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I'm doing Hartshorne exercise 3.1 of chapter 3, which asks one to prove that if $X$ is a Noetherian scheme, then $X_{red}$ affine implies $X$ affine. To do this, I'm trying to show that the higher cohomology of any coherent sheaf $\mathscr{F}$ on $X$ vanishes. Letting $\mathscr{N}$ be the sheaf of nilpotents on $X$, we have the filtration $$ \mathscr{F} \supset \mathscr{N} \cdot \mathscr{F} \supset \mathscr{N}^2 \cdot \mathscr{F} \supset \dots $$ and the quotients $\mathscr{N}^{i-1} \scr{F}/\mathscr{N}^i \mathscr{F}$ naturally have an action of $\mathscr{O}_X/\mathscr{N} = \mathscr{O}_{X_{red}}$. I think I can conclude what I want if I can show that these quotient sheaves are actually coherent on $X_{red}$, but I'm having trouble with this verification. I've tried to use the fact that quotients/pullbacks of coherent sheaves are coherent and I suspect that $\mathscr{N}^{i-1} \scr{F}/\mathscr{N}^i \mathscr{F}$ is the pullback under the natural map $i: X_{red} \to X$ of something like $\mathscr{N}^{i-1} \scr{F}$, but so far I haven't been able to show this.

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You're right, $\mathscr{N}^{i-1} \scr{F}/\mathscr{N}^i \mathscr{F}$ is exactly $i^*\mathscr{N}^{i-1} \scr{F}$. It suffices to verify this affine-locally, so take $X=\operatorname{Spec} A$, $\mathscr{F}=\widetilde{F}$, and $\mathscr{N}=\widetilde{N}$. Then $i^*\mathscr{N}^{i-1}\mathscr{F}\cong (N^iF\otimes_A A/N)^\sim$, and as $R/I\otimes_R M\cong M/IM$, we have $N^{i-1}F\otimes_A A/N\cong N^{i-1}F/N^iF$. Taking the associated sheaf, this gives the claim.

To finish from here, one observes that for any map of noetherian schemes $f:X\to Y$, $f^*$ pulls back coherent sheaves to coherent sheaves and then also that $\mathscr{N}^i\mathscr{F}$ is coherent for any $i$, as both $\mathscr{N}^i$ and $\mathscr{F}$ are coherent.

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  • $\begingroup$ Thanks! I tried the same computation to pullback $\mathscr{N}^{i-1} \mathscr{F} / \mathscr{N}^i \mathscr{F}$ on $X$ to $X_{red}$, and weirdly I get that this pullback is also the sheaf $\mathscr{N}^{i-1} \mathscr{F} / \mathscr{N}^i \mathscr{F}$ on $X_{red}$. Is this correct / is there some intuitive explanation for why this pullback also equals $i^* \mathscr{N}^{i-1} \mathscr{F}$? $\endgroup$
    – Legendre
    Sep 19, 2021 at 14:05
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    $\begingroup$ Yes, that's correct. It shouldn't be so surprising that there are different sheaves which pull back to the same thing under a closed immersion - basically, the difference between the two sheaves happens away from the image of the closed subscheme in $X$. What is kind of interesting here is that $X$ and it's reduction have the same points, so you have to come up with a slightly funny interpretation of what "away from the image" means. $\endgroup$
    – KReiser
    Sep 19, 2021 at 17:08

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