Let $\pi : X \to S$ be a morphism of schemes. Is the $\mathcal{O}_X$-module $\mathcal{O}_X \otimes_{\pi^{-1} \mathcal{O}_S} \mathcal{O}_X$ quasi-coherent? Here, $\mathcal{O}_X$ acts on the tensor product on the right (or on the left, doesn't matter).

Probably it suffices to look at the affine case. But even then it is quite challenging and I don't know an answer.


I edited the answer because I found a counterexample to my reasoning. I am very sorry.

Let $X = Spec(k[x])$, $S = Spec(k)$ with the obvious map between spectra. I claim that $\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X}$ cannot be quasi-coherent, at least if we equip it with the left action. (Or the right one, by the symmetry of the argument.)

Indeed, let $p = (x), q = (0)$ be points of $X$. We have isomorphisms

$(\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X}) _{p} \simeq k[x] _{(p)} \otimes _{k} k[x] _{(p)} \simeq \bigoplus _{e_{i} \in I} k[x] _{(p)}$

of $k[x]$-modules, where $e_{i} \in I$ is some basis of $k[x] _{(p)}$ over $k$. (This is the basis of right summand, since we are using the left action.)

Assume by contradiction that $(\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X})$ is quasi-coherent. Then the localization map

$((\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X}) _{(p)} ) _{(q)} \rightarrow (\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X}) _{(q)}$

would be an isomorphism. But

$((\mathcal{O} _{X} \otimes _{\pi^{-1} \mathcal{O} _{S}} \mathcal{O} _{X}) _{(p)}) _{(q)} \simeq (k[x] _{(p)} \otimes _{k} k[x] _{(p)}) _{(q)} \simeq (k[x] _{(p)}) _{(q)} \otimes _{k} k[x] _{p}$

and the localization map corresponds to the inclusion

$k[x] _{(q)} \otimes _{k} k[x] _{(p)} \hookrightarrow k[x] _{(q)} \otimes _{k} k[x] _{(q)}$

and this is not an isomorphism.

  • $\begingroup$ Ah, thank you. The argument generalizes and shows that almost never this tensor product is qc. $\endgroup$ – Martin Brandenburg Feb 6 '13 at 22:36

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