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    If $A=Ra$ with $ra=0$ and $(r,s)=(1)$, then $A=sA$ and $A[s]=0$

    P-LHIOI

    Area
    Algebra
    Topics
    Cyclic Groups, Modules, Torsion
    Status
    draft

    Let \(R\) be a commutative ring, let \(A=Ra\) be a cyclic \(R\)-module, and suppose \[ ra=0, \qquad (r,s)=R. \] Show that \[ A=sA \qquad\text{and}\qquad A[s]=\{x\in A:sx=0\}=0. \]

    Solution

    Choose \(u,v\in R\) with \[ ur+vs=1. \] Then \[ a=(ur+vs)a=u(ra)+v(sa)=vsa=s(va), \] so \(a\in sA\). Since \(a\) generates \(A\), this gives \[ A\subseteq sA. \] The reverse inclusion is automatic, hence \[ A=sA. \]

    Now let \(x\in A[s]\). Since \(A=Ra\), write \(x=ta\). Then \[ rx=rta=t(ra)=0, \] and by definition \(sx=0\). Therefore \[ x=(ur+vs)x=u(rx)+v(sx)=0. \] Hence \[ A[s]=0. \]

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