Let \(R\) be a commutative domain, let \(A=Ra\) be a cyclic \(R\)-module, and suppose \[ \operatorname{Ann}_R(a)=(r) \] with \[ r=sk \] for nonzero \(s,k\in R\).
Prove that \[ R/(k)\cong sA \qquad\text{and}\qquad R/(s)\cong A[s], \] where \[ A[s]=\{x\in A:sx=0\}. \]
Solution
1. Define\[ \phi:R\to sA, \qquad x\mapsto sxa. \]Then \(\phi\) is a surjective \(R\)-module homomorphism.
For \(t,x,y\in R\), \[ \phi(tx)=s(tx)a=t(sxa)=t\phi(x) \] and \[ \phi(x+y)=s(x+y)a=sxa+sya=\phi(x)+\phi(y). \] Every element of \(sA\) has the form \(s(ya)=sya=\phi(y)\), so \(\phi\) is surjective.
2. One has\[ \ker\phi=(k). \]
Certainly \(k\in\ker\phi\), because \[ \phi(k)=ska=ra=0. \] Thus \((k)\subseteq\ker\phi\).
Conversely, let \(x\in\ker\phi\). Then \[ sxa=0, \] so \(sx\in\operatorname{Ann}_R(a)=(r)=(sk)\). Hence for some \(t\in R\), \[ sx=tsk=s(tk). \] Since \(R\) is a domain and \(s\ne0\), cancellation gives \[ x=tk\in(k). \] Thus \(\ker\phi=(k)\).
3. Therefore\[ R/(k)\cong sA. \]
Apply the first isomorphism theorem to the surjective map \(\phi\) and use 2.
4. Define\[ \psi:R\to A[s], \qquad x\mapsto kxa. \]Then \(\psi\) is a well-defined surjective \(R\)-module homomorphism.
First, \[ s\psi(x)=skxa=rxa=x(ra)=0, \] so \(\psi(x)\in A[s]\). Linearity is immediate from the module laws.
For surjectivity, let \(y\in A[s]\). Since \(A=Ra\), write \(y=ta\). The condition \(sy=0\) says \[ sta=0, \] so \(st\in\operatorname{Ann}_R(a)=(r)=(sk)\). Thus \[ st=usk \] for some \(u\in R\). Since \(R\) is a domain and \(s\ne0\), \[ t=uk. \] Therefore \[ y=ta=uka=\psi(u). \]
5. One has\[ \ker\psi=(s). \]
Since \[ \psi(s)=ksa=r a=0, \] we get \((s)\subseteq\ker\psi\).
Conversely, if \(x\in\ker\psi\), then \[ kxa=0, \] so \(kx\in(r)=(sk)\). Hence \[ kx=usk=k(us) \] for some \(u\in R\). Since \(R\) is a domain and \(k\ne0\), cancellation gives \[ x=us\in(s). \] Thus \(\ker\psi=(s)\).
6. Therefore\[ R/(s)\cong A[s]. \]
Apply the first isomorphism theorem to \(\psi\) and use 4 and 5.