1. Prove that a UFD is normal.Proof: Let A be a UFD, K its field of fractions. Given x ∈ K, we can writex = a/b with a ∈ A and b ∈ A \ {0}, and the factorizations of a and b not havingany prime in common (by canceling any common primes). Then if x is integral overA, we havexn+ cn−1xn−1+ . . . + c1x + c0= 0for some n and ci∈ A. Clearing denominators givesan+ cn−1an−1b + . . . + c1abn−1+ c0bm= 0,which shows that b divides an. But then any prime in b’s factorization would alsobe in an’s, and hence in a’s; we avoid a contradiction only if b’s factorization hasno primes, i.e. b is a unit, which means x ∈ A.2. What is the normalization of A = k[x, y]/(y2− x3)? Here k is a field.Answer: First we give an injection of A into k[t]: define ϕ : A → k[t] by x 7→ t2,y 7→ t3(which is well-defined since (t3)2− (t2)3= 0). We will show injectivity byexhibiting a collection of elements of A which span A and whose images under ϕ arelinearly independent (as a bonus, this will simultaneously show that our collectionis in fact a basis). The desired collection consists of thexiyjfor i ≥ 0 and j = 0 or 1.These span A: indeed, certainly the xiyjfor i, j ≥ 0 span, and we can reduce thej below 2 by successively applying the relation y2= x3. And their images, namely1 together with the tnfor n ≥ 2, are indeed linearly independent.So A is isomorphic to its image A0= ϕ(A) (which, explicitly, is the span of 1 andthe xnfor n ≥ 2). Let’s think about the normalization of A0. The field of fractionsof A0is k(t), the same as that of k[t], since t = t3/t2. And k[t] is integral over A0,since t2− t2= 0. And k[t] is a UFD, hence integrally closed by the first problem.Thus k[t] is the normalization of A0; that is, the normalization of A0is obtained byadjoining the element t3/t2to A0inside its field of fractions. Translating via theabove isomorphism, we see that the normalization of A is obtained by adjoiningy/x inside the field of fractions, and that the result is just a polynomial ring in y/x.3. Let I be an ideal in a ring A. Is Ann(I/I2) = Ann(I) + I?Answer: No. Many of of you gave the following good example: let A =Q∞i=1Z/2Z, and I be the ideal ⊕∞i=1Z/2Z. Then I = I2, so Ann(I/I2) = A;but Ann(I) = 0, so Ann(I) + I = I 6= A.In this example, I is not finitely generated. Even if I is finitely generated, wedon’t necessarily have equality; Tiankai and Nike gave the following example: letA = k[x, y, z]/(y2− xz, x2− yz) and I = (x, y). Then z ∈ Ann(I/I2), but the claimis that z 6∈ Ann(I) + I. This claim is not easy to check, because it can be tricky toget a handle on modding out by y2− xz and x2− yz. There is a technique one canuse, that of Groebner bases, which I won’t explain here, but maybe will in sectionsome day.12The reason the finitely generated case is different is that for I finitely generated,one necessarily has that the radicals of Ann(I/I2) and of Ann(I) + I coincide. Youmight not understand this argument yet, but you soon will:Indeed, It suffices to see that V (Ann(I/I2)) = V (Ann(I) + I) = V (Ann(I)) ∩V (I). Since I = Ann(A/I) and I/I2is isomorphic to I ⊗AA/I (see the nextproblem), this will follow from the following more general claim: let M and Nbe finitely generated modules over A. Then V (Ann(M ⊗AN)) = V (Ann(M)) ∩V (Ann(N)).For this I first claim that p 6∈ V (Ann(M )) if and only if Mp= 0. For “if”, notethat the condition implies that, m1, . . . , mnbeing generators, there are a1, . . . , an∈A \ p with aimi= 0. Then the product of the aiwill be in Ann(M ) but not p.“Only if” is easier, and I leave it to you.Thus what we need to prove is that (M ⊗AN)p= 0 if and only if either Mp= 0or Np= 0. But (and here by = I mean canonical isomorphisms of Ap-modules) wehave(M ⊗AN)p= (M ⊗AN) ⊗AAp= (M ⊗AAp) ⊗Ap(N ⊗AAp) = Mp⊗ApNp,so the claim follows from 2.3 below (certainly if M is finitely generated over A,so is Mpover Ap). Here the second = follows from a general claim made in theproof of that exercise, and the other =’s are instances of the same general factMp= M ⊗AAp, which I invite you to check.2.2 Let A be a ring, I an ideal, and M an A-module. Show that (A/I) ⊗AM isisomorphic to M/IM.Proof: Consider the short exact sequence0 → I → A → A/I → 0.Tensoring with M is right exact, so we get an exact sequenceI ⊗ M → A ⊗ M → (A/I) ⊗ M → 0.This shows that (A/I) ⊗ M is isomorphic to A ⊗ M modulo the image of I ⊗ M →A ⊗ M . However, I claim that A ⊗ M is isomorphic to M via a ⊗ m 7→ am. Indeed,this is certainly bilinear, so gives a good map from the tensor product A ⊗ M ;and m 7→ 1 ⊗ m is an obvious inverse. So (A/I) ⊗ M is isomorphic to M mod-ulo the image of I ⊗M → A⊗M → M, which is IM if you just remember the maps.2.3 Let A be a local ring, M and N finitely generated A-modules. Prove that ifM ⊗AN = 0, then M = 0 or N = 0.Proof: First I claim that if A is an arbitrary ring and B an A-algebra, and Mand N are arbitrary A-modules, we have a canonical isomorphism of B-modules(M ⊗AN) ⊗AB = (M ⊗AB) ⊗B(N ⊗AB). As a lemma, I claim that M ⊗AB hasthe following universal property: for any B-module P , the sets HomB(M ⊗AB, P )and HomA(M, P ) are in canonical one-to-one correspondence (we can consider Pas an A-module through the map A → B: have a ∈ A act through its image).Indeed, to f : M ⊗AB → P we can associate an M → P by m 7→ f(m ⊗ 1), and tog : M → P we can associate M ⊗AB → P by m ⊗ b 7→ bg(m), and these are easilyseen to be inverse. Then the following chain of canonical bijections shows that the3two modules we want to be isomorphic satisfy the same universal property:HomB((M ⊗AN) ⊗AB, P ) = HomA(M ⊗AN, P )= HomA(M, HomA(N, P ))= HomA(M, HomB(N ⊗AB, P ))= HomB(M ⊗AB, HomB(N ⊗AB, P ))= HomB((M ⊗AB) ⊗B(N ⊗AB), P ).(To get the explicit isomorphism and its inverse from this argument, one shouldtake P = (M ⊗AN) ⊗AB in the above and trace id(M⊗AN)⊗ABthrough; then dothe reverse with P = (M ⊗AB) ⊗B(M ⊗AB) and id(M⊗AB)⊗B(M⊗AB)).Now we apply this in our situation with B = A/m. We see that M ⊗AN = 0implies, on tensoring with B, that M/mM ⊗A/mN/mN = 0 (we have also used theprevious exercise). But M/mM and N/mN …