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 $M\simeq M\otimes_AA/I$?
 A: Hint
1. $(a+I)m:=am$. it is well defined because  $I\subseteq \operatorname{Ann}(M)$.
2. $M/IM\simeq M\otimes_AA/I$ 
A: With this answer, I mereley want to stress this point: If $R$ is a ring, then a $R$ module structure on an abelian group $M$ can be thought of as a ring homomorphism
$$R → \mathrm{End}(M).$$

For example, if $M$ is an $A$-module, associate the ring homomorphism
$$Ψ_A\colon A → \mathrm{End}(M),~a↦ (a~·),$$
where $(a~·)$ denotes scalar multiplication by $a$. From this perspective,
$$\mathrm{ann}(M) = \ker Ψ_A,$$
so for any ideal $I ⊂ \mathrm{ann}(M)$, there’s a ring homomorphism
$$Ψ_{A/I}\colon A/I → \mathrm{End}(M),~[a]_I ↦ Ψ_A(a).$$
Now it’s easy to check that the map
$$A/I×M → M,~(α,m) ↦ Ψ_{A/I}(α)(m),$$
defines a scalar multiplication (which is just given by $[a]_I·m = am$).
So from this perspective, the first claim is merely an application of the universal property of the kernel of a map/universal property of factor rings.
A: 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.
