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I am trying to find an example of a ring $R$ in which $x^4=x$ with an additional condition that there exist a non trivial element which is not an idempotent. (A stronger question would be to find an example of such rings where no nontrivial idempotents exist)

The example of a finite ring is the field of order 4.

By a theorem of Jacobson it is known that they are commutative. It can be easily shown that they have characteristic 2 and $x^2+x$ is an idempotent for all $x$ etc.

So I am expecting an example of such rings but not able to find one or prove that they are all finite.

Note : Here ring has the multiplicative identity.

Here one can see about the Theorem of Jacobson.

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For the weaker question, just take a product of the field of order 4 with infinitely many copies of $\mathbb{Z}/(2)$. A non-idempotent element of $\mathbb{F}_4$ with any sequence of zeroes and ones behind it remains a non-idempotent and the result is an infinite ring.

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For the stronger one I was able to prove that no such infinite ring exist. That is If $R$ is a ring in which $x^4= x$ for all $x$ and $x^2 \ne x$ for all $x \ne 0,1$ then $R$ is a finite field .

It can be seen that $-x=x^4=x$ so $R$ has char 2. Also $(x^2+x)^2 = x^2+x$ for all $x$.

This question shows that $R$ is commutative.

Since there are no nontrivial idempotents $x^2+x$ is either $0$ or $1$. For $x\ne 0,1$ if $x^2+x=0$ then $x$ is an idempotent which is not possible. Thus $x^2+x=1$ for all $x$ different from 0 and 1. Thus $x(x+1)=1$ and it follows that all non zero element have inverses. So $R$ is a field.

Since the polynomial $x^4-x$ can have utmost $4$ roots we conclude that $R$ is a field of order utmost $4$. The only possibilities are field of order 2 and 4.

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    $\begingroup$ How do you know that $R$ is commutative? Is the "theorem of Jacobson" mentioned above (it's not clear to me what that theorem states, exactly)? $\endgroup$
    – Stephen
    Aug 7, 2022 at 19:46
  • $\begingroup$ @Stephen I have added a an appropriate link to a question which shows that $R$ is commutative. Here you can see about Jacobson's theorem which I mentioned. $\endgroup$ Aug 7, 2022 at 19:52
  • $\begingroup$ It's a classic exercise in ring theory to show that the ring $R$ is commutative particularly because of it's appearance in Topics in Algebra by Herstein. $\endgroup$ Aug 7, 2022 at 19:56

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