An $R$ module and $S$ module that cannot be an $R$-$S$ bimodule In connection with this question: When is a module over $R$ and $S$ an $R \otimes S$-module?
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.
 A: 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
$$ \begin{aligned}
((0,1)\otimes \sqrt 2) \cdot \sqrt{2} &= ((1,1)\otimes \sqrt 2)\cdot ((0,1)\otimes 1)\cdot \sqrt{2} \\ 
&= \sqrt2 \cdot ( (0,1)\cdot (0,1) ) \\
&= \sqrt2 \cdot \sqrt2\\
&= 2
\end{aligned} $$
A: 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 $\mathbb Q(\sqrt2)$.
