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Suppose $X=X_1\cup\cdots\cup X_n$ is an irreducible decomposition of variety(scheme). Suppose each $X_i$ is projective, then is $X$ projective? Does this hold for proper case? The local ring at the intersection became unclear to me, I am not sure if the valuation criteria can still be verified at the intersection points? (I am reading the definition of a stable curve, which says a complete curve, or projective curve has some properties, I am not clear projectiveness means to treat each component or the whole can be embedded in some projective space?)

Michael Hardy
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  • I do not understand what you want to verify in the local ring of intersection points. However, for stable curves $C$, any power $\omega_C^{\otimes m}$ of the dualizing line bundle is very ample (for $m\geq 3$), so you do not need an embedding for each component. – Brenin May 06 '14 at 16:06
  • Is there a reference for the property or it is very easy? I am not familiar with the dualizing sheaf on a curve in this case..But your answer seems nice! –  May 06 '14 at 16:17
  • A proof is in Deligne-Mumford The irreducibility of the space of curves of given genus, Theorem 1.2. I think that paper contains nothing "very easy". But if you forget about "$m\geq 3$" and just want to show ampleness of $\omega_C$, then I think it is a bit easier. – Brenin May 06 '14 at 16:32

3 Answers3

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If each $X_i$ is proper, then $X$ is. (Just check the definition directly; you don't need to use the valuative criterion.)

Matt E
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Ok, we can have the following counter-example: Take two copies of $\mathbb{P}^3$, pick $L,C$ be disjoint line and conic in the first copy, pick $L',L''$ be disjoint lines in the second copy, then glue $L,L'$, glue $C,L''$. The glued union is projective, because the existence of ample line bundle would imply $1=2$ by intersection theory.

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I think I have a proof that $X$ must be projective if it's a curve over a field. If each component is projective, then they're also proper, so $X$ is proper (Matt's answer). Then by this question, $X$ must be projective.

ggg
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