# Thread: [SOLVED] finitely many idempotents

1. ## [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 .

2. Originally Posted by xixi
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.

3. Originally Posted by NonCommAlg
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!

4. Originally Posted by xixi
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.