0
$\begingroup$

Let $Y$ be a reduced noetherian $1$-dimensional scheme such that the normalization morphism $f:X \longrightarrow Y$ is finite. Let $g:Y\longrightarrow Z$ be a finite flat morphism, where $Z$ is a connected (1-dimensional) Dedekind scheme.

Suppose that the morphism $g\circ f$ from $X$ to $Z$ is etale.

Question. Is the morphism $g:Y\longrightarrow Z$ etale?

Remark. The hypothesis on the dimension is not necessary probably.

Remark. One may assume $Y$ to be integral.

Remark. To assure that $f$ is finite one may suppose that $Y$ is excellent.

Interesting cases one may consider are $Z\subset \mathrm{Spec} \mathbf{Z}$ or $Z\subset \mathbf{P}^1_k$ non-empty open.

$\endgroup$

1 Answer 1

1
$\begingroup$

If $g$ is étale, then $Y$ is regular because $Z$ is regular. Thus $X=Y$. So the anwser to your question is yes if and only if $X=Y$.

$\endgroup$
5
  • $\begingroup$ If $Y$ is not normal then can the hypothesis of the question ever be true? i.e. does there ever exist $Z$ and some map $Y\to Z$ as in the question such that the induced map $X\to Z$ is etale with $X$ the normalization? $\endgroup$ Jul 27, 2011 at 17:43
  • 2
    $\begingroup$ @Kevin: Yes, start with an étale double cover $X\to Z$ with a fibre consisting of two points. Construct $Y$ as the nodal "curve" with these two points identified. $\endgroup$ Jul 27, 2011 at 19:00
  • 1
    $\begingroup$ @Kevin: Yes: $Y=$ union of the two diagonals in the plane, $Z=$ the $x$-axis, map= the $x$-projection. $\endgroup$ Jul 27, 2011 at 19:03
  • 1
    $\begingroup$ @Torsten: You beat me by 3 minutes on this one. $\endgroup$ Jul 27, 2011 at 19:05
  • $\begingroup$ Very nice -- thanks to both of you. $\endgroup$ Jul 28, 2011 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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