Sorry for my bad English.
I'm confusing in Mumford's "Abelian varieties"new edition section 17 p.153.
I have this circumstance;
Let $X$ be an abelian variety, $\cal{L}$be an ample divisor,and ${\cal L}\cong {\cal O}_X(D)$where $D$ is an effective divisor on $X$. And $F\subset X$ be a finite subgroup and $\pi: X\to X/F$.
Now $D$ do not have any multiple component (i.e. multiplicity of all prime divisor of $D$ is 1), and there is an effective divisor $D_1$ on $X/F$ which is also no multiple component and $\pi^{*}D_1=D. $
Moreover $\dim H^0(X, {\cal L})> \dim H^0(X/F, {\cal O}_{X/F}(D_1))$
Then he says as follows;
Since the set of all divisors $D_1$ such that ${\cal L}\cong \pi^*({\cal O}_{X/F}(D_1))$ fall into a finite set of linear equivalence classes, this proves that all sections $s\in \Gamma(X, {\cal L})$ either define multiple divisors, or lie in one of a finite number of lower-dimensional subspaces $\pi^*\Gamma(X/F, {\cal O}_{X/F}(D_1))$. This is contradiction.
But I can't understand what is contradiction. Please help me thanks.
I think you don't understand the argument, let me try to repeat it, he start with an effective divisor $D$ such that $L=O(D)$, having this $D$ is equivalent to having a global section $s\in \Gamma(L)$, then he prove that if $D$ has no multiplicity, $D$ should be a pullback of a divisor $D_1$, so the $L$ should be the pulback of $L'=O(D_1)$, and the section $s$ should be the image of $s_1$ in the associted map $H^0(X/F,L')\to H^0(X,L)$.
so we have a decomposition for the vector space $H^0(X,L)$. it is the union of the image of maps $\phi:H^0(X/F,L')\to H^0(X,L)$ for a $L'$ such that $L=\pi^*(L')$ and the sections $s$ such that the associted $D$ has multiplicity, but if you decompose a vector space into union of subvector space, one of them should be the whole space!
this is the contradiction, because by the dimension argument $Im\,\phi$ can not be the whole space, and in the first paragraph mumford shows that there is at least one effective $D$ without multiplicity.
(if you are not comfortable with this proof, there are other proofs of this fact in more "modern" notes about abelian varieties, but the heart of all of them is the relation between global sections of $L$ and effective divisors $O(D)=L$)