# Krull-Schmidt Analogue for Complete / Graded Rings

Over the ring $\mathbb{Z}$, all finitely-generated modules decompose uniquely as a direct sum of indecomposable submodules; that's the Krull-Schmidt theorem.

I'm given to understand that if a (commutative, Noetherian) ring $R$ is $\mathbb{N}$-graded over a field $k$ (and the degree zero part of $R$ is equal to $k$), then $R$ satisfies this same Krull-Schmidt condition. I'm told that the same holds if $R$ is a complete local ring over a field.

On the other hand, I can't find a good reference for either of these facts. (I can find references to the statements of both facts, but I dislike the notion of citing an unsupported assertion...) So: can anyone point me to a good proof of a Krull-Schmidt theorem for graded or complete local rings? Thanks!

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You should look at Atiyah's aricle "On the Krull-Schmidt theorem with applications to sheaves", freely available here. He gives fairly general conditions on an exact category for the Krull-Remak-Schmidt property to hold.
One of the examples he gives as an application of his criterion is that of coherent sheaves on a projective variety, which translated back into the language of algebra should give the result you request about graded algebras .

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Dear Georges, can you say a few more words on how to deduce the graded ring case from Atiyah's article? I was only able to link it to the category of graded modules over $R$. –  Hailong Dao Oct 4 '11 at 15:08
Dear @Hailong, as you can check in the last words of my post my answer only addressed the case of graded algebras (since Graham had already treated the case of complete local rings, but not that of graded rings). I apologize if my phrasing was ambiguous and I thank you for allowing me to clarify this point. –  Georges Elencwajg Oct 4 '11 at 17:44
Dear Georges, thank you for the clarification. My trouble was more in the sense that there are non-graded modules. Perhaps I was missing something? –  Hailong Dao Oct 4 '11 at 22:31