Let $R$ be a conmutative ring (with unit $1$) and $A$ an associative $R$-algebra (with unit $1_{A}$). Let $A^{e}=A\otimes_{R}A^{opp}$ the eveloping algebra of $A$. Let $M$ be an $A$-bimodule. I want to prove that the following operation is a well defined structure of left $A^{e}$-module over $M$:
$$(a\otimes b)\cdot x=axb$$
for all $a,b\in A$ and $x\in M$. How can I make a rigorous proof?
A convenient way to prove it is to use these two:
For an $R$-algebra $A$, an $A$-module structure on am $R$-module $M$ is equivalent an map of $R$-algebras $A\to \text{End}_R(M)$
The universal property of tensor products- a map of $R$ modules $A\otimes_RB \to N$ is equivalent to an $R$-bilinear map $A\times B\to N$
Combining these you can show it defines a $R$-bilinear map $A\times A^\text{op} \to End(M)$ to get well-definedness of $A^e \to End(M)$ then check its multiplicative