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Assume that $f : X \rightarrow Y$ is a morphism of schemes. Then prove that $f$ is a closed immersion if-f there is an affine cover of $Y$ say $\{ U_i \}$, such that the induced scheme morphisms $f^{-1}(U_i) \rightarrow U_i$, is a closed immersion $ \forall \thinspace i \in I$.

(The above is an exercise from Vakil's notes)

  • the $\Rightarrow$ is doable. The other direction I cannot prove. –  Dec 20 '16 at 22:19
  • What is your definition of a closed immersion? That there exists a quasi-coherent sheaf $\mathcal{I}$ on $Y$ such that $f_{\ast}\mathcal{O}_X\cong\mathcal{O}_Y/\mathcal{I}$? – Servaes Dec 20 '16 at 22:26
  • $f(X)$ is a closed subset of $Y$, and the corresponding sheaf morphism surjective. –  Dec 20 '16 at 22:32
  • Ok... so for starters, are you able to show that $f(X)$ is closed in $Y$ when $f(U_i)$ is closed in $U_i$ for all $i$? – Servaes Dec 20 '16 at 22:38
  • also doable, yes, my problem is the construction of the sheaf, going from local data to global... although you mean $f(f^{-1}(U_i))$, open in $U_i$ I guess. –  Dec 20 '16 at 22:39
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    @mayer_vietoris Recall that a morphism of sheaves is surjective if and only if it is surjective on stalks. Since any point lies in an affine neighborhood, you can prove what you want by using that a ring homomorphism is surjective if and only if it is surjective after localizing at each prime. – Takumi Murayama Dec 21 '16 at 04:00

1 Answers1

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Lemma/Definition 1: Let $\operatorname{Spec} B \rightarrow \operatorname{Spec} A$ be a morphism of affine schemes induced from $A \rightarrow B$, this is a closed immersion iff $A \rightarrow B$ is a surjective ring homomorphism.

Let us use the Affine communication lemma, 5.3.2, pg. 158. We define property $P$ by

Definition 2: $\operatorname{Spec} A \hookrightarrow X$ has property $P$ iff the induced $\operatorname{Spec} B = \pi^{-1}\operatorname{Spec} A \rightarrow \operatorname{Spec} A$ is a closed embedding.

Now we check the conidtions.

(i) We show if $ \varphi:A \rightarrow B$ is a surjection, then $A_f \rightarrow B_{\varphi f}=B \otimes A_{f}$ is a surjection for all $f \in A$. This follows as $\otimes A_f$ is right exact.

(ii) We show if $A_{f_i} \rightarrow B_{\varphi f_i}=B \otimes A_{f_i}$ is surjective for all $f_i$ with $\langle f_1, \ldots, f_n \rangle =A$, then $A \rightarrow B$ is surjective. This then follows from the the two general propositions below.

Affine communication lemma then implies the property holds for all affine open.

Prop 1: If $M \rightarrow N \rightarrow P$ is an exact sequence of $A$ modules, then it is exact iff it is so when localize at all maximal ideals.

Proof: This follows from combining two observations:

a) $M=0$ iff $M_{m}=0$ for all maximal ideals of $A$.

b) localization is exact. So homology of the chain complex of $A$-modules commutes with localization.

Prop 2. $M_m=0$ for all $m$ maximal ideal of $A$ iff $M_{f_i}=0$ for all $i$ where $\langle f_1, \ldots, f_n\rangle =A$.

Proof: One direction follows from a) of Prop 1. Let $m$ be any maximal ideal, there must exist some $f_i$ not in $m$. Thus $M_m \cong (M_{f_i})_m =0$.

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