MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Thanks in advance.

share|cite|improve this question
up vote 10 down vote accepted

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.

share|cite|improve this answer

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).

share|cite|improve this answer

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.