If $A$ is an $R$-module, I am having difficulty proving that $A$ is also a well-defined $R/ann(A)$-module with $(r+ann(A))a=ra$.
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If you assume that the operation from $R/Ann(A) \times A$ to $ A$ is well defined, then A will be an $R/Ann(A) $ module (it follows from the fact that $A$ is an $R $ module, do you see why?).
Now, you need to prove that the operation is well defined. Take two elements $r_1 , r_2 \in R$ that lie in the same equivalence class in $R/Ann(A)$. Then we must have $r_1 = r_2 + x$, where $x \in Ann(A) $.
Thus, if $a \in A$, $r_1 a =( r_2 + x)a = r_2 a$, so the operation is indeed well defined.
Mauro
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More generally, a $R$-module $A$ is also an $R/I$-module in the natural way defined above, iff $IA=0.$ Certainly $\operatorname{Ann}(A)A=0$, hence we can let $I=\operatorname{Ann}(A)$.
MooS
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Do you know any examples of an $R$-module $A$ and an ideal $I$ such that $IA=0$, but $I\neq\text{Ann}(A)$? – Alex Mathers Apr 28 '16 at 02:56
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$A=\mathbb Z/2\mathbb Z$, $I=4\mathbb Z$. – MooS Apr 28 '16 at 13:24
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$\newcommand{\Ann}{\operatorname{Ann}}$ The point here is that you need to see you can take two representatives of the same class, $r+\Ann(A)=s+\Ann(A)$ and see that their product by $a$ is the same. The important fact to note here is that for an ideal $I$, $r+I=s+I$ if and only if $r-s\in I$.
So in our case, $r+\Ann(A)=s+\Ann(A)$ means $r-s\in\Ann(A)$, so $(r-s)a=0$, i.e. $ra=sa$; therefore
$$(r+\Ann(A))a=ra=sa=(s+\Ann(A))a$$
and the operation is well-defined.
Note another way to see this is to see that for an ideal $I$
$$r+I=s+I\iff r-s\in I\iff r=s+x\text{ for some $x\in I$}$$
The third of these equivalences is what Mauro used in his answer, it works equally well. Whichever makes most sense to you is the one you should use. Also, if you didn't know this fact then you should try to prove it for yourself (it's not too bad, I promise!)
Alex Mathers
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