# Checking that a sheaf is coherent in the proof of Hartshorne exercise III.3.1

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.

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.
• 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 $i^* \mathscr{N}^{i-1} \mathscr{F}$? Sep 19, 2021 at 14:05
• 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. Sep 19, 2021 at 17:08