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Let $S$ be a noetherian scheme and $X \rightarrow S$ be an affine morphism of schemes. Consider the diagonal morphism $\Delta: X \rightarrow X \times_S X$. If $\Delta (X)$ is the closed subset of $X \times_S X$, then one can look at the open embedding

$j: U \rightarrow X \times_S X$

of the open complement of $\Delta(X)$.

Has $j$ a chance to be itself an affine morphism of schemes? Or what additional hypotheses would one need to get this property?

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2 Answers 2

up vote 4 down vote accepted

First notice that $j: U\to S$ is affine if and only if $U\to X\times_S X$ is affine (to check this, one can suppose $S$ is affine, then use the facts that $X\times_S X$ is affine and $U$ is open in $X\times_S X$).

So we are reduced to see whether $U$ is an affine scheme when $S$, and hence $X$ are affine. It is well known that then the complementary $\Delta(X)$ of $U$ in $X\times_S X$ is purely 1-codimensional. This is true essentially only when $X\to S$ has relative dimension $1$ (for reasonable schemes). In particular, if $X$ is any algebraic variety of dimension $d>1$, then $U$ can't be affine.

Claim: Let $C$ be a smooth projective geometrically connected curve of genus $g$ over a field $k$, let $X\subset C$ be the complementary of $r$ points with $r+1-g >0$. Then $U$ is affine.

Proof. We can suppose $k$ is algebraically closed (the affiness can be checked over any field extension). Let $D$ be the divisor $D=C\setminus X$ and $$H=D\times C+C\times D + \Delta(C).$$ Then $U=C\times C\setminus H$. It is enough to show that $H$ is an ample divisor on the smooth projective surface $C\times C$. To do this, we will use Nakai-Moishezon criterion (see Hartshorne, Theorem V.1.10).

It is easy to see that $(D\times C)^2=0$ (because $D \sim D'$ with the support of $D'$ disjoint from that of $D$), $(D\times C).(C\times D)=r^2$, $(D\times C).\Delta(C)=r$, and $$\Delta(C)^2=\deg O_{C\times C}(\Delta(C))|_{\Delta(C)}=\deg \Omega_{C/k}^{-1}=2-2g.$$ Thus $H^2=2(r^2+r+1-g)>0$.

Let $\Gamma$ be an irreducible curve on $C\times C$. If $\Gamma\ne \Delta(C)$, it is easy to see that $H.\Gamma>0$. On the other hand, $H.\Delta(C)=2r+2-2g=2(r+1-g)>0$ and we are done.

I didn't check the details because it is time to sleep, but I believe the proof is essentially correct.

EDIT

(1). In the above claim, we can remove the condition $r+1-g>0$:

Let $C$ be a smooth projective connected curve over a field $k$, let $X$ be an affine open subset of $C$. Then $U$ is affine.

The proof is the same as above, but consider $H=n (D\times C+ C\times D)+\Delta(C)$ for $n > g-1$. The same proof shows that $H$ is ample.

(2). Let $S$ be noetherian, and let $C\to S$ be a smooth projective morphism with one dimensional connected fibers, let $D$ be a closed subset of $C$ finite surjective over $S$. Let $X=C\setminus D$. Then $U$ is affine.

Proof. One can see that $D$ is a relative Cartier divisor on $C$ (because $D_s$ is a Cartier divisor for all $s\in S$). So the $H$ defined as above is a relative Cartier divisor on $C\times_S C$. We showed that $H_s$ is ample for all $s$. This implies that $H$ is relatively ample for $C\times_S C\to S$ (EGV III.4.7.1). So $U$ is affine.

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Der QiL, it is perfectly correct, thanks a lot. What I don't see is if one can extract from this the claim for the relative situation, say $X$ the complement of the zero section of an elliptic curve $C$ over $S$. –  Cyril Jun 13 '12 at 7:28
    
@Cyril, I will add some details. –  user18119 Jun 13 '12 at 22:17
    
Thanks for the edit. Is it clear that the complement of a relatively ample divisor is affine relative $S$ or is there a reference for this? One knows of course that this is true fiberwise, but I think affineness is not a fiberwise property. –  Cyril Jun 14 '12 at 6:03
1  
@Cyril, yes you are right. However affineness is local on the target, so we can always work with ''small'' $S$. Here $H$ is the trace of some hyperplane $L$ of some projective space $P/S$. As $H_s$ is ample for every $s\in S$, $L_s$ is a hyperplane in $P_s$. So in a small open neighborhood of $s$, $L$ is defined by a homogeneous polynomial of degree 1 whose coefficients generate the unit ideal in the base ring. Therefore $P\setminus L$ (and thus $C\times_S \setminus H$ is affine). Hope this helps. –  user18119 Jun 14 '12 at 21:15
    
Thanks for clarifying this point. –  Cyril Jun 15 '12 at 5:41

(EDIT: deleted wrong answer, but left up for those reading comments- I had said affine two-space times itself over a line, but see below).

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or easier, affine four space minus $(x, y, x, y)$ isn't affine (i.e. the base is a point and the scheme is a plane). –  user29743 Jun 12 '12 at 14:40
    
Dear countinghaus, I can't follow your argumentation. The set of points $(x,y,x,w)$ with $w\neq y$ looks to me like the complement of the (hyper)plane $w=y$ in affine three space $\mathbb A^3$ with coordinates $x,y,w$. Such a complement is affine (and actually isomorphic to $\mathbb A^2 \times \mathbb G_m$). –  Georges Elencwajg Jun 12 '12 at 17:45

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