Let \(R\) be a PID and let \(A\) be a cyclic \(R\)-module of nonzero order \(r\in R\).
Show that if \(s\) is relatively prime to \(r\), then \(sA = A\) and \(A[s] = 0\).
If \(s\) divides \(r\), so \(sk = r\), then \(sA \cong R/(k)\) and \(A[s] \cong R/(s)\).
Solution
Choose a generator \(a\in A\) with annihilator \[ \operatorname{Ann}_R(a)=(r). \] Thus \(A\cong R/(r)\) and \(ra=0\).
1. If \((s,r)=1\), then \(sA=A\).
By Bézout’s identity there are \(u,v\in R\) such that \[ us+vr=1. \] For any \(x\in A\), one has \(rx=0\) because \(A=Ra\) and \(ra=0\). Hence \[ x=(us+vr)x=s(ux)+v(rx)=s(ux)\in sA. \] Thus \(A\subseteq sA\), while \(sA\subseteq A\) is immediate. Therefore \(sA=A\).
2. If \((s,r)=1\), then \(A[s]=0\).
Let \(x\in A[s]\), so \(sx=0\). With \(u,v\) as in 1, \[ x=(us+vr)x=u(sx)+v(rx)=0. \] Thus the only element annihilated by \(s\) is \(0\).
Now suppose \(s\mid r\), say \(r=sk\).
3. There is an isomorphism\[ sA\cong R/(k). \]
Define \[ \phi:R\longrightarrow sA, \qquad t\longmapsto tsa. \] Every element of \(sA\) has the form \(s(xa)=xsa\), so \(\phi\) is surjective. Moreover, \[ t\in\ker\phi \iff tsa=0 \iff ts\in(r)=(sk). \] Thus \(ts=usk\) for some \(u\in R\). Since \(R\) is a domain and \(s\ne0\), cancellation gives \(t=uk\), so \(t\in(k)\). Conversely, if \(t=uk\), then \(tsa=u(sk)a=ura=0\), so \(t\in\ker\phi\). Hence \(\ker\phi=(k)\), and the first isomorphism theorem gives \[ sA\cong R/(k). \]
4. There is an isomorphism\[ A[s]\cong R/(s). \]
Define \[ \psi:R\longrightarrow A[s], \qquad t\longmapsto tka. \] The image lies in \(A[s]\) because \[ s(tka)=t(sk)a=tra=0. \]
To prove surjectivity, let \(x=ta\in A[s]\). Then \(sta=0\), so \[ st\in(r)=(sk). \] Hence \(st=usk\) for some \(u\in R\), and cancellation of the nonzero element \(s\) gives \(t=uk\). Thus \(x=uka=\psi(u)\).
Finally, \[ t\in\ker\psi \iff tka=0 \iff tk\in(sk). \] Since \(k\ne0\), cancellation gives \(t\in(s)\). Conversely \((s)\subseteq\ker\psi\). Therefore \(\ker\psi=(s)\), and the first isomorphism theorem yields \[ A[s]\cong R/(s). \]