We have the famous classification of rings satisfying $a^2=a$ (for each element $a$) in terms of Stone spaces, via $X \mapsto C(X,\mathbb{F}\_2)$. Similarly, rings satisfying $a^3=a$ are classified by pairs of Stone spaces via $(X,Y) \mapsto C(X,\mathbb{F}\_2) \times C(Y,\mathbb{F}\_3)$. (This kind of classification works for all rings with $a^n=a$ where $n$ is such that every prime power $q$ with $q-1|n-1$ is a prime number. Do these $n$ have a concise description or a name?)


**Question.** How to classify the rings satisfying $a^4=a$ for each element $a$?


They [are](http://ysharifi.wordpress.com/2012/04/19/rings-satisfying-x4-x-are-commutative/) commutative and embed into some product of copies of $\mathbb{F}\_2$ or $\mathbb{F}\_4$. A typical example would be $\{f \in C(X,\mathbb{F}\_4) : f(E) \subseteq \mathbb{F}\_2\}$ for some Stone space $X$ and a closed subset $E$. But because of $\mathrm{Aut}(\mathbb{F}\_4)=\mathbb{Z}/2$ there cannot be a purely topological classification.


I suspect that the sheaf cohomology group $H^1(X \setminus E,\mathbb{Z}/2)$ might enter here. In fact, if $A$ is a ring satisfying $a^4=a$, then $E:=\{\mathfrak{p} : A/\mathfrak{p} \cong \mathbb{F}\_2\}=V(a^2-a : a \in A)$ is a closed subset of the Stone space $X:=\mathrm{Spec}(A)$ and for every compact open subset $V$ of $X \setminus E$ we have an isomorphism of sheaves $\mathcal{O}\_X |\_{V} \cong \underline{\mathbb{F}\_4}|\_V$, because the sheaf of isomorphisms between them is a $\mathbb{Z}/2$-torsor and $H^1(V,\mathbb{Z}/2)=0$. But the potential failure of $\mathcal{O}\_X |\_{X \setminus E} \cong \underline{\mathbb{F}\_4}|\_{X \setminus E}$ should be encoded in $H^1(X \setminus E,\mathbb{Z}/2)$.



Classification of rings satisfying $a^4=a$
Let $\sigma$ denote the nontrivial automorphism of $\mathbb{F}\_4$ and put $C=\{1,\sigma\}$. Let $\mathcal{R}$ denote the category of rings in which $a^4=a$, and let $\mathcal{X}$ denote the category of Stone spaces with an action of $C$. There is a functor $F\colon\mathcal{X}\to\mathcal{R}$ given by $F(X)=\text{Map}\_C(X,\mathbb{F}\_4)$, and there is a functor $G\colon\mathcal{R}\to\mathcal{X}$ given by $G(R)=\text{Rings}(R,\mathbb{F}\_4)$ (with the obvious Zariski-type topology). There are evident natural maps $R\to FG(R)$ and $X\to GF(X)$. I have not checked that these are isomorphisms, but I would be surprised if they were not.


