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.


1 Answer 1


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 $X=Y$.

  • $\begingroup$ 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

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