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Section 5.6 of EGAII is dedicated to Chow's lemma. I am having a hard time following an early step of the proof.

The version of Chow's lemma in the text assumes that $X$ is a separated $S$-scheme of finite type where $S$ is either noetherian or else more generally that $S$ is quasi-compact and $X$ has a finite number of irreducible components. In these cases it is conlcuded that

(i) there exists a quasi-projective $S$-scheme and a surjective, projective morphism $f:X'\to X$,

(ii) that there is an open $U\subseteq X$ such that $f|_{f^{-1}U}:f^{-1}(U)\to U$ is an isomorphism.

The step I am confused about is the one when the proof first reduces to the case when $X$ is irreducible. To do this it says to assumes that Chow's lemma was proved in the irreducible case, and thus taking $X_i$ to be the irreducible components, and $f_i:X'_i\to X_i$ satisfying the conditions of Chow's lemma, we have that if $X'$ is the disjoint union of $X_i'$ and $f$ the morphism from $X'\to X$ which restricts to $f_i$ on $X_i'$, then $f:X'\to X$ is a witness to Chow's lemma.

However, what confuses me here is how we can take the individual $X_i$. What is the scheme structure on them? If we take the reduced structure (which is what the proof seems to be saying to do), condition (ii) will not necessarily hold? Is there a way to fix this?

I notice that the Stacks project goes with a different approach for the proof and they seem to need that $S$ is actually noetherian. Is this way to fix it?

Thanks for any help

Atticus Christensen
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  • I want to write an answer but I'm a bit pressed for time—for now, there is a detailed proof in Görtz/Wedhorn, Theorem 13.100. The idea is to construct an open subset of each irreducible component $X_i$ and use the scheme structure on the scheme-theoretic closure of that open set. Görtz/Wedhorn don't discuss how to construct such an open set, though; for that it looks like the argument in Tag 02O2 in the Stacks Project works. You're right that that part of the Stacks Project assumes noetherianity, but you can do the proof without that assumption, or… – Takumi Murayama Aug 29 '15 at 01:46
  • … even obtain the non-noetherian statement using absolute Noetherian approximation—see Tag 01ZZ in the Stacks Project. – Takumi Murayama Aug 29 '15 at 01:47
  • The stacks project basically assumes quasiseparted in addition to the weaker hypothesis (if I'm not mistaken), which still is not the generality EGA claims. Is the EGA proof salvageable even in the noetherian case? – Atticus Christensen Aug 29 '15 at 02:37
  • This blasted scheme vs prescheme business of EGA confused me. I see the statement of Chow's lemma claimed in EGA holds. I am stlil curious however if the proof via reducing to irreducible case is salvageable – Atticus Christensen Aug 29 '15 at 02:46

1 Answers1

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I think the following works. In the noetherian case, you can just take $V_i = U_i$ below.

Let $S$ be qcqs, let $X \to S$ be separated and of finite type, and let $X = X_1 \cup \cdots \cup X_r$ be a decomposition of $X$ into finitely many irreducible components. Let $U_i = X \setminus \bigcup_{i \ne j} X_j$; note this is a non-empty open subset of $X$ contained in $X_i$ that does not meet any other components, and that $U_i$ contains the generic point $\eta_i$ of $X_i$. Further, let $V_i$ be an affine open subset of $U_i$ containing the generic point $\eta_i \in X_i$. Then, $V_i$ is open in $X$, hence has a canonical open subscheme structure, and $V_i \hookrightarrow X$ is quasi-compact (using the quasi-separatedness of $X$). By [Stacks, Tag 01R8], the scheme-theoretic closure of $V_i$ is therefore a closed subscheme of $X$ with underlying topological space $X_i$.

Takumi Murayama
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  • Ah yes. I see. This then works for the full generality that EGA claims as it has enough assumptions to keep scheme-theoretic closures to behave well. – Atticus Christensen Aug 29 '15 at 12:20