I am trying to think of an explicit example of a morphism $\varphi: X\longrightarrow Y$ of schemes for which the kernel sheaf $ker(\mathcal{O}_Y\longrightarrow \varphi_*\mathcal{O}_X)$ is not quasi coherent. Please help!

Here is what I tried: I understand that since an open immersion $\varphi:U\longrightarrow X$ need not be quasi compact (even though separated), $\varphi_*\mathcal{O}_U$ need not be quasi coherent and so does the kernel. So, I am considering the following example: $X=Spec k[x_1, x_2,……]$ and the open immersion $U= X\setminus \{0\}\hookrightarrow X$ because U is not quasi compact. But I haven’t been able to show that the kernel sheaf of this open immersion is not quasi coherent. Is this example even correct?


  • $\begingroup$ Here is a question where an example of a morphism $f:X\to Y$ with $f_*\mathcal{O}_X$ not quasicoherent is constructed. This should give you the required kernel. $\endgroup$
    – KReiser
    Jan 12 at 19:29
  • $\begingroup$ Your example looks incorrect, because the restriction maps on your $\mathcal{O}_X$ are all injective (since your ring is a domain). I think that the example in the thread @KReiser links to suffers from the same issue, but maybe not the one that the answerer themselves link to. $\endgroup$
    – Aphelli
    Jan 12 at 21:30
  • $\begingroup$ @KReiser Even though $f_*\mathcal{O}_X$ is not quasi coherent, the kernel for all the restrictions are zero and so the kernel sheaf is quasi coherent (as also mentioned by @Aphelli). $\endgroup$
    – Sam
    Jan 12 at 22:54
  • $\begingroup$ Uhoh! My apologies for the link which does not do what I thought it would. $\endgroup$
    – KReiser
    Jan 12 at 23:20

1 Answer 1


Let $Y=\operatorname{Spec} k[x]$ and let $X=\coprod_{n>0} \operatorname{Spec} k[x]/(x^n)$ where the map $X\to Y$ is given on each component by the spectrum of the reduction homomorphism $k[x]\to k[x]/(x^n)$. On global sections, the kernel of the the sheaf map $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is zero: for any polynomial, it vanishes modulo all $x^n$ iff it is the zero polynomial. On the other hand, on $D(x)$, the kernel is all of $\mathcal{O}_Y(D(x))\cong k[x]_x$, as $(f_*\mathcal{O}_X)(D(x))=0$ since $f^{-1}(D(x))$ is the empty set. This shows the kernel cannot be quasi-coherent: a quasi-coherent sheaf on an affine scheme with no global sections must actually be the zero sheaf.

  • $\begingroup$ The scheme $X$ is not affine, because it is not quasi-compact. $\endgroup$ Jan 13 at 7:54
  • $\begingroup$ @DamianRössler yes, that's correct. Forgive me, but I'm not sure why you've left the comment - the problem doesn't require that $X$ is affine and I don't claim that $X$ is affine. Is it just a general warning to future readers, or is there a specific point you're making? $\endgroup$
    – KReiser
    Jan 13 at 13:16
  • $\begingroup$ Sorry, somehow I thought $X$ was the target. So all is fine. $\endgroup$ Jan 13 at 18:57

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