Ideal contained in annihilator problem Let $M$ be an $A$-module, and $I\subseteq \operatorname{Ann}(M)$ be an ideal. Why do we can endow $M$ in a natural way with the structure of an $A/I$-module, and why do $M\simeq M\otimes_AA/I$?

Note that $M$ being an $A$-module is equivalent to $M$ being an abelian group and giving a ring map $\varphi:A\rightarrow \mbox{End}(M)$. Then $\mbox{Ann}(M)=ker(\varphi)$ and so if $I\subset \mbox{Ann}(M)$ our ring map factors through the quotient $A/I\rightarrow \mbox{End}(M)$. This is the natural structure.

Second point: Look at the multiplication map $M\otimes A/I\rightarrow M$ with $m\otimes [a] \mapsto am$. Why is it well-defined? Proof that it is an isomorphism.