# Alterations factor as modification + finite map

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?

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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.

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This is Stein factorization Hartshorne, III.11.5.

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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 – HNuer Mar 1 '11 at 10:56
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). – Karl Schwede Mar 1 '11 at 13:20
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. – HNuer Mar 1 '11 at 13:53

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)?

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Ah - it seems we wrote almost in parallel :-) – Sebastian Petersen Mar 1 '11 at 13:52
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. :) – HNuer Mar 1 '11 at 14:10
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). – Karl Schwede Mar 1 '11 at 15:32
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 – HNuer Mar 1 '11 at 20:33
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. – Karl Schwede Mar 10 '11 at 13:35