# An example of infinite ring such that $x^4=x$ for all $x$

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.

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.

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.

• 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)? Commented Aug 7, 2022 at 19:46
• @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. Commented Aug 7, 2022 at 19:52
• 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. Commented Aug 7, 2022 at 19:56