Why is $M \otimes_R N$ not an $R$ module? My professor said in lecture that for $M$ a $(A, R)$ bimodule and $N$ a $(R, B)$ bimodule that $M \otimes_R N$ is not an $R$ module anymore but an abelian group and that it is naturally an $(A, B)$ bimodule. I can see how it is an $(A, B)$ bimodule but I think that it is also an $R$ module. Did I misunderstand my professor? We can still extend scalar multiplication. 
 A: The only obvious possibility for a left action of $R$ on $M \otimes_R N$ would be by the formula
$$ r \cdot (m \otimes n) = m \otimes rn $$
but in general that's not well-defined; if it were, then if $r,s \in R$, you'd have
$$m \otimes rsn = r \cdot (m \otimes sn) = r \cdot (ms \otimes n) = ms \otimes rn = m \otimes srn$$

You can do this for a commutative ring, but that's nothing new — if $R$ is commutative then any right $R$-module module $M$ can also be considered as an $(R,R)$-module (the left action being $r \cdot m = mr$), so the left $R$-action you propose is just the one described by "an abelian group that it is naturally a left $A$-module" with $A=R$.
A: If $R$ is a commutative ring, then you are right that $M \otimes_R N$ is still an $R$-module via $r(m \otimes n) = mr \otimes n = m \otimes rn $ and you check that this is a left $R$-action. I guess that this is what you have in mind.
If, however, $R$ is not commutative, there is no meaningful way to get a left $R$-action. The one above does not work: $(r_1r_2)(m \otimes n) = mr_1r_2 \otimes n$, but $(r_1(r_2)(m \otimes n)) = mr_2r_1 \otimes n$, but one should have $(r_1r_2)(m \otimes n) = (r_1(r_2)(m \otimes n))$.
I hope this helps.
By the way, the same also applies to the $Hom_R(.,.)$ Functor.
A: In general, $M\otimes N$ is not an $R$-module in any way at all —it is not that one has overlooked a way to do it.
For example, let $R=M_n(k)$, the matrix ring over a field of ome size $n>1$. Then the vector space $V$ of row vector of size $n$ is a right $R$-module in the obvious way, and the vector space $W$ of column vector is a left $R$-module, and $V\otimes_RW$ is not an $R$-module in any possible way.
Indeed, the vector space $V\otimes W$ is $1$-dimensional and if the field $k$ is finite then there is no $R$-module structure on it at all.
