Finite rings $R$ in which $x^{25}=x$ holds

Question

I want to classify finite rings $R$ in which $x^{25}=x$ for all $x\in R$.

I know Jacobson's Theorem that if $x^n=x$ for all $x\in$ then $R$ is commutative. I don't know how to show the Theorem for the special case $n=25$ by elementary methods.

Furthermore, after showing that $R$ is commutative, is there any Theorem similar to Wedderburn's Theorem to conclude that $R$ is a field?

Answer 1

Hint: $\forall x,y \in R$, we have $(xy)^{25} = xy$. But $x=x^{25}$ and $y = y^{25}$. Then $(xy)^{25} = x^{25}y ^{25}$. It is enough to show that $xy = yx$.

Answer 2

Your ring is

1. commutative, by a famous theorem of Jacobson, as mentioned in the comments
2. reduced, (clearly)
3. Finite by assumption, hence it is a semisimple, and therefore a product of finitely many fields (Wedderburn), and
4. All the fields satisfy $x^{24}=1$ on their nonzero elements, which limits the cardinality of each field appearing to at most $25$.

If you run through the cardinalities of finite fields between $2$ and $25$, you'll find that $A=\{2, 3, 4, 5, 7, 9, 13, 25\}$ all will work, while $8, 11, 16, 17, 19$ and $23$ will not. If $q$ is the size of the field, then the nonzero elements are a cyclc group of order $q-1$, and that cyclic group must satisfy $x^{24}=1$, which means $q-1|24$.

So in summary, this proves that any finite product of finite fields with orders in $A$ will satisfy $x^{25}=x$, and conversely every such ring is such a product.

So you see, it is only a field in precisely one of these cases...