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I am reading Kempf's book "Algebraic varieties". On page 55 the author considers a sheaf of rings ${\mathcal{A}}$ (commutative, unital) on a topological space $X$. An ${\mathcal{A}}$-module ${\mathcal{M}}$ is said to be quasi-coherent if there is an open cover $X=\bigcup X_i$ such that we have exact sequences of ${\mathcal{A}}|_{X_i}$-modules $$ {\mathcal{A}}|_{X_i}^{\oplus J}\to {\mathcal{A}}|_{X_i}^{\oplus I}\to {\mathcal{M}}|_{X_i}\to 0 $$ (where the sets $I$ and $J$ can be infinite).

Let $M$ be an ${\mathcal{A}}(X)$-module. Then we can form an ${\mathcal{A}}$-module $M\otimes_{{\mathcal{A}}(X)} {\mathcal{A}}$ by taking it to be the sheaf associated to the presheaf $$ U\mapsto M\otimes_{{\mathcal{A}}(X)}{\mathcal{A}}(U). $$ The author notices that if $M$ is the cokernel of a homomorphism $$ \psi\colon {\mathcal{A}}(X)^{\oplus J}\to {\mathcal{A}}(X)^{\oplus I}, $$ then we have an exact sequence $$ {\mathcal{A}}(X)^{\oplus J}\to {\mathcal{A}}(X)^{\oplus I}\to M\otimes_{{\mathcal{A}}(X)} {\mathcal{A}} \to 0. $$

Then the author remarks that thus on a Noetherian space $X$, an ${\mathcal{A}}$-module ${\mathcal{M}}$ is quasi-coherent if and only if it locally has the form $M_i\otimes_{{\mathcal{A}}(X_i)} ({\mathcal{A}}|_{X_i})$.

Question. How can one prove the assertion of this last remark? Where does one use the assumption that $X$ is Noetherian?

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  • $\begingroup$ This is also proven in the stacks project (search for quasi-coherent modules), even for more general topological spaces (qcqs variant). $\endgroup$ – Martin Brandenburg Jan 5 '14 at 9:59
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You can always express an $A$ module $M$ as a cokernel of $A^{J}\to A^I$ (take a set of generators indicized by $I$, define $A^I\to M$ and then do the same with the kernel and $J$). So, if $\mathcal{M}$ is locally of the form $M_i\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})$, then $\mathcal{M}$ is quasi-coherent because you can express it locally as

$$\mathcal{A}|_{X_i}^{J}\to\mathcal{A}|_{X_i}^{I}\to M_i\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})=\mathcal{M}|_{X_i}\to 0$$

Here you don't have used the assumption that $X$ is noetherian yet. Now, if $\mathcal{M}$ is quasi-coherent and $X$ is noetherian, you have

$${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}\to {\mathcal{M}}|_{X_i}\to 0$$

Now take $\{a_j\}_{j\in J}$, $\{a_i'\}_{i\in I},$ the canonical basis of $\mathcal{A}(X_i)^{J}$, $\mathcal{A}(X_i)^{I}$ and for all $j$ take the image $b_j\in\mathcal{A}|_{X_i}^{I}$ of $a_j$. Locally, $b_j$ is expressed as a linear combination of only a finite number of $a_i'$, but this is true on the whole $X_i$ because $X$ is noetherian (you have to do this for all $j$'s at the same time, so it's not enough to say that this is true locally without $X$ notherian, this is the crucial point).

So, $\mathcal M|_{X_i}$ is the sheafification of

$$U\mapsto\mathcal{A}(U)^I/\mathcal{A}(U)^J$$ that is $\mathcal{A}(X_i)^I/\mathcal{A}(X_i)^J\otimes_{\mathcal{A}(X_i)}(\mathcal{A}|_{X_i})$.

Edit: as requested, adding more details.

Call $P$ the presheaf $U\mapsto \mathcal{A}(U)^I/\mathcal{A}(U)^J$. You have an obvious map $\phi:P\to \mathcal{M}$, induced by $\mathcal{A}^I_{X_i}\to\mathcal{M}$. $\phi$ is an isomorfism everytime you localize to a point $x\in X_i$: this is precisely the fact that ${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}\to {\mathcal{M}}|_{X_i}\to 0$ is exact. Hence, when you sheafify $P$, you get an isomorphism.

You need the fact about the $b_j$'s because $\mathcal{A}^J(X_i)$ is different from $\mathcal{A}(X_i)^J$ (you have to sheafify) hence it's not always true that ${\mathcal{A}}|_{X_i}^{J}\to {\mathcal{A}}|_{X_i}^{I}$ is induced by ${\mathcal{A}}(X_i)^{J}\to {\mathcal{A}}(X_i)^{I}$.

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  • $\begingroup$ Thank you, Giulio. You write that $(a_j)_{j\in J}$ is the canonical basis of $\mathcal{A}|^J_{X_i}$. Do you mean the canonical basis of $\mathcal{A}(X_i)^J$? $\endgroup$ – Mikhail Borovoi Jan 4 '14 at 16:54
  • $\begingroup$ Dear Giulio, you write: So $\mathcal{M}_{X_i}$ is a sheafification of $U\mapsto \mathcal{A}(U)^I/\mathcal{A}(U)^J$. How did you obtain this? How did you use your assertion about $b_j$ and $a'_i$? Please kindly add details! $\endgroup$ – Mikhail Borovoi Jan 4 '14 at 17:04
  • $\begingroup$ Is it clear now? $\endgroup$ – Giulio Bresciani Jan 4 '14 at 17:38
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    $\begingroup$ Not quite. I know that on a Noetherian topological space the presheaf $U\mapsto \mathcal{A}(U)^I$ is a sheaf, and the same for $J$. However, it seems that your argument with $b_j$ and $a'_i$ proves something more - what? I would be very grateful to you if you write a continuous argument answering my original naive question - not just a series of hints.... Many thanks in advance, $\endgroup$ – Mikhail Borovoi Jan 4 '14 at 18:50
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    $\begingroup$ In general, if $R$ is a ring, $S$ an $R$-algebra and there is an $R$-linear morphism $R^J\to R^I$, this induces a morphism $S^J\to S^I$ and there is a canonical isomorphism $R^I/R^J\otimes_RS\simeq S^I/S^J$. This can be seen directly by checking that $S^I/S^J$ satisfies the universal property of the tensor product. In our case, $R=\mathcal{A}(X_i)$ and $S=\mathcal{A}(U)$. $\endgroup$ – Giulio Bresciani Nov 25 '15 at 13:55

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