What is the module structure here? Let $A$ be a local ring with maximal ideal $\mathfrak{m}$, and let $M$ be an $A$-module.
I want to turn the following object into an $A/\mathfrak{m}$-module:  $$A/\mathfrak{m} \otimes_A M$$
I attempt to multiply in this way:
$$\left( [a],\sum_{i = 1}^n [b_i] \otimes_A m_i \right) \mapsto \sum_{i=1}^n [ab_i] \otimes_A m_i$$
(Here, $[a]$ is the class of $a \in A$ modulo $\mathfrak{m}$.)
Is this idea correct, and why or why not?
I think the fact that $\mathfrak{m} = \text{Ann}(A/\mathfrak{m})$ has something to do with it.
Thanks!
 A: Whenever you have an algebra $A \to B$ and an $A$-module $M$ the tensor product of $A$-modules $B \otimes_A M$ has the structure of a $B$-module. You can construct this cleanly via the $A$-trilinear map
\begin{align*}
B \times B \times M &\to B \otimes_A M\\
 (b, x, m) &\mapsto (bx) \otimes m.
\end{align*}
But your comment about annihilators is good too and ties into this construction: if $I$ is an ideal of $A$ then $(A/I) \otimes_A M \simeq M/(IM)$ and if $I$ is contained in the annihilator of $M$ then $IM = 0$, so $(A/I) \otimes_A M \simeq M$. But maybe it's better to show this without tensor products.
A: I believe this is better understood when non (necessarily) commutative rings are involved.
If $R$ and $S$ are rings, an $R$-$S$-bimodule $_RM_S$ is an abelian group $M$ which is also a left $R$-module and a right $S$-module, so that
$$
(rx)s=r(xs)
$$
for all $r\in R$, $x\in M$ and $s\in S$.
If $M_S$ is a right $S$-module and $N$ is a left $S$-module, then $M\otimes_SN$ has nothing more than the structure of abelian group, generally speaking. However, if $_RM_S$ is a bimodule, then $M\otimes_SN$ can be regarded as a left $R$-module by defining
$$
r(x\otimes y)=(rx)\otimes y
$$
extended by linearity (only with respect to sums). This is well defined, as the map
$$
l_r\colon M\times N\to M\otimes_SN
\qquad
l_r(x,y)=(rx)\otimes y
$$
is balanced, so it induces a group homomorphism $\lambda_r\colon M\otimes_SN\to M\otimes_SN$ such that $\lambda_r(x\otimes y)=(rx)\otimes y$. It is routine to verify that we get a left $R$-module structure on $M\otimes_SN$ by defining $r(x\otimes y)=(rx)\otimes y$ (extended by additive linearity).
When a ring $A$ is commutative, any $A$-module can be regarded as an $A$-$A$-bimodule in an obvious way; it's this feature that allows to regard $M\otimes_AN$ as an $A$-module, using the above setup.
In your case, $A/\mathfrak{m}$ is an $A/\mathfrak{m}$-$A$-bimodule, so $A/\mathfrak{m}\otimes_AM$ becomes an $A/\mathfrak{m}$-module by
$$
[a]([x]\otimes y)=[ax]\otimes y
$$
extended by additive linearity.
