Suppose $X$ is a projective variety, $f\colon X\to Y$ is a finite surjective morphism onto variety $Y$, must $Y$ be a projective variety?
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3$\begingroup$ This is true in the genuine quotient case of a finite group $G$. There must be a down to earth way of seeing this (so this is only a comment), but it follows by GIT: take an ample line bundle $L$ on $X$. Then $Y=\oplus_{k}ProjH^0(X,L^k)^G$, so automatically quasiprojective, and is projective as every point is GIT stable (every point has a non-vanishing $G$-invariant section). $\endgroup$– Ruadhaí DervanCommented Jun 19, 2015 at 12:33
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$\begingroup$ @RuadhaíDervan In your comment above, if we know $Y$ can be written as projective spectrum, is it already projective? Why it is only claimed to be quasi-projective? What is the argument using stability? $\endgroup$– user39380Commented Jun 19, 2015 at 12:53
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2$\begingroup$ Well, the map $X->Y$ is only defined on the locus of points where there is a $G$-invariant nonvanishing section, so a priori the image of $X$ might only be quasi-projective. For the argument about stability, try reading this question: mathoverflow.net/questions/6316/… $\endgroup$– Ruadhaí DervanCommented Jun 19, 2015 at 13:00
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No, that is not true. Let $X$ be $\mathbb{P}^3_k$. Let $g:L\hookrightarrow X$ be a line in $X$. Let $h:C\hookrightarrow X$ be a plane conic in $X$ that is disjoint from $L$ and that contains a $k$-point. Let $i:L\to C$ be an isomorphism of $k$-schemes. Let $f:X\to Y$ be the coproduct of the two morphisms $g$ and $h\circ i$. Then $Y$ is a proper $k$-variety, and $f$ is finite and surjective.
If $\mathcal{L}$ were an ample invertible sheaf on $Y$, then the pullback $f^*\mathcal{L}$ would be an ample invertible sheaf on $X$ whose degree on $L$ equals the degree on $C$. Every invertible sheaf on $\mathbb{P}^3$ is of the form $\mathcal{O}(d)$ for some $d\in \mathbb{Z}$. Only for $d=0$ is the degree on $L$ equals to the degree on $C$. For $d=0$, this invertible sheaf is not ample. Thus $Y$ is not projective.
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$\begingroup$ Is there a reference for the construction of coproduct? $\endgroup$– user39380Commented Jun 19, 2015 at 12:37
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2$\begingroup$ Yes, there are references. In this case it is elementary to construct the quotient by hand. Karl Schwede has an expository note about coproducts. The standard reference is Artin's "Algebraization and Formal Moduli, II", which works in the category of algebraic spaces. Also I just learned from Count Dracula (pseudonym, I assume) of a lovely reference by Daniel Ferrand, "Conducteur, Descente et Pincement". At any rate, in this case it is elementary to check that the coproduct $Y$ is, in fact, a scheme. $\endgroup$ Commented Jun 19, 2015 at 12:39
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$\begingroup$ Morover, $Y$ is proper if and only if such an $X$ exists: en.wikipedia.org/wiki/Chow's_lemma $\endgroup$ Commented Jun 20, 2015 at 5:06