Math 250A, anonymous comments



Tue Sep 7 00:38:41 2004 : Look, the first comment of a new semester!



Tue Sep 7 02:35:26 2004 : Hooray!



Tue Sep 7 08:39:05 2004 : Go faster in 250A.



Tue Sep 7 14:57:06 2004 : I would like to second the comment you repeated in class: I think it would be nice if the lectures went a little faster. Otherwise, they're wonderfully clear! It's very nice to be there.



Tue Sep 7 16:16:22 2004 : I'm fine with the pace. If it went a bit faster, I would not complain, but I think that for the most part, the pace of today's (9/7) lecture was the right pace.



Tue Sep 7 21:55:13 2004 : I agree that the pace could be faster, however, as I am sure is the case with most people, I have seen everything we have done so far in undergrad classes. I think what is covered about groups in undergrad algebra will run out soon and once we get through that people will have a better grasp of how much fater (if any) the class should go.



Wed Sep 8 10:24:01 2004 : I think the pace is manageable although I would prefer you go a little slower.



Wed Sep 8 13:37:33 2004 : I've found (what I hope is) an error in problem 12c. In the formula for constructing a semi-direct product, the book lists the first term as x1 ^ (Psi(h1)x2) where exponentiation is shorthand for conjugation as on p. 26. I messed with this formula for a while with no luck. I checked on the web for semi-direct product and several sites had the formula listed instead as: x1*(Psi(h1)x2) This works much better, so is this what we should actually be using?

Reply: I unassigned the problem and wrote up a solution for a corrected version of the problem. Thanks for catching this error.



Wed Sep 8 22:55:34 2004 : On Prob. 23, is it ok to use part (a) in solutions to (b) and/or (c), even though it isn't assigned?



Thu Sep 9 09:25:54 2004 : I'm not too concerned with the speed of the class, although it might be nice if it were little quicker. However I am worried about people other students attempting to ask "show off" sorts of questions. If someone has a question about a step of a proof or a definition or some such thing, that's fine. When someone wants to propose an alternate idea for a proof or something like that, I find it an inappropriate waste of the classes time. Usually people seem to be doing it so they can look smart or clever too - It isn't helping anyone learn.



Thu Sep 9 22:10:36 2004 :



Thu Sep 9 22:10:42 2004 :



Fri Sep 10 03:31:06 2004 : With the above comments... I agree :)



Fri Sep 10 13:36:44 2004 : To the comment immediately above this one: how far up the list do you mean?



Sun Sep 12 18:22:44 2004 : You tend to use #G to denote the order of a group G. Do you (or does Chu-Wee) mind if we use |G| instead?

Reply: As long as the meaning is clear, I don't think that I care what notation you use for the cardinality (size) of a set. The number sign "#" (now often referred to as the pound key!) is used as least as frequently as the vertical bars.



Mon Sep 13 00:23:49 2004 : Regarding Problem 23(a) -- it was proved in class, so you can probably use it.



Tue Sep 14 01:27:29 2004 : mmmm... Algebra



Tue Sep 14 07:12:10 2004 : Some of the homework problems seem to take much more time to figure out what the question is saying than it does to prove that they are true once you understand. Would it be possible to maybe get some problems that require more thinking and mathematical cleverness? (Though assignment 2 was better than assignment 1 in any case.)



Tue Sep 14 12:28:22 2004 : I don't agree with the comment preceeding this one. I think the statements of the problems are, for the most part, easy to comprehend, but some of the proofs require little explosions of trickyness. So maybe you should just go to office hours more often if you want the homework QUESTIONS "clarified".



Tue Sep 14 14:48:37 2004 : Is the average for each hw going to be computed?

Reply: Yes, I presume that it will be computed, and perhaps even posted to the class web page.

OK, check that. The averages will appear on Chu-Wee's homework page.



Wed Sep 15 10:25:13 2004 : I can usually understand the question without help. But often times it takes me 10 minutes to parse what the question is saying and then 5 minutes to actually solve it. It doesn't help that a lot of the notation in the book is pretty bad. For example, using Gs as the orbit of s and G_s as the stabilizer is ridiculous. Also for the double coset problem $H x H'$ looks a lot like $H \times H'$. (The double coset problem in particular seemed to be an excercise in definition masturbation anyway, which didn't help.) Some of the problems are quite interesting problems. a good example was I.20 - the problem was very simple but it took some playing with the material to work it out.



Wed Sep 15 13:45:24 2004 : it takes me longer than 15 minutes to solve a typical problem. am i the only one (hence screwed)?

Reply: It's unrealistic to think that you can look at a problem in a graduate course and solve the problem in some small number of minutes. When I was a beginning graduate student, I would start working on a problem set pretty much as soon as it was assigned. Some problems I got right away, but others seemed opaque to me for a long time. I'd walk around thinking about the hard problems and could usually do all but one or two problems by the time the problem set was due.



Wed Sep 15 15:56:01 2004 : To the person who feels screwed: don't worry, I think it takes a lot people more than 15 mins to go through the entire process of understanding the question, acquiring the background knowledge, and then actually solving the problem. Some problems are easier than others though. It seems like it takes me forever sometimes! Really, to that 15 min person, no one cares exactly how long it takes you to solve the problems, the purpose of this comments page is not to sound smarter than everyone else and make people feel bad indirectly.



Thu Sep 16 02:33:02 2004 : To the 15-minute man, what doesn't help is your own little "excercise" in masturbation. Keep it in your pants next time, bud.

Reply: I hope that we will address each other in a civil way. What we're trying to do here is learn algebra, not score points off each other.



Thu Sep 16 22:55:14 2004 : I apologize to anyone I may have offended with my comments above. I certainly don't claim to be able to solve all or even most of the problems in 15 minutes in any case. I was mostly just objecting to problems where it takes me twice as long to unconvolute the problems as it does to think up a solution once I understand what I am looking for. This is definitely not true of every problem. In any case, each homework so far, including the current one, has gotten more interesting. Perhaps it is just that with less material it is hard to ask interesting question without adding notation, and Lang seems to choose awkward notation a lot.



Fri Sep 17 00:07:28 2004 : I wish we would go more quickly in lecture. I think that many verifications could be left to the listener that are currently being done on the blackboard. I'll admit I've been wakeful anxious nights hoping "modules" will be up next. I hope they're coming soon?



Fri Sep 17 15:46:51 2004 : On problem 8 on the third homework, we are given a group G whose order is a prime power p^k. I see two equally plausible interpretations for the wording of the problem: 1) The prime p itself does not divide n 2) p may divide n, but p^k does not divide n Which was intended? -PH



Fri Sep 17 18:18:15 2004 : Well if the condition is only: p^r does not divide n, then there's a simple counterexample. Let S_n = S_6, and consider the subgroup G of order 4 generated by the elements (1,2) and (3,4)(5,6). Now 4 does not divide 6, yet none of the elements of {1,2,...,6} is fixed by all permutations in G. -Chu Wee



Fri Sep 17 18:20:57 2004 : By the way, I welcome questions by email, although I'd ask that you be specific when posing a question over email (since I often don't have the text when I'm at home). Questions pertaining to other aspects of algebra are also welcome. Enjoy the course. -Chu Wee



Sun Sep 19 13:11:48 2004 : Could you note somewhere in the homework summary, that the typo on problem #1 has been corrected, so that people know to look for a updated version. Or send out and email or something...



Sun Sep 19 18:51:26 2004 : In question 9, we assume G has three normal subgroups N_1, N_2, and N_3, and that G=(N_i)(N_j). Are we supposed to assume that N_i and N_j are not one of the three normal subgroups? I think Im just confused because we go from using numbers to letters for the normal subgroups.

Reply: The intention is to have i and j be integers in the set {1,2,3}.



Tue Sep 21 20:59:28 2004 : hey professor, i was just wondering where we should have read up to at this point in Lang. what will the exam next week cover? thanks

Reply: The exam covers everything discussed in class through September 23 and the material that has been treated in the problem sets. If I skipped over 10 pages in the book, I'm not going to ask you to a question that depends on your knowing the material that I skipped over. I hope that I won't, anyway.



Tue Sep 21 22:27:46 2004 : I was wondering if you could post the new homeworks _before_ Tuesday. This week's homework looks especially long and it would be nice to get started if i have time. And on that note, it would be very helpful if you could recommend what reading we should be doing along with your lectures. Looking at this week's homework, it seems a daunting task to do the 30 pages of background reading required, plus the 9 problems assigned all in one week. I wish I had been told what reading I needed to do earlier (or prompted by having the assignment up earlier).



Wed Sep 22 20:05:30 2004 : To prof Ribet, do you have any recommendation which book we should read that actually covers the things you lectured on since last Thursday? Or... probably somebody in the class have find any suitable book?

Reply: I was attempting to explain that direct sums and other common constructions in terms of representing functors. The concept is explained on pages 64-65 of our book. A functor is representable if it is isomorphic to a specific kind of functor. My way of saying things is not exactly the same way as Lang's, but it comes down to the same thing. I could certainly explain this in class--and I probably will.



Wed Sep 22 21:05:13 2004 : Yes... I agree I think it would be very nice if you (prof) include some notes what reading we should do to accompany the lecture, it doesn't need to be sth from Lang's, at least we have something to compare if we don't really get some parts of the lecture ---- just a suggestion :)

Reply: Thanks for the suggestion. While I'll try to keep it in mind, I can't promise that I'll always do this.



Thu Sep 23 16:29:55 2004 : It would be very nice to see more examples and applications of the basic algebra we are presented in a more advanced context.

Reply: I agree that it would be nice. I will supply some from time to time. The problem is that not everyone in the class is interested in the same sorts of applications. I believe that people taking this class are taking it for all sorts of reasons; not everyone is on the same page.



Fri Sep 24 14:13:19 2004 : Page 31, second illustration of Proposition 5.3...I think he means pi(tau)(delta), where it says tau(delta) - I'm sure the author of the book would agree that pi(tau)(delta) and tau(delta) don't represent the same thing.



Fri Sep 24 14:14:40 2004 : Ignore that last post (and then this one too!); they do represent the same thing.



Fri Sep 24 20:33:30 2004 : Is the exam going to cover the material in this last problem set? Thanks.

Reply: The exam definitely covers everything that we discussed in class through September 23. It covers material from the homework that was handed in on September 21. I presume that your question pertains to the assignment due October 5, which includes a great variety of problems. Some of these problems would have been reasonable exam problems; I'm thinking especially of the problems that use the Sylow theorems to unravel the structure of a group and the problems involving the symmetric and alternating groups.

When you consider the likelihood that a given problem or type of problem will appear on the exam, try to imagine the kind of exam that you would make up. I'm looking for questions that are representative of the material, that are not too hard or too easy, and that will elicit reasonable answers to which I'll be able to assign numerical grades. Sometimes I've screwed up and have given crazy exam questions, but this doesn't happen very often. (I don't think that it has happened very often, anyway.)



Sun Sep 26 23:04:45 2004 : Will the lowest 2 or three homework grades be dropped?

Reply: In my courses, I often drop the lowest score when I compute homework totals.



Mon Sep 27 12:04:54 2004 : Will we have a review for the midterm?

Reply: No, sorry. We're having one right now in office hours, though.



Wed Sep 29 02:02:53 2004 : Chu Wee here. Just a few comments here and there. Those who were in class today would have witnessed a rather heated exchange between a student and me. Now I'd like to make it clear, that I will never, EVER, entertain a reason like "I'm a graduate student, so it's all obvious" as a means to get an undeserved point (or a half-point in this case). The class is a heterogeneous one: there're undergrad students who've just finished taking Math 113; there're CS grad students who're encountering abstract algebra for the first time; and there're Math grad students. I've no means of distinguishing between them, nor do I intend to - simply because I don't believe in double standards in grading. Now what's blatantly, mind-numbingly obvious to a student, might be tricky for another. To make things worse, during office hours, we (Ken and I) often give hints to the students, sometimes liberally. And I've seen homework answers which basically rehash those hints, without any indication that the student's aware of what's going on. As you might have guessed, some of them try to pass it off as being "obvious". Furthermore I feel it's utterly disgraceful to react with such an outburst to a deduction of half a point. I don't, and will never, claim to be infallible, so discussions are always welcome. But to react with such indignation is totally uncalled for, especially in the midst of a class (albeit during the break). We're adults, yes, no? And would you not prefer civil discourse? One last thing, about the comments on the homework. They're there not because I need to exercise my wrists, not because I've gotten too many free pens from the last Thanksgiving sale, and certainly not because I feel an incessant need to display my meagre understanding of the English language. They're there to indicate where and why I've taken points off. So please, RTFC.



Wed Sep 29 07:48:52 2004 : Chu Wee: I think it was really awful the way you were treated in class yesterday. I don't know where people get the idea they can treat graders like shit. I was wishing I had stepped in to say "f*** off." I hope you realized you had support from other students who realized the irrationality of the situation.



Wed Sep 29 09:42:08 2004 : Could you post a solution for problem 14(b) from HW #2? thanks

Reply: Done.



Wed Sep 29 21:19:11 2004 : Chu-Wee, I agree whole-heartedly with the above post. I think you are an extremely instructive GSI, and am saddened by the unpleasantness of the witnessed episode.



Fri Oct 1 03:43:34 2004 : I think you should let your moustache grow out again.



Fri Oct 1 03:44:04 2004 : Addendum: you look like Hugh Woodin with your moustache.



Sun Oct 3 13:46:00 2004 : I think he looks much more inspired and wily with a mustache than Woodin does. Which means I definitely support growing one.



Sun Oct 3 20:42:16 2004 : the problem with the mustache is that it makes the nose look disproportionately large. Other than that it makes Ribet look like someones uncle in a bad 80's movie- I like it



Mon Oct 4 17:21:50 2004 : From problem 46: Does anyone know what it means for H to be a maximal subgroup? Does that mean H=G?

Reply: It means that H is smaller than G (which it has to be in the situation of the problem) and that there are no subgroups of G that contains H, other than H and G.



Mon Oct 4 22:56:36 2004 : In problem 50(b), are both f and g assumed to be surjective? It doesn't seem to work out otherwise.

Reply: It works out, but you may be trying to prove the wrong thing. (You have to figure out what the author means by a pull-back.) When you construct the fibre product, you have a square with four maps: the two original maps and the two projections from the product that you constructed. What's intended is that if one of the two original maps is surjective, then so is the projection map that is opposite from it in the square.



Tue Oct 5 00:45:11 2004 : Um, at this point, what would the "linear combination" on the grades look like?

Reply: If you're talking about the midterm, then the answer is that the grades range from 3 (one student) to 30 (two students). The median grade was 17. There are many grades between 12 and 19. Only 8 out of 31 students got over 20.



Tue Oct 5 10:09:42 2004 : I think the linear combination question was asking how the various exam and homework scores would be weighted relative to each other.

Reply: Something like: midterms, 15% each; final 40%; homework 30%.



Tue Oct 5 11:35:17 2004 : I don't know if it's up for debate: but what about: midterms 20% each, final 40%, and homework 20%??

Reply: It's not up for debate; on the other hand, I appreciate your input. It seems to me that homework in Math 250A is an integral part of the course. Your assignments being graded carefully by an experienced GSI. Many students believe that exams do not represent their abilities and accomplishments as well as their homework papers.



Tue Oct 5 20:11:04 2004 : Would it be possible if the HW is weighted a bit higher than the total of two midterms? I like the HW and learn a lot from it... but screw up in the midterms... Feels like my brain wasn't working at the midterm time



Wed Oct 6 06:39:59 2004 : 15-15-40-30 Sounds pretty close to right. On the one hand, homework should be weighted enough that people definitely do it. On the other hand it shouldn't be so heavily weighted that missing any given problem is a cause for concern. Also since it seemed to come up on the comments for the previous iteration of the course, I'm pretty strongly opposed to takehome tests for the simple reason that one feels pressured to spend an obscene number of hours on them. Anyway, so far the class has been very well taught, and even though I've supposedly seen most topics, the level of depth and sophistication is well beyond what I've experienced before. The last homework was a bit long, but I think that's mostly because it carried over a lot of problems from the previous week.



Thu Oct 7 21:08:06 2004 : I'm not sure if the grading discussion is still on the table, but here's my two cents: I would much, much prefer take-home exams, especially a take-home final exam, just because the in-class testing environment adds a whole extra dimension of difficulty that has almost nothing to do with one's understanding of the material.



Fri Oct 8 10:31:54 2004 : Yesterday's lecture was great! I love this class.



Fri Oct 8 15:09:19 2004 : as far as grades are concerned, I am of the belief that everyone has bad days, so having a large portion of the grade based on one test, like the final, can result in people's grades dropping drasticly despite having done everything else in the class quite, well. However, it is important to have in-class testing mainlt because it tests people's ability to solve problems quickly and independently, which is an important skill for any mathematician, so test should be part of the grade. Homework, on the other hand, tends to deal with deeper results that would be unreasonable to ask students to deal with in only 80 (or 180) minutes, so homework must be a signiicant part of the grade also. my suggestion is this- compute the grade something like 50% homework and 50% either the two midterms or the final, which ever is higher. Then if someone's "brain is not working" on any given day when we happen to have a test it will not crush their grade. Also this should alleviate some of the pressure that people feel while taking the tests, so in general they should do better.



Fri Oct 8 15:19:42 2004 : I propose that a weighted average of the proposed weighted averages be used when calculating grades.



Fri Oct 8 15:37:22 2004 : we should pick our grades out of a hat, anyone who can pull out a rabbit gets an "A"



Fri Oct 8 17:21:43 2004 : I was just about to write a ridiculously long comment on the grade breakdown myself. But what's the point? Everybody should just worry about being able to do problems and getting the material in one's head. Ribet seems to have been flexible in previous classes. So why wouldn't he continue to be flexible?



Fri Oct 8 20:48:01 2004 : I would also very much prefer take-home exams. I slapped my forehead several times on my way home last Thursday, realizing a simple way to do one of the problems I got wrong and another part of a problem I couldn't figure out during the exam because I became nervous. A take home would be a much fairer demonstration of a student's ability.



Fri Oct 8 21:02:52 2004 : If a result is quoted in the book, but the proof is left to the reader, do we then need to prove the result to use it in the homework or can we quote it.

Reply: This is a judgement call, depending on the exact circumstances. I would think that you should prove the result in most cases.



Sat Oct 9 16:38:34 2004 : I would not prefer a take home final because it would probably end up consuming several days during what may already be a fairly hectic exam period.



Sat Oct 9 19:12:43 2004 : I too prefer a timed final.



Sun Oct 10 22:51:08 2004 : The grading in Math 250A (Fall 2001) seemed to work pretty well. I like that one. In that case, I think we don't need to worry so much about either a take- home or in-class final.



Mon Oct 11 13:20:15 2004 : Yes (another person)



Mon Oct 11 21:04:06 2004 : How do I use the "gothic" font in LaTeX?



Tue Oct 12 00:43:59 2004 : You can use $\mathfrak{Texte}.$



Tue Oct 12 19:41:58 2004 : Any idea when we'll be getting Assignment 4 back?

Reply: Either on Thursday or next Tuesday, according to Chu-Wee.



Wed Oct 13 14:51:01 2004 : Lang doesn't really define exactly what a Dedekind ring is. Are they assumed to have any properties besides the group nature of the fractional ideals? In particular, the definition of a Dedekind Ring on Mathworld states that the ring is commutative. Can we assume this, or does it require proof via the group property?



Wed Oct 13 16:50:35 2004 : One of the conditions of a Dedekind ring is that it's a subring of some field K (see page 88 - I think, the edition I have right now is the old one). Since fields are commutative, so is a Dedekind ring.



Thu Oct 14 14:31:22 2004 : Two mathematical questions: Do all Noetherian domains have to be UFDs? (If not, what's a good counterexample?)

Reply: Any ring that's a finitely generated Z-module is automatically Noetherian. In particular, the ring A=Z[\sqrt{-5}] that we considered in class is Noetherian. It's not a UFD because irreducible elements in UFDs are automatically prime: this follows the definition of "UFD" and is something that I forgot to mention in class today. As you know, the element 3 of A is irredubible but not prime.

I've heard that it's possible to have a Rng (Ring without Identity) which has no maximal ideals. How do you do that?

Reply: Beats me.



Thu Oct 14 16:12:50 2004 : For an example of a Rng without maximal ideals, see http://www.math.bilgi.edu.tr/courses/Algebra%20Questions/Ali/03-04/2003Algebra_I_Final.pdf



Thu Oct 14 16:52:16 2004 : Here is an example of a ring (without 1) which does not have any maximal ideals. Let A be an abelian group, and make it into a ring by setting ab = 0 for all a,b in A. Thus, A is a ring without 1. Thus, any ideal in A is just an additive subgroup of A. So, if A does not have maximal subgroups, then it does not have any maximal ideals. For a concrete example, take A = Q (rational numbers under addition). One can check that A has no maximal subgroups, hence it has no maximal ideals.



Fri Oct 15 17:41:45 2004 : Even though Lang doesn't prove it, can we use the fact that a Dedekind ring is Noetherian?

Reply: Only after you've done Problem 13.



Sat Oct 16 18:16:22 2004 : I cannot parse the definition of the Dedekind ring on page 88. What does it mean that "fractinal ideals form a group under multiplication". A fractional ideal is an additive subgroup of a field. How does one mulitply additive subgroups?

Reply: If A and B are fractional ideals, AB consists of finite sums of products a_ib_i with the a's in A and the b's in B. This multiplication clearly has an identity element: the full ring. The essence of the definition is that for every A there is a B so that AB = the ring.

Most people would not say that the field K is a Dedekind ring. For me, and for most people, a Dedekind ring has to be a proper subring of its quotient field. Thus, e.g., the ring of integers is a Dedekind ring, but its fraction field is not.

In the exercises, it may well be true that "ideal" is occasionally used to mean "fractional ideal". Note that fractional ideals are non-zero, and it's Lang's intention to refer mainly to non-zero ideals. Thus his prime ideals are typically non-zero prime ideals of the Dedekind ring. (One refers to them as "primes.") They turn out to be maximal ideals.



Sat Oct 16 18:19:19 2004 : The definition of the product of two ideals is given in the middle of page 87. "Let a, b be ideals of A. We define ab to be ..." A question that I have: in the exercises, when the author uses "ideal", and not "fractional ideal", does he still mean fractional ideal?



Sat Oct 16 18:23:41 2004 : Oops, I misunderstood. Disregard my question.



Sat Oct 16 20:31:32 2004 : I don't really understand problem # 13, when in the hint, it says, given an ideal A, does it mean an ideal A in field K? But, a field doesn't have a proper non-trivial ideal...???? Anybody help...

Reply: The ideal a is a genuine old (non-zero) ideal in the ring. One wants to prove that it is finitely generated. This will make the ring into a Noetherian ring, one of those objects that we discussed in class on October 14. To prove that a is finitely generated, one has to bring in b, which is an inverse to a in the group of fractional ideals. Unless a is the full ring, b will be (only) a fractional ideal and not an ideal of the ring.

For what it's worth, we can infer from Problem 13 that all fractional ideals of the ring are finitely generated modules over the ring. Indeed, if I is a fractional ideal, there is a non-zero x so that xI is a regular ideal. Take generators of xI and multiply these generators by 1/x. You get generators of I.



Sat Oct 16 21:06:12 2004 : In problem # 13, ideals = fractional ideals, or just "ideal"?



Sat Oct 16 23:53:03 2004 : When the author says "ideal", let's assume he means ideal (since he uses the term "fractional ideal" on some other occasions). -CW



Sun Oct 17 15:13:26 2004 : Is a fractional ideal a subset of the dedekind ring? i.e. is an elt of the fractional ideal of the form a/b, or just a?

Reply: A fractional ideal "of R" (where R is a Dedekind ring sitting inside its quotient field K) is a subset of K of the form xI, where I is a non-zero ideal of R and x is a non-zero element of K. If you write x=n/d, then you can say that the fraction ideal is (1/d)J, where J is the ideal nI of R. So fractional ideals are just ordinary non-zero ideals that are allowed to have denominators. Since they are related so simply to regular old ideals, every statement about fractional ideals can be translated into a statement about ordinary ideals. However, allowing denominators turns out to be a convenient invention.

For perspective, consider the case where R=Z, K=Q. The non-zero ideals of R are in obvious correspondence with the positive integers. The fractional ideals are in correspondence with the positive rational numbers. The group of fractional ideals is same thing as the group of positive rational numbers under multplication.



Sun Oct 17 15:23:38 2004 : Doesn't a dedekind ring have to be an integral domain?

Reply: Yes, because it's sitting inside a field.



Sun Oct 17 19:47:14 2004 : Haha... I think we better postponed the HW. There're lots of confusion about it. I read Hungerford's book, and it seems to me that he used different terminology for Dedekind's Ring (or Domain?). Both Hungerford and Dummit&Foote explains Dedekind's Ring in terms of modules and... phew... I think that's a lot to take.



Sun Oct 17 20:07:56 2004 : It's not advisable to refer to other books for this exercise - doing so would only confuse you further. For one thing, different books have different definitions of Dedekind domains. E.g. the standard definition is that R is a noetherian integrally-closed domain, such that every nonzero prime ideal is maximal, but Lang uses the group structure of the fractional ideals as a definition. Everything you need to know is covered in Chapter 2. -CW

Added comment: The standard definition and Lang's definition are equivalent. In the standard treatment, the Dedekind ring is assumed to have the three properties mentioned by CW and then is shown to have the property of Lang's definition. In Lang's treatment, he starts with a very simple hypothesis; one then shows incrementally that you can deduce the standard hypotheses as consequences. One thing that we are not doing in this homework is to show that a Dedekind domain in the sense of Lang is integrally closed in its fraction field K. This would be a good question for a later assignment.



Sun Oct 17 20:48:09 2004 : Hello Prof. Will you post solutions to homework 5?

Reply: I spent a lot of time this weekend answering students' questions and writing up solutions to assignment 6. I'm not eager to write up solutions to number 5. Is there a volunteer?



Mon Oct 18 18:30:43 2004 : Solutions to HW5 are coming up. By the end of this week. -CW



Mon Oct 18 18:44:16 2004 : By the way here's a tip. Try doing 17a first. In my opinion, that's a very useful result. -CW



Mon Oct 18 18:58:31 2004 : I was wondering if you could give some intuition on number 19. for instance, I'm having a problem seeing what c will do to A. is it enough to show that for some a in A and b in B that ca+b=1? Because if so, i don't see why we can't let b=0 and c=a^-1....



Mon Oct 18 20:11:23 2004 : Please ignore the whiny people! This assignment got me to learn a lot, even if it did have a bit of a baptism by fire approach.



Mon Oct 18 20:26:39 2004 : Suppose A,B are ideals of a Dedekind ring D. Do we define "A and B are relatively prime" iff gcd(A,B)=D? Or is it more complicated?

Reply: Sounds correct.



Tue Oct 19 11:09:25 2004 : Perhaps in the next edition of the book, the author might consider reordering the exercises as follows: 13,17,18,14,15,16,19.



Tue Oct 19 11:56:17 2004 : I think the homeworks are very instructive. I only wish I had more time to devote to them. For grad. students who are trying to balance preparation for logic prelim.'s and a 50-75% gsi-ship, it can be a bit much, however. And it would be nice if the exams were wighted more or if there were more leniency with late homeowrk. Don't mean to whine; I am not necessarily suggesting a change, just "thinking aloud" as I consider dropping the course this time around and retaking it some other semester.



Wed Oct 20 00:34:43 2004 : I find it amusing the Chu-Wee called the problem set we got back today the "algebra problem set from hell" or something like that. Thanks for grading it CW!



Wed Oct 20 10:42:15 2004 : Well if problem set 4 is "algebra problems from hell"...what's problem set 6? "algebra problems from the ninth ring of hell"?...sorry bad humor

Reply: Sorry if HW #6 was hard and/or long. I assigned a series of interesting, interconnected problems. I hope that assignment #7 will be less work and more fun. Problem set 4 was long, but it was really a double problem set.



Wed Oct 20 13:00:47 2004 : I hope nobody got offended. The term "algebra problems from hell" was only my bad attempt at humour. -CW



Wed Oct 20 16:41:56 2004 : It could be that I haven't been approaching the homework problems from the right point of view, but it seems to me we haven't had a chance to use any of the category stuff for anything. Is there any chance you could write some interconnected problems to introduce us gently to using this machinery?

Reply: I've been shying away from talking further about categories because I sensed some resistance to them. Plenty of representable functors have come and gone without my pointing them out explicitly. OK, maybe it's time for round two.



Wed Oct 20 17:56:18 2004 : Yes, I agree with the previous comment. I still don't really understand what Lang means by "the universal property".

Reply: The phrase "universal property" usually means that there's a representable functor around.



Wed Oct 20 21:49:09 2004 : What does it mean for one fractional ideal to be an integral ideal over another? Does it just mean that each element in the first satisfies a monic polynomial with coefficients in the second?

Reply: One can talk about ring extensions that are integral; we can, for example, discuss the possibility that S is integral over R when R is a subring of S. I've never heard of this concept being extended to the situation where R and S are replaced by fractional ideals.

Recall, though, that if I is a fractional ideal of R, we sometimes say that I is an integral ideal if it is contained in R. For example, the fractional ideals of the ring of integers are in 1-1 correspondence with the positive rational numbers. Those fractional ideals that are integral ideals correspond to positive rational numbers that happen to be ordinary integers.



Sat Oct 23 16:34:18 2004 : isn't an object O of a category universal if there is only one element in Mor(A,O) or Mor(A,O) for any other A in the category?



Sat Oct 23 20:25:30 2004 : can somebody tell me what an "integral ideal" is? There doesn't seem to be a definition in the book or on Mathworld.



Sat Oct 23 20:25:57 2004 : oops sorry disregard the above post



Sun Oct 24 11:27:21 2004 : In problem #2 is "gothic o" a Dedekind ring like in #1?

Reply: Yes, #2 is a continuation of #1.



Sun Oct 24 23:08:39 2004 : Any advice on part III part a?



Mon Oct 25 01:17:58 2004 : Advice for the previous commenter: Noetherian induction. Now -- any advice for part b?



Tue Oct 26 14:28:43 2004 : Do we have HW this week?

Reply: Yes.



Tue Oct 26 14:37:01 2004 : Is there homework this week?



Tue Oct 26 17:20:54 2004 : I really wish there was not homework this week... i'd even take a double homework due in 2 weeks like last time.



Tue Oct 26 17:26:09 2004 : Hi. Just a note on the homework. It's not necessary to prove every small detail, but you must show me you're aware of what's going on. For instance, in a recent homework, we have a ring homomorphism f:A->A', and some of you said that because I is an ideal, f(I) is also an ideal. This is not true in general, but in the case of this question, it is - because f turns out to be surjective. All you need to do is write something like "because f is surjective, the image f(I) of an ideal I is also an ideal". -CW



Tue Oct 26 21:32:45 2004 : CW- are you going to compute the hw average for the hw we got back today? I feel like we are starting to take for granted that you do all these wonderful kinds of things for us, but if you could continue to do them it would be much appreciated =)



Thu Oct 28 14:35:41 2004 : Nice lecture, Chu-Wee. It was good to see a lot of the categorical ideas we've been talking about in action.



Thu Oct 28 17:01:41 2004 : Chu-Wee have you decided which problems will be graded on the hw we turned in on Tuesday?



Thu Oct 28 19:13:43 2004 : The one you just turned in? Not yet. I'll grade two problems, and Problem 8 (the extra problem in Lang's book) is definitely one of them. By the way, since your midterm is next week, I promise I'll return both your homeworks next Tuesday. -CW



Fri Oct 29 00:01:27 2004 : will the solutions to HW 5 be posted before the midterm, as Chu-Wee promised?

Reply: They were posted on October 31. Happy Halloween, all!



Sat Oct 30 07:19:32 2004 : CW, maybe it's just diasterous coincidence, but it seems like whenever there is a single problem I can't get, therefore don't turn in, it ends up being one of the problems graded. Is this just coincidence? or do you usually try to grade the hardest problem?



Sat Oct 30 20:58:11 2004 : Didn't we show in class that a finitely generated, torsion-free module is free. Is there some reason this doesn't apply to Ch III prob 11? It seems like this should be immediate.



Sun Oct 31 00:23:42 2004 : No, a torsion-free, finitely-generated module over a principal ideal domain, must be free. But for general rings, there are lots of examples of torsion-free modules which are not free.

Comment: Can you cite a specific example of a finitely generated torsion free module over an entire ring that is not a free module?



Sun Oct 31 01:12:50 2004 : With reference to 3 posts above, I'm truly sorry if my grading scheme caused some of you to be placed lower in the class than what you think you deserve. But bear in mind the following. (1) I always decide my grading scheme before looking at any of the answers. (2) The scores should be reasonably well-distributed. (3) I'm running on a tight schedule here, having to grade 30 homework scripts everyweek, and many of you don't even bother to write coherent solutions. (4) Other than HW7, I don't think there's another homework with one question significantly harder than all others. So you can be reasonably assured that it's nothing personal if the only question that you skipped got graded. Statistically, that'll happen to everybody, and it'll happen to some people many times. I don't know why it had to be you, though it's certainly not my intention. You can probably blame it on the alignment of the stars. -CW



Sun Oct 31 01:43:11 2004 : CW- although I do believe it was unfair to grade or even assign the last problem on HW 7 because it was a question that appears in ch. 7 of the book, I think you're otherwise a great and awesome GSI and grader! Note that I am not the person who commented above about your grading.

Comment by Ribet: I'm amazed that you think that a problem is unfair just because it comes from a chapter whose number is greater than III. The problem is perfectly relevant to the exercises from lower-numbered chapters about Dedekind rings. It uses mostly techniques from the earlier chapters. The one extra concept is that of a ring being integrally closed in its fraction field, which we already saw from a different point of view in the previous week's homework. One does need to know that an element is integral if multiplication by it preserves a non-zero finitely generated module, but this is explained very clearly in the opening pages of Chapter VII.

I could go further afield here and add that one purpose of this course is to give students enough background in algebra that they can analyze and solve algebraic problems when they arise "in nature". When presented with a "real" problem that you want to solve, you may not know in advance what chapters of Lang's book you'll need to consult.



Sun Oct 31 12:12:50 2004 : In problem 9(b), the author does not define the maps from S^-1 M' to S^-1 M, and from S^-1 M to S^-1 M''. I see only one obvious map from S^-1 M' to S^-1 M (map m'/s to f(m')/s where f is the map from M' to M), but I feel like there should be a formal notion that captures the intuitive notion that there is "only one obvious map". What is it?



Sun Oct 31 12:25:31 2004 : RE: Sun Oct 31 12:12:50 2004 - It has to be that map, because it's the only one that will make the appropriate diagram commute.



Sun Oct 31 22:13:41 2004 : What will the midterm be covering?

Reply: Everything through Chu-Wee's lecture last Thursday. Although the exam is cumulative, it will emphasize things not covered on the last exam. In practice, this means that the exam will be about rings and modules but not much about groups (except for abelian groups, which are Z-modules).



Wed Nov 3 10:10:37 2004 : It's really hard for me to study for the midterm knowing that Bush will be our president four more years



Wed Nov 3 13:49:30 2004 : For me, studying for the midterm will be a form of escapism. After I'm done taking the midterm, I guess I'll start thinking about politics again so as to avoid thinking about the grade I'm gonna get.



Wed Nov 3 17:02:35 2004 : I'm trying to find an example of an integral domain R, and a finitely generated R-module M, such that M is projective but not free. Obviously R cannot be principal. I keep getting stuck because I have a hard time imagining what say a Z[x]-module or k[x,y]-module of this form would look like.



Wed Nov 3 17:49:59 2004 : Solution to my question from a few minutes ago: Taking the ideal <1+sqrt(10),2> in the ring Z[sqrt(10)] as a module works. Z[sqrt(10)] is Dedekind. By a bunch of straightforward but fiddly arguments the ideal is nonprincipal, and then it's nonfree. But it is projective because we are in a Dedekind domain.



Wed Nov 3 22:38:25 2004 : Could someone tell me how we represent a p-adic integer? What are the operations like? Is the inverse limit module some kind of direct product with new operations?



Fri Nov 5 11:34:48 2004 : Since nobody has replied to the above question, I'll just repeat what I said in the lecture and add a bit more stuffs. You can think of a p-adic integer as an expression in base p. Positive integers have a finite expression in base p, but p-adic integers extend infinitely to the left. For example, if p=5, the p-adic integer (....22223) = 1/2. This is because if you multiply (....22223) * 2, you get 1. Likewise sqrt(2) exists in the ring of 7-adic integers. The interesting thing about p-adic integers is that you can actually perform analysis on it. For example, to compute sqrt(2) mod 7^4, you can take the Taylor series for sqrt(1+x): 2.sqrt(2) = (1+7)^(1/2) = 1 + (1/2).7 + (1/2)(-1/2)/2.7^2 + (1/2)(-1/2)(-3/2)/6.7^3 + O(7^4). This gives sqrt(2) = 235 mod 7^4. -CW



Sat Nov 6 17:53:12 2004 : How do we multiply elements in the group K_0(o) (Exercise 13(c))?

Reply: Start with the monoid of projective modules: if you have two of them, you can add them by taking their direct sum. (Maybe this answers your question already.) You can make a group out of the situation by taking the free abelian group on the set of isomorphism classes of projective modules and then mod'ing out by relations of the form [x+y]-[x]-[y]; here, [x] is the element of the free abelian group corresponding to the generator x, and + is the direct sum. You make further relations (i.e., make the group smaller) by saying that [x+f] and [y+f] are to be equal if f is free and the two modules x+f and y+f are isomorphic. In the book, look around page 40 and then around page 139.



Sun Nov 7 23:58:36 2004 : In 12b, do we need to show that there is an extension to a K-linear map, or simply assume it exists? Also, does anyone know if 12a (from homework *last* week) is useful in solving 12b or 12c?

Reply: If you're in this course to learn algebra, then you shouldn't consider writing something down in your homework solution without understanding why it's true. Once you know why there's a K-linear map as in the situation of the problem, you can ask yourself whether your solution should include a justification for its existence. My take on this question is that you should say what the map is and why it's defined.



Mon Nov 8 19:39:53 2004 : In question 1 of the additional problems you say: find a beta in O that belongs to P^e_i but not to P^(e_i+1) for i=1,...t. I think this is only necessary for i when e_i does not equal 0. (otherwise beta is the principal generator of A from problem 2)



Mon Nov 8 19:41:37 2004 : How do you put the randomized picture in your webpage? it's awesome!

Reply: It's done with javascript. Type something like "display random image using javascript" into google, and you'll get to a tutorial.



Mon Nov 8 21:00:32 2004 : Um, typo in #18b? Should "Let M be a p-divisible group" be "Let A be a p-divisible group"? Also, what does p-divisible mean?

Reply: There's an obvious typo; T_p(A) should be T_p(M) -- something like that. An abelian group is p-divisible if every element of it has a pth root. This means that the map a |-> pa is surjective.



Tue Nov 9 15:22:10 2004 : chu-wee - the "quick tips" on your 250A page are great!



Tue Nov 9 21:25:57 2004 : What does the distribution of sums of midterm scores look like?

Reply: It looks like the distribution of scores for the first midterm. Check out the histogram.



Tue Nov 9 23:32:02 2004 : when will the homework for next week be posted?

Reply: By 10:46 on Wednesday.



Wed Nov 10 01:51:12 2004 : Chu-Wee, I think your answer key is really awesome!



Wed Nov 10 07:53:11 2004 : I think my message wasn't clear. I was curious as to the distribution over the students who took both midterms of (score of midterm1 + score on midterm2). If people tended to get exactly the same score twice, this would tend to be quite wide, ranging from 8 all the way up to 57. If on the other hand there was no correlation between score, it would be a lot lower standard deviation. I suspect the actual behavior is neither of these extremes though.

Reply: OK, now I get it. Here are the midterm scores. Each of the 29 lines that follow represents one student in the class. The first column represents the score on MT1; the second column represents the score on MT2. It is clear that the two columns are correlated only loosely. Some students did better on one exam than on the other. Those of you who are statistically inclined can analyze the data in a formal way:

3	8
5	17
9	6
11	5
12	11
12	18
12	7
12	8
12	8
14	19
14	21
14	7
15	5
16	17
17	10
17	21
18	11
18	21
19	13
19	23
19	26
19	27
21	15
22	25
22	26
24	17
26	17
30	23
30	25
According to Excel, the correlation coefficient between the two midterms is 0.58.



Wed Nov 10 13:58:10 2004 : chu-we , could you post the hw averages and tips on your website for the last few homeworks? thanks!



Wed Nov 10 19:28:14 2004 : is there any hope for us "bottom dwellers"?

Reply: Of course, there's always hope in the sense that you might be doing better on the HW and might do better on the final than you have done on the two midterms. But let's suppose that you really do come out at the bottom of the course at the end of the semester. What then?

I recommend that you look at the grades that I've given out in past semesters, just so that you see what kind of grader I am. Here are some relevent Web pages: /~ribet/113/, /~ribet/110/f02/, /~ribet/114/grades.html, /~ribet/110/f03/grades.html. When I assign grades to students who have the lowest scores in the class, I try to figure out whether they have really been working. Have they been handing in homework? Have they been coming to class? Have they made an effort to grapple with the material, or have they totally lost contact with the subject?



Fri Nov 12 13:29:50 2004 : Due to popular demand, the average homework scores are back. -CW



Fri Nov 12 22:06:54 2004 : yay to chu-wee! thanks!



Sat Nov 13 00:28:35 2004 : In Problem 7 ought we to assume that q is prime? I think it's safe to say q=p?



Sat Nov 13 12:52:18 2004 : In general the number of elements in a finite field is a power of a prime; thus, q = p^n for some prime p and some positive integer n. But I think we don't need to use this fact in this problem.

Reply: Yes, q is the number of elements of k. It's true that q is some power of a prime number p. The problem could have been phrased "Let k be a finite field. Let q be the number of elements of k...." Note that p, in this problem, is supposed to be the characteristic of k.



Sun Nov 14 11:54:52 2004 : I think some of the homework problems this time around can be done in 15 minutes.



Sun Nov 14 23:30:56 2004 : On 18b, I just have no idea where to start on it. I've looked at it when considering its derivatives and if it might have some combinatorial meaning (X choose r and so on), but so far am completely stumped. Any help?



Mon Nov 15 00:04:42 2004 : I predict the two problems graded on this assignment will be 7 and 18. (If anyone else is interested, we should totally start taking bets.)



Mon Nov 15 00:43:01 2004 : Well, if we were going to make bets, and Chu-Wee decided to participate, he would probably win



Mon Nov 15 02:48:56 2004 : Maybe Chu-Wee would bet wrong in order to throw people off and get people to do the wrong problems. Solutions that don't exist are much easier to grade than ones that do, after all.



Mon Nov 15 11:19:22 2004 : Hmm, so it all boils down to the size of the bets, and the difficulty of the questions. People who're familiar with game theory can probably formulate this in quantitative terms. -CW



Tue Nov 16 00:06:45 2004 : To Chu-Wee, your homepage is amazing!



Tue Nov 16 00:27:27 2004 : The correct name for the font invoked by $\mathfrak{text}$ is ``fraktur''(hence \mathfrak), a German type-face from early printing presses. Someone referred to it as gothic in one f the comments above. Not to be pedantic or anything.... just some trivia.



Tue Nov 16 03:52:13 2004 : Thanks. I said "gothic" because that's what Ribet said in class -- "these gothic-looking letters".

Reply: The fonts are commonly called "gothic" as well as "fraktur." If you look at the last lines of the file amssym.def, which can used to implement these fonts in plain TeX, you'll see the definition \let\goth\frak .



Tue Nov 16 07:57:10 2004 : In question 7b (showing that the polynomial of n variables over k, a finite field, has 0(modp) zeros), are the monomials in the polynomial of the form x^i, where x and i are vectors? Or can the monomials include (or not include) any of the variables x_1,...x_2? I mean, is it really a polynomial of n variables, or a polynomial of a vector x?

Reply: In this problem, a single letter x (or X?) is used as a shorthand for (x_1,...,x_n). We are dealing with a polynomial in n variables. For example, if n=2 and we call the variables x and y instead of x_1 and x_2, then the monomials are expressions x^iy^j.



Tue Nov 16 07:58:03 2004 : obviously in the above comment I meant x_n instead of x_2...



Tue Nov 16 18:09:40 2004 : we seem not to have covered a lot of chapter 4. should we be studying independently on the resultant, mason's thm and abc conjecture, power series, etc? Thanks. It does seem that the pace of the class has really picked up lately...

Reply: I want to discuss chapters V and VI before the class ends. Math 250B probably won't cover any Galois theory, so I want to talk about Galois theory in 250A at least to some extent.



Tue Nov 16 18:16:56 2004 : For 7b, here's another solution. Consider a term of f(x). Since the degree of f is less than n, we can't have a term with all the variables x_1,...,x_n being multiplied together. So for every term in f(x), there is at least some x_j missing. Consider an arbitrary term ai(x_1^i_1)...(x_r^i_r). Suppose it equals c. How many other times can this term equal c as we allow the variables x_1,...,x_n to range over K? Since x_j is not in the term, the number of n-tuples (k_1,....,k_n) such that a_i(x_1^i_1)...(x_r^i_r)=c is a multiple of the order of the field, as x_j can take on any value in the field. So the sum of f(x) over k^(n) is = 0 mod p. Note that this solution doesn't use any of Lang's notational craziness.



Tue Nov 16 20:02:30 2004 : Chu-Wee, the dog running at the top of your 250A page is so cute!



Wed Nov 17 14:54:23 2004 : Hi, Chu-Wee here. I can't find my copy of Lang's algebra. Did anyone take it out of my office by mistake? If so, please return it - it's borrowed from the department and I'll have to pay if it's lost. Thanks.



Thu Nov 18 01:44:57 2004 : Post more pics from office hours! :)

Reply: Let's get more large groups of students in office hours.



Thu Nov 18 17:27:39 2004 : today's lecture was fantastic!!



Mon Nov 22 23:03:13 2004 : Woo hoo thanks for deciding to not assign extra problems, happy thanksgiving!



Tue Nov 23 00:16:26 2004 : Woohoo! Happy Thanksgiving!



Tue Nov 23 22:25:02 2004 : Woo hoo! Woo hoo! Woo hoo! Everybody, lets shout it loud...one....two.... three...WOO HOO!!!!!! Happy thanxgiving!!!!



Tue Nov 23 23:27:42 2004 : Come on, it's just turkey. It's not like it's sex or something. That is... unless... you have some very strange Thanksgiving rituals...



Wed Nov 24 13:20:44 2004 : People say way weirder stuff on this comments page than they did on the Galois theory one...



Thu Nov 25 08:59:00 2004 : Well, Galois, being French, never got to appreciate a good Thanksgiving turkey[suffix deleted].



Thu Nov 25 12:32:23 2004 : Problem 9 on this week's HW seems trivial. Am I missing something, or can we just quote a result from the book?



Thu Nov 25 12:56:07 2004 : Never mind. I misread the problem.



Thu Nov 25 18:10:41 2004 : Hey my comment got edited!

Reply: This is not the NBA. Please keep it civil and clean. Thanks.



Thu Nov 25 22:47:20 2004 : gobble... gobble...



Tue Nov 30 17:47:22 2004 : Could the HW perhaps be due next thursday? It seems to be a pretty legnthy assignment...

Reply: This is OK with Chu-Wee, so it's OK with me. You'll all have to come to my office to pick up your graded assignments, though.



Wed Dec 1 22:37:24 2004 : I can't believe the semester is almost over!

Reply: There's always next semester. I'm teaching the undergraduate linear algebra class in 10 Evans and a Math 191 on cryptography.



Thu Dec 2 05:06:55 2004 : Are we going to do any infinite Galois theory in this class?

Reply: We already have, in the sense that we've proved theorems about Galois extensions that are not necessarily algebraic. The main point that we have not discussed is the issue of identifying those subgroups of a Galois group Gal(K/k) that occur as Galois groups Gal(K/E), where E is a subfield of K that contains k. The subgroups in question are the ones that are closed subgroups (i.e., subgroups that are also closed sets) of Gal(K/k); in this statement, Gal(K/k) is given a standard topology that it inherits as a closed subset of a product of finite groups. (The finite groups are given the discrete topology.) Namely, for each E in K that is a finite Galois extension of k, there's a natural restriction map Gal(K/k)->Gal(E/k). We can take the product of these maps, where the index set is the set of all E; this gives us a map from Gal(K/k) to the product of the finite groups Gal(E/k). This map is injective because an automorphism of K that is the identity on every E is visibly the identity.

I would like to be able to describe the closed subgroups of Gal(K/k) without making any explicit introduction of the topology. If I figure out how to do this, I will tell you about it next week.



Fri Dec 3 12:15:55 2004 : What are the chances that we can drop one HW assignment from our final grade: moderate or close to zero?

Reply: 0.99. I thought that it was a done deal that Chu-Wee would compute homeworks in such a way that your bottom 1 1/2 homeworks would not count. If I remember correctly, he is going to list most of your homework scores twice, except that the extra-long assigments will be listed three times. Let's say that there are 8 shorter assignments and 3 longer ones; then he'll list 25 scores. He'll then erase the three lowest ones, leaving 22 scores. These are between 0 and 10, I think. If so, the maximum homework score would be 220. If homework is worth 30%, it would be convenient to multiply these numbers by 3/22 for use in the final computations.



Sun Dec 5 08:43:35 2004 : So can the homework be turned in thursday or not? The course page still says Tuesday but there's a reply to a comment saying thursday is ok.

Reply: Yes, it can. I changed "December 7" to "December 9" on the relevant line of the course web page.



Sun Dec 5 22:31:20 2004 : What are the chances that we will have the same grading arrangement as Fall 2001 (with the option of keeping or dropping the final): moderate or close to zero?

Reply: The latter.



Mon Dec 6 18:35:32 2004 : In office hours you said we didn't have to do #28 on the hw is this still true?

Reply: Sure, let's take off problem 28.



Tue Dec 7 16:23:24 2004 : Hi. I'll be in office tomorrow (8 Dec, Wed) 10-12. Feel free to drop by if you have any questions. It's 1068 Evans by the way. -CW



Tue Dec 7 20:03:11 2004 : Hey Chu-Wee, would it be OK if I showed up at your office at around 2PM? I have class from 10AM-2PM.

Reply: I'll be around in my office (885 Evans) after 2PM. There's a student coming in to see me at 3, though, so my informal supplementary office hour today can be only from 2 to 3. -kr



Thu Dec 9 16:15:05 2004 : Hey Professor, thanks for a fun semester. I really enjoyed the class.

Reply: You're welcome! Study hard for the exam, everyone. Please note my office hours for next week, by the way.



Sun Dec 12 15:09:53 2004 : How will we know when/how we can pick up our last homework?

Reply: The HW papers should arrive in my office some time around the middle of the day tomorrow. I'll put the math graduate students' papers in their mailboxes on the 9th floor of Evans. Other students will have to get their HW back from me when I'm in my office. I don't expect to be in my office on Monday afternoon (after 12:40 or so) but wouldn't be surprised to find myself in Evans a fair amount of the time on Tuesday.



Sun Dec 12 16:24:45 2004 : Will the final focus on the material after the second midterm or will it be more comprehensive?

Reply: The latter.



Sun Dec 12 20:40:13 2004 : wait. i never really got the rule in english. "the latter" = "the second", right? i think it is the other way around with the french equivalents. that's why i always get confused. or am i wrong? you know french, right prof.?



Sun Dec 12 21:09:11 2004 : "The latter" does mean "the second."

Reply: According to m-w.com, the word "latter" may be used as follows:

of, relating to, or being the second of two groups or things or the last of several groups or things referred to
I don't know what construction in French you're alluding to.



Mon Dec 13 10:04:00 2004 : oed is so much better than m-w!



Mon Dec 13 13:24:41 2004 : Hi everyone. I've finished grading the last homework and handed it to the prof. Unfortunately, I handed it in rather late, and he didn't have time to distribute the homework (sorry, Ken). Anyway, here's what I did to the scores. We have 9 homeworks out of 20. I copied these twice: that gives a max score of 360. For HW 4 and HW 8 (which were out of 30), I first lowered them to 20 then repeat each thrice. That gives 120, and a grand total of 480. Then I dropped the lowest three scores, which gives 420. Also, I'll be out of town starting from tomorrow, but questions will always be welcome. Except that I won't be bringing Lang's algebra with me, so please be specific, i.e. don't ask things like "on page 374, 2nd paragraph, what's M_p?". Argh, what a long post. Anyway, good luck for your exams. It's been hard but fun, I believe. Hopefully, many of you will go on to take 250B (commutative algebra) and become lords of the rings. -CW



Mon Dec 13 22:51:52 2004 : Professor, for your solutions on number 9, there is an alternate and perhaps quicker solution. Let x1,...,xp be the conjugates, and K be the splitting field of the polynomial over k. we want to show that xi is in K: [k(x1):k]=p means that the galois group of K/k has a p-cycle- in particular, the cycle is transitive over the xi. We have by hypothesis that x2 is in k(x1). Take the p-cycle phi such that phi(x1)=x2. Then phi takes k(x1) back to k(x1), but iteration of phi means that xi is in k(x1) for 1?i?p, since (phi^i)(x2) is in k(x1), and we're done.

Reply: I'm trying to understand your argument. It sounds similar to the one that I posted, but maybe it doesn't appeal to Artin's theorem. There's a p-cycle in the Galois group, and we fix it up so it takes x_1 to x_2. It takes k(x_1) to k(x_2), which is the same thing as k(x_1), so it maps k(x_1) to itself. The argument is that every x_i is the image of x_1 under some power of phi, so each x_i is again in k(x_1). Hence k(x_1) contains all the conjugates of x_1; it is the splitting field. Therefore it's Galois over k. OK, good.



Tue Dec 14 01:22:11 2004 : "Lords of the rings" -- haha!



Tue Dec 14 18:33:29 2004 : Yes, that was my argument for number 9.



Tue Dec 14 18:34:44 2004 : The only thing i use in the proof is a Sylow theorm and the fact that the Galois group is embedded in Sp. I don't think i appeal to Artin's theroem- i certainly don't do it directly.

Reply: Noted.



Fri Dec 17 00:46:39 2004 : Will there be the true/false/example question on the final that there was on the midterm? Also, should we be boning up on our category theory? Thanks a lot.

Reply: True/false questions are always possible. They tend to be hard. As for category theory, you have to judge for yourself how important it is as an element of the course.



Sat Dec 18 10:37:26 2004 : I was wondering about how many pages of notes we can bring for the final. It would be nice if we could bring three, so we can recycle our notes from the frist two, and then make a third...

Reply: Heck, a single two-sided page should do it. In thinking about what to put on the page, you organize your thoughts and organize the course.



Sun Dec 19 16:14:31 2004 : Could you please give me an example of a direct summand of a free A-module that's not free? How about a projective A-module that is not free? Thanks!

Reply: Projective modules are exactly those modules that occur as direct summands of free modules. Your two questions are the same question. If A is a Dedekind ring and I and J are fractional ideals of A, then you showed in homework that I+J (direct sum) is isomorphic to A+IJ. Also, you showed that I and J are projective. If J is the inverse of I in the group of fractional ideals, then A+IJ is the free modules A+A. If you can prove that I is not free in some situation, than you have your example.

Suppose that I is a fractional ideal of a Dedekind domain A and that I is free with basis v_k over A. It is clear then that every element of the fraction field K of A can be written uniquely as a K-linear combination of the v_k. Indeed, every elemment of K is of the form i/a where i is in I and a is in A; thus every element of K is a linear combination of the v_k. Further, a K-linear combination of the v_k that is 0 can be multiplied by a denominator to yield an A-linear combination of the v_k that is 0. Thus the v_k form a basis of K over K; this means that there is only one v_k, since K is a 1-dimensional K-vector space. Thus I is generated by a single element over A; i.e., I is a principal ideal. In the book, you should find examples of non-principal ideals in a Dedekind domain. To summarize: these examples yield projective modules that are not free.



Sun Dec 19 18:45:01 2004 : I know this may be a little late in the game, but could you explain the difference between free and torision-free? torsion-free means that no elements have finite order? And free? does it mean that it is the direct sum of some ring? please explain.

Reply: A free module over A is a direct sum of copies of A. A torsion-free module M over A is one with the property that ma=0 if and only if m=0 or a=0; here, m is in M and a is in A.



Mon Dec 20 11:42:13 2004 : I don't recall if any examples were done in detail in Lang, but the fact that Z[sqrt(10)] is Dedekind but not principle is done in detail in Hungerford. (Parts of it are excercises, but it walks you through those parts.) Would it be acceptable on the exam to just write that somethig like that is not Dedekind, or would we be expected to prove it in detail?



Tue Dec 21 02:25:09 2004 : Man. That problem about the irreducibility of x^n+x+p was wicked tough.

Reply: Glad you liked it.



Wed Dec 22 19:25:46 2004 : Will detailed grade info be posted like in Fall 2001?

Reply: It's on the main page now.



Thu Dec 23 17:19:39 2004 : It seems like the 2001 class got better grades than we did.



Thu Dec 23 19:21:09 2004 : To the poster above, if you calculate it out, the average grade is B+/A- just like Fall 2001. There just appears to be more variance.