Lemma 33.45.9. Let $k$ be a field. Let $X$ be proper over $k$. Let $Z \subset X$ be a closed subscheme of dimension $d$. If $\mathcal{L}_1, \ldots , \mathcal{L}_ d$ are ample, then $(\mathcal{L}_1 \cdots \mathcal{L}_ d \cdot Z)$ is positive.

**Proof.**
We will prove this by induction on $d$. The case $d = 0$ follows from Lemma 33.33.3. Assume $d > 0$. By Lemma 33.45.6 we may assume that $Z$ is an integral closed subscheme. In fact, we may replace $X$ by $Z$ and $\mathcal{L}_ i$ by $\mathcal{L}_ i|_ Z$ to reduce to the case $Z = X$ is a proper variety of dimension $d$. By Lemma 33.45.5 we may replace $\mathcal{L}_1$ by a positive tensor power. Thus we may assume there exists a nonzero section $s \in \Gamma (X, \mathcal{L}_1)$ such that $X_ s$ is affine (here we use the definition of ample invertible sheaf, see Properties, Definition 28.26.1). Observe that $X$ is not affine because proper and affine implies finite (Morphisms, Lemma 29.44.11) which contradicts $d > 0$. It follows that $s$ has a nonempty vanishing scheme $Z(s) \subset X$. Since $X$ is a variety, $s$ is a regular section of $\mathcal{L}_1$, so $Z(s)$ is an effective Cartier divisor, thus $Z(s)$ has codimension $1$ in $X$, and hence $Z(s)$ has dimension $d - 1$ (here we use material from Divisors, Sections 31.13, 31.14, and 31.15 and from dimension theory as in Lemma 33.20.3). By Lemma 33.45.8 we have

By induction the right hand side is positive and the proof is complete. $\square$

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