阿蒂亚交换代数导引习题解答

Solutions to Introduction to Commutative Algebra Steven V Sam June 28, 2008 Contents 1Rings and Ideals1 2Modules5 3Rings and Modules of Fractions6 5Integral Dependence and Valuations7 8Artin rings9 9Discrete valuation rings and Dedekind domains10 1Rings and Ideals 1. Exercise. Let x be a nilpotent element of a ring A. Show that 1 + x is a unit of A. Deduce that the sum of a nilpotent element and a unit is a unit. Solution. Let u be a unit and x be nilpotent. Suppose u+x is not a unit. Then u+x m for some maximal ideal m. But m is prime, so x m, which implies u m. Thus, u + x is a unit. Letting u = 1 gives the fi rst statement. 4. Exercise. In the ring Ax, the Jacobson radical is equal to the nilradical. Solution. We want to show that m maximal m = p prime p. Since every maximal ideal is prime, the inclusion is obvious.On the other hand, choose f T m and write f = anxn+ + a0. Then 1 fg is a unit for all g Ax, so by (Ex. 2(i), 1 a0b is a unit for all b A, which means a0is nilpotent. In particular, if g = 1, then 1 + f is a unit, so aiis nilpotent for n i 1. By (Ex. 2(ii), f is nilpotent, which means f T p, giving the other inclusion. by M. F. Atiyah and I. G. MacDonald 1 1RINGS AND IDEALS2 6. Exercise. A ring A is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element e such that e2= e 6= 0). Prove that the nilradical and Jacobson radical of A are equal. Solution.That the nilradical is contained in the Jacobson radical is obvious.If f is not contained in the nilradical, then (f) contains a nonzero idempotent, say fg. We claim 1 fg is not a unit. If it were, then from (1 fg)(1 + fg) = 1 fg, we get 1 + fg = 1, or fg = 0, a contradiction. Thus, 1 fg is not a unit, which means f is not contained in the Jacobson radical. 7. Exercise. Let A be a ring in which every element x satisfi es xn= x for some n 1 (depending on x). Show that every prime ideal in A is maximal. Solution. Let p be a prime ideal. Choose nonzero x A/p. Then there is an n 1 such that xn= x, or x(xn1 1) = 0. Since A/p is an integral domain, xn1= 1. If n = 2, then x = 1; otherwise, the inverse of x is xn2 . Thus, A/p is a fi eld, which means p is a maximal ideal. 8. Exercise. Let A be a ring 6= 0. Show that the set of prime ideals of A has minimal elements with respect to inclusion. Solution. Let (P,) be the set of prime ideals of A where p1 p2if p1 p2. Given any chain C of P, we claim that the intersection p of all prime ideals in C is again a prime ideal. Let xy p. Then xy pifor all pi C. Since each piis prime, either x pior y pi. Either x pifor all pi C or there is a least element p0 that contains x. In the fi rst case, x p, and in the second, we conclude that y p since it must be contained in every prime ideal greater than p0. Thus, every chain has a maximal element, so P has a maximal element. In our case, this maximal element is minimal with respect to inclusion. 9. Exercise. Let a be an ideal 6= (1) in a ring A. Show that a = r(a) a is an intersection of prime ideals. Solution. If a = r(a), then a is an intersection of the prime ideals containing it. Conversely, if a is an intersection of prime ideals, then it must be the intersection of all prime ideals containing it, so a = r(a). 10. Exercise. Let A be a ring, N its nilradical. Show that the following are equivalent: i) A has exactly one prime ideal; ii) every element of A is either a unit or nilpotent; iii) A/N is a fi eld. Solution. i) ii) If A has exactly one prime ideal, then it must be maximal, and is equal to N. If x A is not nilpotent, then x / N, so (x) $ N, which means (x) = A, so x is a unit. ii) iii) If p is an ideal with N $ p, then p must contain a unit, so N is maximal, and A/N is a fi eld. iii) i) Since N is maximal and is the intersection of all prime ideals of A, it must be the only prime ideal of A. 11. Exercise. A ring A is Boolean if x2= x for all x A. In a Boolean ring A, show that i) 2x = 0 for all x A; ii) every prime ideal p is maximal, and A/p is a fi eld with two elements; 1RINGS AND IDEALS3 iii) every fi nitely generated ideal in A is principal. Solution. (a) Note that (2x)2= 2x = 4x2= 4x, so 2x = 4x leads to 2x = 0. (b) The fact that every prime ideal is maximal follows from exercise 7. In A/p, if x 6= 0, then x2= x, and x is invertible, which gives x = 1, so A/p has 2 elements. (c) It is enough to show that if an ideal is generated by 2 elements, then it is principal by using induction. Then we claim (x,y) = (xxy+y). The inclusion is clear, and the identities x(x xy + y) = x and y(x xy + y) = y give the other inclusion. 12. Exercise. A local ring contains no idempotent 6= 0,1. Solution. Let m be the maximal ideal of a local ring A, and let e be an idempotent. If e / m, then e is a unit, and e2= e, which means e = 1. If e m, then (1 e)(1 + e) = 1 e2= 1 e. Since 1 e / m (otherwise 1 m), 1 e is a unit, so we get 1 + e = 1, or e = 0. 15. Exercise. Let A be a ring and let X be the set of all prime ideals of A. For each subset E of A, let V (E) denote the set of all prime ideals of A which contain E. Prove that i) if a is the ideal generated by E, then V (E) = V (a) = V (r(a). ii) V (0) = X, V (1) = . iii) if (Ei)iIis any family of subsets of A, then V iI Ei ! = iI V (Ei). iv) V (a b) = V (ab) = V (a) V (b) for any ideals a,b of A. Solution. (a) That V (E) = V (a) follows from the fact that a is the smallest ideal containing E, so must be contained in any prime ideal containing E. Since r(a) is the intersection of all prime ideals containing a, one gets V (a) V (r(a). Conversely, a r(a), so the other inclusion follows. (b) This is clear from the fact that every ideal contains 0 and no prime ideal contains 1. (c) If p V (SEi), then p contains each Ei, which means p T V (Ei). On the other hand, if p T V (Ei), then p contains each Ei, so p V (SEi). (d) Since a b ab, we get the inclusion V (ab) V (a b). If a prime ideal contains a b, then it must contain either a or b by Prop. 1.11, so this gives V (a b) V (a) V (b). Finally, the inclusions ab a and ab b give V (a) V (ab) and V (b) V (ab), so V (a) V (b) V (ab). 17. Exercise. For each f A, let Xfdenote the complement of V (f) in X = Spec(A). The sets Xfare open. Show that they form a basis of open sets for the Zariski topology, and that i) Xf Xg= Xfg; ii) Xf= f is nilpotent; iii) Xf= X f is a unit; iv) Xf= Xg r(f) = r(g); 1RINGS AND IDEALS4 v) X is quasi-compact (that is, every open covering of X has a fi nite subcovering). vi) More generally, each Xfis quasi-compact. vii) An open subset of X is quasi-compact if and only if it is a fi nite union of sets Xf. Solution. Since V (1) = , every prime ideal is contained in X1. Also, XfXg= X (V (f) V (g), and V (f)V (g) = V (fg) because any element of (f)(g) is of the form afg for a A, so Xf Xg= Xfg, and thus the Xfform a basis of open sets for the Zariski topology. (a) This was done above. (b) Xf= V (f) = X f is contained in the nilradical of A f is nilpotent. (c) Xf= X V (f) = f is not contained in any prime ideal f is a unit. (d) From (Ex. 15(i), V (f) = r(f) and V (g) = r(g), so the equivalence is clear. (e) Each open subset of X is the union of Xfi, so we may assume that the open covering consists of Xfiwhere i I for some indexing set I. Then S iIXfi = X, so T iIV (fi) = = V (SiI(fi) = V (a) where a is the ideal generated by the union. This means that a is the unit ideal, so 1 = P iJ gifi for some fi nite sum, and the set XfiiJ gives a fi nite subcovering of X. (f) Again, we may assume that the open covering of Xfis XfiiIfor some indexing set I. Furthermore, since Xfi Xf= Xfif, we can also assume that Xfi Xfsince if XfifiJ gives a fi nite subcover, then so does XfiiJ. Then the argument becomes the same as the previous part. (g) If an open subset X is quasi-compact, then it is clearly a fi nite union of sets Xf. Conversely, if X is a fi nite union of sets Xf, then let XiiIbe an open covering of X. Then Xi is also an open covering for each Xf , so there is a fi nite subcovering for each. Take the union of these fi nite subcoverings to get a fi nite subcovering for X. 18. Exercise. For psychological reasons it is sometimes convenient to denote a prime ideal of A by a letter such as x or y when thinking of it as a point of X = Spec(A). When thinking of x as a prime ideal of A, we denote it by px(logically, of course, it is the same thing). Show that i) the set x is closed (we say that x is a “closed point”) in Spec(A) pxis maximal; ii) x = V (px); iii) y x px py; iv) X is a T0-space (this means that if x,y are distinct points of X, then either there is a neighborhood of x which does not contain y, or else there is a neighborhood of y which does not contain x). Solution. (a) The closed sets in Spec(A) are of the form V (a) ideal a of A.If x is closed, then V (a) = x implies that px= a and no other prime ideal contains a, so pxis maximal. On the other hand, if pxis maximal, then V (px) = x, which is evidently closed. (b) The closure of x is the intersection of all closed sets containing x. Note that V (px) is closed and contains x. It is the smallest such set for the following reason. If x Y V (px), then Y = V (a) for some ideal a, which means a px, and hence every prime ideal containing pxis in Y , which gives V (px) Y . (c) By defi nition, y V (px) if and only if px py, and x = V (px) from part (b). 2MODULES5 (d) If x 6= y, then either px* pyor py* px . Without loss of generality, assume it is the fi rst case. Then y / V (px) = x. The complement of V (px) is open and contains y, but does not contain x, which means X is a T0-space. 2Modules 1. Exercise. Show that (Z/mZ) Z(Z/nZ) = 0 if m,n are coprime. Solution. Let ab (Z/mZ)Z(Z/nZ). Since m and n are coprime, there is a multiplicative inverse n1in Z/mZ, so a b = na n1b = 0. The generators are thus all 0. 2. Exercise. Let A be a ring, a an ideal, M an A-module. Show that (A/a) AM is isomorphic to M/aM. Solution. Tensoring the exact sequence 0 a A A/a 0 with M gives a M AM A/aM 0. There is a canonical isomorphism aM = aM defi ned by rm 7 rm, and A M = M, which means we really have an exact sequence aM M A/a M 0. The cokernel of the map aM M is M/aM, which gives the desired isomorphism M/aM = A/a M. 3. Exercise. Let A be a local ring, M and N fi nitely generated A-modules. Prove that if M N = 0, then M = 0 or N = 0. Solution. Let m be the maximal ideal of A and let k = A/m be the residue fi eld.Then (M AN) Ak = 0, but this is also equal to (M Ak) k(N Ak). But both M Ak and N Ak are vector spaces, so one of them must be zero, say M Ak. By Proposition 2.8 and (Ex. 2.2), M Ak = 0 implies that M = 0. 4. Exercise. Let Mi(i I) be any family of A-modules, and let M be their direct sum. Prove that M is fl at each Mi is fl at. Solution. The module M is fl at if and only if for all short exact sequences 0 / N0 / N / N00 / 0 , we have that 0 / N0 M / N M / N00 M / 0 is a short exact sequence. Since (LMi) N = L(M i N) for all A-modules N, we get that 0 / L(N0 Mi) / L(N M i) / L(N00 Mi) / 0 is exact. This is exact if and only if it is exact at each direct summand, which gives that each Mi is fl at. 5. Exercise. Let Ax be the ring of polynomials in one indeterminate over a ring A. Prove that Ax is a fl at A-algebra. Solution. We may write Ax = L i0Ai where Ai = A. Since A is a fl at A-module, it follows by (Ex. 2.4) that Ax is also fl at. 8. Exercise. i) If M and N are fl at A-modules, then so is M AN. ii) If B is a fl at A-algebra and N is a fl at B-module, then N is fl at as an A-module. 3RINGS AND MODULES OF FRACTIONS6 Solution. Let f : F G be an injective map of A-modules. Then f 1M: F AM G M is injective, and so is (f 1M)1N: (F AM)AN (GAM)A N. The fi rst part follows from the isomorphism F A(M AN)= (F AM)AN. The second part follows by the fact that any A-module can be made a B-module by tensoring with B. By assumption, this base change is fl at, so injectivity is preserved among A-modules. 9. Exercise. Let 0 M0 M M00 0 be an exact sequence of A-modules. If M0and M00 are fi nitely generated, then so is M. Solution. Without loss of generality, we assume that M/M0= M00, so that the factor module M/M0 is fi nitely generated. Every element of M is of the form m + m0for some m M and m0 M0. Since M0 is fi nitely generated, each coset (thinking of as a subset of M) is fi nitely generated, so we conclude that M is fi nitely generated. 24. Exercise. If M is an A-module, the following are equivalent: i) M is fl at; ii) TorA n(M,N) = 0 for all n 0 and all A-modules N; iii) TorA 1(M,N) = 0 for all A-modules N. Solution. We fi rst show that i) implies ii).Let F be a projective resolution of N.Then tensoring with M is an exact functor, so the resulting complex has trivial homology. That ii) implies iii) is obvious. To see that iii) implies i), let 0 / N0 / N / N00 / 0 be an exact sequence. Then there is a long exact sequence TorA 1(M,N 00) / M AN0 / M AN / M AN00 / 0 , so the injectivity of N0 N is preserved. 25. Exercise. Let 0 N0 N N00 0 be an exact sequence, with N00 fl at. Then N0 is fl at N is fl at. Solution. The long exact sequence induced by Tor is TorA 2(M,N 00) / TorA 1(M,N 0) / TorA 1(M,N) / TorA 1(M,N 00 ) . By condition ii) of (Ex. 2.24), TorA 2(M,N 00) = TorA 1(M,N 00) = 0, so TorA 1(M,N 0) = TorA 1(M,N), which fi nishes the exercise by condition iii) of (Ex. 2.24). 3Rings and Modules of Fractions 1. Exercise. Let S be a multiplicatively closed subset of a ring A, and let M be a fi nitely generated A-module. Prove that S1M = 0 if and only if there exists s S such that sM = 0. Solution. If there exists s S such that sM = 0, then by defi nition, m1/s1= m2/s2for mi M and si S since s(s2m1 s1m2) = 0, so S1M = 0. Conversely, if S1M = 0, then pick a generating set m1,.,mr of M. Then for each mi, we have simi= 0 for some sisince mi/1 = 0/1 in S1M. Then s = s1sr satisfi es sM = 0. 5INTEGRAL DEPENDENCE AND VALUATIONS7 2. Exercise. Let a be an ideal of a ring A, and let S = 1+a. Show that S1a is contained in the Jacobson radical of S1A. Use this result and Nakayamas lemma to give a proof of (2.5) which does not depend on determinants. Solution. Pick x,y a, we wish to show that (1 + x)1y is in the Jacobson radical of S1A. By Proposition 1.9, it is enough to show that for all z S1A, 1 (1 + x)1yz is a unit in S1A. Write z = z1/z2with z1 A, z2 S. Then 1 (1 + x)1yz = (1 + x yz1)/z2(1 + x), which is a unit because 1 + x yz1 1 + a. Now we use this to prove Corollary 2.5. Let M be a fi nitely generated A-module and a an ideal of A such that aM = M. Now let S = 1 + a, then S1aM = S1M implies S1M = 0 by Proposition 2.6, so using (Ex. 3.1), there exists s S such that sM = 0. Such an s is of the form 1 + x for x a, so Corollary 2.5 is proven. 3. Exercise. Let A be a ring, let S and T be two multiplicatively closed subsets of A, and let U be the image of T in S1A. Show that the rings (ST)1A and U1(S1A) are isomorphic. Solution. Consider the composition A S1A U1(S1A) and the map A (ST)1A. Since U1(S1A) has elements of ST inverted, we get a map (ST)1A U1(S1A), and similarly, since the elements of U are inverted in (ST)1A, we also have a map U1(S1A) (ST)1A. These two maps must be two-sided inverses of one another by the universal property of localization, so (ST)1A= U1(S1A). 4. Exercise. Let f : A B be a homomorphism of rings and let S be a multiplicatively closed subset of A. Let T = f(S). Show that S1B and T1B are isomorphic as S1A-modules. Solution. The canonical maps B S1B and B T1B are A-linear, and we can use an argument similar to (Ex. 3.3) to fi nd maps S1B T1B and T1B S1B which are inverses of one another. 5. Exercise. Let A be a ring. Suppose that, for each prime ideal p, the local ring Aphas no nilpotent element 6= 0. Show that A has no nilpotent element 6= 0. If each Apis an integral domain, is A necessarily an integral domain? Solution. Suppose x A is nilpotent, so that xn= 0 for some n. Then the image of x in each localization Apis also nilpotent, so for each prime ideal p, there exists an element s A p such that sx = 0. The ideal generated by all such s cannot be proper, or else it is contained in a maximal ideal m, but there is some s / m by hypothesis, hence they generate the unit ideal. Using a relation a1s1+arsr, we get x = (a1s1+arsr) = 0, so A has no nilpotents either. However, let A = C C. Then A has two prime ideals 0 C and C 0. The localization at both is a copy of C, so we have an example of ring whose localizations are all domains, but is itself not a domain. 5Integral Dependence and Valuations 1. Exercise. Let f : A B be an integral homomorphism of rings. Show that f: SpecB SpecA is a closed mapping, i.e., that it maps closed sets to closed sets. Solution. First assume that f is injective. Let V (I) B be a closed subset of B. Then the image f(V (I) is the set f(A) p | p I. We need to show that there exists an ideal J A such that f(A) p J if and only if p I. By Theorem 5.11, J = f(A) I will work. 5INTEGRAL DEPENDENCE AND VALUATIONS8 For general f : A B, let A0= A/kerf. We have a factorization A A0 B, which gives the maps SpecB SpecA0 SpecA. We know from above that SpecB SpecA0is closed, so we need to show that SpecA0 SpecA is a closed map. Given a closed subset V (I) SpecA0, its image in SpecA is V (kerf), so we are done. 2. Exercise. Let A be a subring of a ring B such that B is integral over A,