Global sections of multiples of a divisor Let $D$ be an integral divisor on a smooth projective variety $X$. Consider the multiples $mD$ of $D$ for $m\geq 0$. Clearly, $h^0(X,mD) = 1$ for $m = 0$.
Is there any example where $h^0(X,mD) = 0$ for $1\leq m\leq \overline{m}$, and $h^0(X,mD) = c$ for all $m\geq \overline{m}+1$, where $c$ is a constant?
 A: If $c = 2$, let $D',D'' \in H^0(X,kD)$ be different divisors. Then $2D',D'+D'',2D''$ are three different divisors in $|2kD|$. Similarly, the case $c > 2$ is impossible. So, assume $c = 1$.
Let $D' \in H^0(X,kD)$ and $D'' \in H^0(X,(k+1)D)$ (with $k > 1$). Then $(k+1)D'$ and $kD''$ are different divisors in $H^0(X,k(k+1)D)$. So, the case $c = 1$ is also impossible.
A: Not except in the trivial case $c =1$, $\overline{m}=0$.
Assume for contradiction that $D_1$ and $D_2$ are two effective divisors, with $D_1$ equivalent to $m_1 D$ and $D_2$ equivalent to $m_2D$, and $$m_2 D_1 \neq m_1 D_2 $$
Then there exists an irreducible closed subset of $X$ of codimension $1$, call it $Y$, which $m_2 D_1$ and $m_1 D_2$ contain to different multiplicities. So the divisors $$ c m_2 D_1 , (c-1) m_2 D_1 + m_1 D_2, \dots, c m_1 D_2$$ are all equivalent to $m_1m_2 D$, hence all are the divisors of sections $s_0,\dots, s_c \in H^0 ( X, m_1m_2 D)$, and all contain $Y$ to different multiplicities, so the sections $s_0,\dots,s_c$ vanish to different orders at $Y$.
It follows that $s_0, \dots, s_c$ are linearly independent, as any linear relation of them would have to express a leading term which vanishes to some order at $Y$ as a linear combination of functions which vanish to higher order at $Y$, which is impossible. This contradicts the assumption that $\dim H^0(X, m_1m_2 D) \leq c$.
Thus, for any divisors $D_1, D_2$ equivalent to multiples $m_1 D$, $m_2 D$, respectively, we have $m_2 D_1 = m_2 D_1$. In particular, there is at most one divisor equivalent to $m D$ for each $m$, so $c=1$.
Taking $m_1$ and $m_2$ sufficiently large that there indeed exist such $D_1, D_2$, and choosing $m_1$ and $m_2$ relatively prime so that there exist $a,b$ with $am_1 + bm_2 = 1$, we see that $a D_1 + b D_2$ is equivalent to $D$, and is effective because $m_1 ( a D_1 + b D_2) = D_1$, which is effective, so $H^0 (X, D) \neq 0$, and thus $\overline{m}=0$.
