2
$\begingroup$

I'm learning about de Jong's theory of resolution of singularities and the following fact is used numerous times: an alteration of varieties $h: X \rightarrow Y$ factors as $X \xrightarrow{\pi} Z \xrightarrow{f} Y$, where $\pi$ is a modification and $f$ is a finite morphism. From an exercise in Hartshorne I know I can find a dense open subset in $U \subset Y$ such that the induced morphism $h^{-1}(U) \rightarrow U$ is finite, but I don't know where to go from there. I thought to blow-up the respective closed sets away from these open subsets, but then I'd be moving away from my original varieties. I get the sense this is a known fact. Anyone know a simple proof?

$\endgroup$

3 Answers 3

4
$\begingroup$

As Sandor points out, this is Stein factorization. Let $X$, $Y$ be varieties over a field $K$. Let $h: X\to Y$ be a proper morphism. Then $h_*(O_X)$ is coherent and $Y':=Spec(h_*(O_X))$ (cf. EGA II.1.3 for the definition of $Spec$ of a sheaf of quasicoherent algebras) is finite over $Y$. Consider the Stein factorization $$h:X\buildrel h' \over\longrightarrow Y'\buildrel g\over \longrightarrow Y$$ of $h$. (It is true by Zariski's connectedness theorem (cf. EGA III.4.3.1 - III.4.3.4) that $h'$ has non-empty geometrically connected fibres and $g$ is finite, but this will not be needed here).

Now assume that $h$ is an alteration, i.e. that $h$ is surjective and that there exists a nonempty open subscheme $U\subset Y$ such that $h^{-1}(U)\to U$ is finite. Replacing $U$ by a smaller set we may assume that $U=Spec(A)$ is affine. Define $U'=g^{-1}(U)$ and $V=h^{-1}(U)$. Then $V=Spec(B)$ for some $A$-algebra $B$, because $h$ is affine as a finite morphism. We thus have $h_*(O_X)(U)=B$. Hence it follows from the definition of the sheaf Spec that $U'=Spec(B)$ and the restriction $h': V\to U'$ is an isomorphism. Hence $h': X\to Y$ is a modification.

Note that this argumentation does not use the full strength of Zariski's connectedness theorem in an essential way here. Most of the things are in a sense "a tautology". The only somewhat deeper fact entering is the fact that $h_*(O_X)$ is coherent, because of the properness assumption.

Concerning your comment, the following is clearly true: If $f: X\to Y$ is a finite morphism and $f_*(O_X)=O_Y$, then $f$ is an isomorphism.

$\endgroup$
4
$\begingroup$

This is Stein factorization Hartshorne, III.11.5.

$\endgroup$
3
  • $\begingroup$ What definition are you using for modification? I'm using it to mean a proper, birational morphism, and alteration to mean a proper, surjective, generically finite. Stein factorization seems only to be for projective morphisms. Also, why does having connected fibers imply birationality? Thanks $\endgroup$
    – HNuer
    Mar 1, 2011 at 10:56
  • 1
    $\begingroup$ HNuer, see EGA III, Section 4.3 for Stein factorization in greater generality (ie, proper morphism), but Hartshorne's proof will work for you, I think he only uses projective because he only proved the "projective" case of a few results earlier. Now, connected fibers doesn't always guarantee birationality, but in your particular case it is birational. This follows from the actual construction of the intermediate scheme $Z$ in this case. In particular, it is clear that both $Z$ is isomorphic to $X$ at the generic points of $Z$ (see the proof). $\endgroup$ Mar 1, 2011 at 13:20
  • $\begingroup$ Thanks, I looked up Stein factorization afterward and found it in its generality after I posted that question. Unfortunately I didn't see any of these answers until after I figured out the rest on my own. $\endgroup$
    – HNuer
    Mar 1, 2011 at 13:53
1
$\begingroup$

Thanks to Sándor for the idea of Stein factorization. By Stein factorization as proven in EGA III, 4.3.3 for proper morphisms we have that $\phi$ (I've switched the direction of the morphism) factors as $Y \xrightarrow{\pi} Z \xrightarrow{f} X$ with $\pi$ a morphism with connected fibres and $f$ a finite morphism. By [Hart] II.4.8(e) $\pi$ must be proper. First we note that it's enough to assume that $\pi$ is surjective since it's proper and thus has closed image. Moreover since closed immersions are proper then $\pi$ would remain proper. Also the image of an irreducible scheme is irreducible, so we may assume $Z$ is irreducible, and in fact integral. Now consider the generic fiber of $f$, i.e. $f^{-1}(\eta)$, where $\eta$ is the generic point of $X$. On the one hand we have $\phi^{-1}(\eta)$ and $f^{-1}(\eta)$ are finite sets since they are generically finite and finite, respectively. On the other hand, the fiber of $\pi$ over each one of the finite points in $f^{-1}(\eta)$ must be connected and also finite. Thus the fiber over each one of these points is at most a point. Since now $\pi$ is a surjective morphism between integral schemes we get that $\pi(\zeta)=\nu$, where $\zeta,\nu$ are the generic points of $Y,Z$ respectively, and moreover $\pi^{-1}(\nu)=\zeta$. From the proof of [Hart] Exercise II.3.7, we get that the function fields of $Y$ and $Z$ are equal, and thus indeed $\pi$ is birational. Sound good (albeit wordy)?

$\endgroup$
5
  • $\begingroup$ Ah - it seems we wrote almost in parallel :-) $\endgroup$ Mar 1, 2011 at 13:52
  • $\begingroup$ Yeah, the second after I posted I saw yours and realized that they were the same when you unwind the definition. Thanks for the help anyway, though. :) $\endgroup$
    – HNuer
    Mar 1, 2011 at 14:10
  • 2
    $\begingroup$ Be careful, the exercise you mentioned above does NOT imply that the function fields of $Y$ and $Z$ are equal. Consider for example the Frobenius map (it's the identity on points, but raises sections to their $p$th powers). Basti's proof is right, it uses the construction of the Stein factorization (sheafy-spec). $\endgroup$ Mar 1, 2011 at 15:32
  • $\begingroup$ The proof in that exercise, following Hartshorne's hint, first has you show that the function field of one variety is a finite field extension of the other and the index of the extension is precisely the number of points in the generic fiber, in this case we've shown it's one. I believe that does do the trick, but please tell me where I went wrong. Thanks $\endgroup$
    – HNuer
    Mar 1, 2011 at 20:33
  • $\begingroup$ HNuer, sorry, I should have responded to this earlier. Suppose that $k$ is an algebraically closed field of characteristic $p$, then consider the extension $k[x] \subseteq k^{1/p}[x^{1/p}] \cong k[x^{1/p}]$. The induced map of specs gives us exactly one point in each fiber. In particular, the points of $\text{Spec} k[x]$ are the ideals $(x - a)$ for $a \in k$ and $(0)$. $(0)$ has exactly one point above it, $(0)$. $(x-a)$ has exactly one point above it, $(x^{1/p} - a^{1/p})$. The argument you are giving works for varieties in characteristic zero. $\endgroup$ Mar 10, 2011 at 13:35

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.