# Cohen Macaulay modules over Cohen Macaulay rings by Y. Yoshino

By Y. Yoshino

The aim of those notes is to give an explanation for intimately a few themes at the intersection of commutative algebra, illustration concept and singularity idea. they're in accordance with lectures given in Tokyo, but in addition comprise new examine. it's the first cohesive account of the world and should supply an invaluable synthesis of contemporary examine for algebraists.

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**Example text**

That is, L is isomorphic to a direct summand of M. Hence there are only a finite number of such L. | We close this chapter by noting the following significant result of Auslander. 22) THEOREM. (Auslander [7]) / / R has only a finite number of isomorphism classes of indecomposable CM modules, then R is an isolated singularity. 2), to prove that £(R) admits AR sequences. To see this, let M be an indecomposable CM module which is not free. 13). To do this, let F be the kernel of the natural epimorphism ( , M) —• 5 M and let {JVi, N2,..

Syzygy of Honift(syz tr(M), R) up to free summands. 1) shows that Hom^(tr(M), R) is the second syzygy of M. ) Therefore M is isomorphic to the (d — 2)-th syzygy of Hom^(syz^ tr(M), R) up to free summands. As a consequence, we obtain isomorphisms M ~ s y z ^ H o m ^ s y z * tr(M), R)) ^ , R)) 0 syz rf - 2 (Hom il (y, R)), where ~ stands for the isomorphism up to free summands. As remarked before, neither module in the last term is null, so we see that M is actually decomposed. This contradiction proves the lemma.

We will see that this is actually an object in mod(<£) if M is in <£. 15) LEMMA. For any CM module M, the functor Hom#( mod(Q. That is, Hom^f , M) is a finitely presented functor. , M) is an object in PROOF: Consider a free cover of M : F -^ M —• 0. It is sufficient to show that the cokernel of ( ,g) is the functor Hom#( , M). To see this, it is enough to prove that the image of ( , q) coincides with $*( , M). ) So the lemma follows from the following fact which is obvious from the definition. 1) For a homomorphism / € (X, M) with X G €, if / passes through a free module, then it passes through F.