TITLE: Why an R-module is an R/I-module precisely when it is annihilated by I?
QUESTION [9 upvotes]: Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. Prove that the formula $(r + I) \cdot m = r \cdot m$ for $r + I \in R/I$, and $m \in M$ makes $M$ into an $R/I$-module if and only if $i \cdot m = 0$ for all $i \in I, m \in M$.

I don't quite understand why we must have $i \cdot m = 0$ instead of $I \cdot m \in M$. This is for proving the only if part.

REPLY [6 votes]: A $R$-module is nothing else but a ring morphism $R \to End(M)$. By the homomorphism theorem, this induces a ring morphism $R/I \to End(M)$ iff $IM=0$.