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

Solution 1:

No, your answer to part (a) is not correct.

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.

Solution 2:

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.

Solution 3:

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 $M$.