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.
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.