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Let $R$ and $S$ be regular local rings, and $f: R\rightarrow S$ a surjection that induces an isomorphism on tangent spaces. Is $f$ necessarily an isomorphism?

I believe the answer should be yes, based on how this setup is used in a paper, but I don't see why.

The motivation is to show that the completion of the local ring of a certain scheme is a power series ring.

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1 Answer

up vote 2 down vote accepted

We can write $S=R/I$ and assume that $I\neq 0$. Since the tangent spaces are isomorphic they must have the same dimension. As QiL pointed out in the comments, we have $\dim R=\dim S$. Since $R$ is a domain and $I\neq 0$, we have $\dim S<\dim R$, a contradiction. So $I=0$.

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Thanks; I think I've gleaned enough from your answer to settle my question. I don't quite follow your argument though: what exactly is the contradiction with $I \subset m^2$? EDIT: OK thanks, just didn't understand the definition of regular parameter. – Tony Jan 5 at 21:02
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Note that if $R$ and $S$ are integral domains of the same (finite) dimension, any surjection $R\to S$ must be an isomorphism by comparing the dimensions. – QiL'8 Jan 5 at 21:10
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Dear YACP: Yes, I mean once we know that $R$ and $S$ have the same dimension, the conclusion is immediate because of the above general fact. – QiL'8 Jan 5 at 21:17
@QiL Yes, sure, you are perfectly right! – YACP Jan 5 at 21:25

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