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Zariski's Main Theorem (EGA IV, Thm 8.12.6): Suppose $Y$ is a quasi-compact and quasi-separated scheme, and $f:X\to Y$ is quasi-finite, separated, and finitely presented. Then $f$ factors as $X\xrightarrow{g} Z\xrightarrow{h} Y$, where $g$ is an open immersion and $h$ is finite.

Is there a canonical choice for the factorization $f=h\circ g$, at least under some circumstances?

For example, suppose $f$ factors as $X\to U\to Y$, where $X\to U$ is finite étale and $U\to Y$ is a Stein open immersion (i.e. the pushforward of $\mathcal O_U$ is $\mathcal O_Y$). Then I'm pretty sure the Stein factorization $X\to \mathit{Spec}_Y(f_*\mathcal O_X)\to Y$ witnesses Zariksi's Main Theorem (i.e. is an open immersion followed by a finite map).

In general, when does the Stein factorization witness ZMT? In the cases where it fails to witness ZMT (e.g. $X$ finite over an affine open in $Y$), is there some other canonical witness?

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I suspect that the answer is no (or at least that no such factorization is know), if only because such a factorization seems like it would provide us with an easier proof of ZMT. However, I'm certainly not an expert, so don't take my word for it. The above is just speculation. –  Harry Gindi Feb 18 '11 at 21:55
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Alternative question: Is there a initial factorization, or a terminal factorization? –  Martin Brandenburg Feb 18 '11 at 21:55
    
I would be happy with a non-initial non-terminal factorization that has a nice description, but I guess a terminal factorization is most likely to fit the bill. An initial factorization can't exist because you can always replace $Z$ by $Z\sqcup Y$ and it will still have an open immersion from $X$ and a finite map to $Y$. Similarly, you can throw in random irreducible components, keeping $Z$ connected. –  Anton Geraschenko Feb 18 '11 at 22:48
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A trivial observation: $Z$ is then the normalization of $Y$ in the field of rational functions $K(X)$. This also implies that $X$ is normal. –  Qing Liu Feb 20 '11 at 16:24
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Yes, because $X\to Z$ is then birational and quasi-finite to a normal noetherian scheme. By a variant of Zarisk's main theorem, this implies that the morphism is an open immersion. Note however that one has to suppose that $Y$ is e.g. excellent to ensure that $Z$ is finite over $Y$. If $X$ is not normal, but the ''anormal'' locus $A$ of $X$ is finite over $Y$, then there is a canonical (initial) choice for $Z$: glue $X\setminus A$ with the normalization of $Y\ f(A)$ in $K(X)$ (equivalently: take a factorisation $Z'$, and then normalize $Z'\setminus A$ which is open in $Z'$). –  Qing Liu Feb 20 '11 at 22:25

2 Answers 2

up vote 7 down vote accepted

I think an initial object exists if you work with integral excellent schemes (maybe integral is not really necessarily, but then require that $X$ be schematically dense in $Z$).

So suppose $X, Y$ are integral and excellent. Consider all possible factorizations $X\to Z_{\alpha} \to Y$ with $Z_{\alpha}$ integral. Then $K(Z_{\alpha})=K(X)$. For any pair $Z_{\alpha}, Z_{\beta}$, the closure $Z_{\gamma}$ of $X$ in $Z_{\alpha}\times_Y Z_{\beta}$ gives a factorization $X\to Z_{\gamma}\to Y$ with $Z_{\gamma}$ dominating $Z_{\alpha}$ and $Z_{\beta}$, finite over $Y$, and $X\to Z_{\gamma}$ is an open immersion (one checks that $X\to Z_{\gamma}$ is an immersion, hence open in some closed subscheme $F$, but $X$ is birational to $Z_{\gamma}$, so $F=Z_{\gamma}$). Thus we can consider the projective limite $Z$ of the $Z_{\alpha}$'s.

By construction $Z$ is affine and integral over $Y$. As $Z_{\alpha}$ is dominated by the normalization $\widetilde{Y}$ of $Y$ in $K(X)$ and $\widetilde{Y}$ is finite over $Y$ by excellent hypothesis, $Z$ is finite over $Y$. It remains to see that the canonical map $X\to Z$ is an open immersion. This property is local over $Y$. So we suppose $Y$ is affine. Cover $X$ by principal affine open subsets $D(h)$'s of some $Z_{\alpha_0}$. Then $D(h) \to D_Z(h)$ is a closed immersion because $D(h)\to D_{Z_{\alpha_0}}(h)$ is, and it is birational, so it is an isomorphism and we are done.

It would interesting to compute explicitely the projective limite in some concrete situations. For exemple, consider a surface $S$, finite over $\mathbb A^2_{\mathbb C}$, with non-normal locus $\Delta$. Let $X$ be an open subset of $S$ with $\Delta\cap X$ non-empty and not equal to $\Delta$. The inclusion $X\to S=Z_{\alpha_0}$ is a factorisation. But what is the $Z$ constructed as above ?

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Very nice. It seems natural to try to construct this $Z$ "from the top down" starting with $\tilde Y$. We get an open immersion $\tilde X\to \tilde Y$. $X$ is the quotient of $\tilde X$ by some equivalence relation $R\subseteq \tilde X\times_X\tilde X$. We can take the closure of this relation in $\tilde Y\times_X \tilde Y$ and try to quotient by it. Stopping short of the closure of $R$ will probably produce something non-separated, so if this quotient exists, it's probably the $Z$ we're looking for. In your example, I think this gives $Z=S$. –  Anton Geraschenko Feb 21 '11 at 18:31
    
It depends on $S$. Start with such an $S$, and ''pinch'' a closed point of $S\setminus X$. Then you get a new surface $S_1$ containing $X$ as an open subset. Use Noether's normalization lemma to find a finite morphism from $S_1$ to $\mathbb A^2_{\mathbb C}$. Then the new surface $S_1$ has similar properties to $S$, but it is not equal to the maximal $Z$ because it is strictly dominated by $S$. –  Qing Liu Feb 21 '11 at 22:50
    
Oh ... by a pinch you have in mind something like $k[x^iy^j|i>1$ or $j>1]$. I hadn't realized that it was possible to make higher codimension non-normal loci in this way. For some reason I always thought of non-normality as a codimension 1 phenomenon. –  Anton Geraschenko Feb 22 '11 at 18:45
    
Whatever this $Z$ is, it probably deserves to be called "the normalization of $S$ along $S\smallsetminus X$." Perhaps there is a sensible algebraic notion of taking the "partial integral closure of a ring along an ideal $I$" so that the induced map is an isomorphism away from $V(I)$. –  Anton Geraschenko Feb 22 '11 at 19:28
    
Yes this is a good question. Maybe this kind of partial normalization is done somewhere. What I meant by pinch is something like your example. If $B$ is an $K$-algebra with a rational point corresponding to a maximal ideal $m$, then I consider $K+m^N$ for some integer $N\ge 2$. The inclusion $K+m^N \subset B$ is finite (the quotient is a finite dimensional $K$-vector space) and induces an isomorphism outside of the rational point. –  Qing Liu Feb 22 '11 at 20:55

I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_*\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_*\mathcal O_X$ is finite over $\mathcal O_Y$.

Some comments: one should note that in general, the quasi-coherent algebra $f_*\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_*\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_*\mathcal O_X=h_*\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$ with complementary in $Z$ of codimension at least 2. But I don't have a general criterion.

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