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.

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.

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.

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.

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

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.

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.

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

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?

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.

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

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.

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.

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).

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.

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.

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.

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.

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.

Reply: Nothing like that, sorry.

Reply: Everything that we covered through last week.

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.)

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

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.

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 /~ribet/office_hours.txt, which is the "office hours" link on the course web page.

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.

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"

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.

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.

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.

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

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.

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.

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.

Reply: Noted.

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.

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.

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

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.

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.

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.

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....

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 /~ribet/114/hw5_ans.pdf for details on doing this problem.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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!".

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.

Reply: ??????

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].

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.

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.

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.

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.

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

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.

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.

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?

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.

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].

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?

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.

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.

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."

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.

Added March 31: I wrote up some comments on this problem.

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.

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.

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.

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

Reply: It's the best book.

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

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

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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

Reply: Who made the comments and where did you read them?

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.

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.

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

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.

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!

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.

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.

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.

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.

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.