Again I'm stuck on a problem in Hartshorne. The situation is as follows: Let $X=\text{Spec}(A)$ be an affine, non-singular scheme which is finite of some field $k$. Let also $\mathcal{F}$ be a coherent sheaf on $X$. Let $X'$ be an infinitesimal extension of $X$ by $\mathcal{F}$. This means $X'$ has a sheaf of ideals $\mathcal{J}$ such that $\mathcal{J^2}=0$ and there is an isomorphism $(X',\mathcal{O}_{X'}/\mathcal{J})\cong (X,\mathcal{O}_X)$ under which $\mathcal{J}$ (which is a $\mathcal{O}_{X'}/\mathcal{J}$ module) corresponds to $\mathcal{F}$. We have to show that the extension is trivial meaning $\mathcal{O}_{X'}=\mathcal{O}_X\oplus \mathcal{F}$.

My idea was to try and show that $X'$ is affine. If it is we could apply global sections functor to the exact sequence $$0\to \mathcal{J}\to \mathcal{O}_{X'}\to i_*\mathcal{O}_X\to 0$$ of quasi-cogerent $\mathcal{O}_{X'}$-modules (where $i:X\to X'$ is a homeomorphism) to get an exact sequence $$0\to J\to B\to A\to 0$$ of $B$-modules where $J=\Gamma(\mathcal{J})$ and $B=\Gamma(\mathcal{O}_{X'})$. We could then use the infinitesimal lifting property from the previous problem to see that $B$ is the trivial extension of $A$ and $J$.

If anyone has an idea on how to show $X'$ is affine or want to share another approach to this problem I would be very grateful as always=)

  • 1
    $\begingroup$ (Sorry for posting a wrong answer at first.) I found that this fact ($X$ affine implies $X'$ affine) is written in EGA, I 5.1.9. The proof uses the vanishing theorem for the cohomology of quasi-coherent sheaves on affine schemes. $\endgroup$ Oct 7, 2021 at 3:18
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    $\begingroup$ Similarly to Jun Koizumi's comment above, Hartshorne has this fact that $X$ affine implies $X'$ affine as exercise III.3.1. Personally, I feel proving this fact without cohomology to be a little silly - one can do it by tracing through all the statements if you really want to, but I don't think one really learns anything past an appreciation for the machinery of cohomology for organizing all of these details. $\endgroup$
    – KReiser
    Oct 7, 2021 at 4:41
  • $\begingroup$ Thank you for your answers. In that case I will move past this for now and maybe come back to it once I've read a little bit more about cohomology. $\endgroup$
    – budwarrior
    Oct 7, 2021 at 14:00


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