Lemma 17.13.3. Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces. Assume $i$ is a homeomorphism onto a closed subset of $X$ and that $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Let $\mathcal{F}$ be an $\mathcal{O}_ Z$-module. Then $i_*\mathcal{F}$ is of finite type if and only if $\mathcal{F}$ is of finite type.
Proof. Suppose that $\mathcal{F}$ is of finite type. Pick $x \in X$. If $x \not\in Z$, then $i_*\mathcal{F}$ is zero in a neighbourhood of $x$ and hence finitely generated in a neighbourhood of $x$. If $x = i(z)$, then choose an open neighbourhood $z \in V \subset Z$ and sections $s_1, \ldots , s_ n \in \mathcal{F}(V)$ which generate $\mathcal{F}$ over $V$. Write $V = Z \cap U$ for some open $U \subset X$. Note that $U$ is a neighbourhood of $x$. Clearly the sections $s_ i$ give sections $s_ i$ of $i_*\mathcal{F}$ over $U$. The resulting map
\[ \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ U \longrightarrow i_*\mathcal{F}|_ U \]
is surjective by inspection of what it does on stalks (here we use that $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective). Hence $i_*\mathcal{F}$ is of finite type.
Conversely, suppose that $i_*\mathcal{F}$ is of finite type. Choose $z \in Z$. Set $x = i(z)$. By assumption there exists an open neighbourhood $U \subset X$ of $x$, and sections $s_1, \ldots , s_ n \in (i_*\mathcal{F})(U)$ which generate $i_*\mathcal{F}$ over $U$. Set $V = Z \cap U$. By definition of $i_*$ the sections $s_ i$ correspond to sections $s_ i$ of $\mathcal{F}$ over $V$. The resulting map
\[ \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ V \longrightarrow \mathcal{F}|_ V \]
is surjective by inspection of what it does on stalks. Hence $\mathcal{F}$ is of finite type. $\square$
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