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    Schur's lemma: endomorphisms of a simple module are zero or isomorphisms

    P-XNN6S

    Area
    Algebra
    Topics
    Homomorphisms, Modules, Semisimplicity
    Status
    draft

    Let \(A\) be a simple module and let \[ \phi:A\to A \] be a module endomorphism. Prove that either \(\phi=0\) or \(\phi\) is an isomorphism.

    Solution

    Both \[ \ker\phi \quad\text{and}\quad \operatorname{im}\phi \] are submodules of \(A\).

    Since \(A\) is simple, its only submodules are \(0\) and \(A\).

    If \(\operatorname{im}\phi=0\), then \(\phi=0\).

    Otherwise \[ \operatorname{im}\phi=A, \] so \(\phi\) is surjective. In this case \(\phi\ne0\), hence \[ \ker\phi\ne A. \] Simplicity therefore forces \[ \ker\phi=0. \] Thus \(\phi\) is injective and surjective, hence an isomorphism.

    This is Schur’s lemma in module form.

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