Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am doing some practice questions for my exam and I would appreciate help in solving this problem:

$D,E$ are Cartier divisors on a nonsingular projective surface $X$.
(1) If $D\equiv 0$ show that dim $H^0(X,sD)\leq 1$ for all $s\geq 1$.
(2) If $D$ is effective, $D\neq 0$ and $D\equiv E$ show that $H^0(X,tE)=0$ for all $t\leq -1$.
(3) Is (2) true if $E=D$ and $X$ is nonsingular affine variety of positive dimension?

Here is my workings for (1) and (2): are they correct?
By definition, $D\equiv 0$ means $D.C=0.C=0$ for any curve $C$.
(Note: The remainder of this portion was pointed out to be wrong)
Since I can easily construct a line intersecting with $D$, this shows that $D=0$. Then clearly $sD=0$ for all $s\geq 1$. I know that dim $H^0(X,sD)-1=$ dim$|sD|$, so I can try to show that dim $|sD|\leq 0$. Since $|sD|=\lbrace D'\geq 0|D'\sim sD\rbrace$, therefore $D'\sim 0$ and $D'=0$. This shows that $|sD|=\lbrace 0\rbrace$ and hence dim $|sD|=0$.

For (2), since $D>0$ and $t\geq -1$, hence deg $tE<0$.
$H^0(X,tE)=\lbrace f\in k(X)\setminus\lbrace 0\rbrace|tE+div(f)\geq 0\rbrace\cup\lbrace 0\rbrace$.
Assuming such an $f$ exists, then $tE+div(f)\geq 0$.
But degree of $div(f)=0$, hence this is a contradiction.

Not sure about (3)...

Thanks for reading!

share|improve this question
1  
"This shows that $D=0$." This isn't true. You can have divisors numerically equivalent to $0$, but which are not linearly equivalent to $0$. For example, consider the canonical divisor on an Enriques surface. –  Matt Nov 25 '12 at 17:47
    
@Matt Thanks for the response. I will update my working if I manage to think of a new solution. Please feel free to make any more comments. –  Yong Hao Ng Nov 25 '12 at 18:10

1 Answer 1

up vote 3 down vote accepted

(1) If $h^0(sD)\ge 1$, then there exists a rational function $f$ such that $\mathrm{div}(f)+sD=D'\ge 0$. If $D'=0$, then $sD$ is principal and $h^0=1$. Suppose $D'>0$. Let $L$ be a line, then $$0<\deg D'=D'.L=\mathrm{div}(f).L+sD.L=0,$$ contradiction.

(2) Your proof is correct.

(3) On an affine variety $V$, a coherent sheaf $\mathscr F$ is zero if and only if $H^0(V, \mathscr F)=0$. As $O_X(tE)$ is an invertible sheaf, it is never zero unless $X$ is empty. So $H^0(X, tE)\ne 0$ if $X$ is affine and non-empty.

share|improve this answer
    
May I ask why "deg$D'=D'.L$" and "div$(f).L=0$"? Namely, if degree of divisor is $d$, does it work like a degree $d$ curve and hence Bezout's theorem applies? –  Yong Hao Ng Nov 26 '12 at 2:52
    
@YongHaoNg: how is the degree defined in your class ? –  user18119 Nov 26 '12 at 10:26
    
It is defined as $\sum n_i$ where $D=\sum n_iD_i$ and $D_i$ are prime divisors. –  Yong Hao Ng Nov 26 '12 at 10:29
    
$D_i$ are points on $X$ and $D$ is a divisor on $X$. –  Yong Hao Ng Nov 26 '12 at 10:34
    
OK (though $D_i$ should not be a point !). So just say $D'.L>0$ (take a line $L$ which doesn' contain any irreducible component of $D'$. For the second equality, write $f$ as quotient $P/Q$ of two homogeneous polynomials of same degree $d$. Then $\mathrm{div}(f)=[Z(P)]-[Z(Q)]$, so $\mathrm{div}(f).L=[Z(P)].L-[Z(Q)].L=d-d=0$ by Bézout. –  user18119 Nov 26 '12 at 10:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.