AtiyahMacDonald习题全解_3

48 . embedded isomorphically in Mn, which is itself a Noetherian module (by corollary 6.4). As a submodule of a Noetherian module, A/a will be Noetherian itself, as desired. The statement collapses if we replace the Noetherian condition by the Artinian one. For example, let A = Z and M = G be the subgroup of Q − Z consisting of all elements with absolute value equal to 1/pn, n ≥ 0 (of course p is some fixed prime number, as in the book’s example). As shown in the book, G is an Artinian Z-module, and its annihilator obviously equals 0. If the previous statement was still true in the case of Artinian modules, then it would yield that Z = A/Ann(M) is Artinian, which is false. 6.5 If Y is a subspace of X with the induced topology, then any set U open in Y is of the form G∩ Y , where G is a set open in X. Therefore, any ascending chain {Un}n∈N of sets open in Y is is of the form {Gn ∪Y }n∈N, where {Gn}n∈N is an ascending chain of sets open in X. Since X is Noetherian, {Gn}n∈N is stationary, hence so will {Un}n∈N be. Therefore, any subspace of X is Noetherian. Assume that X is Noetherian, but not quasi-compact (the definition of the book seems to be that of usual compactness). Then, there is an open cover C = {Ci}i∈I of X, such that no finite subcover of C covers X. Let G1 be an arbitrary element of C. Since G1 doesn’t cover X, there is x ∈ X such that x /∈ G1, but x ∈ Ci for some i ∈ I; let G2 = G1 ∪ Ci. Since G2 fails to cover X, there is x′ ∈ Cj ⊂ X with x′ /∈ G2; let G3 = G2 ∪ Cj , etc. In this fashion, we construct an ascending chain of open sets {Gn}n∈N that is not stationary, contrary to the assumption that X is Noetherian. Therefore, X is quasi-compact. 6.6 We will follow an unorthodox order in the proof of the equivalences: (i) ⇒ (iii) By the previous exercise, every subspace of X will be Noetherian and thus quasi-compact (by the previous exercise again). (iii) ⇒ (ii) O.K. (ii) ⇒ (i) Let C = {Gn}n∈N be an ascending chain of open subsets of X. Then, since the subspace Y = ⋃ n∈N Gn of X is quasi-compact and has C as an open cover, there will be a finite subcover {Gin}1≤n≤N of Y . But, if i = max{i1, i2, . . . , iN}, then Y = ⋃ 1≤n≤N Gin = Gi. This Gi is clearly an upper bound of C = {Gn}n∈N, which shows that X is Noetherian. 6.7 We know that the maximal irreducible subspaces {Yi}i∈I are closed and cover X, by chapter 1, exercise 20, (iii). If we assume that the intersection of all the Yi is non-empty (let x be some point of X contained in it), then any neighbourhood of any point of X intersects any neighbourhood of x, therefore we conclude that X = {x}. In this case the statement is vacuously true. In the case ∩i∈IYi = ∅ the set {X−Yi}i∈I is an open cover of X and this necessarily has a finite subcover (since X is Noetherian and thus quasi-compact) and this in turn yields a presentation of X as a finite union of irreducible and closed maximal subspaces. We deduce that the set of irreducible components of a Noetherian space (which are V (pi), i ∈ I and the pi are the minimal prime ideals of A) is finite. 6.8. 49 6.8 Assume that {V (an)}n∈N is a decreasing sequence of closed sets in Spec(A) (the an are assumed radicals of ideals of A; this assumption leads to no loss of generality, since V (a) = V (r(a))). Since V is an inclusion- reversing bijection from the set of all radicals of A to the set of all closed subsets of Spec(A), this chain yields an increasing chain {an}n∈N which must stabilise by the Noether condition on A. This implies that the initial chain stabilises too and therefore Spec(A) is a Noetherian topological space. The converse, however, is not true. Let A = k[x1, x2, . . .] be a polynomial ring over a ring A with infinitely many variables adjoined and let a = (x1, x2, . . .) be the ideal generated by the squares of all the variables. Then, A˜ = A/a is not Noetherian (because A is not), but note that R(A˜) = (x1, x1, . . .) thus Spec(A˜) = Spec(A˜/R) (by chapter 1, exercise 21, (iv)) and this is a point, hence it’s trivially Noetherian. 6.9 If A is a Noetherian ring, then Spec(A) is a Noetherian topological space (with the standard Zariski topology) by exercise 8. Therefore, by exercise 7, Spec(A) will have a finite number of irreducible subspaces, which are exactly the minimal prime ideals of A by chapter 1, exercise 20, (iv). 6.10 Supp(M) is closed since it equals V (Ann(M)) by chapter 3, exercise 19, (v). Now let {Supp(M)∩V (ai)}i∈I be a decreasing sequence of closed subsets of Supp(M). By chapter 3, exercise 19, we see that Supp(M)∩ai = V (Ann(M)∩V (ai) = V (Ann(M)∪V (ai) = V (bi) (where bi is the ideal generated by Ann(M)∩V (ai)). We may assume that bi is a radical, since V (e) = V (r(e)) for all ideals e of A. Since V is an inclusion-reversing bijection between the radicals of A and the closed subsets of Spec(A), we obtain an increasing sequence {bi}i∈I of ideals of A. Letting pi be the minimal prime ideal that contains bi gives rise to an increasing sequence Mpi of submodules of M, which stabilises because M is Noetherian. Therefore the initial sequence must stabilise too and this completes the proof that Supp(M) is a Noetherian topological space. 6.11 By chapter 5, exercise 10, (i) f∗ : Spec(B) −→ Spec(A) always has the going-up property if it’s a closed mapping. If Spec(B) is a Noetherian space, then the converse is also true. Indeed, let V (p) ⊆ Spec(B); then, by the equivalent condition (c) of chapter 5, exercise 10, we have that V (q) ⊆ f∗(V (p)), where q is merely the restriction of p in A. Then, we would like to show that the map is also injective, so that the closed set V (p) is mapped to a closed set. If the inclusion was strict, then we would have the infinite strictly descending chain (the pi arise from the going-up property): f∗−1(V (q)) ⊇ V (p) ⊇ V (p1) ⊇ V (p2) ⊇ . . . of closed sets in Spec(B) (note that f∗ is always continuous, hence the preimage of any closed set is closed), a contradiction by the fact that Spec(B) is Noetherian. Therefore f∗ is closed and this completes the proof. 6.12 Before the actual proof, note that given ideals p and q of A, V (p) = V (p) if and only if p = r(p) = r(q) = q. Now, if {pn}n∈N is an ascending chain of prime ideals in A, then {V (pn)}n∈N is a descending chain of closed sets in Spec(A) and since this space is assumed Noetherian, the sequence {V (pn)}n∈N must be stationary, hence so will the sequence {pn}n∈N be. The converse follows in the same fashion, since V (p) ⊇ V (q) ⇔ p ⊆ q for any two prime ideals p and q of A. CHAPTER 7 Noetherian Rings. 7.1 We note that Σ has maximal ideals and this follows by a typical Zorn’s Lemma argument. Given such a maximal ideal a, assume that x, y /∈ a, but xy ∈ a. Then, b = a + (x) strictly contains a, therefore it must be finitely generated; say b = a0 + (x), where a0 is also finitely generated. Note that a + (x) = a0 + (x) implies a0 ⊆ a. We claim it also implies a = a0+x(a : x). Indeed, the right hand side is obviously contained in the left hand side and conversely, given any a ∈ a, a+ xt ∈ a0 + (x) for every t ∈ A. But then, there are a0 ∈ a0, k ∈ A such that a = a0 + x(k − t), and, since x(k − t) ∈ a, k − t ∈ (a : x), therefore the left hand side is contained in the right hand side too. This shows that the desired equality holds. Since (a : x) strictly contains a it must be finitely generated, therefore a = a0+x(a : x) is also finitely generated, a contradiction. Thus a is prime. As a corollary to the above, we observe that a ring in which every prime ideal is finitely generated is Noetherian (I.S. Cohen). 7.2 Assume that f = a0 + a1x+ a2x2 + · · · ∈ A[[x]] is nilpotent; we will show that all its coefficients an, n ≥ 0 are nilpotent. For that, we just need to show that given any prime ideal p and any coefficient an, an ∈ p. Indeed, let A = A/p be the ring obtained by reduction of A modulo p; since A is an integral domain, so will A[[x]] be (this is trivial to check). In particular, A[[x]] will contain no nonzero nilpotent elements. The natural projection A� A lifts naturally to a projection A[[x]]� A[[x]] and since f is nilpotent, f will also be nilpotent, which yields f = 0 by the above. Hence every coefficient an of f gets mapped to 0 = 0 + p in A, which implies an ∈ p, as desired. Conversely, if all the coefficients of f ∈ A[[x]] are nilpotent, and A is a Noetherian ring, then f is also necessarily nilpotent. By the Noetherian condition on A, there is a positive integer k such that Rk = 0, where R is the nilradical of A. We easily then see that fk = 0, hence f is nilpotent. 7.3 We have the following: (i) ⇒ (ii) If a ∩ S = ∅, then the restriction of S−1a = a, hence we may just put x = 1. Otherwise, (S−1a)c = A, thus x = 0 satisfies the given condition. (ii) ⇒ (iii) Assume otherwise; in particular, if S = {xn}n≥0, then S ∩ a = ∅. Therefore, (a : y) = (S−1a)c = a for some y = xm. But then, xkm = yk = 1 for some k ∈ N, and the sequence terminates, which is absurd. (iii) ⇒ (i) We may assume that a = 0, passing on to A/a if necessary, and then we may repeat the proof of lemma 7.12 since all the chains appearing in that proof are annihilators, and Ann(z) = (0 : z), for all z ∈ A. 51 52 . 7.4 We have the following: (i) This ring is isomorphic to the ring of rational functions C(t), hence it’s not Noetherian. (ii) This set is not even a ring (it doesn’t contain the 0 power series). (iii) This ring is isomorphic to the ring of all rational functions C(t), hence it’s not Noetherian. (iv) This ring is isomorphic to C[z], hence it’s Noetherian. (v) This ring is isomorphic to C[z, w], hence it’s Noetherian. 7.5 Since B is integral over BG (any x ∈ B is a root of ∏σ∈Σ(x− σ(x)) ∈ BG[x]), proposition 7.4 implies that BG is a finitely generated A-generated algebra. 7.6 If the characteristic of K is 0, then Z ⊂ Q ⊆ K, and since K is finitely generated over Z, it will be so over Q, therefore proposition 7.8 yields that Q is finitely generated over Z, an absurdity. Therefore, the characteristic of K equals some prime p and K is a finitely generated algebra over Z/pZ, which, by the Nullstellensatz, implies that K is a field. In particular, it’s a finite field, as desired. 7.7 An immediate corollary of the Nullstellensatz is that if h is a polynomial that vanishes everywhere an irreducible polynomial p does, then p|h. This fact implies that the variety X is well defined as the zero locus of the irreducible polynomials {fαi}α∈I0 and by the same argument there cannot be more then d such polynomials, where d is the minimal of their degrees. In particular, the variety is defined by a finite number of polynomials. 7.8 Indeed, if A[x] is Noetherian, then so is A. Let a1 ⊆ a2 ⊆ · · · ⊆ an ⊆ . . . be any ascending chain of ideals in A. This induces an ascending chain a1[x] ⊆ a2[x] ⊆ · · · ⊆ an[x] ⊆ . . . in A[x] (the ai[x] have the obvious meaning here). This chain stabilises, by the Noetherian condition on A[x]; say man = an+1 = . . . . Then, if there were y ∈ an+1 such that y /∈ an, we would have y ∈ an+1 (the constant polynomial equal to that value), but y /∈ an, a contradiction. This completes the proof. 7.9 Let a be a n ideal of A and let m1,m2, . . . ,mr be the maximal ideals that contain a. Let x0 be a non-zero element of a and let m1,m2, . . .mr+s be the maximal ideals that contain x0. Since mr+1, r+ 2, . . . ,mr+s do not contain a, there exist xj ∈ a, 1 ≤ j ≤ s, such that xj /∈ mj . Since each Amj is Noetherian, the extension Amja of a in Amj is finitely generated; let xi, x2, . . . , xt be the elements of a whose images generate Amia, 1 ≤ i ≤ r. Let a0 = (x0, . . . , xt); we observe that a0 and a have the same extension in Am for al maximal ideals m (since they do in the finite number of ideals m1, . . . ,mr+s and their common extension is the whole of the ring Am in every other ideal m), therefore, by proposition 3.8 we deduce a = a0; in particular, a is finitely generated and since we chose an arbitrary ideal, A is necessarily Noetherian. 7.10. 53 7.10 In chapter 2, exercise 6, we deduced that M [x] ' M ⊗A A[x]. Since M and A[x] are Noetherian A- modules (by assumption and Hilbert’s Basis Theorem respectively), M ⊕ A[x] will be a Noetherian A- module. We also observe that there is a surjective map M ⊕ A[x] � M ⊗A A[x] (the natural projection (m,a(x)) 7→ (m⊗A a(x))), hence M ⊗A A[x] will also be Noetherian by Proposition 7.1. This completes the proof. 7.11 It’s not necessary that A is Noetherian. Consider, for example, A = ∏∞ i=1 Z/2Z, a direct product of infinitely many copies of the field F = Z/2Z = {0, 1}. The strictly ascending chain of ideals 0 ⊂ F× 0 ⊂ F× F× 0 . . . shows that A is not Noetherian. It’s also evident that every element of A is idempotent (A is Boolean). Given any prime ideal p of A, Ap is a local integral domain with maximal ideal pp. We claim that pp = 0; indeed, given any non-zero element x ∈ Ap, we have x(1 − x) = 0, therefore 1 − x = 0, a conclusion that contradicts the primality of p. Therefore, pp = 0 and Ap is a field (which is Noetherian) for every prime ideal p of A. 7.12 By exercise 1 (the Cohen criterion), we may examine just examine ascending chains of prime ideals in A. Let p1 ⊆ p2 ⊆ · · · ⊆ pn ⊆ . . . be one. Since the induced map f∗ : Spec(B) −→ Spec(A) is surjective (by faithful flatness), the above chain gives rise to a chain q1 ⊆ q2 ⊆ · · · ⊆ qn ⊆ . . . , where pi = f∗(qi). Since the latter chain terminates (B is assumed Noetherian), the former must too. Thus A is Noetherian, as desired. 7.13 The fibre of f∗ at p ∈ Spec(B) is of course Spec(k(p)⊗A B) = Spec(Bp/pBp). Since f is of finite type and A is Noetherian, B will also be Noetherian, hence so will Bp/pBp be. This means that Spec(Bp/pBp) will be a Noetherian subspace of B, as desired. Nullstellensatz, strong form 7.14 We will essentially repeat the hint of the book; it constitutes a full proof. It is of course clear that r(a) ⊆ I(V ). Conversely, if f /∈ r(a), then there is a prime ideal p that contains r(a) but not f . Let f be the image of f under the natural projection map A −→ A/p and let C = Bf = B[1/f ]. If m is a maximal ideal of C, then C/m ' k (by the corollary to proposition 7.9; note that C is a finitely generated k-algebra, therefore the conditions of the lemma are satisfied). The images xi, 1 ≤ i ≤ n in C of the generators ti, 1 ≤ i ≤ n define a point x = (x1, x2, . . . , xn) ∈ kn which belong to the variety V , but f(x) 6= 0. Therefore, we also have I(V ) ⊆ r(a). This completes the proof that r(a) = I(V ), as desired. 54 . 7.15 Under the conditions of the problem, we have the following: (i) ⇒ (ii) Since A is flat over itself, so is An = ⊕ni=1A. (ii) ⇒ (iii) Let i : m −→ A be inclusion, which is, in particular, injective. If M is flat, then the map i⊗A id : m⊗AM −→ A⊗AM will also be injective. (iii) ⇒ (iv) If M is flat, then the exact sequence 0 −→ m −→ A −→ k −→ 0 gives rise to the exact sequence 0 −→ m⊗AM −→ A⊗M −→ k ⊗AM −→ 0, hence by definition, TorA1 (k,M) = 0. (iv) ⇒ (i) For this last part, we merely reproduce the book’s hint; it constitutes a full proof. Let x1, x2, . . . , xn be the elements of M whose images in M/mM are a basis for this vector space (we consider, of course, M/mM as a module over the field A/m). By (2.8), the xi generate M over A. Let F be the free module An with canonical basis e1, e2, . . . , en and define φ : F −→ M by φ(ei) = xi; let E be the kernel of this map. Then, the exact sequence 0 −→ E −→ F −→M −→ 0 yields, by the given condition, the exact sequence 0 −→ k ⊗A E −→ k ⊗A F 1⊗Aφ−→ k ⊗AM −→ 0, where k⊗AF, k⊗AM are vector spaces of the same dimension over k. It follows that 1⊗Aφ is an isomorphism, hence its kernel k ⊗A E must vanish. Since E is finitely generated, as a submodule of the Noetherian space F , and A is a local space, we deduce by chapter 2, exercise 3, that E = 0. This implies that M is isomorphic to a free module, hence it’s free, as desired. 7.16 Under the conditions of the problem, we have the following: (i) ⇒ (ii) If M is a flat A-module, then, since flatness is a local property, Mp is a flat Ap-module for all prime ideals p of A. But since Ap is a local ring, exercise 15 implies that Mp is a free A-module. (ii) ⇒ (iii) O.K. (iii) ⇒ (i) Since Ap is a local ring, the given condition is equivalent to flatness of Mp for all maximal ideals m of A, which is equivalent toMm being flat for all maximal ideals m by exercise 15, hence toM being flat since flatness is a local property. We conclude with the following charming epigram: ”flat = locally free”. 7.17 The proof that every submodule of a Noetherian module has primary decomposition follows in exactly the same fashion that propositions (7.11) and (7.12) follow; the proofs are in the book. 7.18 We have the following: (i) ⇔ (ii) This equivalence follows from proposition 7.17. (ii) ⇔ (iii) Consider the mapping φx : M � xM given by m 7→ xm, which yields that M/p = M/Ann(x) ' xM , a submodule of M . Conversely, if A/p is isomorphic to a submodule N of M , then there is an injection φ : A/p −→M . The element x = φ(1) of M satisfies Ann(x) = p. For the least part, we use induction on the number n of generators of M . If n = 0, then M = 0 and the result holds vacuously, so assume that n > 0 and the result holds for modules with less than n generators. Let p be a prime belonging to 0. There’s a submodule N ofM such that A/p 'M/N (since A is Noetherian, 7.19. 55 therefore so is a/p), hence 0 ⊂ N ⊂ M and N has fewer generators than M ; now the inductive hypothesis completes the proof that there is a chain of submodules 0 =M0 ⊂M1 ⊂ · · · ⊂Mr =M, in which each quotient Mi/Mi−1 is of the form A/pi, where pi is a prime ideal of A. 7.19 Any such decomposition of a is a minimal decomposition of primary ideals (since A is Noetherian), hence the uniqueness of the set of associated ideals of a. 7.20 We have the following: (i) Since F contains all open subsets of X and is closed under complements, it also contains all closed subsets of X. Therefore, we may equivalently describe F as the minimal subspace of X that contains all open and closed subsets of X and is closed under finite intersections and unions. Therefore, any element of F will necessarily occur as the intersection of the union of a finite number of open sets with the union of a finite number of closed sets; hence any E ∈ F can be written as E = U ∩C, where U,C are open and closed in X respectively. Conversely, it’s obvious that any set of the above form will belong to F . (ii) If E contains an open subset O of X, then given any x ∈ X and any open neighbourhood N around x, we will have E ∩ N 6= ∅, by the irreducibility of X. Therefore, there is an element of E in any open neighbourhood of X, or, equivalently, E is dense in X. Conversely, if E ∈ F is dense in X, and E = U ∩C (we keep the previous notation, of course), then given any open subset V of X, U ∩ V 6= ∅ and V ∩ U is open in X. Therefore may restrict our attention to open subsets V contained in U . Assume that there is no open set inside E. We can assume V * C; otherwise we obtain a contradiction. But, if x ∈ (X − C) ∩ V , we have x ∈ N such that N ⊆ X − C (since C is closed) and N ′ = U ∩N which will have empty intersection with E, a contradiction since N ′ is a neighbourhood of x and E is dense. Therefore, E contains at least one open set of X. 7.21 Let E ∈ F . Then, E ∩ X0 is a constructible set in the irreducible space X0 and therefore exercise 2(ii) implies that E ∩X0 6= X0 or otherwise E ∩X0 contains a non-empty open subset of X0. For the converse, we use the hint. Assume that E /∈ F ; this implies that the collection G of closed subsets X ′ ⊂ X such that E ∩X ′ /∈ F is non-empty (it contains X) and thus has a minimal element X0, since X is Noetherian. Let X = Y1 ∪ · · · ∪ Yl, where each Yi is irreducible. Then, E ∩X0 = (E ∩ Y1) ∪ · · · ∪ (E ∩ Yk), and by the minimality of X0 each E∩Yi b