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Lazarsfeld said in his book (Positivity in Algebraic Geometry I, page 126, Example 2.1.15) that if $f:X\to Y$ is an algebraic fibre space then $X$ is normal implies that $Y$ is also normal.

His definition of algebraic fibre space is given in Definition 2.1.11: An algebraic fibre space is a surjective projective morphism $f:X\to Y$ of reduced and irreducible varieties such that $f_*\mathcal O_X=\mathcal O_Y$.

To prove the claim in the example, first take the normalization $\nu:Y'\to Y$. Then the normality of $X$ implies that $X\to Y$ factors through $\nu$ so that we have $X\to Y'\to Y$, where $\nu$ is a finite morphism. How to see $\nu$ has to be an isomorphism? E.g., $Y$ is a cuspidal curve $y^2=x^3$.

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Fei Hu
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1 Answers1

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OK, I think I'm making this WAY too hard, since he just says this off-hand. So, hopefully someone can come and give a much better answer. I think I've fallen into a possibly red herring ZMT approach.

Consider our factorization

$$X\xrightarrow{g}Y'\xrightarrow{\nu}Y$$

Note that $g$ is proper, $Y'$ is Noetherian, normal, and integral, $X$ is integral, and that the geometric fiber of $g$ is geometrically integral. The only thing which is non-obvious is the last statement. But, this amounts to the statement that $K(Y)=K(Y')$ is algebraically closed in $K(X)$, but this follows from Example 2.1.12 in Lazarsfeld. Thus, we may conclude (see here) that $g_\ast\mathcal{O}_X=\mathcal{O}_{Y'}$.

Thus, we deduce that

$$\nu_\ast(\mathcal{O}_{Y'})=\nu_\ast(g_\ast\mathcal{O}_X))=(\nu\circ g)_\ast\mathcal{O}_X=f_\ast\mathcal{O}_X=\mathcal{O}_Y$$

but since $\nu$ was the normalization map, this clearly implies that $\nu$ is an isomorphism, so that $Y$ is normal.

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Alex Youcis
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  • Thanks! Alex. The point is pseudoconnected or $\mathcal O$-connected is stronger than geometrically connected of fibre. – Fei Hu Feb 26 '16 at 08:59
  • @HuFei No problem. I'm not totally sure what you're second comment was with respect to. Were you satisfied with this answer, and were just commenting on it, or did you think that I missed something? Thanks! – Alex Youcis Feb 26 '16 at 09:03
  • I think the crux of the matter is exactly where you say "clearly": namely, showing that if the pushforward by a finite map (e.g. normalisation) takes the structure sheaf to the structure sheaf, the map must be an iso. That takes a little argument with an affine cover. – Nefertiti Feb 26 '16 at 09:51
  • @Nefertiti Really? You know that $\nu$ is affine. So, if you take any affine $\text{Spec}(A)\subseteq Y$ you have that $\nu^{-1}(\text{Spec}(A))=\text{Spec}(A')$. Then, the map $A\to A'$ is just the isomorphism $\mathcal{O}Y(\text{Spec}(A))\to\nu\ast\mathcal{O}_{Y'}(\text{Spec}(A'))$, and so $\nu$ is an isomorphism. By crux, I guess you mean important, not difficult? – Alex Youcis Feb 26 '16 at 10:00
  • Of course, I meant to write $\nu_\ast\mathcal{O}_{Y'}(\text{Spec}(A))$ above. – Alex Youcis Feb 26 '16 at 10:11
  • @AlexYoucis You answered my query. At the begining, I was confused on the normalization of a cuspidal curve. Since this is bijective, so if I just consider the geometric fibre, there is no contradiction. – Fei Hu Feb 26 '16 at 10:31
  • @HuFei Good. :) Yeah, that was also something I was considering too, since it was obvious that $\nu$ had to have connected fibers which, of course, is not enough. I think the key is the result I posted you. In particular, we think about $\mathcal{O}$-connectedness as `basically being connected fibers' by ZMT, but of course, the point is that this is false if the target is non-normal, as we see above. – Alex Youcis Feb 26 '16 at 10:32
  • @AlexYoucis and by definition $\nu_*\mathcal O_{Y'}(\mathrm{Spec} A)=\mathcal O_{Y'}(\nu^{-1}(\mathrm{Spec} A))=\mathcal O_{Y'}(\mathrm{Spec} A')$ – Fei Hu Feb 26 '16 at 10:37
  • @HuFei Yeah, that's correct of course. I made a typo above originally. :) – Alex Youcis Feb 26 '16 at 10:38
  • @AlexYoucis ;-) – Fei Hu Feb 26 '16 at 10:39
  • Sure, I meant that was the point that actually seemed to be confusing the OP. But maybe I am butting in unnecessarily. – Nefertiti Feb 26 '16 at 11:56