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For a homomorphism of rings $R \to S$, the following are equivalent:

a) $(-) \otimes_R S : \mathrm{Mod}(R) \to \mathrm{Mod}(S)$ reflects isomorphisms

b) $R \to S$ satisfies effective descent with respect to modules.

c) $R \to S$ is pure: For every $R$-module $M$ the natural map $M \to M \otimes_R S$ is a monomorphism.

For a reference, see "Descent Theory for Schemes" by Bachuki Mesablishvili; this work generalizes Grothendieck's descent theory for quasi-coherent modules.

Question. Let $R$ be a ring. Is there a noetherian ring $S$ together with a ring homomorphism $R \to S$ satisfying the equivalent properties above?

I don't know which one of (a) or (c) are more easy to verify. I've also included (b) to illustrate the strength of the condition, but also the geometric significance: Is every (affine) scheme "built up" out of noetherian schemes with respect to the "topology" of effective descent mophisms for quasi-coherent modules?

An affirmative answer would settle another problem in my work. On the other hand, I have no idea how to construct a reasonable noetherian ring $S$ over $R$ at all. If $R$ is a domain (which you may assume for convenience), then $S = \mathrm{Quot}(R)$ is an example, but of course it does not satisfy the condition.

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up vote 8 down vote accepted

I am afraid that this is not true for any non-noetherian ring.

Let $R \to S$ be a descent extension, and assume that $S$ is noetherian. Let $I$ be an ideal of $R$; there will be a finitely generated ideal $J \subseteq I$ of $R$ such that $JS = IS$. Hence $R/I$ and $R/J$ become isomorphic when tensored with $S$, so they are isomorphic are $R$-modules. So $I = J$. Hence $R$ is noetherian.

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Thanks Angelo! I have no idea why I didn't see this obvious proof, especially after having written down the same for faithfully flat $R \to S$. –  Martin Brandenburg Nov 27 '11 at 17:35
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