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In connection with this question: Modules and tensor products

Question: For two commutative rings $R$ and $S$ (with unity), is there an abelian group $M$ which has $R$ module and $S$ module structures, but never has an $R$, $S$ bimodule structure? (All structures unital.)

It may always be that a bimodule struture is possible, using the free product mentioned on that page, but I'm not familiar enough with the construction.

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up vote 8 down vote accepted

Consider the abelian group $M=\mathbb Q^2$ and let $R=\mathrm{End}_{\mathbb Z}(M)$ be its endomorphism ring. $R$ is in a natural way a $\mathbb Q$-algebra and in fact it is isomorphic to $M_2(\mathbb Q)$, the algebra of $2\times 2$ rational matrices.

Then $M$ is both a left and a right $R$-module, but it is not an $(R,R)$-bimodule in any way.

Indeed, such a structure would give us a morphism of rings $\Phi:R\otimes R^{\mathrm{op}}\to\mathrm{End}_{\mathbb {Z}}(M)$ such that $\Phi(r\otimes s)(m)=rms$ for all $r$, $s\in R$ and all $m\in M$. Since $R\otimes R^{\mathrm{op}}$ is a simple algebra, the map $\Phi$ must be injective, but then we have $16=\dim_{\mathbb Q}R\otimes R\leq\dim_{\mathbb Q}\mathrm{End}_{\mathbb {Z}}(M)=4$, which is absurd.

A similar idea works if we restrict to commutative rings:

For example, the fields $R=\mathbb Q(\sqrt2)$ and $S=\mathbb Q(\sqrt 3)$ are both isomorphic to $M=\mathbb Q^2$ as abelian groups, so $M$ can be made into a left $R$-module and into a left $S$-module. But it cannot be made into an $R\otimes S$-module in any way. Indeed, such a structure would give a morphism of rings $R\otimes S\to M_2(\mathbb Q)$. One can check that $R\otimes S$ is a field, so the map must be injective. Since its domain and codomain have the same dimension over $\mathbb Q$, the map is moreover an isomorphism. This is absurd, because its domain is commutative and its range is not.

We can recast this last example in terms of pure linear algebra: it says that there are not two matrices $A$, $B\in M_2(\mathbb Q)$ which commute and such that $A^2=2I$ and $B^2=3I$; these matrices correspond to the action of $\sqrt2$ and $\sqrt3$ on $\mathbb Q^2$ for a hypothetical $(R,S)$-bimodule structure on $M$. In these terms we can argue as follows: $A$ becomes diagonalizable over $\mathbb Q(\sqrt2)$ with distinct eigenvalues, and since $B$ commutes with $A$, $B$ itself must be diagonalizable over $\mathbb Q(\sqrt2)$: this is impossible, because the egenvalues of $B$ are square roots of $3$ and there is no such thing in $\mathbb Q(\sqrt2)$.

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$R$ and $S$ are assumed to be commutative in the question. –  Cihan Apr 25 '12 at 20:01
    
Ah! ${}{}{}{}{}$ –  Mariano Suárez-Alvarez Apr 25 '12 at 20:02
    
Thanks, it's an example to have around anyway! –  rschwieb Apr 25 '12 at 20:13
    
The new example has a few finesses I really like! –  rschwieb Apr 25 '12 at 22:08
    
@MarianoSuárez-Alvarez Sorry to bother Mariano, but, did you have time to read it? –  Pedro Tamaroff Apr 25 '12 at 23:13
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Take $R = \mathbb Q^2$ and $S = \mathbb Q[\sqrt 2]$. As abelian group, they are isomorphic, let call $M$ the common group strucure. (Take $(1, \sqrt 2)$ as a basis for S).

Then $M$ is a $R$-module and a $S$-module. However, it is not a $R\otimes S$-module such that $(r\otimes 1)\cdot m = r\cdot m$ and $(1\otimes s)\cdot m = s\cdot m$ — which I think this is a minimal compatibility condition.)

Indeed, if it were, then on the one hand $$ \begin{aligned} ((0,1)\otimes \sqrt 2) \cdot \sqrt{2} &= ((0,1)\otimes 1)\cdot ((1,1)\otimes \sqrt 2)\cdot \sqrt{2} \\ &= (0,1)\cdot( \sqrt{2} \cdot \sqrt2)\\ &= (0,1)\cdot(2, 0) \\ &= 0 \end{aligned} $$

But on the other hand $$ \begin{aligned} ((0,1)\otimes \sqrt 2) \cdot \sqrt{2} &= ((1,1)\otimes \sqrt 2)\cdot ((0,1)\otimes 1)\cdot \sqrt{2} \\ &= \sqrt2 \cdot ( (0,1)\cdot (0,1) ) \\ &= \sqrt2 \cdot \sqrt2\\ &= 2 \end{aligned} $$

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I think the question asks for an abelian group which admits "some" $R$-module structure and "some" $S$-module structure but "no" $R \otimes S$-module structure. There is no compatibility requirement. In particular there should be no ring homomorphisms between $R$ and $S$ because otherwise restriction of scalars gives a bimodule structure on $M$. In your example there is a ring homomorphism $\mathbb{Q}^2 \rightarrow \mathbb{Q}[\sqrt{2}]$ given by projection. –  Cihan Apr 25 '12 at 20:19
    
Yes, I would like to see (if possible) a module with no $R\otimes S$ module structure, despite having an $R$ structure and $S$ structure separately. –  rschwieb Apr 25 '12 at 20:25
    
I cannot deny ! I guess the proof of Mariano can be adapted to my example, but there is no point, his is very good. –  Lierre Apr 25 '12 at 22:55
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