Tensor product of two $R$-modules $R$ a ring. Given $M$ a right $R$-module and N a left $R$-module, we define the tensor product
$$M \otimes_R N$$
where (among other properties) $mr \otimes n = m\otimes rn$ holds for all $r \in R, m \in M, n \in N$.
My question is probably trivial but why do we care so much about $M$ being a right module and $N$ being a left one? Couldn't we just define something where $rm \otimes n = m\otimes rn$ holds, in the case they are both left $R$-modules? Can someone give me an example of why this is actually important?
 A: Your definition $rm \otimes n = m\otimes rn$ would  work when $R$ is commutative, since in which case left $R$ modules are right $R^{op}=R$ modules. 
But in case of noncommutative rings $rm \otimes n = m\otimes rn$  does not work, since this would lead to $m \otimes (rs).n = m\otimes (sr).n$ in the following way  $$m \otimes (rs).n=m \otimes r.(s.n)=r.m \otimes s.n=s.(r.m) \otimes n=(sr).m \otimes n=m\otimes (sr).n$$
A: I would say that this is because the tensor product is made to study bilinear forms, and that bilinear forms are usually asked to satisfy $b(mr,n)=b(m,rn)$.  
More precisely, recall that a bilinear form $b:M\times N\to X$, where $X$ is any abelian group, is a map satisfying 


*

*$b(m_1+m_2, n_1+n_2) = b(m_1,n_1)+b(m_1,n_2)+b(m_2,n_1)+b(m_2,n_2)$, and

*$b(mr,n)=b(m,rn)$,


for all $m,m_1,m_2\in M$, $n,n_1,n_2\in N$ and $r\in R$.
The tensor product of $M$ and $N$ is then defined to be a kind of repository of "universal expressions involving bilinear forms".  This is made precise in its universal property: if $t:M\times N \to M\otimes N$ is the defining map, then for any bilinear map $b:M\times N\to X$, there is a unique morphism of abelian groups $\overline{b}:M\otimes N \to X$ such that $\overline{b}t=b$.

Now, you could very well decide that you are not interested in bilinear maps, but instead in maps $c:M\times N\to X$ (with $M$ and $N$ now both left modules) satisfying


*$c(m_1+m_2, n_1+n_2) = c(m_1,n_1)+c(m_1,n_2)+c(m_2,n_1)+c(m_2,n_2)$, and

*$c(rm,n)=c(m,rn)$,


for all $m,m_1,m_2\in M$, $n,n_1,n_2\in N$ and $r\in R$.  Note that property 4 is not the same as 2.
You could then define a new object, say $M\otimes' N$, with the universal property that it comes with a map $s:M\times N\to M\otimes' N$ satisfying properties 3 and 4, and such that whenever you have a map $c:M\times N\to X$ satisfying 3 and 4, there is a unique morphism of abelian groups $\overline{c}:M\otimes' N \to X$ such that $\overline{c}s=c$.
You could then construct such an object as you would the tensor product; in short, its elements would be linear combinations of symbols of the form $m\otimes' n$, subject to linearity in $m$ and $n$, and to $rm\otimes' n = m\otimes' rn$.

So I guess a short answer to your question would be: the property you want could be imposed, but it is not the one we choose to study, because bilinear forms.
