Timeline for Which rings are subrings of matrix rings?
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Nov 10, 2009 at 19:46 | comment | added | Graham Leuschke | Right. Conversely, if M is an MCM T-module, then it is an MCM A-module since T is finite over A; as A is regular, every MCM A-module is free over A by Auslander-Buchsbaum. Since every T-endomorphism of M is also an A-endomorphism, we have T \subseteq \End_T(M) \subseteq \End_A(M), so T is a matrix ring. | |
| Nov 10, 2009 at 13:03 | comment | added | David E Speyer | Let A be local. As I understand it, an MCM module would mean a T-module M for which there is a regular sequence of length n? That's definitely true if T is a subring of a matrix ring: If T is contained in the m x m matrices then A^m is a T-module and, since it is free as an A-module, the generators of the maximal idea of A give a regular sequence. | |
| Nov 10, 2009 at 3:06 | comment | added | Graham Leuschke | I haven't yet thought at all about whether this is equivalent to T having a MCM module, but just wanted to point out that (a) any local ring containing a field admits a MCM module by work of Hochster in the 70s; (b) in dimensions 1 and 2 existence of MCM modules is trivial; and (c) MCM modules exist in dimension 3 (mixed characteristic) by work of Heitmann from ~2002. | |
| Nov 9, 2009 at 23:17 | comment | added | Kevin Buzzard | I have now independently heard that the case of $A$ regular is "open and difficult". If T is finite as an A-module then (and I don't understand what I'm writing here) "this is precisely the issue of whether T has a finitely generated maximal Cohen-Macaulay module. This is true in dim < 3 and not known in dimension 3 in any characteristic, although it is known to be true in many instances. " | |
| Nov 9, 2009 at 20:41 | history | answered | David E Speyer | CC BY-SA 2.5 |