Extending a module over an integral domain to a vector space over its field of fractions Let $A$ be a commutative unital ring with field of fractions $K$. Regarding $K$ as an $A$-module, we have that for each $A$-module $M$, $K\otimes_A M$ is an $A$-module and therefore an abelian group. I need to make sense of $K\otimes_A M$ as a vector space over $K$, a bit like how the complexification of a real vector space is a complex vector space constructed using tensor products.
So it must be necessary to construct a map $K\times(K\otimes_A M)\to K\otimes_A M$. I know that maps of tensor products are usually constructed from bilinear ones on the usual product. We have an $A$-scalar multiplication $A\times(K\otimes_A M)\to K\otimes_A M$ and an $A$-bilinear map $(a,m)\mapsto a\otimes_A m:A\times M\to A\otimes_A M$, but where do I go from here?
 A: I will write simply “ring” for a unital (but not necessarily commutative) ring.
See this for the more general notion of tensor product of modules over a ring.
Given an (additive) abelian group $A$, and a ring $S$, a map $S \times A \to A$ that makes $A$ a $S$-module is “equivalent” to a ring homomorphism $S \to \operatorname{End}_\mathbb{Z}(A)$.
More precisely, given $m \colon S \times A \to A $, define $S \to \operatorname{End}_\mathbb{Z}(A)$ by $s \mapsto m(s,\_)$; and given $r \colon S \to \operatorname{End}_\mathbb{Z}(A)$, define $S \times A \to A$ by $(s,a) \mapsto r(s)(a)$.
It is easy to check that these two correspondences are mutually inverses.
Using the above, I will prove that if $R$ and $S$ are two rings, $M$ a $(S,R)$-bimodule, and $N$ a left $R$-module, then the abelian group $M \otimes_R N$ is a left $S$-module such that
$$
\forall s \in S,\, \forall m \in M,\, \forall n \in N \quad s(m \otimes n) = (sm) \otimes n.
$$
Indeed, given $s \in S$, the $R$-balanced mapping
$$
\mu_s \colon M \times N \to M \otimes_R N, \quad (m,n) \mapsto (sm) \otimes n
$$
induces a group homomorphism $\tilde \mu_s \colon M \otimes_R N \to M \otimes_R N$ such that $\tilde \mu_s \circ \otimes = \mu_s$. It is easy to prove that the mapping $s \mapsto \tilde \mu_s$ is a ring homomorphism $S \to \operatorname{End}_\mathbb{Z}(M \otimes_R N)$, as desired.
