# Example of a morphism of schemes whose kernel sheaf is not quasi coherent

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?

Thanks.

• 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. Jan 12 at 19:29
• 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. Jan 12 at 21:30
• @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).
– Sam
Jan 12 at 22:54
• Uhoh! My apologies for the link which does not do what I thought it would. Jan 12 at 23:20

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.
• The scheme $X$ is not affine, because it is not quasi-compact. Jan 13 at 7:54
• @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? Jan 13 at 13:16
• Sorry, somehow I thought $X$ was the target. So all is fine. Jan 13 at 18:57