Why $(r+(p))(a+pA) = ra+pA$ holds?

42 Views Asked by At

Let $R$ be a principal ideal domain, $A$ a left $R$-module, and $p\in R$ a prime. Let $pA=\{pa\mid a\in A\}$ and $A[p] = \{a\in A\mid pa=0\}$. I am trying to prove $A/pA$ is a vector space over $R/(p)$, and $A[p]$ is a vector space over $R/(p)$. In order to finish my proof I need to show following: for $r\in R$

$(r+(p))(a+pA) = ra+pA$ where $a\in A$.

$(r+(p))a=ra$ where $a\in A[p].$

For the first part $(r+(p))(a+pA)=ra+rpA+(p)a+(p)pA=ra+prA+(p)(a+pA)=ra+pA+(p)(a+pA).$

How can we conclude $(p)(a+pA)=0?$ And also for the second part why $(p)a=0?$