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This question is based on question $3.11$ from chapter $2$ of Hartshorne, found on page $92$.

Part $a)$ of said question asks to show that closed immersions are stable under base extension. In other words, if $f:Y \rightarrow X$ is a closed immersions and $g:X'\rightarrow X$ a morphism of schemes, then the induced map $f':Z=Y\times_X X' \rightarrow X'$ is also a closed immersion.

My question is:

How can we prove this without using that a closed immersion is affine?

I ask because although I can reduce to the case where $X'$ and $X$ are affine fairly easily, I have not been able to find a proof anywhere (including on related questions such as this one on the site) that doesn't use part $b)$ of the question in Hartshorne, namely that if $X = \operatorname{Spec}A$ then $Y$ is affine, and in fact is the closed subscheme determined by some ideal of $A$. Without this, I can't see how to reduce to the affine case since even if $X$ is affine, the restriction of $f$ to an arbitrary affine piece of $Y$ need not be a closed immersion.

Although part $b)$ doesn't rely on part $a)$ and we therefore could use it to prove part $a)$, I can't help but think that if this is what Hartshorne had in mind then he would have switched the order around. This is the first time that I've had to use a later exercise to prove an earlier one and would like to avoid it if at all possible.

Of course it's pretty much impossible to prove that such a proof doesn't exist. I would however accept an answer from someone who feels that they have enough experience to say that if such a proof exists they would most likely have seen it, or who can provide some reasonably convincing heuristic argument that it can't be done.

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  • $\begingroup$ You might be able to use a more concrete description of the stalks at points of $X \times Y$ as detailed here. But I think you would have to show some other stuff in order to apply this and I don't see how to do that without this "affine" proof, which I really do think is the "right" one. $\endgroup$ – Hoot Mar 23 '15 at 8:17
  • $\begingroup$ The Stacks Project shows this in the generality of locally ringed spaces using quasi-coherent sheaves of ideals; then again this might not be satisfying since it uses material from II.5 in Hartshorne. Also, I hope my answer on the linked question was not in vain; sorry it had taken so long! stacks.math.columbia.edu/tag/01JU $\endgroup$ – Takumi Murayama Mar 25 '15 at 5:42
  • $\begingroup$ @TakumiMurayama Thanks for the link, but yes, I would prefer an elementary answer. I think that your answer was good, but it did ultimately end up using $3.11 b)$, which I wanted to avoid. This is why I was badgering you so much about it, because you didn't explicitly mention that part of the question in your pre-edit answer so I had hoped that you had found a way around it. $\endgroup$ – Tom Oldfield Mar 25 '15 at 10:19
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As is usual with questions involving base change, one approach is to try to phrase everything in terms of universal properties, i.e., to ask "what functor does $f:Y\to X$ represent?" In this case, the answer is $$\text{Schemes}_{/X}\to \text{Sets}$$ defined by sending $h:A\to X$ to the set of $\mathcal{O}_X$-module homomorphisms $\mathcal{O}_X/\mathcal{I}_Y\to h_*\mathcal{O}_A$, where $\mathcal{I}_Y$ is the ideal sheaf of $Y$. In terms of universal properties, this says $f:Y\to X$ is characterized up to unique isomorphism by the fact that given any $h:A\to X$ such that $h^\sharp:\mathcal{O}_X\to h_*\mathcal{O}_A$ factors through $\mathcal{O}_X/\mathcal{I}_Y$, there exists a unique homomorphism $\phi:A\to Y$ such that $h = f\circ \phi$.

Once this has been established, we can check that $f':Z\to X'$ satisfies this property. Given $h:A\to X'$ such that $h^\sharp$ factors through $\mathcal{O}_{X'}/\mathcal{I}_Z$, we have that $(g\circ h)^\sharp$ factors through $\mathcal{O}_X/\mathcal{I}_Y$ by commutativity of the pullback square. As $f:Y\to X$ is a closed immersion, there is a unique $\phi:A\to Z$ such that $g\circ h = f\circ \phi$. By the universal property of base extension, there exists a unique $\psi:A\to Z$ such that $f'\circ \psi = h$ and $\mathrm{pr}_1\circ \psi = \phi$. This $\psi$ shows that $f'$ satisfies the above "universal property of closed immersions".

Of course, it remains to show that the claims in the first paragraph are correct. I won't give too many details, but the idea is that given $h:A\to X$ with $h^\sharp$ factoring through $\mathcal{O}_X/\mathcal{I}_Y$, one first shows that $h(A)\subseteq f(Y)$, giving a continuous map $\phi:A\to Y$, and then obtains $\phi^\sharp:\mathcal{O}_Y\to \phi_*\mathcal{O}_A$ by adjunction from $\mathcal{O}_X/\mathcal{I}_Y = f_*\mathcal{O}_Y \to h_*\mathcal{O}_A$, using the facts that (1) $f^*f_*\mathcal{O}_Y\to \mathcal{O}_Y$ is an isomorphism in the case of closed immersions and (2) $h_*\mathcal{O}_A \cong f_*\phi_*\mathcal{O}_A$ canonically.

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  • $\begingroup$ Thank you for the interesting answer. Sorry it took me so long to reply, I have only recently started to understand the link between ideal sheaves and closed subschemes, so I couldn't make much sense of your answer at the time. I am still reluctant to accept the answer since I would ideally like to see a proof that only uses the properties of schemes outlined in Hartshorne before the exercise is set, but this is a really nice alternative proof nonetheless. $\endgroup$ – Tom Oldfield Apr 22 '15 at 11:46
  • $\begingroup$ @TomOldfield Fair enough :) $\endgroup$ – Jan Ladislav Dussek Apr 22 '15 at 17:28

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