In Hartshorne, a closed immersion of schemes is defined to be a scheme morphism $\Phi \colon Y \to X$ such that $\Phi$ is a homeomorphism onto $\Phi(Y)$, $\Phi(Y)$ is closed in $|X|$ and $$ (*) \hspace{1cm}\Phi^\# \colon \mathcal{O}_X \to \Phi_* \mathcal{O}_Y \text{ is surjective.} $$

Can I replace $(*)$ by the condition that $\Phi^\#_U \colon \mathcal{O}_X(U)\colon \to \mathcal{O}_Y(\Phi^{-1}(U))$ is surjective for all affine open subsets $U \subseteq X$, or does this lead to another definition?

  • $\begingroup$ Is this really stronger? If you have a closed immersion $i:Y \to X$ and $U=\mathrm{spec}(A)$ is an open affine in $X$, then $i^{-1}(U)$ is an open affine too and of the form $\mathrm{spec}(A/{\mathfrak a})$. $\endgroup$ – Jürgen Böhm Mar 18 '15 at 16:15
  • $\begingroup$ That would make sense to me, but I'm not sure how my condition implies the surjectivity of $\mathcal{O}_X \to \Phi_*\mathcal{O}_Y$. Can I restrict myself to considering affine open subsets when forming the colimit in the stalk definition? $\endgroup$ – legacytron Mar 18 '15 at 16:16
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    $\begingroup$ You can, yes. (For quasicoherent sheaves, this is II.5.1(b) of Hartshorne, but in any case the affine open sets form a basis for the Zariski topology, so you can always take the colimit with respect to them only.) $\endgroup$ – Jake Levinson Mar 18 '15 at 16:26

It is equivalent. We have a short exact sequence of sheaves

$$0 \to I_Y \to \mathcal{O}_X \to \Phi_*\mathcal{O}_Y \to 0,$$

where $I_Y$ is the ideal sheaf of $Y$. This is quasicoherent on $X$, so if we restrict to an affine open set $U \subset X$, the associated sequence of global sections

$$0 \to I_Y(U) \to \mathcal{O}_X(U) \to \Phi_*\mathcal{O}_Y(U) \to 0$$

is again exact. (See Hartshorne II.5.6 and II.5.9.)


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