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Math Help - [SOLVED] finitely many idempotents

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    [SOLVED] finitely many idempotents

    If R is a reduced ring and has only finitely many idempotents then prove that R is a direct product of a finite number of indecomposable zero-dimensional reduced rings i.e. fields .
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    Quote Originally Posted by xixi View Post
    If R is a reduced ring and has only finitely many idempotents then prove that R is a direct product of a finite number of indecomposable zero-dimensional reduced rings i.e. fields .
    first of all we need our ring to be commutative. second of all, the claim is basically false and R=\mathbb{Z} is a counter-example. certainly there's something missing in your question.
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    Quote Originally Posted by NonCommAlg View Post
    first of all we need our ring to be commutative. second of all, the claim is basically false and R=\mathbb{Z} is a counter-example. certainly there's something missing in your question.
    Yes , R is commutative & I have missed this assumption that every principal ideal of R is an idempotent ideal , sorry!
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    Quote Originally Posted by xixi View Post
    Yes , R is commutative & I have missed this assumption that every principal ideal of R is an idempotent ideal , sorry!
    but then we wouldn't need the condition "R is reduced" anymore because it would be a result of the conditions "R is commutative" and "every principal ideal of R is an idempotent". here's why:

    suppose x^2=0. then, since Rx=Rx^2=\{0\}, we'll get x=0, i.e. R is reduced. anyway, i'll assume the problem is this:

    Claim: let R be a commutative ring with 1. if the number of idempotents of R is finite and every principal ideal of R is an idempotent ideal, then R is a direct product of finitely many fields.

    Proof of the claim:

    1) J(R)=\{0\}, where J(R) is the Jacobson radical of R.

    Proof: let x \in J(R). since Rx=R x^2, we have x=yx^2, for some y \in R. thus x(1-yx)=0 and hence x=0 because 1-yx is a unit of R.

    2) every principal ideal of R is generated by an idempotent.

    Proof: let I=Rx. since Rx=Rx^2, we have x=yx^2 for some y \in R. then xy=e is an idempotent and it's easy to see that I=Re.

    3) R is artinian.

    Proof: since the number of idempotents of R is finite, the number of principal ideals of R must also be finite, by 2). so the number of ideals of R is finite and thus R is artinian.

    4) R is a direct product of finitely many fields.

    Proof: by 1) and 3), R is a commutative semisimple ring and therefore, by Wedderburn-Artin theorem, R is a direct product of finitely many fields.
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