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.



Reference request, direct summand conjecture in dimension 2
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. 


