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!