# Is any quasi-projective variety isomorphic to a closed subvariety of a product of a projective space and an affine space?

Let $k$ be an algebraically closed field. Let $\mathbb{P}^n$ be a projective space over $k$. Let $\mathbb{A}^m$ be an affine space over $k$. Is any quasi-projective variety isomorphic to a closed subvariety of $\mathbb{P}^n\times \mathbb{A}^m$ for some $n, m$?

-

Edit Remove some false statements on the general situation.

Counterexample. Let $X=\mathbb P^2\setminus \{ \text{one closed point}\}$ over a field $k$. Suppose $X$ is closed in a $P\times A$ with $P$ projective and $A$ affine. Composing with projection then gives a morphism $p: X\to A$. As $A$ is affine, this morphism corresponds to a homomorphism $O(A)\to O(X)=k$ of $k$-algebras and is given by $X\to \mathrm{Spec}(O(X))\to A$. Therefore $p$ is constant and $X$ is a closed subscheme of $P\times \{ p(X) \}$ which is projective. But $X$ is not projective !

More generally, there is a notion of anti-affine varieties which don't have non-constant morphisms to any affine variety. Any non projective anti-affine variety is a counterexample to your question.

Remark The properties : being a closed variety of a product $\mathbb P^n\times \mathbb A^m$; projective over an affine varietiy; affine over a projective are clearly equivalent. And the examples in the answers show that this is a strictly stronger condition than being quasi-projective.

There is one way (probably useless) to characterize varieties $X$ which are embeddable in a $\mathbb P^n\times \mathbb A^m$: namey, $O_X(X)$ is finitely generated $k$-algebra and the canonical morphism $X\to \mathrm{Spec}(O(X))$ is projective. The only non-trivial point is the "only if" part. Suppose $X$ can be embedded in some $\mathbb P^n\times \mathbb A^m$. Composing with the projection to $\mathbb A^m$, we get a projective morphism $f: X\to \mathbb A^n_k$. By the theorem of direct image $O(X)$ is finite over $O(\mathbb A^n)$. So this is a finitely generated $k$-algebra. The morphism $f$ factorizes through $X\to\mathrm{Spec}(O(X))\to \mathbb A^m$. This implies that the first morphism is projective.

This gives another way to provide examples of quasi-projective varieties which are not embeddable into $\mathbb P^n\times \mathbb A^m$: it suffices that $O(X)$ is not f.g. over $k$. Such $X$ exist with $X$ quasi-affine (Nagata, related to Hilbert's 14th Problem).

-
For the notion of anti-affine, google with Brion and anti-affine. – user18119 Nov 26 '12 at 13:28
Could you explain why $O(X) = k$ and $X$ is not projective? – Makoto Kato Nov 28 '12 at 19:10
Write $\mathbb P^2$ as union of three copies $U_0$, $U_1$, $U_2$ of $\mathbb A^2$. If $f\in O(X)$ and $X=\mathbb P^2\setminus \{ x_0\}$, then $f\in O(U_i\setminus {x_0})=O(U_i)$. Hence $f\in O(\mathbb P^2)=k$. Finally $X$ is not projective because otherwise it would be proper, hence closed in $\mathbb P^2$. – user18119 Nov 28 '12 at 23:01

The variety $X=\mathbb A ^2 \setminus \lbrace 0\rbrace$ cannot be embedded as a closed subvariety $X\subset \mathbb{P}^n\times \mathbb{A}^m$.

Indeed, if it were the restriction $p:X\to \mathbb A^m$ of the projection $\mathbb{P}^n\times \mathbb{A}^m \to \mathbb{A}^m$ would be proper so that the fibers would be complete subvarieties of $X\subset \mathbb A^2$, hence necessarily finite subsets.
But then $p$ being proper with finite fibers would be finite and thus an affine morphism .
Then, by the definition of affine morphism, $X$ would be affine since $\mathbb A^m$ is: but this is an absurd conclusion.

Reference : I used Corollary 12.89 in Görtz-Wedhorn's Algebraic Geometry, which says that it is equivalent for a morphism to be finite or quasi-finite and proper or affine and proper.

-
Dear Georges, I don't have the book. Could you give us a sketch or a hint of the proof that a proper morphism with a finite fibres is finite? I know a proof that a proper morphism between affine schemes is finite. – Makoto Kato Nov 29 '12 at 11:06
Dear Makoto, you can find a proof that a quasi-finite proper morphism is finite in EGA $III_1$, Proposition (4.4.2), page 136 (It is a consequence of Zariski's Main Theorem, a rather difficult result). All of EGA is online; the volume you need is here. – Georges Elencwajg Nov 29 '12 at 13:05
Thanks! So it needs ZMT. I suspected it a bit. – Makoto Kato Nov 29 '12 at 13:24

No. Let $X$ be the variety $\mathbf{A}^1-\{0\}$. Suppose that $X$ is a closed subvariety of $\mathbf{P}^n\times \mathbf{A}^m$. The projection of $\mathbf{P}^n\times \mathbf{A}^m$ on $\mathbf{A}^m$ is a projective morphism. Thus, there is a projective morphism from $\mathbf{A}^1-\{0\}$ to $\mathbf{A}^m$. Since $\mathbf{A}^1-\{0\}$ is affine this implies that $\mathbf{A}^1-\{0\}$ is also affine. No contradiction here! But....

You should replace $\mathbf{A}^1-\{0\}$ by $\mathbf{A}^2-\{0\}$ in the above. You will obtain with the same reasoning that there is a finite morphism from $\mathbf{A}^2-\{0\}$ to some affine space. This would imply that $\mathbf{A}^2-\{0\}$ is affine. Contradiction.

Here are some ideas on how to show "finite" = "quasi-finite + proper". I wrote this in a hurry but I think it will help the OP get on the right track. Let me show that finite implies quasi-finite and proper.

Finite morphisms are affine. So to check that finite implies quasi-finite you can consider a finite morphism of rings $A\to B$. Let $p$ be a prime ideal of $A$. You want to show that there are only finitely many prime ideals $q$ in $B$ such that $q\cap A$ is $p$. This is equivalent to showing that Spec $B\to$ Spec $A$ is quasi-finite. To show this claim you only need the definition of what a finite ring morphism is. To show that "finite" implies "proper" you first note again that "finite" implies "affine". Then you note that "affine" implies "separated". Furthermore, "finite" clearly implies "finite type" (by definition). So you have to check that finite morphisms are "universally closed". Since "finite" is "invariant under base change" it suffices to check that a finite morphism is closed. To check that a finite morphism is closed you can reason locally. So you have $A\to B$ a finite ring morphism. Take a closed of Spec $B$. This corresponds to an ideal $I$. You can reduce to the case $I=(0)$ by replacing $B$ with $B/I$. Now, what does it mean for Spec $B \to$ Spec $A$ to be closed?

The other implication is a bit more difficult. It suffices to show that a proper morphism of affine schemes is finite. This can be found in Liu's book on algebraic geometry. I don't have the book with me now so you'll have to look at the relevant chapters on "proper" morphisms and "finite" morphisms.

-
Dear Bordo, your assertion that necessarily $m=1$ is incorrect. Actually the finite morphism $\mathbb A^1∖{0}→\mathbb A^2:z\mapsto (z,1/z)$ embeds $\mathbb A^1∖{0}$ as a closed subvariety of $\mathbb A^2=\mathbb P^0\times \mathbb A^2$ – Georges Elencwajg Nov 26 '12 at 12:31
Thank you Georges. I will leave edit the post accordingly. – Bordo Nov 26 '12 at 13:03
Could you explain why a projective morphism $\mathbf{A}^2-\{0\} \rightarrow \mathbf{A}^m$ is finite? – Makoto Kato Nov 26 '12 at 15:39
It's the same argument as Georges gives. This map is proper and hence the fibres are complete subvarieties of $\mathbf{A}^2$. Thus, this map is quasi-finite. And then you use that "finite" = "proper +quasi-finite". – Bordo Nov 27 '12 at 8:39
@Bordo The point of your proof is that "finite" = "proper + quasi-finite". I would like to know a proof of it. – Makoto Kato Nov 27 '12 at 22:11