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Let $f : X \to Y$ be a finite morphism of connected, reduced, pure-dimensional projective schemes of equal dimension satisfying $f^*\omega_Y \cong \omega_X$. What can be said about $f$ in this case? Is $f$ surjective? Etale? I am failing to come up with examples showing otherwise. One can assume that $\omega_Y$ is an invertible sheaf.

To clarify, I'm interested in the geometric consequences the condition $f^*\omega_Y \cong \omega_X$ imposes.

Mellon
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  • If $Y$ is smooth over a field, then such a map is etale. I assume your interest is in the situation where $X$ and $Y$ are Gorenstein but not smooth? – David E Speyer Jul 27 '17 at 13:59
  • Yes, that is correct. Would you mind sharing your proof or reference on this? – Mellon Jul 27 '17 at 16:33
  • A simple counterexample is an embedding of a point. – Sasha Jul 28 '17 at 19:45
  • @Sasha they need to be equidimensional schemes, so I don't think embedding a point will be a counterexample. – Andrew Jul 28 '17 at 22:09
  • @Andrew A point is equidimensional, isn't it? – Sasha Jul 29 '17 at 07:48
  • By equidimensional I simply meant "of equal dimension". I have updated the question. – Mellon Jul 29 '17 at 08:43
  • Would you please explain $\omega_\text{space}$? Algebraic geometry isn't my field of expertise, but if $\omega_\text{space}$ is a 2 form, then this looks like a condition stipulated in symplectic geometry. – Chickenmancer Jul 31 '17 at 17:32
  • @Checkenmancer $\omega_X$ is the dualizing sheaf of $X$. Look up e.g. Hartshorne for a reference. – Mellon Jul 31 '17 at 21:16

1 Answers1

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In the case you describe we can use the higher dimensional analogues of the Riemann-Hurwitz theorem to see that the ramification locus of the morphism will be empty. So the morphism will be unramified, but not necessarily etale or surjective. Let $f:\mathbb{P}^2\coprod\mathbb{P}^1\rightarrow\mathbb{P}^3$ be a morphism embedding the first factor as a hyperplane H in $\mathbb{P}^3$ and the second factor as a line in H. Clearly the morphism is finite and unramified, so by Riemann-Hurwitz we see $f^*\omega_{\mathbb{P}^3}=\omega_{\mathbb{P}^2\coprod\mathbb{P}^1}$. This morphism obviously not surjective, and is not flat since it is not an open immersion, and so it is not etale either.

Ajacket
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  • I don't think that $f^*\omega_{\mathbb{P}^3}$ is isomorphic to $\omega_{\mathbb{P}^2 \sqcup \mathbb{P}^2}$. Is the condition of being dominant overlooked in your application of the Riemann-Hurwitz formula? In any case, my question specifies that $X$ and $Y$ should be puredimensional, connected and of equal dimension. – Mellon Aug 04 '17 at 10:49