1. ## Action over quotient

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

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

thanks!

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

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

thanks!
$\displaystyle J$ 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 $\displaystyle s_1 + J=s_2 + J.$ then $\displaystyle s_1 -s_2 \in J.$ thus by the definition

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