Author: William A. Stein
Math 129: Topics in Number Theory

# Math 129: Virtual Office Hours

## Email me a question!

Since I post the answers here, everyone benefits.

 On Monday 02 May 2005 10:33 pm, you wrote: > in second problem do you mean Norm_L/K(a^(1/n)) instead of Norm_L/K(a) since a > is already in K? so the problem should read as "\gothic p is unramified if it > is coprime to n and a". It should be Norm_{K/Q}(a). The conclusion might also be true under the weaker hypothesis that \p is coprime to n*a, but you don't have to prove this stronger statement to get full credit. > Btw do we need the nth roots of unity to be in K? It is not necessary to have the nth roots of unity be in K in order to consider the field L = K(a^(1/n)). The field Q(zeta_n) is unramified outside n, so if you replace L by L(zeta_n), then the resulting field can only have additional ramification at primes dividing n. Thus for the purposes of this problem, without loss of generality you may assume that K contains the n-th roots of unity.  > For 1(a) of this problem set, when you say to "write down the > Riemann-Roch Theorem," I assume this means the statement of the theorem > with no proof; is this correct? Yes, this is correct.  > I'd like to do a project related to algebraic combinatorics. Doing something on Groebner basis might work. One very exciting paper is this one by Faugere.  On Wednesday 16 March 2005 06:52 pm: > i'm using eisenstein's criterion in 4.b to prove irreducibility of the > polynomial x^2n+2, a result we learned in math 122. i remember the proof > has something to do with looking at the factorization of f modulo p and i > suspect i could come up with it if i have to, but i was wondering if it > suffices to just cite this result. It suffices to just cite this result. Thanks for asking. > i suppose while i'm asking i should also ask the same question of the > primitive element theorem (i.e. that every number field K/Q is given by > some irreducible f). You can also just cite this result.  On Wednesday 16 March 2005 07:35 pm: > My question concerns the optimism with which to approach the map > \Phi_L:C_L\rightarrow C_K. Is it possible that the map is not > well-defined to begin with? My trouble is that if it were not > well-defined, you would't > have asked part b, and, more importantly, wouldn't have made the little > note after the problem. But, on the other hand, I do have some reasons to > believe it is not always well-defined...(although I have to recheck my > arguments) I'm not going to give this away, except when I said "be optimistic" I only meant that as a general strategy --- I'm not guaranteeing that in all cases a proof is what is required. > Also, when you say "Prove or give a counterexample," do you really mean > "prove or disprove"? EVEN if the map is well-defined, I still believe I > can prove that sometimes \Phi_L is not a group homomorphism. Because there > are other ways to disprove it, for > example, by proving that a counterexample exists, without explicitely > finding it (well, I can _explicitely_ find K and L, but not the ideal that > messes things up). I consider giving an L and a K and proving that there's an ideal that "messes things up" as giving a counterexample. You don't have to give every detail of the counterexample, just details of a proof that it is one.  > In 3(c) we are asked to "prove that there is a number field L such that > \Phi_L is the 0 map; i.e., \Phi_L sends every element of C_K to the > identity of C_L." But \Phi_L is a map from C_L to C_K. Did you mean that > \Phi_L sends every element of C_L to the identity of C_K? There is a typo in 3(c). The map you should consider is the one from C_K to C_L.  On Wednesday 16 March 2005 03:10 pm: > (for problem 3) i just wanted to check that I*OL is the set of all finite > sums of products ab where a \in I and b \in OL. at first i thought it was > just the set of products, but that doesn't appear to be closed under > addition. thus I*OL is the smallest ideal containing I in OL. am i > correct? Yes, you are correct.  On Wednesday 16 March 2005 12:00 pm: > In example 7.2.6, we show that disc(O_K)=5, where K=Q( sqrt{5} ). There > are 2 complex imbedding of K, thus the sign of the discriminant should be > negative according to the second problem on the midterm. > > Is this a typo? No. Perhaps you're confused about the meaning of complex embedding. It's an embedding that doesn't have image in the real numbers.  On Wednesday 16 March 2005 07:31: > I guess the last email that I sent has problems in it. But I know that the > map p--> p \cap O_K must span all the primes of O_K. I saw it in a > book(before the exam) You are completely right, and this is very easy, and thanks for pointing this out! If \wp is a prime of O_K, let q_1, ..., q_n be the primes of O_L that lie over \wp, so the q_i are the primes that appear in the factorization of \wp*O_K. Then each q_i intersected with O_K is a prime ideal (since the inverse image of a prime is prime), and that prime ideal must be \wp, since it contains \wp.  On Wednesday 16 March 2005 12:58: > thanks...just one more. i'm unclear on what you mean by the sign of the > discriminant. i know that the discriminant is an integer so it's > perfectly natural for it to have a sign in the \pm 1 sense, but i don't > recall ever hearing a special mention of this sign so i wanted to check > and make sure my interpretation is correct. maybe i'm being misled by the > point value for the problem, which suggests that it's easier than number > 1, though the solution for 1 was much more obvious to me... If the discriminant is negative, then the sign is -1, and if the discriminant is positive, then the sign is +1.  On Tuesday 15 March 2005 09:14 pm: > two questions (sorry for all the e-mails). > 1. for problem 1 is R a ring with unity? Yes, it's a commutative ring with unity. > 2. for portions of solutions that closely mimic proofs given in the book, > should we be concerned with plagarism? obviously if i'm basing a proof > off one that you give, they will be similar, but it's also easier to just > adopt your notation then deliberately come up with something new. No, it's perfectly fine. In fact, in your answer you can just say what modifications are needed for your solution and reference the book freely.  On Tuesday 15 March 2005 08:17 pm: > "We gave examples in class to show that the ring of integers of a > number field need not be generated by one element. Is the ring of > integers always generated by at most two elements?" > > I'm guessing that you meant to write the following: > > "We gave examples in class to show that *an ideal in an order* need > not be generated by *two* elements. Is *such an ideal* always > generated by at most *three* elements?" > > Or am I very confused? Well at least ene of us is very confused. Here's why I think it's you. > We gave examples in class to show that the ring of integers of > number field need not be generated by one element. The example I'm referring is Dedekind's example of an essential discriminant divisor, i.e., a cubic field such that O_K =/= Z[alpha] for any alpha in O_K, so the ring of integers is *not* generated by one element. The question then arises, can you always generate with at most 2? This question is reminiscent of the ideal generators question, but it's actually an entirely different question. [...] and he responds: "Ah, OK. I guess I was confused about the definition of "generated". You mean "generated as a ring over Z"."  On Tuesday 15 March 2005 07:55 pm: > The class > group is defined as > the fractional ideals of the ring of integers as a group modulo the > principal ideals. How > do you take modulo of a group? Can I assume our operation is > multiplication, so > two fractional ideals are equivalent if one is a principal ideal times > the other via the definition > we supplied while we were doing unique factorization? Yes.  > In question 3 of the midterm, do we need to show that \Psi_L and > \Phi_L are well-defined (for Psi_L, that if I and J are equivalent > modulo the principle ideals of K, then (I*O_L) and (J*O_L) are > equivilent modulo the principle ideals of L; and for Phi_L, that if I > and J are equivalent modulo the principle ideals of L, then (I cap > O_K) and (J cap O_K) are equivalent modulo the principle ideals of K)? Part of the problem is either to prove that they are well defined (if they are), or give a counterexample. Thanks for clarifying this. > And for "Challenge Question 1" on the 129 web site, isn't the ring of > integers generated by the single element "1"? Unless you mean > generated as a Z-module, in which case the number of elements required > is the degree of the extension over Q. The question is whether or not every order is generated by AT MOST 2 elements. To give a counterexample, means giving an order such that the minimal number of generators is >= 3, as a ring.  Page 11: should it be Agarwal (not Agrawal)? Also, I think technically it should be finitely-generated abelian groups according to general English rules of hyphenation, http://owl.english.purdue.edu/handouts/grammar/g_hyphen.html but this is probably not the most important rule to follow.  Hey Professor Stein, a few items from the early part of the book: page 7: this book *may* be freely... Also, you may not care about this, but technically "the book" is the subject of this sentence so I think "without requiring you to obtain..." implies that the book would be the one doing the requiring... either way it might be better reworded "More precisely, you may freely redistribute, copy, or even sell this book without obtaining written permission from me." Acknowledgments: his name is spelled Peter Behroozi. two missing commas: after David Escott and after Andrew Ostergaard. Chapter 1: 1.1, last bullet should probably read "Basics"  > Mistake in Homework 5, Problem 2: > > In class we showed that any equivalence class of ideals contain an integral > ideal of norm atmost 2^{r+s}*1/v*M*sqrt(d_K). In our homework problem we have > M=n^(-n). So in order to obtain Minkowski's inequality we must have v=2^(r-s)* > PI^s/n!. so in part b v should be 2^r (2PI)^s D_{r,s}(1) instead of D_{r,s}(n) I agree. Here are more details, mainly for the virtual office hours page. We showed that every ideal is equivalent to one with norm 2^{r+s} sqrt(|d_K|) * M / v the Minkowski bound is (4/pi)^s*sqrt(|D_K|) n!*n^{-n}n, and M=n^{-n}, so it seems to me that v should be 2^{r+s} sqrt(|d_K|) * n^{-n} / ((4/pi)^s*sqrt(|d_K|) n!*n^{-n}) = 2^{r+s} * (pi/4)^s / n! = 2^{r-s} * pi^s / n!.  On Friday 11 March 2005 10:13 am: > In problem 2 of homework 5, in the definition of S, is the x_{v+s} > (under the square root) supposed to be squared? Yes.  On Wednesday 09 March 2005: > I have a little question about a hint for problem 5. Is it supposed to be > a conceptual or a computational problem? I mean, I know *in theory* how to > do it with a computer. Namely, I come up with good enough extension, say > of degree 4, Q(a), > whose ring of integers is not Z[a], I take an ideal I of Z[a], which I > generate by 3-4 elements, say. Next, I ask MAGMA for a basis of this ideal > in the order as a free Z--module (I figured out that command; if the > order is the maximal one, I can ask for two-element generaators also, but > not if the order is proper). Next, I write down two general elements of I > and check if the generators for I belong to the ideal generated by those > two elements. This condition is equivalent to a system of 4 equations to > be solved in integers. So, if I figure out elements of the ideal such that > this system does not have a solution, I will be done. If the computer does > not produce an answer, I change the ideal. If I try this a couple of times > and does not work, I change the extension. I've decided I'll give you some hints, in case you haven't solved the problem yet: (1) Prove that if I is an ideal in a ring R, and I can be generated by n elements, then I/I^2 can be generated by <= n elements as an R/I-module. This is useful, e.g., if you choose I to be maximal, then R/I is a field, and the minimal number of generators of I/I^2 is exactly the dimension of I/I^2 as an R/I-vector space. If you're in the context of our problem, then R/I will be a finite field and I/I^2 will be a finite set. (2) We proved that if I is an ideal in the ring of integers O_K, then and if m is *any* integer in I, then there's an alpha in O_K such that I = (m, alpha). From this it follows that for any integer m the quotient ring O_K/(m) is a principal ideal ring. Thus you might want to look for an order O in O_K such that O/(m) is not a principal ideal ring, then lift a non-principal ideal to O. (3) There are other orders besides Z[a]. For example, O = Z + m*O_K.  On Monday 07 March 2005 09:54 pm: > I have a little question about a hint for problem 5. Is it supposed to be > a conceptual or a computational problem? I mean, I know *in theory* how to > do it with a computer. Namely, I come up with good enough extension, say It's a *conceptual* problem. A computer will be of little use, but you could probably solve it with just some paper. > Or, is it something we can construct using some sort of general arguments/ > reasoning? Yes, *you* can.  On Sunday 06 March 2005 03:49 pm: > By now I'm quite comfortable generating simple extensions of the field of > rational numbers in Magma. However, is it possible to generate extensions > by more than one variable? For example, how would I create the field > Q(sqrt(2), sqrt(3))? > > I have tried generating K = Q(sqrt(2)) and then generating L = K(sqrt(3)), > but I can't seem to force L to be interpreted as an extension of Q. Magma is extremely good at relative extensions, and can turn them into absolute extensions. Here's an example. [email protected]:~$magma ... > R := PolynomialRing(RationalField()); > K := NumberField(x^2-2); > S := PolynomialRing(K); > L := NumberField(y^2-3); > L; Number Field with defining polynomial y^2 - 3 over K > AbsoluteField(L); Number Field with defining polynomial x^4 - 10*x^2 + 1 over the Rational Field  On Wednesday 02 March 2005 09:11 pm: > I have a question about Problem 5b. Technically, we defined the ring of > integers only for number fields, not for arbitrary extensions of Q. This > is in a strict sense. By ring of integers, do you mean what one thinks you > mean, namely, all elements in Q(...) that satisfy a monic polynomial with > integer coefficients? Or, do you mean the ring of integers of Q(a), where > a is a root of x^5+7x+1. I think it is unlikely that you mean the second > thing, but I wish you did:)) (othrwise, I have to search hard to find the I meant the ring of integers of Q(a), where a is a root of x^5+7x+1. Sorry for the confusion.  On Tuesday 01 March 2005 10:19 pm: > On problem set 3 (the one due Thursday), problem 5(b) asks for the ring of > integers of Q(x^5+7x+1). Should this actually be Q(x)/(x^5+7x+1)? Yes, it should be Q[x]/(x^5+7x+1). Note the square brackets. Thanks for the correction, and also your comments in class today!  On Friday 25 February 2005 09:23 pm: > Namely, let$P$be a nonzero prime ideal. Then, according to definitions, > the set$M=\{ \{0\}, P\}$is not closed because we cannot find I with > M=V(I). Thus, I guess a condition should be added in part b for the finite > sets: either to be \{ (0) \} or not to contain (0). I just wanted to make > sure > we change part b, not the definition of the topology on X. You're right. I'll update the problem again.  On Friday 25 February 2005 09:36 am: > I may have missed this in class, but in problem 2 of homework 3, I > assume that$I\$ is allowed to vary through all integral ideals, not > just prime ones. Is this correct? You're right. I've posted a new version of the assignment here: http://modular.math.washington.edu/129-05/homework/3/