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What's the easiest (by which I mean uses the least fancy machinery) proof of the direct summand conjecture in dimension 2?

Recall that the direct summand conjecture says that:

Conjecture (Hochster): If $R$ is a regular ring and $S$ is a module finite integral extension, then $R \to S$ splits as a map of $R$-modules.

It is trivial in characteristic zero (via the trace map) and not that hard in characteristic $p > 0$ using Frobenius-type methods. In mixed characteristic it is known up to dimension 3.

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Are you asking for the easiest proof? –  Manny Reyes Oct 5 '10 at 17:17
    
Yes. I am asking for the easiest proof. I fixed the typo above. –  Karl Schwede Oct 5 '10 at 20:58

1 Answer 1

up vote 5 down vote accepted

You may assume that $R,S$ are complete and $S$ is a domain. Now take the integral closure $T$ of $S$, which is $S$-finite. Since we are in dimension $2$, $T$ is maximal Cohen-Macaulay module over $R$, so $T$ is $R$-free. Thus the composition map $R\to T$ splits (as it takes $1$ to $1$) whence the map $R\to S$ splits.

The moral of this is that existence of small Cohen-Macaulay modules implies a lot of things, and you can get that for free in dimension $2$ via integral closure.

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May be Mel's CBMS note has a reference? Embarrassingly, I do not own a copy (-:. –  Hailong Dao Oct 5 '10 at 17:58
    
I looked in Mel's CBMS book before asking, but I didn't see this there. That's a very nice proof, thanks! Do you know of any other approaches? –  Karl Schwede Oct 5 '10 at 20:45
    
Karl, the other way I know is via the monomial conjecture, the reduction to that case is elementary: you need that the extension is cyclically pure, and it's enough to use ideals gen. by powers of elements in a s.o.p. In dimension 2, monomial conjecture amounts to showing $x^ty^t \notin (x^{t+1},y^{t+1})$ for all $t>0$, $x,y$ s.o.p. Probably that has an elementary proof. –  Hailong Dao Oct 6 '10 at 2:39
    
Long, thanks. That's very helpful. –  Karl Schwede Oct 6 '10 at 4:56

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