Proposition 19.4.7632B1323B741

Proof. Step 1. We first show that there is $\Delta \in \mathcal{K}_ n$ such that $\Delta _ k(L)\subseteq \Delta $. For this purpose, using 19.3.3, we may assume that $\nu $ is induced by an admissible flag $Y_{\bullet }$ on $X$. We may further assume that each $Y_ i$ is smooth.

Fix $s\in \mathrm{H}^0(X,L^ k)^{\times }$ for some $k\in \mathbb {Z}_{>0}$. Assume that $s\neq 0$. We need to show that for each $i=1,\ldots ,n$, $\nu (s)_ i\leq Ck$ for some constant $C>0$, independent of the choices of $k$ and $s$.

Fix an ample divisor $H$ on $X$. Take a large enough integer $b_1>0$ such that

Then $\nu (s)_1\leq b_1 k$. Next take a large enough integer $b_2$ such that, for every $a\in [0,b_1]$,

After removing the first vanishing order of $s$ along $Y_1$, the resulting section on $Y_1$ is a section of $k((L-aY_1)|_{Y_1})$ for some $a\in [0,b_1]$. Hence $\nu (s)_2\leq b_2k$. Continuing in this manner and choosing the constants uniformly over the compact ranges of the preceding normalized vanishing orders, we conclude that $\nu (s)_ i/k$ is bounded for each $i$.

Here the expressions involving $L-aY_1$ are understood numerically. At the $i$-th step the previous normalized vanishing orders lie in a compact polytope, and the intersection number is continuous in these parameters; hence the constants $b_ i$ can be chosen uniformly.

Step 2. Observe that $\Gamma (L)$ is clearly a semigroup. It remains to show that $\Gamma (L)$ generates $\mathbb {Z}^{n+1}$ as an Abelian group.

For this purpose, take two very ample divisors $A$ and $B$ so that $L=\mathcal{O}_ X(A-B)$. After choosing $A$ and $B$ ample enough, we may guarantee that there exist sections $s_0\in \mathrm{H}^0(X,A)$, $t_ i\in \mathrm{H}^0(X,B)$ for $i=0,\dots ,n$ such that

and $\nu (t_ i)$ is the $i$-th unit vector $e_ i\in \mathbb {R}^ n$ for $i=1,\ldots ,n$.

Since $L$ is big, we can find $m_0>0$ such that for any $m\geq m_0$, we can find an effective divisor $F_ m$ on $X$ linearly equivalent to $mL-B$. Let $f_ m=\nu ([F_ m])$. Then

Since $(m+1)L$ is linearly equivalent to $A+F_ m$, we have

It follows that $\Gamma (L)$ generates $\mathbb {Z}^{n+1}$. $\square$