Math Help - Action over quotient

1. Action over quotient

Let $A$ a conmutative ring, $M$ an $A$-module and $J= \{ a \in A: am=0 \forall m \in M \}$

Prove that $J$ is a subring of $A$ and that $M$ is an $R/J$-module with action $(s+J)m=sm$.

thanks!

2. Originally Posted by roporte
Let $A$ a conmutative ring, $M$ an $A$-module and $J= \{ a \in A: am=0 \forall m \in M \}$

Prove that $J$ is a subring of $A$ and that $M$ is an $R/J$-module with action $(s+J)m=sm$.

thanks!
$J$ is actually more than just a subring ... it's an ideal of A! prove it! (J is called the annihilator of M in A) to show that M has a structure of an R/J module, we only need to show that the defined

action is basically well-defined, because other properties of modules will be directly transfered from R-module structure of M. so suppose $s_1 + J=s_2 + J.$ then $s_1 -s_2 \in J.$ thus by the definition

of J, we have: $(s_1-s_2)m=0, \ \forall m \in M.$ hence: $(s_1+J)m=s_1m=s_2m=(s_2 + J)m.$ Q.E.D.