# How singular can the Stein factorization of a proper map between smooth varieties be?

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?

• 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. Mar 27, 2015 at 12:37
• @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.
– jmc
Mar 27, 2015 at 12:55

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 $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.

• 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.
– jmc
Mar 26, 2015 at 15:12

$\hat{X}$ 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.

• 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.
– jmc
Mar 26, 2015 at 12:41
• @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. Mar 26, 2015 at 14:12
• Moreover, I don't see why $X\to\hat{X}\to Y$ is the Stein factorization of $X\to Y$. Mar 26, 2015 at 15:01
• @HNuer: Yes, of course, $X \to Y$ is only generically finite, but I guess this is precisely what was asked. Mar 26, 2015 at 20:29
• @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. Mar 26, 2015 at 20:31

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).

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.”