Finite morphism inducing an isomorphism away from codimension 2 to a strict henselian ring of dimension 2 satisfying $S_2$ is an isomorphism?

Question

Let $R$ be a strictly henselian local ring of dimension 2, satisfying Serre's condition $S_2$. Let $X = \text{Spec }R$, and let $f : Y\rightarrow X$ be a finite morphism inducing an isomorphism over the complement of the unique closed point of $X$. Is $f$ an isomorphism?

Is this true in higher dimensions as well (possibly replacing $S_2$ with $S_n$?)

I know this is true if we also assume $R$ normal (so that it also satisfies Serre's condition $R_1$).

References would be appreciated!

Answer 1

This is not quite true as stated, even for normal rings rings $R$. For example let $R = k[[x,y]]$ and $S = k[[x,y]] \times k$ and the map $R$ to $S$ sends $x,y$ to themselves in the first coordinate and sends $x, y$ to zero in the second.

But I think this is basically the only thing that can go wrong. Say now that $Y = \mathrm{Spec} S$, and so we are just looking at a finite inclusion of Noetherian rings $R \subseteq S$ with $R$ local satisfying $S_2$.

We have a map from $S$ to the $R$-module $S$-2-ification (in this case $S' = \Gamma(U, f_* O_Y)$ where $U$ is the punctured Spec), call it $S'$. It follows immediately that $S' = R$ since $R$ is $S_2$ (a map between two S2 modules that is an isomorphism in codim 1 is an isomorphism, see for example Hartshorne: Generalized divisors on Gorenstein rings).

In particular, we have a ring map $S \to R$ such that the composition $R \to S \to R$ is an isomorphism. Viewing $S$ as an $R$-module, this means that $S = R \oplus M$ for some $R$-module $M$. Furthermore, this $M$ must be supported at the origin of $M$ (since it goes away if we work away from the origin). Now we know $M$ is killed by a power of $m_R$, so we can construct an ideal in $S$, $I$ the set of things that kill $m_R^n S$ for some $n \gg 0$. Since $R$ is Serre-2, $I \cap R = 0$, hence we have an injection $R \to S/I$, and this must indeed be an isomorphism.

In other words, as long as $H^0_m(S) = 0$ (here $H^0_m(S)$ is the submodule of $S$ killed by a power of $m$, also known as the zeroth local cohomology / cohomology with support at $m$), $R = S$.

This doesn't require any additional hypothesis in higher dimensions.