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For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite).

This question is related to that one Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding

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    $\begingroup$ As user74230 suggested in an answer below, I assume you mean $\otimes = \otimes_R$? Then there is a paper by Goodearl --- I am traveling and don't remember a more precise reference --- that studies more generally the map $M \otimes_R \prod_i N_i \to \prod_i(M\otimes N_i)$. My memory is that it is always injective when $R$ is Noetherian, but at that level of generality injectivity can fail when $R$ is not Neotherian. Actually, I think the failure is witnessed by modules that are isomorphic to $R^X$ for some $X$. $\endgroup$
    – Theo Johnson-Freyd
    Mar 19, 2015 at 21:32
  • $\begingroup$ Yes, I mean $\otimes = \otimes_R$. Thank you for the reference. $\endgroup$
    – Duchamp Gérard H. E.
    Mar 19, 2015 at 23:26
  • $\begingroup$ I found the reference and think it is : K. R. GOODEARL, DISTRIBUTING TENSOR PRODUCT OVER DIRECT PRODUCT, PACIFIC JOURNAL OF MATHEMATICS Vol. 43, No. 1, 1972 $\endgroup$
    – Duchamp Gérard H. E.
    Mar 19, 2015 at 23:36
  • $\begingroup$ That's the one. $\endgroup$
    – Theo Johnson-Freyd
    Mar 20, 2015 at 23:40

1 Answer 1

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The key point is to generalize the problem in order to make more effective the use of limits. Consider more generally for any $R$-module $R$ the natural map $$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$ given by $m \otimes (r_y) \mapsto (r_y m)$. In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective.

If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$. Hence, it suffices to treat the case when $M$ is finitely generated. In case $M$ is finitely presented then $f_{M,Y}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy). So this gives an affirmative answer when $R$ is noetherian.

Are you interested in non-noetherian $R$?

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  • $\begingroup$ Thank you for your answer, I find it convincing and will redo the details. My primary interest was for $R$ being the ring of analytic (complex) functions on an open domain (which I think is not noetherian). $\endgroup$
    – Duchamp Gérard H. E.
    Mar 19, 2015 at 15:25
  • $\begingroup$ (a) I do not understand (infinite) products as direct but rather as inverse limits can you tell me more ? (b) The ring of entire functions $\mathbb{C}\rightarrow \mathbb{C}$ is unfortunately not noetherian (take the ideals $I_k$ of functions vanishing on $\{k,k+1,k+2,\cdots\}$, it is strictly increasing) ... but (c) I greatly appreciate your generalization of the ``field case'', thanks. $\endgroup$
    – Duchamp Gérard H. E.
    Mar 19, 2015 at 16:51
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    $\begingroup$ @DuchampGérardH.E.: Yes, such rings are not noetherian, and certainly the direct products are not direct limits. But that is not relevant, since I put the problem in the wider generality where the left factor is an arbitrary $R$-module, so it becomes meaningful to consider expressing it as a direct limit of finitely generated submodules. That being said, I wrote my answer specifically without even invoking the notion of direct limit, so I'm not sure what causes you to raise question (a) in your comment. $\endgroup$
    – user74230
    Mar 20, 2015 at 1:29
  • $\begingroup$ Well, I choosed my words carefully : my first reflex is to see products as inverse and not direct limits. I thought you might have a ``hidden trick'' to consider direct limits here a $\endgroup$
    – Duchamp Gérard H. E.
    Mar 20, 2015 at 5:27
  • $\begingroup$ Considering my title as a context, I wanted you to tell me more which is done now. I however maintain my edit for the sake of clarity (for an external reader) and accept your contribution as answer (in which I learned a lot). $\endgroup$
    – Duchamp Gérard H. E.
    Mar 20, 2015 at 5:49

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