Lemma 28.26.14. Let $X$ be a scheme. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $i : X' \to X$ be a morphism of schemes. Assume at least one of the following conditions holds

1. $i$ is a quasi-compact immersion,

2. $X'$ is quasi-compact and $i$ is an immersion,

3. $i$ is quasi-compact and induces a homeomorphism between $X'$ and $i(X')$,

4. $X'$ is quasi-compact and $i$ induces a homeomorphism between $X'$ and $i(X')$.

Then $i^*\mathcal{L}$ is ample on $X'$.

Proof. Observe that in cases (1) and (3) the scheme $X'$ is quasi-compact as $X$ is quasi-compact by Definition 28.26.1. Thus it suffices to prove (2) and (4). Since (2) is a special case of (4) it suffices to prove (4).

Assume condition (4) holds. For $s \in \Gamma (X, \mathcal{L}^{\otimes d})$ denote $s' = i^*s$ the pullback of $s$ to $X'$. Note that $s'$ is a section of $(i^*\mathcal{L})^{\otimes d}$. By Proposition 28.26.13 the opens $X_ s$, for $s \in \Gamma (X, \mathcal{L}^{\otimes d})$, form a basis for the topology on $X$. Since $X'_{s'} = i^{-1}(X_ s)$ and since $X' \to i(X')$ is a homeomorphism, we conclude the opens $X'_{s'}$ form a basis for the topology of $X'$. Hence $i^*\mathcal{L}$ is ample by Proposition 28.26.13. $\square$

## Comments (2)

Comment #5468 by Dat Pham on

Can some explain to me why do we need the condition $i(X')\subseteq X$ is locally closed? I think knowing that $i$ induces a homeomorphism between $X'$ and $i(X')$ is already sufficient to deduce that the open sets $X'_{s'}=X'\cap X_s$ form a basis for the topology of $X'$.

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