Sun Jan 4 15:39:40 2004 : The book stores don't seem to have textbook information for this class.. so they have no _Galois Theory_ books yet. Also, do you know if the differences between the second and third edition are significant, for the purposes of this class? -t

Wed Jan 21 07:42:52 2004 : Hi... Concerning the material in yesterday's lecture... those A+B and then Aw+Bw^2 and then Aw^2+Bw... well, the explanations made sense to me. Nonetheless, i wonder (just curious) if the explanations given in the class was a more "intuitive" approach to this problem? it doesn't look to me like a mathematical explanations. i read the book and the author seemed to give the same sort of explanation. is it because the author want to give a more intuitive approach rather than a more mathematical one? thanx.

Reply: We looked at a cubic equation y^3+py+q=0 and wanted to find the roots. We did something totally unmotivated to do this. First, we decided to look for solutions of the form u^(1/3) + v^(1/3). Further we decided to look only at u and v that satisfied the two equations u+v = -q and u^(1/3)v^(1/3) = -p/3. We did really prove, however, that if u^(1/3) and v^(1/3) satisfy these equations, then u^(1/3) + v^(1/3) is a solution to the original equation.

Wed Jan 21 11:16:47 2004 : will you put lecture schedule in web so we can know whats coming up in the next class?

Reply: When I teach a course, I generally follow the outline of the book that we're using. For the first few lectures, I will be discussing some material at the beginning of Stewart's book that I view as review from Math 113. I'll do this quickly, and you might find that I'm not following the book's outline. Once we get to new material, however, it should be very easy to see what's coming next: just look at the next section of the book. If I deviate seriously from the book's plan, or if I plan to discuss topics that are not in the book, I'll let people know in advance.

Wed Jan 21 20:36:47 2004 : On page 8 of the textbook he starts with the equation t^3 + a*t^2 + b*t + c = 0 and makes the substitution t=y-a/3 to get the new equation y^3 + p*y + q = 0, and claims that p=(a^2 - 2*a^3 + 3b)/3 and q= (2*a^3-9ab + 27c)/27. I get that p should be (3b-a^2)/3.

Reply: If f(t) = t^3 + a*t^2 + b*t + c and g(y) = f(y- a/3), then the coefficient of y in g(y) is g'(0), which is f'(-a/3). When I differentiate f and plug in -a/3 for t, I get a^2/3 - 2a^2/3 + b, which is the same thing that you get.

We need to keep track of errors in the book. (I'll send Stewart a list of typos and errors at the end of the semester.) Please post them or send them to me by e-mail when you find them.

Thu Jan 22 07:19:38 2004 : I'm puzzled about the setup for exercise 1.11 on page 15 of the text. First, does "n zeros and ones" mean n each of zeros and ones (total of 2n), or a combined total of n zeros and ones, or something else? Second, does a unique "arrangement" mean a unique ordering, so that for example, 0111 is different from 1110, or does "arrangement" mean something else?

Non-reply: I won't be able to comment until I'm re-united with my copy of the textbook. (I'm at home; it's at school.) I hope that Stewart's discussion of the Conway sequence will be sufficiently illuminating that I'll be able to figure out what's going on. You might want to look at http://www.ocf.berkeley.edu/~stoll/answer.html, which might inspire you. (This page was written by Cliff Stoll, who's a good person to meet. See especially the Acme Klein Bottle Home Page.) I was surprised to see an early morning comment, by the way. Aren't students supposed to sleep in?

OK, scratch all that. I guessed incorrectly that the problem pertained to the sequence at the top of page 18, which comes after problem 1.11. This is a Math 55 kind of problem. I understand it as follows. You have n symbols strung in a row. Each symbol is a 0 or 1. As the problem says, the 1's that occur (if any) have to be in clumps of size bigger than 2. The number P(n) is the number of ways of doing this. For n<3, P(n)=1 because you can have only 0's. We have P(3) = 2 because we can write 000 or 111. We have P(4) = 4 because we can write 0000, 1111, 0111 and 1110. I counted P(5)=7: More precisely, there is 1 way of writing a string of length 5 that uses no 1's and 1 way that uses 5 1's. There are 3 ways that have 3 1's, namely 11100, 01110 and 00111. There are 2 ways that have 4 1's, namely 01111 and 11110. The challenge here is to see how to express P(n) in terms of smaller P(i) in some inductive way. Note that it's true that P(5) = 2P(4) - P(3) + P(1) = 8-2+1 and that P(4) = 2P(3)-P(2)+P(0) = 4, so it looks like the formula is correct.

Thu Jan 22 10:15:14 2004 : in page 9 of text-book, when it tries to solve y^3 + 3y -36 = 0, i think the number "-18" in y terms has to be just "18" with no negative sign. Is that a typo? this is because q = -36, and so, using Cardano's teh negative sign cancels out. If that's so, then in homework problem #4, we must have 18 instead of -18?

Reply: Yes, you are completely right. This is a typo to keep track of. If you compute the numbers numerically, you'll see that the cube root of (18 + the square root of 325) is a tad more than 3; it's roughly 3.30278. The cube root of (18 minus the square root of 325) is around -0.30278.

Thu Jan 22 10:49:34 2004 : Near the top of page 9 of the text, there is a statement, "Because u+v=q", but earlier on page 8, formula 1.5 gives u+v=-q. If q is not zero, I don't see how these both can be true.

Reply: It's clear that the "u+v=q" on line 3 of page 9 was intended to be "u+v=-q". If you compare the displayed formulas for u and for v on that page, you'll see that the right-hand sides sum to -q. Let's make sure that this typo gets on our list.

Thu Jan 22 14:28:05 2004 : the same question concerning homework #11 as above. i don't quite understand the wordings for this problem.

Reply: I replied to the question around the same time that you wrote your comment and didn't see your question until after I wrote my reply. When we send the author our list of errata at the end of the semester, we should suggest that he give further explanation along the lines of my reply. Including the calculation of the first few P(n) would probably be very helpful to readers of the book.

Thu Jan 22 21:21:10 2004 : I'm a bit confused how to proceed on the HW problems. While algebra, i'm sure, works for 1.4, would it not be alllowed to show that 3 is a solution for the equation, that Cardano's formula yields the messy sum, that that sum yields all the answers, so one must be three? It seems that you DON'T want that, but i wanted to be sure. Additionaly, do you want us to use Cardano's formula to get the messy sum/equation in 1.7 or just accept it and simply plug in the substitution given at the beginning of the problem? And finally, is it ok that we write in math sentences as opossed to english ones in our HW? i.e., can we use a lot of symbols and abbreviations (that are of course widely accepted as ok)? thanks a lot

Reply: This is a multi-part question! For 1.4, it's pretty clear that the intent is to prove the displayed identity (for well chosen values of the cube roots) without any direct reference to Cardano's formula or cubic equations. I did this problem by working backwards from the desired answer. Namely, if A and B are the two cube roots, then A+B is supposed to be 3, and A^3+B^3 is clearly 36. This gave me two equations in the two unknowns A and B, which is solved without much sweat. I then had simple Bombelli-style formulas for A and B. These formulas were unproved, because I obtained them from the assumption "A+B=3", but it was easy to derive them directly and then complete the problem by noting that the sum of A and B is 3, as required. Exercise 1.7 seems to be a straightforward issue because Bombelli is telling you what the cube roots look like and what's to be proved is that they can be chosen so as to sum up to 4 -- no big deal. It doesn't seem to be that the solution to this problem requires any reference to Cardano at all. I don't know what a "math sentence" is. I definitely prefer words to symbols.

Fri Jan 23 15:58:59 2004 : On page 7, the last equation of section 1.4.2, where did the 1/2 come from? Did he use the little talked about "multiply the right side by 1/2" rule? When you said there would be errors in the book, I assumed you meant once in a while. I'm amazed at how many errors there are! One editorial comment (not exactly a typo). At the top of page 32, he says "This is a generalization of the Remainder Theorem, in which f is assumed to be linear." That sentence is just confusing. The Remainder Theorem (on page 26) does not have a poly. f. He apparently is refering to the f in the Division Algorithm which _follows_ that comment. Like C, my brain won't let an author talk about a variable before he's defined it. Greg

Reply: Well, he does say on page 7 that he has the usual quadratic formula "except for a change of notation". The change is that the symbol "1/2" denotes the factor 1 in the displayed equation. :-) I agree that it's disheartening to find misprints. My hope is that the misprint rate will go down as we get into Galois theory proper. I do agree with your comment about the top of page 32. It would have been much better to write "This is a generalization of the Remainder theorem, which pertains to division by a linear polynomial".

Sat Jan 24 15:36:25 2004 : I was slightly amused to read Stewart's little blurb about the omission of commas after displayed formulas (page xii) -- just last night in reading a fluid dynamics textbook (Kundu and Cohen, page 46), I encountered the convention that a displayed formula followed by a comma indicates an implied spatial derivative! And I, of course, had been confusedly ignoring them as just an element of punctuation. :-S

Reply: Perhaps our author has read "Eats, Shoots & Leaves: The Zero Tolerance Approach to Punctuation" by now.

Sun Jan 25 15:38:31 2004 : I don't suppose you can give us a hint on numbers 4 and 11. In particular, I can find the roots to use in 4 by other means than proof but have no idea where to begin. in actually proving that they work On 11, if it is indeed a math 55 kind of question, for those of us who haven't had math 55, in what direction should we be looking?

Reply: For 1.4, I wrote A and B for the two cube roots that are supposed to sum to 3. I fooled around with the expression "A+B=3" and its cube and ended up with formulas for A and B that did not involve cube roots. Once I had those formulas, which were derived from the end result that I was trying to prove, I was able to prove the formulas directly. Looking at the formulas for A and B, you can see immediately that their sum is 3.

For 1.11, I thought immediately of the Fibonacci numbers, which are defined by a recursive formula like the one that we're trying to prove. The ratio between successive Fibonacci numbers approaches the "Golden Mean," which is roughly 1.68. (It's 1/2 the sum of 1 and the square root of 5.) It's probably good to have those numbers somewhere in the back of your mind. Often you have a sequence of numbers that end up being counted by the Fibonacci numbers. For example, suppose I walk up a flight of n steps, taking either one or two steps at a time. In how many ways can I do this? The answer is a Fibonacci number (maybe even the nth). The key to seeing this is to show that the numbers in question satisfy the same recursive relation as the Fibonacci numbers. Namely, when I start walking up the stairs, I can begin by taking either 1 step or 2 steps. If I take 1 step, I have n-1 steps left to go, and the number of ways of continuing is S(n-1), where S(n) is the number of ways of walking up n stairs. If I begin by taking 2 steps, there are n-2 left, so I can continue in S(n-2) ways. We get that S(n)=S(n-1)+ S(n-2). For problem 11, you have to count the number of ways of making a string of n letters out of 0's and 1's; the strings are required to obey the rule that 1's come in groups of 3 or more. It was helpful to me to write P(n) as the sum of the number of strings that begin with 0 and the number that begin with 1. The number that begin with 0 is clearly P(n-1) because there are n-1 spaces to fill in after an initial 0 and the requirements for the n-1 spaces are exactly those counted by P(n-1). Strings that begin with 1 have to begin with 111 because of the rule about clumps of 3 or more. The aim is to count those and to relate the number of them to P(n-1), P(n-2),....

Sun Jan 25 17:51:24 2004 : Hi... i read your comment 'bout how to do 1.4.: does that involve solving a cubic polynomial? B/c i was hopping that i can just prove A+B = 3 without having to deal with any cubics - just through algebraic manipulation. Thnx.

Reply: Just manipulate, don't solve. Good luck.

Sun Jan 25 17:58:23 2004 : Hi in one of the examples on page 3 it says " The set of all polynomials in ï, with integer coefficients, is a subring of C but not a subfield" I was wondering what ï meant in this context.

Reply: The example talks aboout polynomials in π, which is intended to be the complex number 3.1415926535...., i.e., the ratio of the circumference of a circle to its diameter.

Sun Jan 25 20:53:08 2004 : Thanks for the advice on A+B=3. It's sometimes very funny how such a deceptively simple piece of advice can solve a much more complex-looking matter.

Mon Jan 26 19:27:38 2004 : Thanks for the hint about 11. what a nice way to think about the problem.

Mon Jan 26 21:45:36 2004 : On page 26 the expression for q(t) is wrong: the degree of q(t) should be n-1 and the constant term should be a1.

Reply: You're certainly right about the degree, and I have no reason to doubt what you say about the constant term. This needs to be added to the typo/errata list. We'll just direct the author to this page and ask him to go through the stuff that we point out.

Here's another thing to correct: On page 14, problem 1.3 (which I didn't assign) talks about the "prime factorization" of a rational number. Presumably the author intends the rational number to be positive here, but that isn't stated a priori. The prime factorization of a negative number would involve the "sign" -1, and the prime factorization of 0 is undefined.

Here's something else that I want to point out; it's not really a misprint, but I think that it's something in the book that should be changed. The author has a non-standard view of what an irreducible polynomial is: On page 36, he says that a polynomial over a subring R of C is irreducible if it cannot be factored as a product of polynomials of lower degree. He says explicitly that all constant polynomials are irreducible. Even if R is a field, most people insist that irreducible polynomials be non-constant, for the same reason that 1 is deemed not to be a prime number. (It would mess up statements about the uniqueness of factorization.) For most authors, an irreducible polynomial over R would be the same thing as an irreducible element of the ring R[X]; recall (from my lecture last Thursday) that an irreducible element of a ring is an element that is neither 0 nor a unit but which cannot be factored non-trivially.

Mon Jan 26 22:38:53 2004 : About problem 4, I did the problem using your hint and it does work out (big surprise!). However, now that I think about it again, I'm not entirely sure that everything that I did was mathematically "legal." The reason is the fact that I am using the result of the proof in the proof itself when I assumed A+B=3 (this is what the problem is asking me to proof). However, without doing so I will not be able to find formulas for A and B that doesn't involve cube roots. Am I doing something wrong? Help...

Reply: The logic of the solution is as follows. You suppose first that A+B=3 and use this assumption to figure out what A and B would have to be. You then prove directly that A and B actually have the values that you calculated. Here, you no longer use the assumption. Finally, you look at the values of A and B and observe (big surprise) that the two numbers add up to 3. You've thus proved the relation that you wanted.

Tue Jan 27 19:13:00 2004 : well this might be a "stupid" question, but i do forget what "up to isomorphism" means...

Tue Jan 27 19:53:34 2004 : An isomorphism is a one-to-one and onto function (bijection?) between two mathematical structures such that the behavior of the structures is respected. For example, an isomorphism f between two groups G and H is a function that assigns an element h=f(g) of H to each element g of G such that, for any two elements g_1 and g_2 in G, the following holds: f(g_1 * g_2) = f(g_1)*f(g_2), and, if f(g_1)=f(g_2) then g_1=g_2. If two groups are isomorphic, then they are really "the same," except the elements might have different "names." Thus, when counting groups (or any other sort of structure with a sense of isomorphism), one usually considers isomorphic groups to be equivalent, and only count these equivalence classes. It's another way of saying that you don't count renamings or rearrangements as being substantially different.

Tue Jan 27 22:01:29 2004 : random question, but why after submitting a comment are we sent to your 110 site? thanks

Reply: Thanks for pointing this out. Fixed, I hope.

Wed Jan 28 08:14:57 2004 : In the new homework #2 assignment, first problem, I don't see that the equation D(n)=D(n+1)-D(n-1) is true for all n. I calculate D(1)=0, D(2)=-1, D(3)=-1, D(4)=0, which gives me D(4)-D(2)=0-(-1)=1, which does not equal D(3). Am I miscalculating here?

Wed Jan 28 08:35:05 2004 : I think the correct formula for the HW problem should be: D(n) = D(n+1) + D(n-1), so instead of doing subtraction, we have to do addition. Am I right?

Reply: Yes, thanks. I fixed the formula. As I said in class, the formula is intended to say that each term is the sum its two neighbors.

Wed Jan 28 10:31:58 2004 : Is there any possibility of turning in late homework?

Reply: That's basically between you and the grader. At the present time, there are solutions to the homework up on the class web site. I suspect that the grader will be reluctant to accept any further papers.

Wed Jan 28 22:17:12 2004 : who is thr reader for this course?

Sat Jan 31 18:00:57 2004 : I'm working on proving D(n)=D(n+1)+D(n-1) by working backwards. This gives me 2P(n)-F(n+2)=2P(n+1)-F(n+3)+2P(n-1)-F(n+1). Now it seems to me that we have to consider the F's and P's separately in this equality, as I know of no relations between them. So -F(n+2)=-F(n+3)-F(n+1), or F(n+2)=F(n+3)+F(n+1). This rearranges to F(n+3)=F(n+2)-F(n+1), but the Fibonacci sequence is F(n+3)=F(n+2)+F(n+1). So F(n+1)=0?!? I must be messing up the algebra somewhere, but it seems fairly straight forward and I've gone through it again and again! Is there some hidden relation between F & P that I haven't seen? Where did I go wrong??

Reply: You seem to be saying two things: (1) there is no relation between F and P because you don't know of any; (2) if two functions are unrelated, any relation between them is really a pair of identities. I don't know if you can make (2) precise mathematically, but it seems to me that (1) is false because the whole problem is about establishing a certain relation between P and F. The relation is that the quantity D(n) satisfies the simple "difference equation" D(n) = D(n+1)+D(n-1).

As you more or less have explained, the relation that you have to prove may be written

(a certain expression involving P-values) = (a certain expression involving F-values).
You can probably prove it by induction, using the recursion formulas that you know, one for each of the two functions.

Sat Jan 31 19:26:48 2004 : Do we need to prove that polynomial functions over C are continuous, or can we just assume it?

Reply: If you say that polynomials are continuous, the reader is very unlikely to challenge your assertion as needing proof. In this algebra course, we assume that students have some background in calculus and "advanced calculus" (now called real analysis, usually). Of course, I hope that you know how to prove that polynomials are continuous.

Sun Feb 1 18:30:40 2004 : Professor, i think i can prove that your formula does NOT work where D(n)= D(n+1)+D(n-1). in particular, we get -[F(n+3)+F(n+1)]= -F(n+2) by expansion working backwards, but that leaves us to prove that 2P(n)=2p(n+1)+2p(n-1). from there it's not hard to get iff statements to 0=p(n)+p(n-3), an obvious contradiction... so where did i go wrong? or does the formula not work?

Sun Feb 1 18:38:01 2004 : nevermind, i was WRONG, sorry

Mon Feb 2 23:59:55 2004 : Hey, this is Emily, the grader for this class. I wanted to let you all know that I'll be posting comments on homework sets on my webpage, which you can find at http://math.berkeley.edu/~eep/114 I'm going post to this page when many students make the same sort of mistake on a problem. I'll try to clarify the mistake and indicate how to fix it, or give a solution that avoids the problem. If you'd like me to clarify a comment from your homework, or you want to see the solution to a problem that Ribet doesn't write up, please email me and ask. I'm eep AT math DOT berkeley DOT edu.

Tue Feb 3 01:57:56 2004 : Professor, I'm trying to solve the limit problem to number one using the method that you told us in office hours, i.e. by combining the two limits and manipulating the ratios. I ended up with lim n->infinity (2P(n+1)/2P(n) - F(n+3)/F(n+2))=0. I recognized that if you were to combine the numerator then it'll be D(n+1) and the denominator will be D(n), but I know I have to take the common denominator. I have tried three pages worth of manipulating those P(n)s and F(n)s and still cannot find a numerator that's either bounded or converges. Can you give me more hint? Please?

Reply: You have to figure out what happens to a fraction A/B if you change A by a little bit and also change B by a little bit. Suppose that the new A and B are A+a and B+b, where a and b are small. Write (A+a)/(B+b) - A/B in such a way that you see that it is small. Now apply your general insight in the case where A is F(n+3), say, and B is F(n+2).

Tue Feb 3 14:07:13 2004 : Ahhhh, it makes sense. Thank you, professor.

Thu Feb 5 18:18:21 2004 : Hi--I have a question about class today. In the beginning, when you were proving that f(t+1) is irreducible because of Eisenstein, you expanded f(t+1) to be t^(p-1) + a_(p-2)t^(p-2) + ... + (a_1)t + a_0, and concluded: a_0 = f(0) = 1^(p-1) + 1^(p-2) + ... + 1 + 1 = p. Did you mean to say f(0)? a_0 = 1 + ... + 1 = p because if we set _t_ to zero, those 1's are left over, right? but f(0), when we're talking about f(t+1), = 1, doesn't it? (and f(0) = 1 if we're talking about f(t) also.)
(testing html :) )
Also, four insignificant errors in the book:
1) p. 35 he says "we end this chapter by ..." he means we end this _section_.
2) p. 42 (where he's giving an example of the f(t+1) thing, he says "see Lemma 13 for details"--what the hell is Lemma 13? Is that that multiply right side by 1/2 rule? I can't find any lemmas whatsoever that have a 1 or a 3 in their number.
3) (perusing ahead) Fig. 7.2 on p. 77 is wrong. He says p_1 and p_2 are the centers of those circles, but in the figure they are not. (Looks like they are about 1/2 the distance from the center--Lemma 13?)
4) Finally, p. 90, he says, "chapters 15 and 18 derive..." I think he means "Chapters 14 and 15."
It's amazing that someone who spent half (1/2!) a page in the preface explaining his use of punctuation wouldn't have gone over his proofs a little more carefully! --Greg

Reply: Thanks for the pointers to these glitches in the book. As I've said before, we'll send everything to the author at the end of the semester. Concerning your first comment about f(0), I seem to have written down "f(0)" when I needed to write down "the value of f(t+1) at t=0," which is f(1). Thanks for calling attention to this. When I started lecturing yesterday (Thursday), my mind was addled by my weekly donut-induced sugar rush.

Fri Feb 6 00:04:36 2004 : Hi, today you talk what Q(a) looks like when "a" is algebraic over Q. My question is: will Q(a) look the same if "a" is transcendental over Q?

Reply: If a is transcendental, the field Q(a) has infinite dimension as a Q-vector space. It's totally different in character from the fields that we studied on Thursday in class. See page 62 of Stewart's book for some discussion. We'll talk about this in class next week.

Fri Feb 6 12:23:50 2004 : Are we going to cover Syllow thm in this class? I am looking forward to see it.

Reply: My intention has been to lecture on it. Once we do the essentials of Galois theory (which links up group theory and field theory), we will want to apply Galois theory in various ways. For certain applications, it really helps to know the Sylow theorems.

Sun Feb 8 11:53:32 2004 : Regarding homework problem 3.12, for the function phi(n) may we assume the two formulas is problem 3.9 or should we prove those two results as well. Thansks.

Reply: In doing 3.12, you can assume that you know a formula for phi(n). The standard formula for phi(n) is often proved in Math 55 and in Math 113. I'll explain in class on Tuesday how we know the formula. In essence, I'll be doing problem 3.9 for you.

Sun Feb 8 13:23:47 2004 : can we use theorems proved in class on the hw, specifically the one that t^(p-1)+...+t+1 is irreducible over Q where p is a prime? Also, to what extent do the inverse polynomials in the hw on ch. 4 have to do with determinant of certain matrices? nothing at all? thanks

Reply: My answer to your first question is that you should judge from the context whether a given HW question comes before or after the proof that the t^(p-1)+...+t+1 is irreducible. If the question asks you to prove that t^4+t^3+t^2+t+1 is irreducible, it would be against the spirit of the problem to answer simply that this is a special case of a polynomial that was treated in class. A good answer might be to begin with that statement but then show how the method used in class works in the specific case of the problem. Concerning the second question, my inclination is to say that the problems don't have any obvious relation to determinants but that I wouldn't be totally surprised if someone found a way to do the problems by making use of matrices. It's a theorem, for example, that sums and products of algebraic numbers are again algebraic. The usual proof of this theorem involves some standard facts about determinants.

Sun Feb 8 14:59:42 2004 : So is the homework still due tuesday? or has there been a change because of the material we have not covered in class yet?

Reply: The material on the Euler function is something everyone should know by the end of a 2-semester sequence in algebra. On the other hand, it would be going overboard to delay the homework just because we didn't establish the formulas of exercise 3.9. I'll explain on Tuesday why they're true. For the mean time, take them for granted as something that will be known by the time that the homework is graded!

Sun Feb 8 15:11:13 2004 : Question: on problem 3.6, when the textbook says to factor the polynomials from 3.5 into irreduceables, does that mean irreduceables over Q? Because if the polynomial is irreduceable over Q to begin with, then there isn't really anything to do.

Reply: The context makes it pretty clear that the author wants you to factor each polynomial into irreducibles over Q. If a polynomial happens already to be irreducible, there is nothing further to do.

Sun Feb 8 16:06:42 2004 : Question on 3.8: When they ask you to find zeros of polynomials over Q,R,and C and the polynomial happens to be reducible over Q and all the roots are in Q, do we have to find all the other factors over R,C? I understand if the roots are in Q they are automatically in R and C, since Q Reply: If a polynomial over Q happens to split completely into linear factors, then it can't split further over larger fields. For polynomials like this, you can kick back and relax once you've done the work over Q. All polynomials factor completely over C, but only some factor completely over R; few factor completely over Q.

Sun Feb 8 17:00:16 2004 : i'm having a really bad time trying to find the inverses- it seems like multiplying what we want to find an inverse of with something that has generic coefficients and solve for those coefficients may be an extremely messy business- is that what we need to do?

Reply: I can imagine two approaches. The first is to do out in these concrete examples the general proof that we gave in class for the proposition that 1/f(a) can be written as a polynomial in a whenever a is algebraic and f(a) is non-zero. The technique involved finding the polynomial of smallest degree that is satisfied by a. The second approach is to build on the idea that you can write the inverse of, say, 1+2i or 1+sqrt(5) by rationalizing the denominator. When there are more complicated irrational expressions, you can imagine having to rationalize two or more times in succession. I'm deliberately answering without doing the problem myself; once I can do the problem, then it's hard for me to give out a hint without pretty much saying how to do the problem.

Sun Feb 8 19:02:06 2004 : Professor, can you give us hints for 3.12 and 3.15? For 3.12 I cannot find the relationship between the solution for n and n+1. And for 3.15 I figured out part of the thm when n is a prime number but for the non-prime number I cannot figured out the formula. Help? Thanks.

Reply: For 3.12, you have to show that phi(n) gets big as n gets big. It might be helpful to paraphrase the question as follows: for each m, phi(n) is less than m only for finitely many n. Since you have a formula for phi(n), this should be manageable. As far as 3.15 goes, there seems to be an evident pattern, and the issue is to justify it. It might help to notice that a is prime to a number n is and only if n-a is prime to n.

Sun Feb 8 21:25:17 2004 : It seems like the homework has been rather computational so far. I know it's only been three weeks, and we're just now entering fresh material. And I also know that a bit of computation is good and healthy - to keep us on our feet. But will the homework get better? That is, will we not have problems that simply make us verify horribly messy (high school) algebra? Thanks.

Reply: The homework has seemed plenty theoretical to me so far. I'd describe it as a mix between computation and theory. I suspect that the homework will gradually get more theoretical. In my opinion, by the way, there's nothing wrong with a concrete problem if it makes you understand general concepts better.

Sun Feb 8 23:01:50 2004 : do you drop any homeworks at the end of the semester when considering hw grades? thanks

Reply: I always have dropped the lowest homework score. One can argue that this policy hurts certain students, namely the ones who have 14 good scores instead of 13. However, students have always wanted the "drop a score" policy.

Sun Feb 8 23:05:42 2004 : i'm trying to do prob. 3.16, but, honestly, i don't understand the wording of the problems? Can u pharaphrase it for me? Thanx.

Reply: Find all positive integers n such that g^2-1 is divisible by n for all g relatively prime to n.

Sun Feb 8 23:19:06 2004 : Do you have any helpful hints for 3.16? It seems that the natural thing to do would be to derive a contradiction whereby having every element of the group squared being equal to 1 would contradict something but I don't know what the something is. There doesn't seem to be any natural reason why this should be true, and yet I vaguely remember from 113 that there is something special about groups of order 24 but don't know what it is.

Reply: If you take a prime number p, for example, it's very hard for all the squares mod p to be 1. For p>2, we saw in class that there are (p-1)/2 different squares, so p has to be 3 if there's only one square. If n is a possibly composite number such that g^2 is 1 mod n for all g prime to n, you should find that n is seriously constrained. It might help to use the Chinese Remainder Theorem here.

Mon Feb 9 11:27:53 2004 : I just wonder: Is Math H113 = Math 113 + Math 114?

Reply: H113 probably differs quite a bit from year to year. What gets done depends on the inclination of the instructor and the students. When I taught the course last year, we didn't do any field theory of Galois theory, but we did do the Sylow theorems. My class had quite a lot of group theory and a bit of ring theory. There weren't a lot of fields poking around.

Mon Feb 9 11:33:03 2004 : I just wonder: Is Math H113 = Math 113 + Math 114?

Mon Feb 9 17:16:15 2004 : No, Math H113 = H * Math 113. The H can be factored out of the "Math."

Mon Feb 9 18:35:42 2004 : A small error in the book: On p. 45, between Def. 3.26 and Lemma 3.27, it is written, "When K=R and we draw a graph, as in Figure 1.1, ..." I think Stewart means to reference Figure 3.1 on p. 44 instead.

Reply: Keep finding the misprints. We'll bring in the author toward the end of the course.

Tue Feb 10 04:01:11 2004 : Dear Professor, In problem 3.12 is the upper bound for possible values of n the number of possible products recieved from multiplying (by any number of iterations) the numbers amongst and between the two sets: 1) all primes which do not divide m and are less than m, 2)all primes which divide m and all their powers up to exponents which do not produce a number greater than m? The multiplication part is important so I will delineate further. Example: using this "multiplication" given (a,b,c) we have the possiblities: a,b,c,ab,ac,bc,abc. So the upper bound (with no reference to specifics from our problem) would be 7. Thanks for your time.

Reply: I think that you should just get a crude bound for n's that can possibly have phi(n) = m and that you shouldn't worry about any refined analysis. You can probably see in various ways that phi(n) can never be an odd number bigger than 1. Does that fit into your framework?

Tue Feb 10 05:40:24 2004 : i'm stuck on #3.16. it seems to me like a more number theoretic question. Where should the Chinesse remainder thm comes in? i tried to do this problem again since 4 a.m. yet results in nothing.

Reply: Let n be a positive integer. Say that (*) holds for n if g^2=1 mod n for all integers g that are prime to n. We were discussing this problem in my office yesterday and seemed to think that it was helpful to know that (*) for n implies (*) for all divisors of n. Equivalently, if (*) fails for n, it fails for all multiples of n. To see this relationship, it is helpful to think about the Chinese Remainder Theorem.

Tue Feb 10 07:49:21 2004 : hmm... I actually got the answer, at last, but without using Chinesse Rem thm at all. Prof., do you mind to post the answer for this prob. after we turned in the hw today? I think this is the question of lots of interest? Thanx.

Reply: I'll be happy to post a solution. You can also post yours -- or e-mail it to me.

Tue Feb 10 23:28:14 2004 : Did someone find a way to solve 3.16 without the explicit use of the chinese remainder theorem?

Wed Feb 11 13:08:15 2004 : The way I solved 3.16. is like this (check if I didn't do any mistakes here). I first showed that n cannot possibly have any prime divisors greater than 3. To do so, suppose, n has prime factors p1, p2, ... , pm, all are greater than 3. Now, divide it into 2 cases: (1) 3 divides n -> x = 2(p1.p2....pm)+3 is relatively prime to n. But x^2 doesn't equal 1 mod n. (2) 3 doesn't divide n -> x = (p1.p2....pm)-2 is relatively prime to n. But, again, x^2 not equal 1 mod n. Hence, n cannot have any prime factors greater than 3. We're left with n = 2^a. 3^b. It's easy to check that a has to be less than 4 and b must be 1.

Reply: Inspired by the comment above, I'd like to present an even shorter proof of 3.16. Suppose that the property (*) holds for n. Write n as a power of 2 times an odd number m. Let g = m+2. Then g is odd, and is prime to m, so it has no common factor with n. Consider g^2, which is supposed to be 1 mod n. It's 1 mod m, in particular. On the other hand, it's clearly 2^2 = 4 mod m. Hence 1 mod m is 4 mod m, which means that m divides 3. Thus n is either a power of 2 or 3 times a power of 2. In either case, 5 is prime to n, so that 5^2 is 1 mod n. Since 5^2 is 1 mod n, 24 = 5^2-1 is 0 mod n; thus n divides 24.

Thu Feb 12 09:40:09 2004 : Hi... this question has nothing to do with math 114. it's about math 250. Is it basically a more advanced version of math 113 and 114 or it studies totally diff subjects on Group theory? For students contemplating of doing a grad. degree in math, which one is a better course to take: math 250A or 202A? And, from your experience, how much more work usually a student has to devote in order to succeed in grad course? Thanx.

Reply: I think it's true that 250A treats a lot of the same topics that are covered in 113 and 114. It does it faster (in one semester instead of two) and more deeply. In Math 250A, you see some material that tends to get skipped over in 113; an example might be the structure theorem for finitely generated abelian groups. I don't think that there's any clear choice between 250A and 202A -- it depends on whether you like algebra or analysis more. Either course is a good introduction to what graduate school is like when you first start out. (Once grad students have taken a few basic courses, they move on to more advanced topics and start working on research problems.) It's hard for me to guess how much time our undergraduates spend on courses like 250A and 202A. The best way to find out is to ask around in the common room.

Thu Feb 12 22:16:42 2004 : I still don't quite get why phi(p^k)=p^k-p^(k-1) for prime p. Why are there p^(k-1) numbers that are not relatively prime to p^k? And for problem 4.9, He meant p,q in Q[t1,....,tn], right? Is it enough to show that the set is a field by showing for p/q not 0 there is inverse of p/q in the set? Thank you.

Fri Feb 13 00:26:46 2004 : Consider a sequence of integers: 1,2,3,...,p^k. Now, gcd(n,p^k) not equal to 1 iff p|n. So, the only possible n's are p,2p,3p,...,(p^k-1)p. This gives a total of p^k-1 of them. Consequently, phi(p^k) = p^k - p^k-1. Hope this helpful.

Fri Feb 13 13:42:08 2004 : I'm not sure what 4.9 is asking me to do. I might be getting lost in the translation. I'm understanding that p(x_1, ... x_n) is a polynomial, each term of which is a coefficient from c and zero or more elements x_i, that is, p(x_1, ... x_n) = a_0,...,0 + a_1,0,...,0 x_1 + a_n,...,n x_1 ... x_n. Is this right? I think I'm ok so far, but now I get a little lost. Is n the cardinality of X, so that all of the elements of x are represented in each polynomial (likely with some zero coefficients?) If so, what is q(y_1,...,y_n)? Are the x_i's distinct from the y_i's? Also, what's meant by C(X)? Each of the field extensions mentioned in the text appears to be a subfield of C? Are we extending the abstraction? Are the members of X not members of C? If I understand the question, I can show how each rational expression described is a member of C(X). Intuitively, I can see that this describes all of C(X). But I'm having trouble conceiving of a methodology for proving that nothing else can be a part of C(X). Thanks.

Reply: Um, yeah, I see the difficulty. It's 99.9% apparent to me that the exercise meant you to describe Q(X) and not C(X). Misprint alert! In this short chapter, the author discusses subfields of C that are generated by subsets X of C. He focuses in his exposition on the situation where X has exactly one element. The problem asks for a description of Q(X) in the general case. When you see what's going on, you'll realize that the problem doesn't have deep content. It asks for a characterization of Q(X) that's analogous to the description of the span of a subset X of a vector space V; the span of X is the set of all finite linear combinations sum a_ix_i where the x_i are taken from X. Here you have polynomial expressions, and quotients of them, instead of linear combinations. Hope this helps. Too bad about the misprint(s).

Sat Feb 14 09:44:16 2004 : This is related to problem 5.1...is there a clear cut way to tell if the extension is simple or not? The book wasn't so terribly clear about this...

Reply: The intent of the problem is for you to investigate the situation without a clear idea ahead of time whether or not the extension is simple. If you think that it's simple, you might try to find a single a (= alpha) in the field such that Q(a) is the whole field. Perhaps the sum of the two square roots might work? If you suspect that the extension is not simple, try to imagine a strategy that would prove this. For example, if you can prove that the degree of the extension is 4 and that every Q(a) has degree 1 or 2 over Q, then you have a proof. I say all this because it's a common situation in mathematics that you come to a problem and don't know what the answer will be. By day, you try to prove that it's simple. By night, you try to prove that it isn't. One day or one night, you realize that you've solved the problem.

Sat Feb 14 17:10:36 2004 : I think I am confusing notation. When we speak of the extension K(t); does it neccarily look like p + qt where p and q are in the field K? Also, I understand the definitions of algebraic and simple extension separately but not the definition of a simple algebraic extension. Also What happens when we consider the case where K is the set of rational functions over C? In particular, in the proof of therem 5.3: the set of rational expression expressions K(t) is a simple transcendental extension of the subfied K of C. For the simple extension K(t):k if p is a polynomial over K s.t. P(t) = 0 why is p neccesarily o? thank you.

Reply: If K is a field and t is understood to be a variable, K(t) denotes the field whose elements are fractions p(t)/q(t), where p and q are polynomials and q is non-zero. You start with K[t], the polynomial ring over K with the variable t. This ring is an infinite-dimensional K-vector space; a basis is the set of monomials t^i with i=0,1,.... The ring K[t] is an integral domain, but not a field. The field K(t) is the field of fractions of K[t]. Constructing it from K[t] is like constructing Q from the ring of integers Z.

A simple extension is an extension L of K that is generated by one element of L. This means that there is an a in L such that L=K(a). An algebraic extension L/K is one for which each a in L satisfies an algebraic equation over K. A simple algebraic extension is of the form K(a), where a is algebraic over K. This is the kind of extension that we have talked about in class; we've done this a fair bit. You know that all elements of K(a) are actually polynomials in a (with coefficients taken from K).

We can consider C(t); that's a simple extension of C that is not algebraic. We can also consider K(t) when K is a field that is not necessarily C. For example, K might be a subfield of C.

I don't think that Theorem 5.3 has any deep content. The assertion is that K(t) is not an algebraic extension of K. Well, it's enough to see that t satisfies no polynomial with coefficients in K (other than 0). This is true by the constructions of K[t] and K(t). At a very low level, polynomials in t are constructed so that a polynomial is not the zero polynomial unless its coefficients are all 0. If t satisfied a polynomial with coefficients in K, the opposite would be true.

Sun Feb 15 14:04:29 2004 : Hey professor. i was wondering if Q((sqrt(5)+1)/2)=Q(sqrt(5)), not by isomorphism but literally. thanks

Reply: Sure, these are the same thing. You can express each as a (very simple) polynomial in the other, so each number is a member of the field generated by the other. The equality of the two fields follows immediately.

Sun Feb 15 16:59:19 2004 : Professor, about problem 5.1, if I happen to find a polynomial with degree 4, and prove that it is the min. poly., how does this prove that the extenstion is not simple? Also, how do you know that all Q(a) over Q have degree 1 or 2, would this help to see that the extension is not simple? Thanks!

Reply: First of all, understand that I was speaking hypothetically about a line of attack on the problem. I tried to evoke a situation where you could prove that the extension has degree 4 but that each subextension Q(a)/Q has degree 1 or 2. If you know that this is true, then you may conclude that the full extension is not of the form Q(a). In the question that you have posed, you're talking about an irreducible polynomial of degree 4. That would be more in the direction of proving that the extension is in fact simple. Indeed, if the minimal polynomial of a is of degree 4 and if a is contained in a field of degree at most 4, then the field is generated by a and is therefore simple.

Sun Feb 15 17:38:10 2004 : p(x1,x2,...,xn)= some polynomial over Q in n variables in problem 4.9, right? so then p in Q[t] is not true; instead p is in Q[t1,...,tn]. is this right? is there yet another misprint in this book?

Reply: The polynomial ring in n variables t_1,...,t_n, over a field K is denoted K[t_1,...,t_n]. Since this is a cumbersome notation, the ring K[t_1,...,t_n] is sometimes given the nickname K[t]. I don't know whether the notation was used somewhere in the book before this problem appears. If not, then it should have been explained in the statement of the problem.

Sun Feb 15 17:47:13 2004 : hey professor. will there be a review scheduled for the midterm and will we have a practice midterm/review sheet? thanks

Mon Feb 16 14:11:05 2004 : What materials will be covered in midterm #1?

Reply: Everything that we covered through last week.

Mon Feb 16 22:46:22 2004 : On question 5.3, can we consider the polynomials t^2-2 and t^2-4t+2 to be "equal", because they are equivalent mod 4? If so, it would automatically follow that the two extensions are isomorphic, so I feel like Im missing something here

Reply: The two polynomials are clearly not equal in any literal sense. I don't know what sort of "equality" you have in mind. If two polynomials happen to be congruent mod 4, you wouldn't normally expect them to define isomorphic extensions of Q. For example, t^2+1 defines Q(i)=Q(sqrt(-1)) but t^2-3 defines Q(sqrt(3)); these two fields are clearly not isomorphic. (The integer 3 is a square in the second field but not in the first.)

Mon Feb 16 22:46:46 2004 : hi professor. on the exam, are questions like solving the cubic and quartic equations like in the begginning of the book going to be fair game? thanks

Reply: I don't see why not. The exam covers what we have done.

Mon Feb 16 23:15:10 2004 : Yup, i have the same question regarding prob. 5.3. it seems clear to me that the two are isomorphic. I feel like missing sth too. Prof., do you mind to talk about these "isomorphic" stuff on class tom? I think the book is doing fine when it talks about another subject but everytime it starts talking about isomorphism between two field, or even more: between two extensions, the book doesn't seem to me to take the subject too seriously. At best it often only says something superficial comment in a hope that the reader will just understand everything right away. In my math 113 class, every isom. relationships were given rigorous treatment but here, i think i miss sth. Thanx.

Reply: Because this is a "second course" in abstract algebra, it is reasonable for the textbook to assume a certain sophistication of its readership that might be absent in students at the start of Math 113. Sometimes providing too much in the way of detail can obscure the main point. I disagree that the book is failing to take isomorphism questions seriously. On the other hand, I will be happy to establish various isomorphisms in class today. For example, we can do Problem 5.3 if you like.

Tue Feb 17 04:09:20 2004 : Will you have any additional office hours this week, since we missed Monday and have an exam on Thursday? Also, can you post your regular office hours please?

Reply: Those students who came to class on Thursday asked for additional office hours this week. I proposed Tuesday from 10 to 11:30. My regular office hours are those that were discussed in class a couple of weeks ago. These are posted at http://math.berkeley.edu/~ribet/office_hours.txt, which is the "office hours" link on the course web page.

Tue Feb 17 10:06:31 2004 : Question about context. Ex. 4.10.g says every subfield of C (complex) contains Q (rationals). I thought, no, Z_5 does not contain Q. Yet Prop 4.4 "proves" every subfield of C contains Q. (Erroneously, it seems to me--what if 1+1+...+1 = n = 0?) Where does the confusion lie? I am guessing the context of this question is limited in some way that forgoes the possibility of finite fields, but I don't see that stated explicitly. There's a slight disconnect between the book and the lectures, because in class we talk a lot about what Stewart would call "abstract" fields, which he is going to such great lengths to avoid. It sometimes makes the context of a question difficult for me to discern. Thanks.

Reply: A subfield of C is a field that is contained in C. It lives within C. Its field operations (+,*) are those of C. The field with 5 elements is not a subfield of C. In C, it is impossible to add up a bunch of 1's and get 0.

Tue Feb 17 10:18:12 2004 : By the way, it's really annoying when Stewart says, as on p. 53, "For a proof, see Exercise 4.9" then you look at Ex. 4.9 and it says "Prove that C(X)..." He means "for a proof, see your paper after you figure it out for yourself." :)

Reply: As I've said many times, we'll ask the author to read this page and take your remarks into account. I agree that it would have been better to write something like "The proof is left as an exercise, namely exercise 4.9"

Tue Feb 17 10:25:25 2004 : Fn (i.e.) Zn is not a subfield of Z but rather the set of equivalence classes of Z mod n. In other words, Zn= Z/nZ as the professor loves to say it. looking at this structure, it doesn't seem to really be a subfield of C but rather a totally different animal using the same elements for naming as C. Also, just as Z5 could be called {0,1,2,3,4} it seems it could also be called {5,6,7,8,9} without much of a problem b/c 0=5 and so one. so Z5 doesn't really have a unique naming in Z... which is kinda iffy if you want to play the subfield game with C. but that's just my opinion.

Reply: In a field F, the multiplicative identity ought to be called "1" because you want to be able to write 1*x = x for every x. If you add 1 to itself, your natural impulse is to write 2 for the result. If you take the additive inverse of 2, you call it -2. What goes on is that you can construct, for each integer i, an element of F that you'll want to call i. The construction gives you a function Z->F, namely i |-> the element of F called i. This function is a "canonical" (God-given) homomorphism of rings. When it is injective, one says that F has characteristic 0. For example, it is injective if F is a subfield of C, so those fields have characteristic 0. Some fields have non-zero (positive) characteristic. Finite fields are examples. Not all fields of positive characteristic are finite, however.

Tue Feb 17 11:31:54 2004 : About 5.1, didn't we just do the equivalent problem last week, #4.4? Doesn't proving 1, sqrt(5), sqrt(7), sqrt(35) are linearly independent prove that Q(sqrt(5), sqrt(7)) is not a simple extension? (I thought I sent this, but it didn't show up--sorry if it's repeated)

Reply: To say that Q(sqrt(5), sqrt(7)) is a simple extension means that there's some polynomial expression in the two square roots, say a, such that each square root is a polynomial in a. Because we're talking about polynomial expressions and not linear expressions, the questions are not obviously equivalent to each other.

Wed Feb 18 21:33:39 2004 : This is Ribet. A student wrote to me this evening as follows:
I just found two big errors in the book: on p. 60, his Lemma (5.8) about congruences is completely wrong. He says that if a_1 cong. a_2 (mod m) and b_1 cong. b_2 (mod m) then a_1 + a_2 cong. b_1 + b_2 (mod m) ! which is easily shown to be false. He means that a_1 + b_1 cong. a_2 + b_2 (mod m).

His proof tries to prove his erroneous statement, but by the extreme brevity so many math books are fond of, he completely glosses over that the second equality in his proof also isn't true. (Does this happen a lot in mathematical articles?)

His second claim is also false. He says that also, a_1*a_2 cong. b_1*b_2 (mod m). He means to say that a_1*b_1 cong. a_2*b_2 (mod m) In this case he actually got the proof right, because he proved the correct statement rather than the one he stated.

Wed Feb 18 22:27:53 2004 : Here's an e-mail exchange that I had with a student this morning:

Hi Professor--About simple extensions of Q that we were talking about in class yesterday, you started by extending Q with the sqrt(2), Q(sqrt(2)):Q. Then you extended that with sqrt(3), Q(sqrt(2),sqrt(3)):Q(sqrt(2)). The degree of each of those extensions is 2, and the degree of [Q(sqrt(2),sqrt(3)):Q]=4. But then you showed there exists a such that [Q(sqrt(2),sqrt(3)):Q(a)]=1, and [Q(a):Q]=4. Is all that right so far? Can you explain more about what is going on here? What does it mean that you can extend the field in two small steps, or extend the field in one big step?

I think that I must have been trying to prove, for well chosen a, that Q(a)=Q(sqrt(2),sqrt(3)). One way to do this is to take a in such that Q(a) has the same degree over Q as Q(sqrt(2),sqrt(3)). Because the relative degree [Q(sqrt(2),sqrt(3)):Q(a)] will be 1, you'll conclude that Q(a)=Q(sqrt(2),sqrt(3)). To know that the two fields have the same degree, you have to calculate the two degrees. The degree of Q(a) is the degree of an irreducible polynomial satisfied by a. The other degree is at most 4 because each of sqrt(2), sqrt(3) has degree 2 over Q; I must have been explaining that.

Is it the case that if you take any simple extension b where [Q(b):Q]=n, and n is not prime, that you can find intermediate fields for each factor of n? (in some way the converse of the tower law?) For instance, if [Q(b):Q]=6, then there must be another extension c s.t. [Q(c):Q]=3 and [Q(b):Q(c)]=2? And for that matter, another extension d s.t. [Q(d):Q]=2 and [Q(b):Q(d)]=3? Is it that those extensions must exist if there's an extension of degree 6? And yet even so, Q(b):Q is simple, even though in some way it is, I don't know, "compound"?

This is a natural guess, but it turns out to be false. Galois theory turns the question into one involving groups and subgroups, and the fact of the matter is that a finite group G need not contain subgroups whose orders are arbitrary divisors of the order of G.

The following question was asked in class, but could you please say a little more, then, about what is _not_ a simple extension? Any "compound" extension with degree < infinity is simple (and algebraic)? But an extension of two non-algebraic numbers is not simple (Like Q(pi,e):Q), and yet all simple transcendental extensions are isomorphic? Well, that's weird!

If you have fields E and F, with E contained in F, you can consider the extension E(a_1,...,a_n) when a_1,...,a_n are in F. If the a's are all transcendental over E and satisfy no algebraic relations among them, then the extension is not simple. If the a's are all algebraic and E has characteristic 0 (e.g., if E and F are subfields of C), then there's a theorem that the extension is simple. If E has characteristic p, where p is a prime, then there is no such theorem. Sometimes the extension is not simple.

Ken Ribet

Wed Feb 18 22:47:45 2004 : i don't get why if dim of F over R is finite then F must equal C?

Reply: In the context, F was an extension of C and we can consider [F:C] and [F:R]. The tower law tells is that the latter number is twice the former number. This means that both degrees are finite if either one is finite and that we have [F:R] = 2[F:C] when the numbers are finite. The essential point here is that C has no finite field extensions over than C itself! If F/C is finite, then every a in F is algebraic over C. Its minimal polynomial is an irreducible polynomial over C. Because C is algebraically closed, that polynomial has degree 1. Hence C(a)/C has degree 1, so that C(a)=C. We conclude that a is in C. Since all elements of F are in C, F=C.

Wed Feb 18 23:11:43 2004 : hi professor ribet and class, i have two questions about the lecture on tuesday: 1. at the end of the period i thought you asked whether we could prove a theorem that if we have some a_1,...,a_n all algebraic in C then the degree of Q(a_1,...,a_n) is finite and less than or equal to [Q(a_1) : Q]*...*[Q(a_n) : Q] (i'm hoping i quoted this right, i probably should've written it down). anyway, i just wanted to make sure that saying a number is algebraic in C as in this case is equivalent to saying it's algebraic over Q. i only ask because if you just meant algebraic over C then you could just take a_1 = pi and have [Q(pi) : Q] = infinity.

Reply: If I said that a number is "algebraic in C," I presumably meant that the number was a complex algebraic number. By definition, a complex number is algebraic if it is algebraic over Q.

2. earlier in the lecture you asked us to prove that if we have some F containing C where [F : R] is finite then F = C. i was wondering if it would suffice to say that this implies [F : C] is finite which means that every element a of F is algebraic over C which means any such element has a minimal polynomial in C, but the only irreducible polynomials in C have degree < 2 which means a satisfies a linear polynomial in C which means a is in C so F = C.

Reply: I think that you asked the same question that I answered a couple of minutes ago. It sounds as if you understand what's going on.

i apologize if either or both of my recollections of your questions are incorrect, as they likely are. also, even though i'm sure you've already written the test and won't be changing it before tomorrow, finding roots of cubics and quartics doesn't seem particularly interesting so i'll cross my fingers that we don't have to deal with them tomorrow.

Reply: We didn't do any quartic equations in class, so you should probably infer that they're not likely to be on the exam. Cubic equations were discussed in class, however.

Wed Feb 18 23:12:12 2004 : i thought the answer to that is that we know that something is an algebraic extension of something else iff the dimension of that extension is finite. But the only algebraic extension that is finite over R is C because by the fundamental theorem of algebra, any polynomial in R has roots over C. (i think the statement equivalent to this is that C is algebraicly closed). it so happens that the degree of C over R is 2. So if the extension over R is finite and the element adjoined (or one of them) is not in R, then the extension is algebraic which means the element must be in C which means the elements adjoined are in C which implies that the degree is 2. So we can also say the degree is finite iff the degree is less than or equal to 2. i think that solves the problem- am i right, professor?

Wed Feb 18 23:14:39 2004 : Could you restate the Chinese remainder theorem as an isomorphism again? because i'm trying to find it and most sources don't write is that way... thanks

Reply: Write Z_n for the ring of integers mod n. Suppose that n and m are relatively prime. Then the natural map Z_{nm} -> Z_n x Z_m is an isomorphism of rings. This isomorphism induces an isomorphism of groups of units. The group of units of Z_n, call it U(n), consists of numbers mod n that are prime to n. It's a group under multiplication. The unit groups of Z_{nm} is U(nm). The unit group of a product of rings is the product of the two groups of units. We emerge with an isomorphism of groups U(nm) = U(n) x U(m) when n and m are relatively prime.

Thu Feb 19 00:56:03 2004 : on page 43, stewart talks about reducing mod p but then talks about Z[t] --> Zn[t], and never says that n is prime. in his examples, n is prime. he might want to specify that, because all he talks about is reducing mod n in the sentences on the page. just one more typo in the long list of them.

Thu Feb 19 10:35:24 2004 : Hi-- Another small typo. p. 3, Example 1.2.4, he says we will prove something in "Theorem 5." There is no "Theorem 5." (I didn't look for what theorem he means--I'm still looking for Lemma 13.)

Thu Feb 19 17:55:24 2004 : Yay the midterm is over. Hope everyone has a good weekend!

Thu Feb 19 18:07:02 2004 : This midterm: is it easy or difficult?

Thu Feb 19 18:17:57 2004 : I, at least, didn't have quite enough time. 10 more minutes would have been enough.

Thu Feb 19 18:42:33 2004 : questions about the test. How did question 1 relate to what we've been learning? Was it just a general question about gcd, congruences, and so on? Or did it have some significance in fields, extensions, etc, that I missed? Secondly, on the last one, did we need to identify which solution came out to 1, 0, and -1? Or was it enough just to show what the three formulae were?

Thu Feb 19 19:59:10 2004 : Ups... i thought in problem #4 we only need to identify whether the given solution using Cardano's is related to y = -1 or 1 or 0????? That's how i read the problem

Fri Feb 20 12:30:19 2004 : I think it was reasonable but like the other person said there wasn't enough time. 10 more minutes would have been enough.

Fri Feb 20 18:28:24 2004 : I have a question on problem 6.9. I figured out for the case where the extentions L/K and M/L are finite, and for the case where M/L is infinite but L/K is finite, but for the rest(M/L finite, L/K infinite, and the case where both of them are infinite), I'm not so sure how to approach the problem. Are there any hints?

Reply: You're trying to prove that every element a of M satisfies an algebraic equation with coefficients in K. It's enough to show that a lies in an extension of K that has finite degree over K. See if you can find such an extension.

Sun Feb 22 14:42:24 2004 : Isn't problem 6.8 false as stated? What if the polynomial p is linear? (Something in the smaller field will automatically be in the bigger field. But the degree of such a p would of course be 1, so trivially it would be relatively prime to the other quantity.) Otherwise (i.e., if p is "bigger") I think the problem (and my proof) is ok.

Reply: You're right. The statement is certainly not correct in the case where p is a linear polynomial. In this case, p always has a root in K, so it has a root in L. Further, the degree of p, which is 1, is coprime to every positive integer in this situation. (Two integers are coprime when their gcd, or hcf if you prefer, is equal to 1.) Another correction for the next printing or edition.

Here's a minor correction: on line -10 on page 109, the author refers to "Theorem 16". The intended reference is to Theorem 5.16 on page 63.

Sun Feb 22 16:29:09 2004 : to show that an infinite set of numbers are linear independent, is it enough to show that any arbitrary finite subset of them are? thanks

Reply: Yes; that's how linear independence is defined.

Sun Feb 22 21:36:34 2004 : what does it mean for a liner transformation to be singular?

Reply: If one has a linear transformation T:V->V, where V is a finite-dimensuional vector space, one says that T is non-singular if it is invertible and singular if it isn't. The transformation T is invertible if and only if its determinant is non-zero (and there are other equivalent conditions as well). Thus T is singular if and only if det(T)=0, i.e., if and only if 0 is an eigenvalue for T.

Mon Feb 23 00:27:11 2004 : Professor, for problem 6.8, can you give an example?

Reply: Take the polynomial p(x)=x^3-x-1 over K=Q. This is an irreducible polynomial of degree 3. If L/K is an extension of degree prime to 3, then p(x) has no roots in L. For example, p has no roots in a quartic extension of Q.

A cleaner and stronger statement of the problem might go as follows: Let p be an irreducible polynomial over K and let L be an extension of K in which p as a root. Then the degree of p divides the degree of the extension L/K. In this statement, the degree of p can be 1; it's not a special case.

A related statement, whose truth we can explore later on, is the following one. Take an irreducible polynomial p over K and an extension L of K. Then p remains irreducible over L if the degree of p is prime to the degree of the extension L/K.

Mon Feb 23 09:06:12 2004 : Prof., can u possibly give more hint for problem 6.9.? My question is the same as the person's above

Mon Feb 23 10:09:53 2004 : For 6.9, if we have algebraic extensions M-L and L-K, and an element a is in M, then by definition of algebraic, we have a polynomial over L that has a as a root. the trick is to show that all of those coefficients of said polynomial are either in K or are experessible as a (finite?) sum of elements in K. But wouldn't this be true in the extension of K when you adjoin all of those coefficients, since we can show that this extension is finite and is a subset or L? or is this reasoning wrong? thanks a lot.

Mon Feb 23 15:15:49 2004 : For problem 6.14 we are trying to show that Q(sqrt6, sqrt10, sqrt15)= 4, not 8. In the book it is proved that Q(sqrt 2, sqrt 3, sqrt 5) has degree 8, and it is obvious that the former field is contained in the latter field. Is it sufficient to show that [Q(sqrt 2, sqrt 3, sqrt 5): Q(sqrt 6, sqrt 10, sqrt 15)] is 2? If so how would we go about doing this, is it enough to find an element in the field with degree 8 that isnt in the field we are looking at?

Mon Feb 23 16:19:13 2004 : Professor, I was the one who asked the question about 6.14 but I figured it out so thanks anyway.

Mon Feb 23 17:35:31 2004 : Professor, would it be possible for you to put up another chart, but this time with the last 4 digits of our sid's next to our scores?

Reply: This is a bit over the top. You're going to get back the exams tomorrow. If you can't wait until then, send me e-mail and I'll let you know your score.

Mon Feb 23 18:53:58 2004 : If we wish to use well-known results, such as, for example, the fact that the square root of a prime number is irrational and that the sum of a rational number and an irrational one is irrational, do we have to prove these things, or can we just cite them as given? I'd like to use some of these ideas in one of this week's problems, and I can prove them if necessary, but it's more work. What is the verdict?

Reply: The basic answer is that you should use your judgement. You might perhaps be able to allude to the essence of the proofs of the indicated assertions without going into them in gory detail. For example, the square root of a prime p is irrational because t^2-p is an Eisenstein polynomial and therefore irreducible over Q. (In first courses, one proves that the square root of 2 is irrational with a long argument, but we can do it in a flash by appealing to a result that we have studied.) Hope this helps....

Tue Feb 24 01:00:25 2004 : just a quick question for everyone: is there a reason why nobody has asked a question about 6.11? did he do this in class and i just happened to be zoning out? am i just a complete idiot and everyone has managed to tackle the problem? seriously, totally clueless.

Reply: We had some discussion of this problem in office hours yesterday. My suggestion was to prove the result about linear independence by proving something stronger, namely: if p_1,...,p_t are distinct primes and if a_1,...,a_t are the square roots of the p's, then the field Q(a_1,..,a_t) has degree 2^t over Q. The aim is to prove this by induction on t, using the tower law. You'd have to show that p_t is not a square in the field generated by a_1,...,a_{t-1}. This seems like a good strategy, but we didn't do the problem to the end, so I am not completely sure that it works. Once you know the result about the degree, you'll see that some obvious set of generators with 2^t elements has got to be linearly independent over Q. This set includes the numbers a_1,...,a_t; that's why the degree result would be stronger than the desired linear independence result. See http://math.berkeley.edu/~ribet/114/hw5_ans.pdf for details on doing this problem.

Tue Feb 24 14:04:45 2004 : Which problems are you going to grade this week? Thank you.

Tue Feb 24 23:59:20 2004 : from the Grader: I intend to grade questions 7, 9, 11, and 12 from this homework set (time permitting.)

Wed Feb 25 14:43:58 2004 : This is Ribet. Here's a comment about the proof of Proposition 9.14, which is stated on page 113 and proved on pages 113 and 114: In the previous lemma (9.13), the author proves that f is separable if and only if the hcf of f and its derivative f' (or Df) is 1 in Sigma[t]. In 9.14, he says that f is separable when it is irreducible over K, and the reason is that f and f' can have no non-trivial common factor in K[t]. We need an extra piece of information, which we discussed in class last week or the week before. It's that two polynomials in K[t] that have hcf=1 in K[t] continue to have hcf=1 in extensions of K[t]. This is proved by invoking Euclid. Namely, if f and g have no common factors over K, we can find polynomials u(t) and v(t) in K[t] so that 1 = uf + vg. It follows from this equation that a polynomial that divides both f and g in Sigma[t] (for example) has to divide 1 in Sigma[t] and therefore must be a constant.

Reply: I'm replying to my own comment. In fact, lemma 9.13 proves that f is separable if and only if hcf(f,f')=1 in K[t]. I think so, anyway. I concluded that there is simply a misprint in the statement of the lemma and that Sigma should be K.

Thu Feb 26 10:26:14 2004 : two more very small corrections: Page 92, in example 3 he means "as in example 6.8. In the first line of text on page 93, the second "is" should be deleted.

Thu Feb 26 10:38:57 2004 : Hi Professor--I think I finally figured something out that has been confusing me for weeks. Can you verify this for me, please? The words "variable" and "indeterminate" have two meanings in our context. First is the indeterminate of the polynomial, the obvious one: t^4 + 3t^3 + 2--the t's there are the indeterminate of the polynomial. But Stewart uses the word in a different way as well. When he talks about an extension like Q(x_1, x_2, x_3), I thought he was talking about a function with three variables, like x_1^3 + (x_1^2)x_2 + x_2(x_3^2) + x_3. But now I realize what he is trying to talk about when he says Q(x_1, x_2, x_3) is like a _generic_ representation of a field extended by three elements--Q(sqrt(2),sqrt(3),sqrt(5)) for example. He wants to talk about _any_ field extension with three elements. Am I right? Or was I right the first time? These two possible meanings are vastly different (and if my new reading is correct, I think it's drastically misleading to call the x's "indeterminate"). I guess here's my basic question. Are we EVER talking about functions with more than one variable? (I am feeling this is a really retarded question, one I should have known a long time ago, which is why I'm writing from an internet cafe so you can't trace my IP. :) )

Reply: There are no retarded questions.

The question is interesting enough if there is a single variable, so let's talk about Q(x). The notation "Q(.)" is used in two ways. If t or x is understood to be an indeterminant, the Q(t) or Q(x) refers to the field of rational functions in the variable (t or x) over Q. If a (or alpha...) is a complex number, then Q(a) is the smallest subfield of C containing a. If you read, say Q(r), you have to decide from the context whether r has a pre-defined meaning in a field or is to be viewed as a "new variable". Needless to say, you can refer in general to "Q(r) where r is a complex number" where r isn't a specific complex number that you've singled out yet.

This issue is subtle enough that I'd prefer to discuss it in office hours so that you can show me the passages in the book where constructions like Q(x_1, x_2, x_3) are used. I don't know exactly what you're referring to.

Fri Feb 27 11:46:34 2004 : 85 more errors in the book (some minor proofreading, others big). p. 94: the right-hand margin is messed up. p. 94 he refers to exercise 7.1. I don't think that's the one he meant. p. 95 is a perfect example of my confusion about notation. He says L=C(t_1, ..., t_n) It looks to me that t_1,...,t_n are fixed numbers, not variables. But then he means Q, not C. But he uses this notation consistently (extending C) which makes me think he means variables (as on p. 53). p. 98 The box at the end of the proof of 8.11 is on the wrong line. p. 98 should read PROOF OF THEOREM 8.10. p. 99 he writes S_n\A_n. Did he mean S_n/A_n? (don't know). p. 99 He means Lemma 8.11 (not 11). p. 100: "More precisely is not a complete sentence. Needs a comma, not semicolon. p. 100 he says "See Exercise 15.14." That's a really interesting exercise--see for yourself. p. 100, he says "prop. 8.9 shows..." I don't think he means 8.9 (but maybe he does).

Here's my favorite one: p. 101: "Since k is prime to p there exist integers q, l such that kp + ql = 1." That important algorithm is attributed to Euclid's lesser-know younger brother Billy. But in this case he needs Euclid's algorithm, not Billy's: there exist integers q, l s.t. kq + pl = 1. p. 103. He refers to exercise 8.6--I think he means a different one. --Greg

Fri Feb 27 16:14:36 2004 : A couple more: p. 109, he refers to Theorem 16. It's not obvious to me which one he meant. p. 111, the proof at the "end" of the proof of Th. 9.9 is after only the forward direction of the iff proof. The box should be on p. 112, at the end of the entire proof. Toward the bottom of p. 111, a block of text starts with "Furthermore..." What he is saying is either unnecessary or wrong, I'm not sure which. He already said K is a subset of K(theta_1). It looks to me he should not have mentioned K subset K(theta_1), but I'm afraid there is some other subset relation he meant to point out but didn't. p. 112 he says "substituting in (8.1). He means 9.1.

Sat Feb 28 18:43:05 2004 : I have a question on 6.16 (b). How are we gonna go about showing that J_beta is only finitely distinct for different betas? The only ways I could think of how to do this was to show that J_beta has only finitely many different set of bases as a vector space over K, or to show that there are finitely many minimal polynomials that are distinct for K(alpha1+beta*alpha2). Am I on the right track?

Reply: In this part of the problem, you're trying to show that L/K is simple if there are only finitely many M between L and K. You start by assuming that there are only finitely many M and want to prove that L is generated by a single beta over K. The first part of the argument is part (a), in which you show that L/K is a finite extension. It follows from this that there is a finite subset of L that generates L over K. (Take, for example, a vector space basis of L over K.) Thus L is generated by n elements, for some n. You want to prove that L is generated by a single element, and you do this by induction on n. The first step of the induction argument -- part b -- is to show that if L is generated by two elements, then it's generated by a single, well chosen, element a_1 + ba_2. (I'm writing in roman characters instead of in Greek.) The reason that only finitely many different fields J_b (or J_beta) can occur is because that is your hypothesis.

Sun Feb 29 23:42:15 2004 : probl #8.1. looks trivial to me. Hmm... i'm afraid if i actually missed sth here. And prob. #8.2, what does it mean by "elementary symmetric poly"?

Reply: I don't claim that 8.1 is very hard. Triviality is in the eye of the beholder. As far as the elementary symmetric polynomials go, they're defined in chapter 8. The elementary symmetric polynomials in three variables a, b, and c are: a+b+c, abc and ab + ac + bc. These are, up to sign, the coefficients of the polynomial (x-a)(x-b)(x-c).

A bonus problem for readers of this comment page: simplify the product with 26 terms (x-a)(x-b)(x-c)...(x-z).

Mon Mar 1 00:19:37 2004 : Professor, I have some questions on the true false for chapter 6. There is one that says "Every extension of R is finite". It is obvious that the extension C of R is finite, but what else is there? If we adjoin the indeterminate t, is that an infinite extension? Also, there is another that says "Every vector space is isomorphic to the vector space corresponding to some field extension". Im not sure exactly what this means, can you clarify? I know that two vector spaces are isomorphic if they have the same dimension, but I dont even know if this is relevant. Thanks!

Reply: I agree that R(t) is an infinite field extension of R. A finite extension of R is isomorphic either to R or to C.

Two vector spaces are isomorphic if and only if they have the same dimension. The "Every vector space..." sentence means (in my interpretation) the following: If K is a field and V is a vector space over K, then there is a field extension F:K such that F and V have equal dimensions as K-vector spaces.

Mon Mar 1 01:09:27 2004 : Hi Professor--another question about 6.16(b). I understand everything you said in your last post, and everything written in the book until the very last line (basically, the part we are actually supposed to show). I understand why there can only be finitely many J_b (only finitely many M, by assumption), but I don't understand how it then follows that one of those J_b must be L itself. Why couldn't all of the J_b be strict subfields of L?

Reply: Well, this is basically what you have to figure out. I'd try the following type of argument. Because K is infinite (by assumption), there are infinitely many b. Also, there are finitely many J_b. Thus there have to be two different b's for which the J_b are the same; say J_b=J_c. In this situation, you might be able to fool around and show that both generators a_1 and a_2 are in J_b, which will show that J_b is all of L=K(a_1,a_2). Give it a shot.

Mon Mar 1 12:52:57 2004 : Comment on: "A bonus problem for readers of this comment page: simplify the product with 26 terms (x-a)(x-b)(x-c)...(x-z)." That's funny. For a second, I had the robotic urge to multiply out.

Mon Mar 1 21:20:03 2004 : on 6.15, i've played your game of taking a_n=c+d(a_(n-1)), where c and d are in K(a_1,...,a_n-2). but it is very hard to prove that c and d are nonzero. can you give me some insight into this? i this seems harder than the answer you provided in the pdf that went along with 6.11 cuz we don't have much of a "rationality" thing going on here, in a concrete way. thanks a lot

Mon Mar 1 22:01:45 2004 : Can you give any hints on the part of 6.16d where we have to prove that m_M=m_M' implies M=M'. Thanx.

Reply: Presumably, the technique is to prove that M is generated by the coefficients of m_M. Let N be the field generated by these coefficients over K. Then N is inside M. If N is smaller than M, the tower law probably gets violated.

Mon Mar 1 22:16:57 2004 : concerning prob 8.2., aren't all the eq. given in the problem already symmetry? i.e. the first one: a^2 + b^2 + c^2

Reply: I believe that this was answered by a student comment below. As the student said, the point is to write the polynomials in the problem, e.g., a^2+b^2+c^2, as polynomials in the three symmetric expressions a+b+c, ab+bc+ac and abc. This illustrates the general theorem that symmetric expressions in n variables are polynomials in the elementary (or standard) such expressions.

Mon Mar 1 23:14:44 2004 : for 6.16(d), if M = Q, L = Q(sqrt(2)), we have: m_M(t)= x^2 - 2, and m(t)= x - sqrt(2).......so, what we actually have is: m divides m_M, not the other way around?????????

Mon Mar 1 23:50:37 2004 : on 8.5, i've stared and stared at it, but fleshing it out is really, really, messy. is there intuition i'm missing? ack... thanks

Reply: I don't know how to think about these identities intuitively. I was able to establish them on the board yesterday after a long struggle. One of the students who witnessed this was heard to mutter "This sucks!" That's what Serge Lang calls "grassroots feedback". Maybe a good way to do this problem is by induction on the degree of the polynomial. Suppose you pass from one degree to the next. You have to multiply the polynomial by some linear factor (t-b). When you do this, lambda_j changes in a transparent way: a term b^j gets added. How do the coefficients a_i of the polynomial change? The formula is simple and involves only two terms. This might be the start of an extremely efficient argument.

Tue Mar 2 05:44:43 2004 : i think my intuition about most of this homework has been horrid, but here are my best answers to some of the questions that have so far gone unanswered: 1. the equations in problem 8.2 all seem to be symmetric with respect to swapping alphas, betas and gammas, but they are not written in terms of elementary symmetric polynomials. in fact, it seems to me that you wouldn't be able to write them out in terms of elementary symmetric polynomials if they weren't symmetric in the first place. regardless, our job is to put them in terms of the three elementary symmetries. 2. for the example given with M=Q and L=Q(sqrt(2)), it seems as though you have your definitions mixed up. m is meant to be a polynomial over K=Q, so m(t) = t-sqrt(2) is invalid, since sqrt(2) is not in K. perhaps you read that m is supposed to be a polynomial over L.

Tue Mar 2 05:44:54 2004 : 3. as far as the game that involves a_n=c+d(a_n-1), i'm not sure what your attempted argument is but it seems that if you squared both sides you would get a_n^2=c^2 + (d(a_n-1))^2 +2cd(a_n-1), and by defintion everything but 2cd(a_n-1) should be in K(a_1,...,a_n-2), so either c or d must equal 0. other than that, the professors last comment for 6.16(b) is extremely useful, 8.5 is giving me a bastard of a time as well, and while 8.1 didn't seem altogether complicated, i found some difficulty in being completely rigorous about it, so i wouldn't say it's exactly trivial, though perhaps simple to some. p.s. bonus question to anyone really, really bored who reads this board (and to professor ribet as well): regarding the product (x-a)(x-b)...(x-z), what is the lowest power of x for which the coefficient (when fully expanded into a sum) contains a summand which is a word in the english language (when using the optimal permutation of letters)? somehow i doubt i worded that properly, but oh well.

Tue Mar 2 14:24:56 2004 : Hi prof, do you mind to post the sol for prbl. 6.15, 6.16 and 8.1? i think i get the answer but i want to see how you would actually solve the problem. Thanx.

Tue Mar 2 14:50:06 2004 : which problems are you going to grade this week? thanks

Tue Mar 2 22:22:14 2004 : The easy problems looked pretty challenging this week: I think you should grade those.

Tue Mar 2 22:49:01 2004 : This is Emily, the grader: I am going to grade problems 16 and 17 from Chapter 6, and 1 and 5 from Chapter 8.

Wed Mar 3 22:47:52 2004 : for Emily: So far your comments have been helpful. If you could not just criticize, but also make suggestions for writing style, then you will be even more helpful. However, I think you take off a bit too many points for mistakes, but that is just a minor complaint. Lastly, I appreciate you posting comments for the homework on your webpage.

Thu Mar 4 00:45:31 2004 : To Emily, our grader: I personally think you are doing an exceptional job. The webpage and it's content are very helpful. The homework comments, especially the phases pointing out the key ideas we skipped over, are also helpful.

Thu Mar 4 11:50:17 2004 : from Emily, the grader: I've changed my mind and I'm going to grade question 8.2 instead of 8.5. Hope you guys don't mind ;)

Thu Mar 4 14:57:53 2004 : hey emily: out of curiosity, how many people did 8.5 anywhere near correctly? could you perhaps post solutions on how you would have done it? thanks a lot. i also think you have done a great job correcting. one thing: could you post the average HW grade for each week? thanks :)

Fri Mar 5 22:51:04 2004 : To Emily: I think you've been doing a wonderful job grading. I disagree with the person above who said you took off too many points, to me it seems that you're pretty lenient about giving us credit in spite of dumb mistakes. I appreciate that. And you're certainly welcome to comment on style if there are things that should be said.

Sat Mar 6 19:11:36 2004 : Emily, I have been planning on telling you as well, I think you're the best grader for a math class I've ever had. I love all your comments. All of them are helpful, and more could only be better. :) Please keep up the good work. Also, please stop counting off so many points when _I_ make mistakes, but continue to grade everyone else in the same way. Thank you.

Sun Mar 7 08:10:20 2004 : This concerns problem 9.8 of the homework, I was just wondering if I could just say that since [Sig:K] is an integer such that 1<=[Sig:K]<=n if f is a polynomial of degree n over K, and so [Sig:K] divides n!. Is that a right direction or no?...Doesn't seem right as it didn't use the hint....

Reply: If you take an irreducible cubic over the rational field, then odds are that its splitting field has degree 6, rather than degree 3, over Q. For a specific example, you can take t^3 -t-1, which we've seen in class a few times. It has exactly one real root (and two non-real roots), so its splitting field must be bigger than Q(a), if a is the real root.

Sun Mar 7 16:55:52 2004 : From Emily, in response to the question about 8.5: Well, as I decided not to grade the problem, I really can't say how many people got it right and how many got it wrong. It's a tough problem, I agree. Anyhow, I don't have time to write up a solution at the moment, but perhaps someone in the class who thinks they have a good solution would like to post it here?

Sun Mar 7 20:42:37 2004 : In 9.8, are we specifying a field K? as in, if i'm running an induction on the degree of f, then the inductive hypothesis will work over both a field K and a field Q, right? just wondering. thanks

Reply: If you work by induction, you should probably phrase the statement to be proved in a way that has n coming first. You could say: "Let n be a positive integer. Suppose that f is a polynomial of degree n over a subfield K of the field of complex numbers. Then the degree over K of the splitting field of f is an integer that divides n!".

Mon Mar 8 22:01:23 2004 : Hi Prof--for 9.5(a) ("is Q(t):Q normal?"), instead of the way we did it in OH today, couldn't we simply use the definition of "algebraic" and "transcendental"? Since "t" is transcendental, it seems the result follows immediately. Thx.

Reply: Using definitions seems like a good idea, but you probably have to prove something. You need to know, for instance, whether there are a non-constant elements of Q(t) that satisfy non-zero polynomials with rational coefficients.

Tue Mar 9 01:12:32 2004 : I'm a little rusty on my arithmetic. What's negative infinity factorial?

Tue Mar 9 04:06:04 2004 : I'm a little lost. I believe Q(t) is the set of all rational expressions p/q, p, q are polynomials in the indeterminant t, with coefficients in t. Is this right? If not, what is it? And what would it mean for a member of Q(t) to split?

Reply: What is distracting about this problem is that the extension of Q is called Q(t). You would do better to think of it as Q(x), where x is an indeterminant. Suppose you want to prove that Q(x)/Q is normal. Then you have to show: Suppose that p(t) is an irreducible polynomial over Q with a root in Q(x). Then p(t) is a product of linear factors over the field Q(x). When you factor p(t) over Q(x), you have to think of p(t), which began life in Q[t], in the larger ring Q(x)[t].

Tue Mar 9 04:06:50 2004 : Sorry. With coefficients in Q.

Tue Mar 9 11:06:14 2004 : I believe that negative infinity factorial comes out to 5.3.

Tue Mar 9 11:15:08 2004 : So, when we're constructing splitting fields, we don't need to worry if the elements we are using to extend the field are multiple, right? If we have a irred. poly of degree 2 over field K, with roots a_1 and a_2, the splitting field is K(a_1, a_2)--we don't need to worry whether a_1 = a_2, right? It might be redundant, but K(a_1) = K(a_1,a_2)?

Reply: The short answer to your question is "yes". However, in the current context, all fields are subfields of C. As we've seen, an irreducible polynomial over a field K has distinct roots when it's split over C. Hence your example is not ideally chosen -- we could never have a_1 = a_2 in the situation that you described. On the other hand, we could have decided to construct the splitting field of (t-1)(t-1)(t-1)(t-1) over K! It's the field gotten by adjoining 1, 1, 1 and 1 to K, which turns out to be K itself.

Wed Mar 10 21:15:27 2004 : Hey professor, could you maybe write up and post 9.6? Thanks a lot

Reply: The problem is to show that quadratic extensions L/K are normal. If L=K(a) and m(t) is the minimal polynomial of a over K, then L is actually the splitting field of m. Indeed, if the roots of m are a and b, then a+b lies in K: it's the negative of the coefficient of t in m. Once L contains a, it contains b as well. Thus L is normal. Extensions of degree > 2 are not normal, in general. We've seen in class some examples of non-normal cubic extensions L/Q. We could take, for instance, L = Q(a), where a is the real cube root of 2. Then L is a subfield of the real field R. The extension L/Q is not normal because t^3-2 does not split over L -- in fact, it doesn't split over R.

Sat Mar 13 13:01:43 2004 : For the problem asking us to find the order of the galois extention of Q(root(5), root(3), root(7)):Q. Since Q(root(5), root(3), root(7)):Q is normal, the order of the galois group is the degree of the extension. The book asks us to argue that the degree of the extention is 8, by using the tower law and showing each subsequent extension has degree 2. However we have proved in two homework problems that for prime numbers pi Q(rootp1, rootp2, rootp3 . . .rootpn):Q has degree 2^n. Can we use the argument or do we need to argue as the book hints? Thank you.

Reply: It seems to me that you probably proved the statement in your previous homework exactly as the book now hints, i.e., by using the tower law and showing that each apparently quadratic layer has degree 2, and not 1. If you are satisfied with your proof on the previous homework and can say in all honesty that you have established the assertion in a previous assignment, then this is surely fine.

Sat Mar 13 13:03:47 2004 : Is there another way to find the galois group of an extension l:k beside looking at all the possible k monomorphisms and deciding which are k-automorphisms? Thank you

Reply: The answer to this question is surely "yes," but it would be helpful if you had a specific extension out on the table. I'm sure that there are lots of situations where you know that L:K is a Galois extension for some simple reason, perhaps because it's given as the splitting field of a polynomial over K, which is a field inside of C. In such situations, the issue isn't to figure out which K-monomorphisms L->C land in L: they all do. The problem is to decide what the Galois group Gal(L:K) actually is. Sometimes you can do this indirectly, but it's hard to say more without seeing an example.

Sat Mar 13 23:10:20 2004 : In one of the hw problems, we are asked to find the normal closure of the extension Q(sqrt2, sqrt3):Q. But isn't this already a normal extension, because it is the splitting field for the polynomial x^4-10x^2+1 right?

Reply: If L:K is normal, it's its own normal closure.

Sun Mar 14 15:37:23 2004 : On the bottom of page 122 for example 10.7.2, we are given 4 candidate Q-automorphisms for the Galois group. However I do not understand why these four automorphisms are the right ones, besides the identity of course. For example, why is p+rk+s(k^2)+t(k^3)+q(k^4) not an automorphism, where k=exp(2ipi/5)?

Reply: The "rk" term means that k^2 is sent to k under the automorphism. But k is (k^2)^3, so k has to be sent to k^3, k^3 has to be sent to k*k^3=k^4, and so on. Everything is determined by the image of k under an automorphism because everything in the field is a polynomial in k with rational coefficients. Automorphisms respect field operations and leave rational numbers alone.

Sun Mar 14 17:32:18 2004 : On page 92 of the book, the author gives 3 Q-automorphisms of Q(sqrt2,sqrt3,sqrt5). But shouldnt there be 8 automorphisms, since the extension is normal and finite, and the degree of the extension is 8? Am I missing something here? Because I believe there are 4 Q-automorphisms for Q(sqrt2, sqrt3) namely the id, f, g, amd fg where f is the map sqrt2-> -sqrt2 and g is the map sqrt3, -sqrt3. Is this wrong?

Reply: He didn't say that the three automorphisms are all there are. He goes on to explain that they generate a group of order 8.

Sun Mar 14 21:40:06 2004 : Professor, is there an easier way to find the normal closure of the extension of Q(a,b):Q (where a=the sqaure root of 2 and b=the cube root of 2) than finding all of the roots of the min. polynomial? Because we already know one of the roots, but it is messy, so it will lead to other messy roots I imagine! Also, I cannot figure out how to find the Galois group of this extension, I think it will have 4 elements, but since the extension is not normal I am not sure. Thanks.

Reply: I don't know how you're thinking about the problem, so it's hard to say what you should do differently. When you say "the min. polynomial," you make me guess that you are trying to view Q(a,b) as a simple extension, perhaps the one generated by a+b. You can think of the normal closure as the splitting field of (t^2-2)(t^3-2). Does this help?

Mon Mar 15 13:00:36 2004 : About the question about example 10.7, I saw it more simply than that. a(z) can be z, z^2, z^3, and z^4. (z=5th root of unity). So for a_2, for instance, a(z) = z^2. So, the map is a_2:p + qz + rz^2 + sz^3 + tz^4 = p + qz^2 + rz^4 + sz^6 + tz^8 = p + qz^2 + rz^4 + sz^1 + tz^3. (The point is that z^8, for instance, is the same as z^3 (mod 5). So those four maps are the only ones that show up by replacing a(z) with z^2, z^3, etc, and taking the powers mod 5.)

Mon Mar 15 13:17:29 2004 : Hi Professor--small question about the language of theorem 10.5. He says that G is a subgroup of the group of automorphisms of K. What is this "group of automorphisms"? What I mean is, I thought we always thought of automorphisms in terms of the field it fixes--"K-automorphisms". I understand why K-autos. of L form a group. It seems like "automorphisms of K" could be more than just the ones that fix a certain subfield. Can you clarify, please? Thanks.

Reply: An automorphism of K is a bijection K->K that respects the field structure of K. The set of automorphisms of K forms a group under composition. If K is a subfield of C (as is typically the case in our course so far), then the rational field Q is a subfield of K. You can see immediately that Q is fixed by all automorphisms of K because the automorphisms are required to take the number 1 to itself and are required to respect addition, subtraction and so on. Thus the group of automorphisms of K is alternatively the group of Q-automorphisms of K.

Mon Mar 15 21:41:49 2004 : on 11.3, are we finding the galois groups of the extensions given in a-d of 11.2, or of their normal extensions?

Reply: Even with my decades of teaching experience, it's hard for me to predict what you'll do here. :-) Reading the problems, I find 11.3 pretty ambiguous. However, because 11.4 is about finding the Galois groups of the normal extensions, I infer that 11.3 is about finding the Galois groups of the extensions themselves. Recall (p. 91) that the Galois group of an extension L:K is the group of automorphisms of L that are the identity on K. When L:K is a finite extension, the order of the group may be considerably smaller than the degree [L:K].

Tue Mar 16 14:52:51 2004 : To the person who made a comment to help on 10.7.2, thank you, this hint really helped me see what was going on, at least in this example.

Sat Mar 20 06:58:29 2004 : page 155, after lemma 15.5: "We can say more..." is clearly false as stated. For example, if we took K=C, the Galois group would be trivial. The reference to §21.6 is relevant only when K=Q.

Wed Mar 24 19:39:42 2004 : Hi for 12.3, I was trying to show that Q(r):Q is the splitting field for the minimal polynomial of r.(where r is gamma) But I got stuck in trying to show that sqrt(2-sqrt(2)) is in Q(r). Do I have use the regular way in which to show that it is a linear combination of the basis for Q(r) over Q?

Reply: I'm away from Berkeley and don't have a copy of the book. Thus I don't know what r is. If you want to show that something is in Q(r), all you have to do is write it as a polynomial in r. Does this help?

Wed Mar 24 23:55:41 2004 : For 12.6 e, what does it mean quotient of the Galois group of L:K? Does it mean it is the kernel of a group homomorphism? Or does it mean a quotient group? I'm not sure what it is...

Reply: If G is a group, the quotients of G are the groups G/N where N is a normal subgroup of G. If h:G->G' is a homomorphism, the image of h is isomorphic to the quotient G/N where N is the kernel of h.

Sat Mar 27 23:01:36 2004 : In 12.3 we're asked to show Q(r):Q is normal with cyclic Galois group G, where r = sqrt(2+sqrt(2)). Is there a theorem we can appeal to to show that G is cyclic, or do we explicitly need to identify the group and a generator of the group?

Reply: Presumably, the extension has degree 4. If it's normal, its Galois group is of order 4. All groups of order 4 are abelian. The non-cyclic ones are "Klein 4 groups": these are groups with 3 subgroups of order 2. If you know, for some reason, that your extension has only one quadratic sub-extension, then the extension has to be cyclic.

Sun Mar 28 22:11:01 2004 : On p. 133, we have the last line before the fundamental theorem of galois theory saying that M is a subset of M*` and H is a subset of H*`, where ` is a dagger in the book. But this doesn't make sense because * maps fields to groups and ` maps groups to fields. so, by notational convention, should we say that M is a subset of M*` or of M`*? i only ask because it's rather critical for a true/false question... (b and c on 12.6)

Reply: Just ignore the symbols at the end of that line on page 133. It's true that the two maps * and † reverse inclusions. This is explained correctly on page 93. In displayed equation (8.2) on page 93, you see what the author was trying to say on page 133, but this is not what it means for the maps to reverse inclusions. In the true/false, one of b and c makes sense and the other one doesn't. If you are presented with a nonsensical statement, you can reply "rubbish" instead of "true" or "false."

Tue Mar 30 00:41:14 2004 : Prof., is there an easier way to do prob. 13.2c and 13.3c? it involves some messy calculations, isn't it? do u mind to post the sol. to this prob after we turn in the hw? thanx.

Reply: I'm reminded of a comment by Woody Allen; I forget whether this is in a film or in one of his books. The question was posed as to whether or not sex is dirty. Allen's answer: "When it's done right"! I think that Galois theory is very much the opposite: if you're doing messy calculations, then you're perhaps not doing it right. I don't want to commit myself too far here because I haven't worked out this example yet. (Lots of students came to my office hour yesterday, but no one asked about t^4 - 3t^2 + 4.) I'd start by letting u = t^2, so that u^2-3u+4=0; the field generated over Q by one root of t^4 - 3t^2 + 4 is then a quadratic extension of Q(sqrt{-7}). My impression is that this quadratic extension is normal over Q; in other words, I believe that [K:Q]=4.

Tue Mar 30 11:23:10 2004 : Hi prof... i'm the one who asked prob. 13.2 and 13.3 above. last time i did some litle miscalculation. i fix it already and, yes, it's simple. But i don't quite understand your comment. u're saying Q(sqrt(-7)) is normal over Q and then u proceed saying that [K:Q] = 4. i suspect, here, u go one step further from Q(sqrt(-7)) to Q(sqrt(sqrt(-7))) and letting the last field be K. it's true that K is also normal over Q. But i don't see why we can use this argument because i had a hard time trying to show that t^4-3t^2+4 has at least one root lying in K. Thanx.

Tue Mar 30 20:27:24 2004 : I'm a little confused by question 14.2 on next week's HW, and wondering if perhaps it has a typo? The question says: "If n>= 5, prove Sn is not soluble using only the simplicity of A5." Should we use the simplicity of An, rather than A5, or am I missing something?

Reply: The point of the question is that knowing that A_5 is simple gives you enough ammunition to prove that S_n is not solvable for n > 4. The book proves that A_n is simple for n > 4, but you can get away with knowing less.

Fri Apr 2 07:47:50 2004 : On page 145, I think he means 1=H0 instead of 1+H0 in the series of normal subgroups for H.

Sun Apr 4 03:36:55 2004 : Comment on your comment: "I don't want to commit myself too far here because I haven't worked out this example yet." Based upon the number of typos we've discovered in the text so far, is it wise to assign us problems that you haven't attempted? You may very well be asking us to prove the impossible.

Reply: When I assign problems, I have a pretty good sense of what's involved in doing them. If there's a mistake in the problem, students are pretty quick to report the mistake. I think that there's a certain value in confronting exercises whose difficulty isn't known in advance, by the way. This is the opposite of the situation that you get in a calculus class, where you're sure that the exercises are similar to the examples that are worked out in the textbook.

Now, here are some comments on the next chapter: Chapter 15. As I mentioned before at some point, the author is off base after lemma 15.5 when he says that a certain Galois group is of order p-1. It might be trivial; this is the case, for example, when K=C. A new comment pertains to the next page, when the author says before Lemma 15.7, "Again we can say more; the Galois group is cyclic of order p...." This doesn't make any sense for various reasons. One reason is that there's no p in the context. Another reason is that the Galois group again could easily be trivial, for example if a=1 or K=C.

Sun Apr 4 11:55:44 2004 : I was watching TV this morning and saw an ad for some mutual fund company. At the end of the ad were the words: "No load+" then "No fees*" (where + was a dagger). I didn't understand it.

Mon Apr 5 19:31:12 2004 : In office hours we talked about finding the conjugacy class of A_5, and while we were looking at the conjugacy class of the 3-cycles we were looking at (123) and trying to define a function s=sigma such that s(1)=a s(2)=b s(3)=c so that we could get all the 3-cycles into one conjugacy class. However I cannot find an even permutation s such that s(123)s^(-1)=(124). In other words, I cannot find any such permutation s when two of the numbers in the cycle on the right hand side "line up" with two of the numbers in (123) i.e. cycles like (124), (153), (523). Even when I make the cycle (124) look like (241) or (412) I still cannot find an even permutation sigma. Does this make sense? Help!

Reply: You want an s that sends 1 to 1, 2 to 2 and 3 to 4. Then 4 can go to 5 or to 3 and 5 is forced to go to 3 or to 5. If 3 goes to 4 and 4 goes to 3, then 5 is fixed and s is a 2-cycle, which is odd. But if 3 goes to 4, 4 goes to 5 and 5 goes to 3, then s is (345), which is even. So take s=(345) and you're golden.

Tue Apr 6 15:21:57 2004 : hi prof, do you mind to post the answer for prob 13.6. thnx.

Reply: Which extension(s) in 13.1 are you interested in? You might want to look at my discussions of the "(c)" extension.

Tue Apr 6 21:39:20 2004 : Hi Professor, I am just curious, why are you using Lang's book for Math 250a next semester? It seems like students hate this book with a passion and many people complain about how poorly written it is.

Wed Apr 7 19:53:13 2004 : Hi professor I was looking at your second midterm from the last time you taught the class, which asks to give a separable extension that is not normal, and a normal extension which is not separable. I was wondering if you could give an example of each? The book does not do a very good job in differentiating between the two, and does not even give an example of a separable extension, so I do not think I have a very good understanding of the whole concept, can you help please? Thanks.

Reply: We have stuck, so far, to extensions inside C. They are all separable. We haven't seen any examples of non-separability.

Wed Apr 7 22:02:17 2004 : are we allowed to use a "cheat sheet" for this exam?

Reply: As I announced in class on Tuesday, the answer is in the affirmative.

Wed Apr 7 23:24:57 2004 : On page 139, the subgroup T should be {1,s^2,t,(s^2)t} not {1,s^2,t,s^2,t} where s=sigma and t=tau. On another note, the book says this is one of the subgroups for D8, whereas my 113 book says it is a subgroup for D4. So is this group D8 or D4? Finally, I was wondering if you could give an example of a Galois extension with degree 114?

Reply: As is said below, some people write D_n for the nth dihdreal group, while other people write D_{2n} for the same thing. I think that D_{2n} is the more usual notation.

Let p be the prime number 229. If K is the splitting field of t^p-1, K:Q is a Galois extension of degree p-1=228. The Galois group of this extension is an abelian group (actually, a cyclic group); the group contains an element of order 2, namely the complex conjugation automorphism c. Let L be the fixed field of {1,c}; this is called the real subfield of the field of 229th roots of unity. The extension L:Q has degree 114.

Thu Apr 8 00:23:27 2004 : D_8 is the same as D_4. The 4 refers to the fact it's a square. The 8 refers to the number of elements in the group. Stewart refers to the general dihedral as D_2n, which makes it more clear what he means. U also should not have that final comma.

Thu Apr 8 00:35:40 2004 : Hi professor on the same page 139 I do not know why the fixed field of the subgroup {1, s^2,st,(s^3)t} is Q(i(sqrt2)). I can find fixed fields when we are looking at subfields like {1,s^2} but when the subgroups involve more than one permutation or compositions of them I get lost and confused.

Thu Apr 8 02:01:42 2004 : It is obvious that S corr. to Q(i), because s, s^2, s^3 all leave i alone. T corr. to Q(sqrt(2))--remember that sqrt(2) is x^2 (x=xi--fourth root of 2). s^2: x-> -x, so (-x)^2 = x. t: x->x, and (s^2)t: x-> -x, so again (-x)^2=x. So x^2 is fixed in T, so the fixed field of T is Q(sqrt(2)). For U, again x^2-> x^2, and i->i, so ix^2->ix^2. st: x->ix, so x^2->(ix)^2=-(x^2). But st: i-> -i, so st: ix^2 -> (-i)(-(x^2))=ix^2. And finally, (s^3)t: x^2-> (-ix)^2= -(x^2), and (s^3)t: i-> -i, so (s^3)t: ix^2-> (-i)(-(x^2))=ix^2. Again, x^2=sqrt(2), so in every permutation, ix^2 is not changed, so Q(i*sqrt(2)) is the fixed field of U.

Thu Apr 8 10:10:36 2004 : It seems like every Galois group of order 4 we see is iso. to Z_2 X Z_2. Are any iso. to Z_4? Can you show us one, please? Thanks.

Reply: Take the splitting field of t^5-1 over Q. The Galois group of this field over Q is the group of invertible integers mod 5. That group is cyclic of order 4; (2 mod 5) is a generator.

Thu Apr 8 10:23:04 2004 : It would be very instructive to hear your thought processes in answering "c" in ch. 13. Even though I understand your write-up of the solution, it would take me 85 years to figure it out on my own.

Reply: I can't remember exactly what I did. I problably first tried to find relations among the roots. Once you know how the roots are related, you see that automorphisms are constrained.

Thu Apr 8 20:19:39 2004 : There is a typo for problem 15.1(b). It reads (sqrt(6)+2(5)^(1/3))/)^4. What should it be?

Reply: Good question!!! I'll see if this problem was in the previous edition; that might be a clue.

OK, it was in the previous edition. The fraction bar is extraneous; just remove it. The quantity should be the fourth power of (sqrt(6) + twice the cube root of 5).

Thu Apr 8 21:09:48 2004 : i thought this one was actually longer

Reply: I think that you're in the minority. This time, there were people who left early; that wasn't true last time. This time, there was less frantic writing at the end. This time, one student told me explicitly while handing in the exam that it was shorter.

Fri Apr 9 02:30:59 2004 : Is Z2 x Z2 isomorphic to the klein 4 group? also, in your test answers you say that sqrt(2) is in the field... how did you find that? or is it a typo?

Reply: Yes, Z/2Z x Z/2Z is the Klein four-group. In problem 2, we have in K a number whose square is 2 (namely a - 2/a), so the field contains a square root of 2.

Fri Apr 9 11:45:47 2004 : What is [R:C]?

Reply: Since [C:R]=2 and [C:R][R:C]=[C:C]=1, one could infer that [R:C] = 1/2. If you don't want to go there, consider [R:C] as undefined.

Sat Apr 10 12:50:26 2004 : hi prof, r u going to drop one or two lowest hw scores? thanx.

Reply: Sure. How about if we drop the lowest score and count the next-lowest score only 50%? Supppose that there are 14 homeworks, each worth 20 points. The maximum possible homework total will then be 250, instead of 280. Since homework is supposed to count 25% of the course grade, this sounds like a convenient situation. Of course, mathematicians can do computations with numbers that are not necessarily round, so we'll be able to figure out your grade no matter what.

Sat Apr 10 16:45:17 2004 : i think that's a good idea.

Sun Apr 11 15:53:20 2004 : what is enough to show that a function has exactly two nonreal roots? can we just say that the graph of the function shows that there are two nonreal roots? also, how should we treat 15.1b, where there is clearly a typo. thanks

Reply: You can find the number of real roots of a polynomial by the methods of Math 1A: figure out where the function is increasing and decreasing by taking the derivative and calculate some strategic values of the function. A graphing calculator may show only n-2 real roots, where n is the degree, but there might be two tiny hidden roots because of some quick oscillation near the x-axis or perhaps because the function doubles back to the x-axis when it's beyond the range of the graph.

Sun Apr 11 16:27:42 2004 : oops, yeah, disregard my 15.1 b question...

Sun Apr 11 18:56:20 2004 : Professor, for problem 15.5 I do not understand how the hint that says let u=t+1/t helps. In other words, I do not see how one would use this substitution to obtain a more manageable polynomial and solve it by radicals. Can you possibly elaborate on this hint? Also, for the last polynomial on problem 15.3, I am trying to show that the polynomial has two nonreal roots, but I am having trouble because I am not quite sure how to show that the polynomial has 5 real roots. By taking the derivative of the polynomial, we get a polynomial of degree 6, but I am not sure how to find the roots of this polynomial. I think there are 4 real roots, which would make the original polynomial have 5 real roots, but I can only see this using a graphing calculator. Help!

Reply: For 15.5, the point must be that a symmetric polynomial in t can be expressed as a polynomial in t + 1/t. The degree-6 polynomial in t that is in the problem can be written as a cubic in t + 1/t. If you solve the cubic, you'll see the possible values of t + 1/t. For each, you can find the possible t by the quadratic formula.

You should look at Sturm's theorem and Decartes Rule of Signs as possible methods for bounding the number of real roots of a polynomial. You can find out about them by searching on google. Look for example at http://www.math.niu.edu/~rusin/known-math/96/sturm.

Mon Apr 12 00:34:10 2004 : for the above comment, the way i solved it was basically seeing what (1+1/t)^3 is and so forth... and then getting some iff equation that was cubic and was symmetric. the substitution would then be pretty immediate and easy. this is also the first time where i found that cardano was actually illuminating in any way...

Mon Apr 12 15:54:24 2004 : In my copy of the book 15.1b reads (sqrt(6)+2*thirdroot(5)/)^4 Since there is nothing after the / and before the last ) it looks like this is a typo. What is the denominator supposed to be?

Reply: There is no denominator. Look up for comments that were posted on April 7 or 8.

Tue Apr 13 10:30:32 2004 : Am I missing something about 15.1, or is it just really easy?

Thu Apr 15 02:45:15 2004 : hi prof., i read some comments concerning the text u're going to use for math 250A next fall. it seems that the book is ranked "five-star" for its "too terse proofs". One commentator said that whenever the author says "it's obvious" it simply means we have to be ready to scribble 4-5 pages to see why it is "obvious". The book, many said, is much better treated as a reference book but not as a book for begining grad stud.

I looked at the readers' comments on amazon and didn't see anything this extreme. The book is clearly intended for students with a strong background in algebra. It's also better for students who have a support system available to them. In the case of our course, students have a professor who can supply examples and details in proofs, fellow students who are excellent, a GSI who will be holding office hours, and so on. The author will be visiting Berkeley during the semester and can be counted on to make guest appearences in class.

Thu Apr 15 20:46:25 2004 : why are you complaining so much? no one is compelling you to take 250a.

Thu Apr 15 22:50:09 2004 : I agree with the above person...

Thu Apr 15 23:02:14 2004 : I think I may be confusing notation. Is it correct to say that over some field K, a being expressible by radicals is the same as saying the extension K(a):K is radical? Thank you.

Reply: To say that a can be expressed in terms of radicals is to say that there is a radical extension of K that contains the field K(a) (or the element a: it's the same thing). However, it's my understanding that a subextension of a radical extension might not be a radical extension. Accordingly, it could conceivably be the case that K(a) is not itself a radical extension of K, even though it's contained in one. Coming up with a concrete example is something that you're supposed to do for homework, I believe.

Thu Apr 15 23:04:21 2004 : For the outside homework problem; it seems like finding the galois group of L:Q where L is the splitting field of a = 6 +sqrt(5) is relatively trivial. Is it possible you meant for L to be the splitting field for the minimal polynomial of sqrt(a). Thank you.

Reply: Thanks for the correction. You were absolutely right about my intention.

Fri Apr 16 13:38:49 2004 : Proof of Lemma 15.4--I understand everything up to the last line, but how does the fact that each alpha_i is radical imply that each beta_ij has to be radical as well? He doesn't even say what the radical sequence he is considering, but I'm guessing it's b_11, b_12,...,b_1r,b_21,...,b_rr?

Reply: Suppose that L is generated over K by some elements a_1,..a_r and that E is the normal closure of L over K. Let s_1,..,s_n be the elements of the Galois group of E over K, i.e., the set of automorphisms of L that are the identity on K. If you list s_1(a_1), s_1(a_2),...,s_1(a_r); s_2(a_1),...,s_2(a_r);...;...s_n(a_r), you will have a list of n*r elements. These elements are by no means asserted to be distinct. A lemma, which I urge you to check, is that these elements s_i(a_j) generate E over K. If the a_j now have the radical property (meaning that some positive power of a_j is in the field generated over K by a_1,...,a_{j-1}), then the list that I just wrote out also has the radical property. Indeed, for each i and j, s_i(a_j) is in the field generated by s_i(a_1),...,s_i(a_{j-1}) and thus, in particular, in the field generated by all elements to the left of s_i(a_j) on the list that I wrote out.

Sat Apr 17 13:45:23 2004 : Can anyone suggest a good starting point for 15.6 (Showing that a subfield of a radical extension need not be radical). I've tried a few examples with no luck, and can't think of any plan of attack. Thanks.

Reply: Suppose that you take a random irreducible cubic polynomial, something like t^3-t-1. Let a be a root and consider Q(a). Is Q(a) a radical extension of Q? Is it contained in a radical extension of Q? Consider Q(a) as a potential example. If it doesn't work, try something different!

Mon Apr 19 10:32:18 2004 : In Example 16.4.8, Stewart says that Z_6 is not a field because some elements (2,3,4) do not have inverses. Fine, but then he says that indeed 2*3 = 0 but 2,3 != 0. Also fine, but I don't see the connection. How does 2*3=0 illuminate the fact that some elements don't have inverses? Thanks.

Reply: Your question is answered at the top of page 166, where the author explains that the product of two non-zero elements of a field is again non-zero. If xy =0 but x and y are non-zero in a field, you can multiply by the inverses of x and y to get the false equation "1=0". You can paraphrase what the author is saying as the observation that fields are integral domains. Since the ring of intgers mod 6 is not an integral domain, it is not a field.

By the way, here's a trivial misprint: on line 1 of page 185, the constant coefficient of g is asserted to be tau^2. The constant coefficient is (-tau)^s.

Mon Apr 19 21:12:37 2004 : On 15.12b, we seem to have a definition of a radical extension as having only finitely many adjoined elements to a field. is that a typo? i.e. is Q with all the sqaure roots of primes adjoined to it a radical extension of Q?

Reply: The author's definition implies that radical extensions are finite. I'm not sure how to definite radical extensions of infinite degree. I don't believe that we need to consider them, however.

Mon Apr 19 22:32:00 2004 : I guess this is more like what I had in mind: suppose we have two extension K(a) and K(b) and we know that they are isomorphic. If a is radical over K, can we conclude that b is also radical over K? If I understand correctly, you're saying this is true because isomorphisms are supposed to preserve common properties, and so the radical property is preserved by the isomorphism. But is there a more direct way to prove it?

Reply: You need an isomorphism between K(a) and K(b) that takes a to be and that is the identity on K. If you have such a thing, then you have a dictionary that says that the two fields K(a) and K(b) are the "same" in such a way that a and b get identified. In this situation, a given power of a is in K if and only if the same power of b is in K. Note that we are in this situation if a and b are roots of the same irreducible polynomial over K.

Mon Apr 19 22:50:08 2004 : I, too, confused if there is any way to prove directly: if K(a) isom with K(b) and a is radical over K then b is radical over K.

Tue Apr 20 10:32:43 2004 : I think this is a clue to 15.6, but the extensions in a radical extension have to be ordered, right? For instance, if you have a = sqrt(1+sqrt(6)) as a root, you have to adjoin sqrt(6) before you adjoin a. Otherwise a^2 would not necessarily be an element of the base field.

Reply: The definition of "radical extension" requires that there be an (ordered) sequence of elements with certain properties. If you change the ordering, the new sequence is unlikely to have the right properties.

Tue Apr 20 11:23:00 2004 : If we have a number and we want to show that no power of it is rational, and we know its minimum polynomial, how could we go about doing that?

Sat Apr 24 21:17:02 2004 : Can we assume that the degree of a polynomial is finite? For example, is the product of (t-alpha) for all alpha a member of the natural numbers a polynomial, or more specifically a member of Q[t]?

Reply: The product that you are alluding to cannot possibly make sense. What would its contant term, be, for example? It would be the infinite product (-1)(-2)(-3)..., which doesn't converge. You can sometimes give a sense to an infinite product of polynomials as an infinite sum of terms a_it^i, where i ranges over the non-negative integers. Such a sum is called a formal power series. In order to say that an infinite product of polynomials is a formal power series, you need to be sure that each coefficient a_i can be computed as a finite arithmetical expression. In the specific example that you gave, the constant term a_0 cannot be computed in a meaningful way.

Sun Apr 25 19:53:23 2004 : What's wrong with my reasoning for 17.6? Suppose an isomorpmhism f:R[t]/ exists? We can define L=R[t]/ as {bt+a:a,b in R} equipped with usual addition, multiplication mod (t^2+1) Now f(-1) = f(i^2) = (f(i))^2 = -1 (b = 0, a = -1) So (f(i))^2 = (bt+a)^2 for some a,b in R = bt^2+2abt+a^2 = -1 mod (t^2+1) So a^2 = -1 => a = i => a not in R, a contradiction Therefore such an isomorphism does not exist. Therefore the fields are not isomorphic.

Reply: I'm just responding to what you have written, without opening a copy of Stewart. You say that the congruence bt^2+2abt+a^2 = -1 mod (t^2+1) implies that a^2=-1. But t^2 is -1 mod t^2+1, and the constant coefficient of t^2 is not -1; as you see, it's 0.

Sun Apr 25 20:18:54 2004 : 17.13 asks for us to find all irreducible quadratics over K and to construct all possible extensions of K by an element with quadratic minimal polynomial. So does this mean after we should find an extension for every minimal polynomial we have found? Also, Im not quite sure what those extensions should look like. For example, x^2+1 is irreducible over Z3, so would Z3(i):Z3 be the appropriate extension, or could we just say the extension is Z3(a):Z3, such that a satisfies the min. poly. x^2+1 over Z3? The books seems to be doing it the second way, but this way seems sort of pointless, am I missing something?

Reply: Again, I'll reply without looking at the book. The number i is a complex number, so it makes no sense to write "Z3(i)". If you want to make a splitting field for t^2+1 over K = Z/3Z, you would normally construct K[t]/(t^2+1). A generator for the extension K:Z/3Z is the element t mod t^2+1. Since this element has square = -1, you might want to call it i. It's an analogue of the complex number i, but it is not equal to the complex number i.

Mon Apr 26 22:50:12 2004 : On 17.10, are we supposed to show that for ANY field of characteristic 2 there exist quardratics whose splitting fields aren't just extensions by square roots of elements of the field, or do we just need to exhibit such a field, along with a sample polynomial?

Reply: You are supposed to a exhibit a quadratic extension E:F, where E and F are fields of characteristic 2, such that E cannot be written as F(sqrt(a)) with a in F.

Tue Apr 27 07:27:38 2004 : hi prof... do you mind to talk a little bit about constructing extensions field in the class? perhaps with example.

Reply: I think that we've done this several times. Please come to an office hour.

Wed Apr 28 10:29:38 2004 : What was the average on the last homework?

Wed Apr 28 11:12:08 2004 : Here's a good typo. On page 239, the author says "Consider the field extension Q(theta.zeta):Q, which turns out to be of degree 10...." The bottom field of the extension should be Q(theta), not Q.

Reply: Not only that, but on page 232, problem 20.5, there's a "GF" in the wrong font.

Thu Apr 29 11:43:19 2004 : Third line in the proof of Thm. 20.1, Stewart refers back to Thm. 16.1, which, I believe...doesn't even exist? And then, on page 184, the (p-r)! term in the binomial coefficient has to be in the denominator.

Reply: Theorem 6.1 on page 67 was apparently intended instead of Theorem 16.1. I agree, of course, about the binomial coefficient.

Fri Apr 30 15:53:17 2004 : Proof of Lemma 20.6 doesn't seem right. He's supposed to find an element g of G that has order e(G) but instead defines g to be the product of a_j's--which are integers...? It looks to me like he wants g to be the product of g_j^(a_j)'s. If not, then how does a_j^(mq)=1?

Fri Apr 30 16:14:49 2004 : Never mind, I'm wrong. I thought by a_j he meant the exponent. At least it makes sense now :)

Mon May 3 23:15:51 2004 : Professor, for problem 20.5, we must show that the Galois correspondence is a bijection. Is this the same as just showing that finite fields of the same size are isomorphic?

Reply: For an extension L:K, the Galois correspondence is the pair of maps * and † that connect up intermediate fields M between K and L with subgroups of the group Gal(L:K) (i.e., the group of automorphisms of L that are the identity on K). The problem is presumably asking you to show that * and † are 1-1 correspondences in the situation of the problem. (I don't have my book open right now but am pretty sure that this is it.)

Tue May 4 00:58:22 2004 : I took the request to show that the Galois correspondance is a bijection to be equivalent to showing that the extension is actually Galois.

Tue May 4 18:48:47 2004 : Nice lecture today.

Reply: It's the end of the semester. I must be slipping.

Tue May 4 23:09:22 2004 : I couldn't really figure out 20.11. Is there any way I can see the solution? It seemed interesting.

Tue May 4 23:26:21 2004 : Here's the solution for 20.11: For a finite field F, consider the set S of all the squares of the elements of S. what is the order of S? Well, consider the group homomorphism Phi: F* --> F* s.t. a --> (a^2). Note that F* is the multiplicative group of F. it is pretty easy to see that this is a group homomorphism. what is the order of Phi(F*)? well, we see that the order of the kernel of Phi is 2 (i.e. 1 and -1), so the order of Phi(F*) is half the order of F*. These are all the squares in F except for 0, so the order of S is half the order of F* plus 1, which is over half the order of F. Now, consider y which is NOT a sqaure in F. Consider S'=y+S. It is pretty apparents that the order of S' is the same as the order of S. But this mean that S' and S have a nontrivial intersection (since they both take up more than half of F), so y is a sum of squares in F, and we're done.

Tue May 4 23:28:16 2004 : Sorry, S'=y-S. and the order of S is 1 plus half the order of F*. note if F=Z2, then the whole problem is trivial, so let the order of F be ?3

Wed May 5 01:39:43 2004 : hi prof, can u post the sol. for prob. 20.11? thanx :)

Reply: The solution is pretty much covered by what students have posted. We have a finite field F and an element a of F and want to write a as x^2 + y^2 with x and y in F. If F has characteristic 2, then a is already a square, so we can write a = x^2 + 0^2 and go away happy. (In a field of characteristic 2, the squaring map is the Frobenius automorphism.) Assume, then, that F has q elements and that q is odd. A key point is that the set of squares in F has (q+1)/2 elements and that this number is more than q/2. As has been explained by student posters, the set of non-zero squares in F has (q-1)/2 elements. Indeed, the squaring map F^* -> F^* is a group homomorphism whose kernel {-1,+1} has order 2, so its image has order (q-1)/2 as claimed. The number (q+1)/2 is gotten by remembering that 0 is a square. Consider the set of squares in F, i.e., the set of all x^2. Consider the set of elements of F of the form a-y^2. These sets each have (q+1)/2 elements; they must overlap because (q+1)/2 + (q+1)/2 is bigger than q, the number of elements of F. Because they overlap, there is an x^2 that is also an a-y^2. The equation x^2 = a-y^2 amounts to the desired relation a = x^2 + y^2.

Wed May 5 03:01:40 2004 : hm... for the above one, this is true except where the field has characteristic 2. then, that homomorphism with have kernel equal to 1 (since 1=-1), so everything is a square.

Wed May 5 10:52:59 2004 : To the person above: that's why prof. argued for the case char=2 separately.

Wed May 5 13:42:10 2004 : i know, i was talking about the solution above the prof.'s....

Wed May 5 16:00:53 2004 : For part of this week's homework, we are supposed to show that d*N(d) can be represented an alternating series in q, using a formula that we derive. However, when I use the formula to calculate N(d) for compound numbers d, I don't get an alternating series. For example: 1*N(1) = q 2*N(2) = q^2 - N(1) = q^2 - q 3*N(3) = q^3 - N(1) = q^3 - q 6*N(6) = q^6 - 3*N(3) - 2*N(2) - N(1) = q^6 - (q^3-q) - (q^2-q) - q = q^6 -q^3 -q^2 + q This isn't strictly an alternating series, because the coefficients on the quadratic and cubic terms have the same sign (right?). I see similar patterns for other compound numbers like 10. Should we ignore the part directly referring to an alternating series, or have I misunderstood what is meant?

Reply: The word "alternating" was included only to underscore that some of the terms have minus signs. It was not intended to imply a strict alternation +/-/+/-/... Perhaps another word or phrase could have been used instead. The formula that follows shows exactly what's going on.

Thu May 6 19:18:35 2004 : Dear Ian Stewart--You may have noticed that lately there haven't been as many corrections to your "book" as there were in the beginning of the semester. Don't take that to mean there are less mistakes, I think you finally just exhausted all of us.

Thu May 6 21:41:50 2004 : That's kind of mean. And it might not be entirely his fault.

Reply: I'm inclined to delete the "Dear Ian Stewart..." entry and the "kind of mean" reply. I told the author about this page and will encourage him to read it carefully after the course is over. I think that our job is to keep track of changes that need to be made and to offer constructive criticism.

A general remark that I'll make is that I was unhappy with the author's decision to prove results only under an unnecessary hypothesis (namely, that all fields in sight are subfields of C) and to re-state the same results in a later chapter with the comment that the proofs given in earlier chapters go through without change. What he says is essentially true, but I don't think that readers encountering Galois theory for the first time are in a good position to verify the author's claims without more guidance. One could actually re-do all the proofs, but that would be boring and time-consuming. I think that it would have been better to state and prove the results abstractly while offering lots of examples involving finite fields and subfields of C.

Sat May 8 15:25:56 2004 : In this weeks hw. Is q supposed to be a prime or is F_q just an arbitrary finite field where q is of the form p^n, for some prime p?

Reply: The problem is pretty clear on this point (lucky for me). It starts off with a finite field and lets q be the number of elements in it. This number is not necessarily a prime number.

Sun May 9 13:35:40 2004 : Professor, in the first problem you've said the q is some number not necessarily prime, but is q equal to p^n? We know how to do this problem for the polynomial (t^q)-t, where q=p^n, but I do not understand how we would do it for t^(q^n)-t where q is not prime. Any hints?

Sun May 9 15:53:43 2004 : the number of elements in any finite field is a power of a prime; thus, q = p^k for some integer k and some prime p.

Mon May 10 23:47:30 2004 : Hi prof... this week hw is very interesting - its very stimulating. Can you write up a solution (just for the first two parts) later on? Thanx. :)

Mon May 10 23:52:48 2004 : Here are some corrections to the book. page 184: He says that (p/r) (p choose r) = (p!/r!)(p - r)!. It should be p!/((r!)(p - r)!). (In other words, (p-r)! should be in the denominator. The paragraph after Definition 17.15, he means example 17.12. On page 185: In the first line, I believe it should be tau^s, not tau^2. In the first equation, the u after the minus sign should be tau, I think. It should read v(u)^p - tau(w(u))^p = 0. page 228: The first line of Theorem 20.2, he means: "let q = p^n ...". Oh, while I'm here, back on page 137, the third off-set line from the bottom should read [K:Q] = 2 * 4 = 8. Instead of a multiplication symbol, there's a decimal point. On page 138, in figure 13.1 the lower right corner of the square should be labeled "-i*xi" (no negative sign in the book). page 144: Example 14.2.2, I believe the symmetric group S_3 is of degree 6, not 3? On page 183, In the multiplication table, he doesn't have the multiplication symbol in the upper left corner of the table. He does other places; I think he should have it there as well.

Reply: Thanks for these corrections. I hope that the author will be grateful to you for your efforts.

Tue May 11 00:09:12 2004 : Hi Prof. Can you help me with couple questions about the book? In Example 17.16 (p. 184) he says let K = K_0(u) where u is transcendental over K_0. K_0 is finite (Z_p). Is K finite? (My thinking is that it is infinite, because there are infinite unique powers of u in K (the def. of transcendental). But K_0 has characteristic p. Is K infinite, but still has characteristic p? If so, what's up with that?)

Reply: The field K is infinite, and you've given the reason. Even though students like to join together the concepts "finite field" and "characteristic p," there are plenty of infinite fields of characteristic p. You've got one in front of you! Another kind of example arises if you piece together all the finite fields of characteristic p. The "union" of these fields has an infinite number of elements. Each element is in GF(p^n) for some n.

Later in the example, he sets tau = v(u)/w(u), where v,w are elements of K_0[u]. I get that. Then he rearranges a little bit, and ends up with v(u)^p - tau(w(u))^p = 0. Where do the exponents p come from, and why? Then he says that the terms of highest degree cannot cancel. Could you help me by saying more explicitly what he is trying to do here? Which terms of highest degree, exactly, and why can't they cancel?

Reply: tau is a root of t^p-u=0. In other words, tau is a pth root of u; tau lies in a splitting field Sigma for t^p-u. The author is explaining that this polynomial is not a separable polynomial. It's definitely a pth power of something: it's (t-tau)^p. The author wants to show that it's irreducible as an element of K[t]. He supposes that it's reducible, i.e., that it factors, and investigates the situation. (We did this in class as well.) He ends up with the conclusion that tau actually lies in K, which means that it's a fraction v(u)/w(u), where v and w are polynomials with coefficients in K_0. Since tau is a pth root of u, this means that v(u)^p/w(u)^p = u. To see that this implausible equation is in fact impossible, you multiply by the denominator and examine the resulting identity of polynomials.

I'm also confused by the proof to Prop. 17.18 (p. 185) I know what the proof is trying to say, but there are so many convolutions and negations in the language I'm getting lost. Could you reword it more simply? Char.=0 I understand. We're showing that f is insep. iff all powers of t are div. by p. Now, f is insep. iff f and Df have a common factor >=1. The next sentence is where I get lost: "If so, then since f is irred. and Df has smaller degree than f, we must have Df = 0." So he's saying that f and Df have a common factor >=1, f is irred, and so Df = 0, but still Df has a common factor with f? Is that possible? In other words, if we had a poly: pt^3 + pt^2 + pt^1 + p, in field Z_p, does that poly both have degree 3 _and_ equal 0?

Reply: The polynomial Df has degree less than the degree of f; I would say that this is true even if Df=0, in which case I would say that the degree of Df is minus infinity. If f and Df have a common factor, then f has to divide Df. Since a non-zero multiple of f has degree at least as big as deg(f), we see that Df must be 0 in the case that f and Df have a common factor. The equation Df=0 is true if and only if f is a polynomial in t^p. The polynomial that you consider, pt^3 + pt^2 + pt^1 + p in field Z_p, is the 0 polynomial. Its degree is not 3; it's either minus infinity or is undefined, depending on your conventions. A polynomial can appear to have large degree but in fact have small degree -- consider the complex polynomial (4^2-17+1)t^2 + (6-2*3)t + 7, which has degree 0, for instance.

Thank you.

Tue May 11 00:35:01 2004 : One small verification, please. In a field F with q elements, the reason that x^q = x is that F\{0} is a cyclical multiplicative group with q-1 elements, and x^(q-1)=1. So multiplying both sides by x, x^q = x. Right? Thanks.

Reply: When x is non-zero, we have x^(q-1)=1, as you said. This follows from Lagrange's theorem because the multiplicative group has order q-1; it's not necessary to know that the group is cyclic. As you say, one gets x^q=x by multiplying by x. When x=0, the equation x^q=x is true as well. Thus it's true for all x.

Tue May 11 00:46:50 2004 : One more small proofreading error. On page 228, thm. 20.2 is an if-and-only-if. In the proof, he puts the proof-ending box after the forward direction. Instead there should be no double space before the word conversely, no box after "f over P," there should be a box instead after |K| = q, and there should be a double space before "Since splitting fields." (Chapman & Hall has saved an awful lot of money by having us proofread their proofs for them (wow, three uses of the word "proof" in one paragraph!). Normal publishers actually pay someone to do this; they have us paying them for the privilege.) I think now I'll drink some 86-proof Jack Daniels, and go to bed.

Reply: My guess as to what happened with this book is that passages from the older edition were re-typed for the new edition and that no one appreciated that this process would introduce new typographical errors. Well, at least the book was inexpensive: the majority of textbook publishers would have charged \$100 for a book of this length.

Tue May 11 10:17:36 2004 : Thank you for your help, professor. In light of what you wrote, one of my corrections was wrong. On page 185, the equation at the top of the page is right as it is. As for the textbook being cheap, I guess that's relative! This has nothing to do with this class, but I have a real problem with publishers charging us even \$50 for a paperback book that contains mostly repackaged 100-year-old information. I think they are taking advantage of students; they have us in a virtually monopolistic situation. What do you think of this idea: I'd like to start a website that archives math (and other) textbooks, and original texts (such as Gauss's Disquisitiones Arithmeticae or Euclid's Elements, and so on)that are past copyright expiration. Students could download them for free, or we would print a copy for a small fee. Here's the hard part. Students have to organize against the outrageous prices of textbooks, and insist their math (and other) departments use cheap texts. I'm pretty sure that might meet with some pretty stiff resistence from the academic community, but what an interesting fight it would be!

Reply: I agree with some of what you say but certainly not with everything that you say. Some comments that I don't agree with: (1) Your statement about "100-year-old information". Math textbooks get written periodically because students' interests and their background changes. Even if the theorems are 150 years old, the expository and pedagogical contribution of the author is recent. (2) Your phrase "virtually monopolistic situation". There are quite a few textbook companies, and different publishers drift in and out of trying to publish for upper-division mathematics students. There aren't enough students to give anyone a big profit, so publishers often lose interest in us. If you look at http://math.berkeley.edu/courses/text.shtml , you'll see a fair number of names in the publisher column. You won't see the name of our publisher at all, by the way.

There are already quite a few math books that can be found on line. If you type "on-line mathematics textbooks" into google, you'll get lots of hits. A recent trend is for publishers to allow authors to retain digital rights to their books. This means that readers will have the option of buying a copy of a book in the store or printing out the same book at home. (Readers can also read the book on their laptops without printing anything.) Although authors don't write math textbooks to get rich, they do put in huge amount of time and effort while they're writing. Authors tend to feel that they deserve modest royalties as partial compensation for their creations.

Tue May 11 17:39:31 2004 : Hi Professor--I agree with everything you said, and I know that some books are better than others, and the authors put a lot of time into them, and what they hope is exactly that theirs will be better than what came before it, and if we don't reward authors with adequate royalties, we might stagnate. But I actually think upper division math classes are a bad example of this problem--I'm thinking more of \$120 for a Calculus or Physics I book, or \$15 for a copy of a novel by Dickens, or even \$6 for a Shakespeare play. And, too often new versions come out _only_ to change the exercises to prevent students from buying used copies. Also, "monopoly" isn't the right word--I can't remember the word for when a group of corporations get together to fix prices, such as OPEC or the drug companies do.

Tue May 11 17:47:14 2004 : Hey Professor, Thanks for a great class.

Tue May 11 19:19:29 2004 : We owe you a class round of applause. From lecture to office hours, thank you for a wonderful class.

Tue May 11 20:22:32 2004 : Respected sir, Trisecting an angle, doubling a cube and squaring a circle are the three famous construction problems in geometry, which have been proved impossible by the new generations of mathamaticians with a straightedge(not marked) and compass alone , . I have made possible the construction problem of trisecting an angle with a straightedge(not marked) and compass alone. I am a student of high school and I have worked out for two months to find out a method for trisecting an angle with a compass and straightedge(not marked). It was only a coincidence that I came to know that the process of trisecting an angle with a compass and straightedge(not marked) is one of the impossible construction problems.I also have logical proof of trisectingan angle. .So sir,kindly suggest me the way through which I could introduce my method of trisecting an angle before mathematicians at international level. Yours faithfully, Manish Garg (e-mail:manish_garg88@rediffmail.com) (Phone No. :241313 Code No. 91-07771) (Postal address: Manish Garg C/O Shri V.D.Garg Mines Rescue Station B/13 Manendragarh, Distt.: Koria, State:Chhattisgarh(INDIA) P.I.N. : 497442)

Reply: You sent me e-mail and I replied on May 9 as follows:

Trisecting an arbitrary angle with a compass and straightedge is impossible. The impossibility is demonstrated in many books. For example, the textbook used in my current course on Galois theory proves this impossibility. If you go to http://math.berkeley.edu/~ribet/114/, which is my course web page, you will find a link to the publisher's page for the book.

Trisections have been proposed by many people over the years. A famous book entitled "A budget of trisections" discusses some of the proposed methods. This book was published by Springer-Verlag in 1987; the author is Underwood Dudley. I got this information from http://www41.homepage.villanova.edu/robert.styer/trisecting%20segment%20revised/references.html.

If you insist that you want to make your method known to mathematicians, I suggest that you post it to the USENET news group sci.math. If you post to this group, you will get feedback.

Wed May 12 12:31:58 2004 : this comments page is very nice because of all the juicy gossip on it.

Wed May 12 17:49:35 2004 : I agree! I'm not even in 114 and I like checking this page now and then because there's so much fun stuff going on.

Thu May 13 17:25:01 2004 : Professor, I think you've discovered the next Ramanujan!

Manish, don't listen to them. It's easy to trisect an angle using only compass and straight-edge (not marked). I figured this out a long time ago. Just fold the paper into three equal sections, which is clearly allowed under the rules and restrictions.

Squaring a circle is harder, because it requires folding the paper, then licking and ripping it.

--Greg

Seriously, the problem has to do with the "transcendence of pi." That's why folding and licking are required.

Thu May 13 17:25:49 2004 : I'm dying to know what the "Garg Mines Rescue Station" is.

Thu May 13 17:30:48 2004 : Manish, our book this semester addresses trisecting an angle explicitly. Here's what it says (page 81): "If you claim a trisection of a general angle using ruler and compasses [Stewart is anal about the '-es'--not sure he's right], ... the above proof shows that you are, therefore, claiming that 3 is a power of 2; in particular, since 3 != 1, you are claiming that 3 IS AN EVEN NUMBER.... Do you really want to go down in history as believing you have proved this?" So be careful before you post to sci.math!

Fri May 14 12:20:12 2004 : Hey, not to ruin the fun, but Id appreciate it if the people who are not in this class do not post about things that are not directly related to the class. I was under the impression that this was an anonymous comments page for MATH 114, not a free for all for random people with random questions.

Sat May 15 19:35:24 2004 : Hi Professor--I asked the other day about infinite fields with characteristic p. Any field, for instance, Z_p(u) with u transcendental over Z_p, has characteristic p but is infinite. Can you explain this more thoroughly? What I specifically don't understand is, u*p != 0, right? It seems to break the definition of characteristic to have an element that does in the field that does not =0 when multiplied by p.

Oh, I see that Fraleigh's definition of "characteristic of a ring" is different from Stewart's def. of characteristic of a field. Fraleigh's is the one I said above, Stewart's involves the prime subfield. What is the confusion? Thanks.

Reply: If F is a field of characteristic p, then F(u) is an infinite field. An infinite subset of F(u) is the sequence of powers of u: 1, u, u^2, u^3, and so on. The product of u with p is 0; it's not a non-zero element of the field.

I know of only one definition of "characteristic". If R is a ring with identity, there's a unique ring map Z->R that maps 1 to 1, 2 to 1+1, etc. The kernel of this map is of the form nZ where n is a non-negative integer. The integer n is unique; it's called the characteristic of R. If R is an integral domain, n is either 0 or a prime number p. If R is a field of characteristic p, then it contains a copy of Z/pZ as its prime field. If R is a field of characteristic 0, it contains a copy of Q as its prime field. The prime field of a field may be characterized as the smallest subfield of the given field.

Sat May 15 23:20:05 2004 : Professor Stewart-- Hi. I have a suggestion for future editions of your book. In this edition, you hardly mention symmetric polynomials and the relationship between roots and coefficients. I read about it in more detail in another book and it has really helped to illuminate for me what is going on with subgroups of the galois group, and their corresponding fixed fields, and why each fixed field is what it is for a given subgroup. Thanks.

Mon May 17 18:48:27 2004 : We didn't cover everything in the book.

Tue May 18 15:32:54 2004 : Yes, we also didn't cover some things that weren't in the book--such as symmetric polynomials. Stewart only mentions them in passing.

Reply: We covered some things that aren't in the book.

Wed May 19 11:40:10 2004 : I know, we talked about symmetric polynomials and roots on the last day. My comment was only a suggestion to Stewart for an improvement for his book. Speaking of which.....

Some more corrections: Chapter 16: On p. 170 he makes reference to exercise 16.2, p. 173 to ex. 16.3, and p. 174 to ex. 16.4. I cannot see a connection with any of those exercises and the topic under discussion. They should be checked (perhaps I'm missing something). On p. 173 he discusses the field of fractions of R[t]. This discussion would be much more helpful the first time he brings up rational expressions, on page 53. On page 252, the proof of Prop. 22.3 states "by theorem 4 and proposition 4 there exists..." Neither th. 4 nor prop. 4 exist.

Fri May 21 10:34:30 2004 : Professor, I agree with the earlier post - somehow between the end of the course and getting our homeworks back you were cheated out of a well-deserved round of applause. So, ClapClapClapClapClapClapClapClapClapClapClapClapClapClapClap Here's one student that learned a lot and thoroughly enjoyed this class.

Fri May 21 15:59:01 2004 : It was a very good test, with lots of fun, interesting problems, I think.

Reply: I expected to get roasted for a hard exam: I perceived that there was too much sighing going on in the room.

Fri May 21 18:34:04 2004 : I for one found it difficult. then again, i didn't get the first one, which is one i really should have gotten. i really loved the class. i'm interested in seeing the answers, especially to the last one.

Fri May 21 23:02:29 2004 : The final was not easy at all. I agree that the problems were interesting, but I can't enjoy doing them under the pressures of a final. But I guess that's just my attitude. A priori, no exam is fair, but Galois Theory is sexy so I guess that makes things even.

Sat May 22 12:12:13 2004 : I remembered to put my name on every page this time. I hope someone notices.

Sat May 22 22:36:30 2004 : I'm wondering if you're still going to let our final be 45%. i know that i did pretty well on the midterms and HW but tanked the final. i don't know how many of my classmates had a similar experience...

Reply: With my system, a final with low scores counts less than a final with high scores. I add in the numbers as if the scores were really well distributed between 0 and 45. If, in reality, everyone scores between 0 and 20, the result will be a final exam that counts not all that much more than a midterm.

Sat May 22 22:45:57 2004 : I did not.

Sun May 23 00:56:06 2004 : i guess... yeah... the final was pretty interesting. i was hoping that we can actually did those problems as hw prob. it would be very great. hahahha... i think i agree... there were too much pressure to do those problems in the final.

Sun May 23 10:18:36 2004 : yes, too much pressure. What we need is a finals system with no pressure. Where the professor says, listen, look at this test for a while, if you feel like answering a few of the questions go ahead, but you know, whatever. My scores would probably actually go up!

Mon May 24 14:13:53 2004 : On the solutions to the final (Problem 2b): 5^2=25 divides 100 so Eisenstein does not apply or am I missing something?

Reply: Thanks for the correction. I fixed this on the solution sheet (I hope).

Mon May 24 17:44:43 2004 : Are you going to post grades on this website with point breakdowns like you have done in the past?

Reply: That sounds like a reasonable idea. I expect to be able to do this some time Wednesday (May 26) and possibly even on Tuesday. The table will be ordered by total grade (a number from 0 to 100) and will show each student's MT scores, homework score (a number between 0 and 1), final exam score, total grade and letter grade. If there are students who signed up for the course on a pass-not pass basis, the letter grade will be converted appropriately. The total grade is the average of the midterm scores, plus the final exam score, plus 25 times the homework score.

I haven't finished grading completely as I write these words, but I can tell you that the final exam scores will be low. You have lots of company if you thought that the exam was hard.

Wed May 26 09:02:41 2004 : Here's something really interesting. This is from the manager of Customer Fulfillment Services at CRC Press: "The editing department has informed me that they do not edit for the accuracy of the author or editors mistakes only that the book is grammatically correct. This book [Galois Theory]was published as a camera ready title meaning the author did all the work we basically just had the book laid out and printed."

Wed May 26 11:34:57 2004 : So maybe he thinks that all those crazy things he said in the book are actually true.