# If $M$ is an $R$-module and $I\subseteq\mathrm{Ann}(M)$ an ideal, then $M$ has a structure of $R/I$-module

Let $M$ be an $R$-module and let $I$ be an ideal of $R$ such that $I$ is a subset of $\mathrm{Ann}(M)$.
Define a product of an element of $R/I$ by an element of $M$ as follows:
$$(r+I)\cdot m=rm.$$
(a) Show that this product is well defined.
(b) Show that this product, together with the sum of elements of $M$, turns $M$ into an $R/I$-module.

Let $r+I,s+I \in R/I$ such that $r+I=s+I$. Then $(r+I)m=(s+I)m$, for all $m \in M$. Then $rm=sm \implies r=s$ (since the inverse of $m$ is in $M$), so the product is well defined.

Is my answer to (a) correct?

How can I use Ann(M) to answer (b)?

#### Solutions Collecting From Web of "If $M$ is an $R$-module and $I\subseteq\mathrm{Ann}(M)$ an ideal, then $M$ has a structure of $R/I$-module"

You need to show that, for any $r,s\in R$ such that $r+I=s+I$, and any $m\in M$, we have
$$(r+I)\cdot m=rm=sm=(s+I)\cdot m.$$
You cannot conclude from this that $r=s$; effectively, that’s the whole point of this construction. You should instead note that
$$r+I=s+I\iff r-s\in I$$
and therefore, if $r+I=s+I$, then $r-s\in\mathrm{Ann}(M)$. What does that tell you about $(r-s)m$ for any $m\in M$? Now note that
$$rm+(r-s)m=sm.$$

Also, the “inverse” of an element of a module makes no sense; a module only has scalar multiplication with element of a ring and itself.

Hint: For a) you are going about it wrong. You want to show that if $s+I=r+I$ then $(s+I)m=(r+I)m$! Instead you want to show that if $r-m\in I$ then $rm=sm$ (try wrtiting this equality as $rm-sm$ and doing some factoring).

For b) this should follow immediately from the fact that $M$ is an $R$-module and the $R/I$-module structure is really the $R$-module structure in disguise.

It still seems to me that the previous answers are not complete.

To show that the multiplication $R/I\times M\to M$ is a well-defined map, you need to show that for all $m,n\in M$, $r,s\in R$, if $m=n$ and $r+I=s+I$, then $(r+I)m=(s+I)n$.

Notice the new $n$, here. However, this is not any harder to prove than the previous answers have explained. You can show that $(r+I)m=(s+I)m$ and then use the fact that $sm=sn$ in $M$.