Let \(A\) be an \(R\)-module (or abelian group), and let \(p \in R\) be a ring element (e.g. a prime integer \(p \in \mathbb{Z}\)). Define the multiplication-by-\(p\) map: \[\phi_p: A \longrightarrow A, \qquad x \longmapsto p x.\] (1) Prove that \(\phi_p\) is an \(R\)-module homomorphism when \(p \in Z(R)\). (2) Define the image submodule \(p A \coloneqq \operatorname{im}(\phi_p)\) and the kernel submodule \(A[p] \coloneqq \ker(\phi_p)\) (\(p\)-torsion submodule). (3) Compute \(p A\), \(A/pA\), and \(A[p]\) for \(A = \mathbb{Z}\), \(A = \mathbb{Z}/n\mathbb{Z}\), and \(A = \mathbb{Q}/\mathbb{Z}\).
Solution
Assume \(p\in Z(R)\). Then for \(r\in R\) and \(x\in A\), \[ \phi_p(rx)=p(rx)=(pr)x=(rp)x=r(px)=r\phi_p(x), \] so \(\phi_p:x\mapsto px\) is \(R\)-linear. Its image and kernel are \[ pA=\{px:x\in A\},\qquad A[p]=\{x\in A:px=0\}. \] Hence \[ 0\longrightarrow A[p]\longrightarrow A\xrightarrow{\,p\,}A \longrightarrow A/pA\longrightarrow0 \] is exact, and the first isomorphism theorem gives \(A/A[p]\cong pA\).
For abelian groups and prime integer \(p\):
If \(A=\mathbb Z\), then \[ pA=p\mathbb Z,\qquad A/pA\cong\mathbb Z/p\mathbb Z,\qquad A[p]=0. \]
If \(A=\mathbb Z/n\mathbb Z\) and \(d=\gcd(p,n)\), then \[ |pA|=n/d,\qquad A/pA\cong\mathbb Z/d\mathbb Z, \qquad A[p]=\langle n/d\rangle\cong\mathbb Z/d\mathbb Z. \] Since \(p\) is prime, \(d\in\{1,p\}\).
If \(A=\mathbb Q/\mathbb Z\), multiplication by \(p\) is surjective, so \[ pA=A,\qquad A/pA=0, \] while \[ A[p]=\left\{\frac{k}{p}+\mathbb Z:0\le k<p\right\}\cong\mathbb Z/p\mathbb Z. \]