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    An irreducible generates any principal ideal containing it in a PID

    E-FODKS

    Area
    Algebra
    Topics
    Factorization, Ideals, Principal Ideal Domains
    Status
    draft

    Let \(R\) be a PID and let \(I=(b)\subsetneq R\). If \(a\in I\) is irreducible, prove that \[ (a)=(b)=I. \]

    Solution

    Since \(a\in(b)\), there is \(r\in R\) such that \[ a=rb. \] Because \(a\) is irreducible, either \(r\) or \(b\) is a unit. But \((b)\) is proper, so \(b\) is not a unit. Hence \(r\) is a unit, and therefore \(a\) and \(b\) are associates. Thus \[ (a)=(b)=I. \]

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