Find all integral domains $R$ such that $x^{4} = x$ for all $x \in R$.

Question

At the moment, I am not even studying ring theory but I was thinking of the following question.

Question: Find all integral domains $R$ such that $x^{4} = x$ for all $x \in R$.

My initial thought was to rewrite the equation as $x(x - 1)(x^{2} + x + 1) = 0$. From this, solutions $x = 0, x = 1$ and the non-trivial solutions to the equation $x^{3} - 1 = 0$ follow, namely $\omega = e^{2\pi i / 3}$ and $\omega^{2}$. Clearly, the trivial ring $R = \{0\}$ (where $0 = 1$) is one such integral domain. Also, $\mathbb{F}_{2}$ clearly satisfies the condition.

Moreover, I have considered the integral domains $\mathbb{F}_{2}[\omega] = \{a + b\omega \mid a, b \in \mathbb{F}_{2}\}$ and $\mathbb{F}_{2}[\omega^{2}] = \{a + b\omega^{2} \mid a, b \in \mathbb{F}_{2}\}$. These rings also satisfy the condition, since we have for example that $(a + b\omega)^{4} = ((a + b\omega)^{2})^{2} = (a^{2} + b^{2}\omega^{2})^{2} = a^{4} + b^{4}\omega^{4} = a + b\omega$ since $\omega^{3} = 1$.

From this point, I am wondering if I have found all integral domains that satisfy the mentioned condition. I would not be sure on how to proceed in showing that there are any others. If anybody could help that would be greatly appreciated.

Answer 1

Let $R$ be an integral domain such that every element of $R$ is a root of $X^4-X\in R[X]$.

1. As shown in this question the number of roots of a polynomial over an integral domain is at most its degree, and so $|R|\leq4$.
2. As shown in this question every finite integral domain is a field.
3. As shown in this question the only fields of such small order are $\Bbb{F}_2$, $\Bbb{F}_3$ and $\Bbb{F}_4$, up to isomorphism.

A quick check shows that only $\Bbb{F}_2$ and $\Bbb{F}_4$ work.

Answer 2

For the bigger picture you may consult this paper, section 2. Let us call a ring an $n$-ring if $x^n = x$ for all $x$. Then clearly every $n$-domain is an $n$-field (just cancel $x \neq 0$ to get $x^{n-1}=1$), and these are exactly the $\mathbb{F}_q$ with $q-1 \mid n-1$. Table 2 on page 9 in the linked paper mentions some examples. For $n = 4$ the only prime powers are $q = 2,4$. When for example $n = 2023$, we get $q=2,4,3,7$.