# Restricting finite morphism to finite etale morphim

If $f:X \rightarrow S$ is locally of finite type, there is unique largest open subset $U$ in $X$ such that $f|U$ is etale.

Suppose $f$ is finite and $U$ is nonempty. Is it true that $f|U$ is finite etale?

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No, because $U\to S$ finite implies that $U\to X$ is finite (at least when $X\to S$ is separated), so $U$ would be closed in $X$.

If you want an example, take a non-trivial morphism from a projective smooth curve $X$ to the projective line over $\mathbb C$.

You might ask whether $U\to f(U)$ (if the latter is open in $S$) is finite, but this is not true either. Consider for example $S=\mathrm{Spec}\mathbb Z$ and $X=\mathrm{Spec}\mathbb Z[t]$ with $t^3+t^2+2t+2=0$. In the fiber above $2$, there is one étale point and one ramified point.

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No. If $f^{-1}(f(U))\neq U$, then $f|_U$ is not proper and hence not finite. This can easily happen if there are unramified points mapping to a branch point. (In other words if there are unramified and ramified points mapping to the same image).

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Here is an example of a finite morphism with no finite etale restriction.

Let $X = S = \mathbb{P}^1_k$ and $f : X \rightarrow S$ is defined by sending $x$ to $x^2$. Then this map is ramified at $0, \infty$. so $U = \mathbb{P}^1_k -\{0,\infty\}$.

Then restriction of $f$ can't be finite and etale. It should be open and closed so, surjective. But $\mathbb{P}^1_k$ has no nontrivial finite etale cover.

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This is not a very good example, because $f|_U$ is finite over its image. – Sándor Kovács Mar 24 '11 at 22:23