Let \(R\) be a commutative ring, let \(A\) be an \(R\)-module, and let \(p\in R\). Define \[ A[p]=\{a\in A:pa=0\}. \] Show that \(A[p]\) is naturally an \(R/(p)\)-module.
Solution
Define \[ (r+(p))\cdot a:=ra, \qquad r\in R,\ a\in A[p]. \]
1. The action is well defined with respect to the residue class of \(r\).
If \(r-r'\in(p)\), write \(r-r'=sp\). Then for \(a\in A[p]\), \[ ra-r'a=(r-r')a=spa=s(pa)=0. \] Hence \(ra=r'a\).
2. The action preserves \(A[p]\).
If \(a\in A[p]\), then \[ p(ra)=r(pa)=0 \] by commutativity of \(R\). Thus \(ra\in A[p]\).
3. The module axioms descend from the \(R\)-module structure on \(A\).
For \(r,s\in R\) and \(a,b\in A[p]\), \[ (r+(p))\cdot(a+b)=ra+rb, \] \[ ((r+s)+(p))\cdot a=ra+sa, \] \[ ((rs)+(p))\cdot a=(r+(p))\cdot((s+(p))\cdot a), \] and \[ (1+(p))\cdot a=a. \]
Therefore \(A[p]\) is naturally an \(R/(p)\)-module. If \((p)\) is maximal, then \(R/(p)\) is a field and \(A[p]\) is consequently a vector space over that field.