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Suppose $M$ is a module over $R$ and $S$, commutative rings with $1$.

Under what conditions is $M$ also a $R \otimes S$-module?

Also, a more general question: How to construct a structure of a $R \otimes S$-module? In other words, when one wants to construct a map from $R \otimes S$ to another ring, one can use the universal property of tensor products. What can one do to have a structure of a $R \otimes S$-module?

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For eaxmple $(r\otimes s)m:=rm$ –  userNaN Apr 25 '12 at 18:40
    
@Norbert: This might be silly but how do I check this is well-defined? –  averageman Apr 25 '12 at 18:42
    
It seems reasonable to ask for $r.(s.m)=s.(r.m)$ for all $r\in R, s\in S$ and $m\in M$, i.e. for $M$ to be a $(R,S)$-bimodule. –  Olivier Bégassat Apr 25 '12 at 18:43
    
Shouldn't you require that the $R\otimes S$ structure respect the $R$- and $S$- structures? i.e., that $(r\otimes 1_S)m = rm$ and $(1_R\otimes s)m = sm$ for all $r\in R$ and $s\in S$? –  Arturo Magidin Apr 25 '12 at 18:45
    
I wound up posting my related question: math.stackexchange.com/questions/136921/… –  rschwieb Apr 25 '12 at 20:11

3 Answers 3

Commutativity is unimportant. If $R, S$ are two rings and $M$ is a (left) module over both of them, then all you know is that $M$ is a (left) module over the free product $R*S$ by the universal property of the free product (a slight misnomer as the free product is actually the coproduct in the category of rings). The free product is very different from the tensor product, and among other things is often noncommutative even if $R, S$ are commutative; for example the free product of $\mathbb{Z}[x]$ with itself is $\mathbb{Z} \langle x, y \rangle$, the ring of noncommutative polynomials in two variables (e.g. the free ring on two elements).

To get an $R \otimes S$-module structure you need to assume that the actions of $R$ and $S$ commute, i.e. that $r(s(m)) = s(r(m))$ for all $r \in R, s \in S$.

To get $R \otimes S$-modules in general, take a tensor product $M \otimes N$ where $M$ is an $R$-module and $N$ is an $S$-module.

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For general rings there is a one-to-one correspondence between $R-S$ bimodule structures on $M$ and the $R\otimes S^{op}$ module structures on $M$.

In your case, when $S$ is commutative, this amounts to an $R\otimes S$ module. I can't immediately think of an example of a module over two commutative rings which cannot be a bimodule over them... maybe this is already addressed by the QY's post earlier.

Update: So, you will be able to get an $R\otimes S$ structure exactly when you can get an $R-S$ bimodule structure. This is indeed more restrictive than merely being a module over the two rings. Mariano Suárez-Alvarez came up with a neat example at An $R$ module and $S$ module that cannot be an $R$-$S$ bimodule demonstrating that it's not always possible to make an $R\otimes S$ module out of $M$.

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The OP does not ask about a bimodule but about a module over $R$ which is also a module over $S$. –  Qiaochu Yuan Apr 25 '12 at 19:56
    
@QiaochuYuan The poster asked about the $R\otimes S$ module structures, and as noted, these correspond to the bimodule strucutres. I have some posters telling me at math.stackexchange.com/questions/136921/… that it is possible to have two such ring structures without an $R\otimes S$ module structure. –  rschwieb Apr 25 '12 at 20:09
    
@QiaochuYuan It seems they are still sorting it out, so there is no verdict yet. If a counterexample surfaces, of course, it will be an example that an $R\otimes S$ module structure is not always possible. –  rschwieb Apr 25 '12 at 20:21
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I think I found a nice way of constructing an $R \otimes S$-module structure.

Notice that the data of an $R$-module structure (for any ring $R$) is given by:

  • an abelian group $M$, and
  • a homomorphism of rings $R \rightarrow End(M)$, where $End(M)$ is the ring of group endomorphisms of $M$ (where the operations are addition and composition).

Having said that, to construct an $R \otimes S$-module structure on $M$, we just need a ring homomorphism from $R \otimes S$ to $End(M)$. This can be constructed using the usual universal property of tensor products.

As an example, let $M$ is an $R$-module and an $S$-module such that the actions of $R$ and $S$ commute. Assume these structures are given by maps $\varphi: R \rightarrow End(M)$ and $\psi: R \rightarrow End(M)$, then we can define a structure of $R \otimes S$-module on $M$ via the ring homomorphism $r \otimes s \mapsto \varphi(r) \circ \psi(s)$.

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As I said in my answer, you only get a map $R \otimes S \to \text{End}(M)$ if the actions of $R$ and $S$ commute with each other. For example, let $R = \mathbb{Z}[x], S = \mathbb{Z}[y]$ and $M = \mathbb{Z}^n$ acted on by two noncommuting integer matrices $A, B$. The homomorphism you attempt to define is from $\mathbb{Z}[x] \otimes \mathbb{Z}[y] = \mathbb{Z}[x, y]$ so the images of $x$ and $y$ need to commute (since $(x \otimes 1)(1 \otimes y) = (1 \otimes y)(x \otimes 1) = x \otimes y$), which they don't by assumption. –  Qiaochu Yuan Apr 27 '12 at 15:11
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The universal property of tensor products is that $R \otimes S$ is the universal ring admitting a map from $R$ and from $S$ whose images commute. The universal property you want is of the free product $R*S$, not the tensor product. (The tensor product of commutative rings has the universal property you want in the category of commutative rings, but $\text{End}(M)$ isn't commutative in general.) –  Qiaochu Yuan Apr 27 '12 at 15:14
    
Yes, I corrected the commutativity thing. The example was flawed (not anymore) but the general fact I stated first is still true... –  averageman Apr 27 '12 at 17:30
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The reason why I chose this answer instead of the one by Yuan is because, in my opinion, this one is clearer than his one since it doesn't free product. –  averageman Apr 30 '12 at 15:19

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