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A little bit of motivation (the question starts below the line): I am studying a proper, generically finite map of varieties $X \to Y$, with $X$ and $Y$ smooth. Since the map is proper, we can use the Stein factorization $X \to \hat{X} \to Y$. Since the composition is generically finite, $X \to \hat{X}$ is birational, and therefore a sequence of blowups. I am currently interested in the other map: $\hat{X} \to Y$. I would like to apply Casnati–Ekedahl's techniques from “Covers of algebraic varieties I” (Journal of alg. geom., 1996). For this, I need $\hat{X} \to Y$ to be Gorenstein. (Since $Y$ is Gorenstein (since it is smooth), this is equivalent with $\hat{X}$ being Gorenstein.) When is this true?

Specifically, in my case $X \to Y$ is the albanese morphism of a smooth projective surface: so $Y$ is an abelian surface, and I am in the situation that the albanese morphism is surjective.


Let $f \colon X \to Y$ be a proper map between two varieties $X$ and $Y$ over a field $k$. Assume $X$ and $Y$ are smooth (and proper, if you want).

Let $\pi \colon X \to \hat{X}$ and $\hat{f} \colon \hat{X} \to Y$ be the Stein factorization ($f = \hat{f} \circ \pi$). Of course, in general $\hat{X}$ is not smooth. However:

Q1: Does $\hat{X}$ have some other nice properties?

I am thinking in the direction of, e.g., Gorenstein or Cohen–Macaulay. If not, does it help if we assume a bit more on $f$? Or, alternatively:

Q2: Under what conditions is $\hat{X}$ Gorenstein?

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    $\begingroup$ You probably know this, but a normal surface is Cohen--Macaulay, by Serre's criterion. So not quite as good as Gorenstein, but not totally awful. $\endgroup$ – tracing Mar 27 '15 at 12:37
  • $\begingroup$ @tracing — Thanks! Actually, I did not know this. I am just starting to learn things about the Gorenstein and Cohen–Macaulay properties. I'll have to look into Serre’s criterion. $\endgroup$ – jmc Mar 27 '15 at 12:55
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$\hat{X}$ can be as bad as you want. For example, take your favorite non-Gorenstein variety $\hat{X}$ in $\mathbb{A}^N$. By Noether Lemma there is a finite morphism $\hat{X} \to \mathbb{A}^n =: Y$. Take $X$ to be a resolution of singularities of $\hat{X}$. Then $X \to Y$ is a quasifinite morphism between smooth varieties.

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  • $\begingroup$ Ouch! It can be nasty indeed. Murphy's law hits again. Well thanks for the fast answer! I guess there is not much that one can do in light of Q2, right? As in: no sensible condition on $f$ will prevent examples like you gave in your answer. $\endgroup$ – jmc Mar 26 '15 at 12:41
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    $\begingroup$ @Sasha, unless I'm using a different definition then you, the morphism $X\to Y$ is only generically finite, not quasi-finite, unless the $X$ is just the normalization of $\hat{X}$. That is, there might be positive dimensional fibers. $\endgroup$ – HNuer Mar 26 '15 at 14:12
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    $\begingroup$ Moreover, I don't see why $X\to\hat{X}\to Y$ is the Stein factorization of $X\to Y$. $\endgroup$ – Laurent Moret-Bailly Mar 26 '15 at 15:01
  • $\begingroup$ @HNuer: Yes, of course, $X \to Y$ is only generically finite, but I guess this is precisely what was asked. $\endgroup$ – Sasha Mar 26 '15 at 20:29
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    $\begingroup$ @LaurentMoret-Bailly: To ensure that $X \to \hat{X}$ has connected fibers it is enough to take $\hat{X}$ to be normal. Then the first map has connected fibers and the second is finite, so it is the Stein factorization by the universal property. Of course, without normality of $\hat{X}$ this may be wrong. $\endgroup$ – Sasha Mar 26 '15 at 20:31
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The only restriction I see is that $\hat{X}$ must be normal (because $X$ is): if $\phi$ is a rational function on (some affine open subscheme of) $\hat{X}$ which is integral over $\mathscr{O}_\hat{X}$, then $\phi\circ\pi$ is integral over $\mathscr{O}_{X}$, hence in $\mathscr{O}_{X}$. In other words, $\phi$ lies in $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.

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  • $\begingroup$ Ok, so when $X$ and $Y$ are surfaces, this might be interesting. Because by your argument, the singular locus $\hat{X}$ is $0$-dimensional, and we can wonder what kind of singularities might occur. $\endgroup$ – jmc Mar 26 '15 at 15:12
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For what it's worth, one can say the following sort of thing.

Since $Y$ is log terminal so is $(\hat{X}, -\mathrm{Ram})$. This doesn't mean much since in the pair, the boundary has a negative coefficients (ie, the singularities of $\hat{X}$ can be arbitrarily bad). But it does say things like:

if $\hat{X}$ has really bad singularities at some points, then $\mathrm{Ram}$ also has really bad singularities at those points too. Another way to say this is if the ramification divisor has mild singularities, then $\hat{X}$ does too.

Note that of course, $K_{\hat{X}} + (-\mathrm{Ram}) \sim f^*(K_Y)$. The right side is Cartier, and thus so is the left. So the pair $(\hat{X}, -\mathrm{Ram})$ is log-Gorenstein (again, this doesn't mean much unless you control the ramification divisor in some sense).

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There are three good answers to this question, and together they have more or less answered what I wanted to know. I find it hard to choose one of them as best, but nevertheless I think this question should have an accepted answer to move it from the unanswered list. Hence a CW-answer summarizing the (in my eyes) most important points made.


  • Laurent Moret-Bailly points out that $\hat{X}$ must be normal.
  • Sasha then says that besides that, it can get as bad as you want. Take a normal subvariety $\hat{X} \subset \mathbb{A}^{N}$. By Noether's lemma we get a finite map $\hat{X} \to \mathbb{A}^{n} = Y$. A resolution of singularities $X \to \hat{X}$ has connected fibres. The composition $X \to Y$ is generically finite.
  • Karl Schwede remarks that the pair $(\hat{X}, -\mathrm{Ram})$ is log-Gorenstein (where $\mathrm{Ram}$ is the ramification divisor). He also states the slogan “if $\hat{X}$ has really bad singularities at some points, then $\mathrm{Ram}$ also has really bad singularities at those points too. Another way to say this is if the ramification divisor has mild singularities, then $\hat{X}$ does too.”
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