Algebra II, Take Home Exam To be submitted on Canvas by 10:00 P.M. on Wednesday, May 6. You may use material from Jacobson and from the class notes in your solutions to these problems. If you use a result from these sources you should cite it (e.g. \The proposition on slide 6 of Lecture 5"). You may not consult with other people or use other references. 1) Let A be a commutative ring in which every element x satisfies xn = x for some n > 1 (depending on x). Show that every prime ideal in A is maximal. Solution. Suppose that P is a prime ideal of R. For x 2 R, let x be the class of x in R=P . Suppose x 6= 0. xn = x for some n > 1 implies xn = x which implies x(xn−1 − 1) = 0. Thus xn−1 = 1 since R=P is an integral domain. Now n > 1 implies x is a unit in R=P , with inverse xn−2. Thus R=P is a field so P is a maximal ideal. 2) Let A be a commutative ring with nilradical Nil(A). Show that the following are equivalent. i) A has exactly one prime ideal. ii) Every element of A is either a unit or a nilpotent. iii) A=Nil(A) is a field. Solution. i) ) ii). Suppose A has exactly one prime ideal P . By Proposition 8, on slide 10, lecture 12, the nilradical Nil(A) is the intersection of the prime ideals of A so Nil(A) = P . Thus every element of P is nilpotent. If x 2 A n P , then x cannot be contained in a maximal ideal of A since P is the unique maximal ideal of A. Thus x is a unit in A. ii) ) iii). Suppose x 2 A. Let x be the class of x in A=Nil(A). Now if x is nilpotent then x = 0 in A=Nil(A) and if x is not nilpotent then x is a unit in A which implies x is a unit in A=Nil(A). Thus A=Nil(A) is a field. iii) ) i). The assumption of iii) implies m = Nil(A) is a maximal ideal of A. By Proposition 8, slide 10 of Lecture 12, Nil(A) = \Q where the intersection is over all prime ideals of A. Thus m ⊂ Q for all prime ideals Q of A. Thus Q = m is the only prime ideal of A since m is maximal. Thus m is the unique prime ideal of A. 3) Let A be a (commutative) local ring, M and N be finitely generated A-modules. Prove that if M ⊗A N = 0 then M = 0 or N = 0. Hint: Apply the Corollary of Slide 6 of Lecture 10 when I = m is the maximal ideal of A. You may also use the following Lemma without proof, but you must explain when you use it. Lemma 0.1. Suppose that ' : A ! B is a homomorphism of commutative rings and E, E0 are two A-modules. Then there is an isomorphism of B-modules 0 ∼ 0 (E ⊗A B) ⊗B (E ⊗A B) = (E ⊗A E ) ⊗A B: A proof of this lemma can be found in Proposition 3, page AII.8.3 of Algebra, by N. Bourbaki. 1 Solution (to Problem 3). Let F = A=m which is a field. Suppose M ⊗A N = 0. Then ∼ 0 = (M ⊗A N) ⊗A A=m = (M ⊗A A=m) ⊗A=m (N ⊗A A=m) by the Lemma. But since F is a field, M ⊗A A=m and N ⊗A A=m are vector spaces over F and thus are free F -modules. M is a finitely generated F -module since the images of a finite set of generators of M as an A-module in M ⊗A A=m are generators of M ⊗A A=m as an A=m-module. Similarly, N ⊗A A=m is a finitely generated A=m-module. Thus both M ⊗A A=m and N ⊗A A=m have finite ranks. Thus 0 = rank((M ⊗A A=m) ⊗A=m (N ⊗A A=m)) = rank(M ⊗A A=m)rank(N ⊗A A=m) by the corollary on slide 20 of Lecture 9. Thus one of M ⊗A A=m or N ⊗A A=m has rank 0 and is thus zero. Without loss of generality, M ⊗A A=m = 0. Now ∼ M ⊗A A=m = M=mM by the corollary of slide 6 if Lecture 10. M=mM = 0 implies mM = M. Since A is a local ring with maximal ideal m (so that m is the Jacobson radical of A) we have that M = 0 by Nakayama's Lemma (Proposition 16, slide 5 of Lecture 13). 4) Let A be a commutative ring, M be a finitely generated A-module and ' : M ! An be a surjective homomorphism. Show that Ker ' is a finitely generated A-module. Solution. We have a short exact sequence of A-modules ' (1) 0 ! Ker ' !ι M ! An ! 0: An is a projective module over A since it is a free module (by the theorem on slide 17 of Lecture 5). Thus (1) is split exact by the theorem on slide 19 of Lecture 5. Thus there is a homomorphism : M ! Ker ' such that ι = idKer ' by the theorem on slide 3 of Lecture 5. Thus is surjective. Since M is a finitely generated A-module its image Ker ' is thus a finitely generated A-module. Explicitely, if y1; : : : ; yk generate M as an A-module, then (y1); : : : ; (yk) generate Ker ' as an A-module. 5) Let A be a commutative ring, M be an A-module and I an ideal of A. Suppose that Mm = 0 for all maximal ideals m which contain I. Prove that M = IM. ∼ Solution. Suppose m is a maximal ideal of A. Then (M=IM)m = Mm=(IM)m by Proposi- tion 20, slide 2 of Lecture 14. Suppose I 6⊂ m. Then there exists y 2 I \ (A n m). Suppose z z 1 yz x 2 Mm. Then x = s for some z 2 M and s 2 A n m and so x = s = y s 2 (IM)m. Thus (IM)m = Mm and so (M=IM)m = 0. Now suppose I ⊂ M. Then Mm = 0 by assumption and so (M=IM)m = 0. Thus (M=IM)m = 0 for all maximal ideals m of A and so M=IM = 0 by Proposition 22, slide 5 of Lecture 14. Thus M = IM. 2.