SOME FINITENESS CONDITIONS IN ALGEBRA ANDREW BAKER Throughout, R will be a commutative and unital ring. However, much of the theory also works for non-commutative rings. Recall that the Jacobson radical of R is the intersection of all the maximal ideals of R, \ J(R) = m: maximal m/R This is an ideal of R which is characterized as the largest ideal I/R for which 8t 2 I 8s 2 R 1 + st 2 R£: 1. Nakayama's lemma Theorem 1.1 (Nakayama's lemma). Let M be a ¯nitely generated R-module and I/R. Suppose that IM = M. (a) There is an element t 2 I for which (1 + t)M = 0. (b) If I ⊆ J(R), then M = 0. Proof. See [4] for the commutative case and also [5] for (b) in the general case. ¤ Corollary 1.2. Suppose that I ⊆ J(R) and N 6 M with M=N ¯nitely generated. If N + IM = M, then N = M. Proof. We have IM=N = M=N, hence M=N = 0. ¤ Here is an important sample application of Nakayama's lemma and its corollary. Proposition 1.3. Suppose that M is a ¯nitely generated R-module and that F is a ¯nitely generated free R-module for which there is an isomorphism F= J(R)F =» M= J(R)M. Then for any factorization F / F= J(R)F ' »= ² ² M / M= J(R)M the map ': F ¡! M is an epimorphism. Proof. We have im ' + J(R)M = M. ¤ This means that construction of `minimal' free covers can be done by working modulo J(R). For example, if R is a local ring with unique maximal ideal m, then M=mM is a ¯nite dimensional vector space over the ¯eld R=m. Given a basis for M=mM, there is a corresponding isomorphism » » Rd=mRd ¡!= (R=m)d ¡!= M=mM which lifts (non-uniquely) to a homomorphism Rd ¡! M which is an epimorphism by Proposition 1.3. If M is projective over R, this implies that M is free of rank equal to dimR=m M=mM. Here is another application. Proposition 1.4. Let S be a commutative ring. Suppose that M is a ¯nitely generated S-module and ': F ¡! M is an S-module epimorphism. Then ' is an isomorphism. Date: 21/04/2009. 1 2 ANDREW BAKER Proof. Let R = S[X] and view M as the R-module obtained by with extending the existing action of S by taking X ¢ m = '(m): Consider the ideal RX / R, and notice that (RX)M = R'M = M. Hence for some f(X) 2 R, (1 ¡ Xf(X))M = (1 ¡ f(X)X)M = 0; from which it follows that ' is invertible with inverse f('). ¤ 2. Noetherian and coherence conditions Let R be a commutative ring. De¯nition 2.1. Let M be an R-module. (a) M is ¯nitely generated if for some n 2 N there is a short exact sequence Rn ¡! M ! 0: (b) M is ¯nitely presented if for some m; n 2 N there is a short exact sequence Rm ¡! Rn ¡! M ! 0; i.e., there is a short exact sequence Rn ¡!¼ M ! 0: with ker ¼ ¯nitely generated. An important result is the following. Proposition 2.2. Suppose that M is ¯nitely presented and that there is an exact sequence 0 ! K ¡! P ¡! M ! 0 with P ¯nitely generated projective. Then K is ¯nitely generated. Proof. This requires Schanuel's Lemma which says that if for an R-module N there are exact sequences 0 ! K ¡! P ¡! N ! 0; 0 ! L ¡! Q ¡! N ! 0; with P and Q projective, then P © L =» K © Q: Given this result, let 0 ! ker ¼ ¡! Rn ¡!¼ M ! 0 be short exact with ker ¼ ¯nitely generated. Then P © ker ¼ =» K © Rn: But since P © ker ¼ is ¯nitely generated, so is K © Rn. Therefore K is ¯nitely generated. ¤ Recall that an R-module M is Noetherian if for every increasing chain of submodules Nk 6 M, ¢ ¢ ¢ ⊆ Nk ⊆ Nk+1 ⊆ ¢ ¢ ¢ there is a k0 such that Nk = Nk0 (k > k0): For such a module, every submodule is ¯nitely generated. The ring R is said to be Noetherian if it is a Noetherian R-module. For such a ring, every ¯nitely generated R-module M is Noetherian, hence it is ¯nitely presented. It is now easy to deduce that M admits a free resolution ¢ ¢ ¢ ¡! Fn ¡! Fn¡1 ¡! ¢ ¢ ¢ ¡! F0 ¡! M ! 0; where each Fn is ¯nitely generated and free. If R is also local with maximal ideal m, we can use Nakayama's lemma to show that such a resolution can be chosen to be minimal in the sense that the boundary maps factor as Fn ¡! mFn¡1 ⊆ Fn¡1; and therefore the complex ¢ ¢ ¢ ¡! R=m ­R Fn ¡! R=m ­R Fn¡1 ¡! ¢ ¢ ¢ ¡! R=m ­R F0 ! 0 has trivial di®erentials and so its homology has Hn(R=m ­R F¤) = R=m ­R Fn: SOME FINITENESS CONDITIONS IN ALGEBRA 3 For a more general ring R, a resolution is minimal if the boundary maps factor as Fn ¡! J(R)Fn¡1 ⊆ Fn¡1; and such minimal resolutions exist for ¯nitely generated modules if R= J(R) is semi-simple. What about more general conditions on a ring forcing similar results? De¯nition 2.3. An R-module M is pseudo-coherent if all its ¯nitely generated submodules are ¯nitely presented. A ¯nitely generated pseudo-coherent module M is coherent. The ring R is coherent if it is a coherent R-module. So an R-module M is coherent if and only if it and all its ¯nitely generated submodules are ¯nitely presented. It is standard that a ¯nitely generated module over a Noetherian ring is coherent. But there are more examples. Theorem 2.4 (see [3, theorem 2.3.3]). Let fR®g®2A be a directed ¯ltered system of coherent rings and suppose that R = colim® R® is flat over each R®. Then R is coherent. In particular, this happens if whenever ® 4 ¯, R¯ is flat over R®. Example 2.5. If S is a Noetherian ring, then the in¯nitely generated polynomial ring S[x1; x2;:::] = S[xi : i > 1] = colim S[x1; : : : ; xn] n is coherent but not Noetherian. Example 2.6. Any localization of a coherent ring R is coherent. In particular, if p /R is a prime ideal, Rp is coherent. Example 2.7. If R is coherent and I/R is a ¯nitely generated ideal, then R=I is coherent. Example 2.8. If R is semi-hereditary then every countably generated polynomial algebra over R is coherent. However, in general, if R is coherent, the polynomial ring R[x] need not be coherent. Coherence satis¯es some useful properties. Proposition 2.9. Let 0 ! M 0 ¡! M ¡! M 00 ! 0 be an exact sequence of R-modules. If two out of M 0; M; M 00 are coherent, the third is. Proposition 2.10. Let ': M ¡! N be a homomorphism between coherent modules. Then ker ', im ' and coker ' are all coherent. Proposition 2.11. If the following triangle of homomorphisms of R-modules is exact ' / L `@ M @@ || @@ || @@ || Ã @ ~|| θ N then if two out of L; M; N are coherent so is the third. We also have Theorem 2.12. The ring R is coherent if and only if every ¯nitely presented R-module is coherent. Over a coherent ring R, the coherent modules admit resolutions by ¯nitely generated free modules. For a coherent local ring there are minimal resolutions of coherent modules; the proof is similar to that in the Noetherian case. References [1] N. Bourbaki, El¶ements¶ de Mathematique, Fasc. XXVII: Alg`ebreCommutative, Hermann (1961). [2] J. M. Cohen, Coherent graded rings and the non-existence of spaces of ¯nite stable homotopy type, Comment. Math. 44 (1969), 217{228. [3] S. Glaz, Commutative Coherent Rings, Lecture Notes in Mathematics 1371 (1989). [4] T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag (1999). [5] , A First Course in Noncommutative Rings, Second edition, Graduate Texts in Mathematics 131, Springer- Verlag (2001). [6] H. Matsumura, Commutative Ring Theory, Cambridge University Press (1989). [7] J-P. Serre, Faisceaux alg¶ebriquescoh¶erents, Ann. of Math. 61 (1955), 197{278. Department of Mathematics, University of Glasgow, Glasgow G12 8QW, Scotland. E-mail address: [email protected] URL: http://www.maths.gla.ac.uk/»ajb.