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Is there an example of a variety over the complex numbers with no embedding into a smooth variety?

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I don't understand the question. What if the variety is singular? Or do you mean embedding of an analytic variety into an algebraic variety? – shenghao Oct 16 '09 at 21:44
@shenghao: I think the OP is not insisting on an open embedding. A singular projective veriety admits an embedding into a nonsingular one! – Qfwfq Apr 27 '10 at 15:26
up vote 6 down vote accepted

Yes, there are even toric varieties. In fact, there are complete toric varieties with trivial picard group (see e.g., Eickelberg "Picard groups of compact toric varieties..." 1993). We have the following simple observations:

1) Any variety which can be embedded into a smooth variety has an ample family of line bundles.

2) A proper variety with trivial picard group cannot have an ample family of line bundles.

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wait, isn't that a proof that it's not projective? – Ben Webster Oct 9 '09 at 0:21
If it does not have any non-trivial line-bundles then, as you say, it cannot be projective. BUT, it can also not have any ample family of line bundles (any such family would be trivial => scheme is quasi-affine which is a contradiction to the properness) and thus cannot be embedded into a smooth variety by 1). – David Rydh Oct 9 '09 at 4:49
Section 2 of the paper "Higher K-theory of Schemes" (or SGA 6) has facts about ample families of line bundles; MR data is at – David Zureick-Brown Oct 13 '09 at 6:10

A really neat well known example is as follows:

Choose a conic $C_1$ and a tangential line $C_2$ in $\mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $\phi:\mathbb{P}^2\rightarrow{X}$ with $C:=\phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $\mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $\mathbb{P}^2$. Note that the degrees of $L'|C_1$ and $L'|C_2$ both coincide with the degree of $L|C$ and are therefore equal. But $L'\cong\mathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'\cong\mathcal{O}$, which is non-effective! A contradiction!

In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme.

In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1\amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism above), and $Z=\mathbb{P}^2$, but the argument is more general provided any finite set of closed points in $Z$ is contained in an affine open set. $X$ will denote the quotient.

Claim: Suppose $j:Y\rightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Y\rightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram (which I don't know how to draw here, but one visualize it easily): $Y\xrightarrow{j}{Z}$, $Y\xrightarrow{g}Y'$, $Y'\rightarrow{X}$, $Z\xrightarrow{h}X$ where $X$ is a scheme, $h$ is finite and induces monomorphisms on coordinate rings and $Y'\rightarrow{X}$ is a closed immersion.

Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $B\subset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $A\rightarrow{A/I}$. Put $A'=j^{-1}(B)$ and $Spec(A')=X$. The claim is clear for $X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$, $X$ would be replaced by $U'=h(U)$ which is an open subset.

Now this guarantees the existence of $X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $X$ that it is actually reduced and irreducible. QED.

I hope this is satisfactory.

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"There would surely exist an effective (and nonzero) line bundle on X" for the following reason: take an open affine neighborhood U on the ambient smooth scheme Y which intersects X. Take a divisor $D_U$ on Y intersecting X, and let D be its closure. Then the restriction of $O_Y(D)$ to X is the required effective nonzero line bundle. But why is it obvious that the quotient X is a scheme? It is an algebraic space by general Artin's results but why a scheme? The usual trick of proving that X is projective (and therefore a scheme) does not work. – VA. Dec 27 '09 at 15:46
@VA: I have edited my answer to include the proof. – Maharana Dec 27 '09 at 21:19
I am almost sure that your Claim is false in general (without assuming existence of an invariant cover, which of course exists in our case). For example, let $Z$ be the affine line with doubled points, $Y=Spec(k[x]/(x^2)\oplus k[y]/(y^2))$, where $x$ and $y$ are the coordinates near the two copies of the doubles point, and $Y'=Spec(k[x,y]/(x^2,xy,y^2))$. – t3suji Dec 27 '09 at 23:04
@t3suji: yes you are right. The correct hypothesis is that $Z$ must be such that any finite set of closed points is contained in an affine open set. This is how your example will not work. I will insert this into my answer now. Thanks! – Maharana Dec 27 '09 at 23:19
I agree that gluing an affine scheme along a closed subscheme gives an affine scheme via a dual construction for rings. And having an invariant affine cover reduces the problem to the affine case. Very nice explicit example! – VA. Dec 28 '09 at 0:35

See example 4.9 in Ravi Vakil's paper , and the references therein.

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