# Classification of rings satisfying $a^4=a$

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 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 $H^1(X \setminus E,\mathbb{Z}/2)$.

• Just a quick note on the commutativity proof: there is a remarkable theorem of Jacobson that if for all $r \in R$ there is $n(r) > 1$ with $r^{n(r)} = r$, then $R$ is commutative, which may serve as a better reference if you are going to consider larger $n = 7, 9, ...$ – dvitek Sep 27 '14 at 14:19
• Yes, I've studied this result and its proof in detail. Probably the shortest and quickest proof ($\approx$ one page) is given in "Elementary proofs of a theorem of Wedderburn and a theorem of Jacobson" by Nagahara and Tominga. – Martin Brandenburg Sep 27 '14 at 14:41
• I am very stupid, or at least forgetful. In the paper "Topological representation of algebras" (1947), which I already read several years ago, Arens and Kaplansky give exactly this classification of $4$-rings in terms of $C_2$-actions. And the proof is very simple, one just extends scalars to $\mathbb{F}_4$, where classification is simple, and then one does (what is nowadays called) descent. – Martin Brandenburg Oct 14 '14 at 14:01

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.

• Hm, $\mathcal{X}$ is equivalent to the category of boolean rings with an action of $C_2$. Is there a direct algebraic way to see that this is $\mathcal{R}$? – Martin Brandenburg Sep 27 '14 at 18:39
• Unfortunately, $G(R)$ is not a Hausdorff space and not totally disconnected. For example, $G(\mathbb{F}_4)$ has two points and carries the indiscrete topology. Nevertheless, we can ask if $R \to \mathrm{Map}_{C_2}(\hom(R,\mathbb{F}_4),\mathbb{F}_4)$ is an isomorphism. It is injective, but I have no idea how to prove surjectivity. – Martin Brandenburg Oct 9 '14 at 10:11
• @MartinBrandenburg: It is a Stone space if you topologize it based on $\mathbb{F}_4$ being discrete, rather than only $\{0\}$ being closed. – Eric Wofsey Oct 9 '14 at 10:17
• @EricWofsey: Yes, because then it is a closed subspace of $\prod_{r \in R} \mathbb{F}_4$. Notice that this is not a Zariski-like topology. – Martin Brandenburg Oct 9 '14 at 10:20
• I think I have found a proof that $R \to F(G(R))$ is an isomorphism, but it's quite long. Should I make this a CW answer? – Martin Brandenburg Oct 13 '14 at 11:22

Here is a proof for one part of Neil Strickland's answer.

If $R$ is a $4$-ring, then $R \to \hom_{C_2}(\hom_{\mathsf{Ring}}(R,\mathbb{F}_4),\mathbb{F}_4)$, $r \mapsto (f \mapsto f(r))$ is an isomorphism.

Proof. If $r \in R$ lies in the kernel, this means that $f(r)=0$ for all $f \in \hom(R,\mathbb{F}_4)$. Hence, $f$ lies in every prime ideal, and therefore in the nilradical, which is zero. This proves that injectivity..

For surjectivity, let $X:=\hom(R,\mathbb{F}_4)$ and let $\alpha : X \to \mathbb{F}_4$ be a $C_2$-equivariant continuous map. Write $\mathbb{F}_4=\{0,1,u,u^2\}$. We may write $\alpha$ as $\alpha = \chi_{Y} + u \chi_{Z} + u^2 \chi_{\sigma Z},$ where $Y = \alpha^{-1}(\{1\})$ and $Z = \alpha^{-1}(\{u\})$ and hence $\sigma Z = \alpha^{-1}(\{u^2\})$. Notice that $Y,Z,\sigma Z$ are disjoint clopen subsets of $X$ and that $\sigma Y = Y$.

In the decomposition of $\alpha$, we have to be a bit careful since $u \chi_Z$ does not lie in $\hom_{C_2}(X,\mathbb{F}_4)$, but $\chi_Y$ and $u \chi_{Z} + u^2 \chi_{\sigma Z}$ do. We show that both functions are in the image.

First, let us treat $\chi_Y$. It suffices to find some idempotent element $r \in R$ such that $Y = \{f \in X : f(r)=1\}$, because then $(f \mapsto f(r))$ equals $\chi_Y$. Consider the map $X \to \mathrm{Spec}(R)$, $f \mapsto \ker(f)$. It is easily seen to be continuous. Since $f$ and $\sigma f$ have the same kernel, we get a continuous map $X/C_2 \to \mathrm{Spec}(R)$, which is clearly bijective. Since $X/C_2$ is compact and $\mathrm{Spec}(R)$ is Hausdorff, the map is a homeomorphism. It follows that there is a 1:1 correspondence between clopen subsets of $X/C_2$ and the clopen subsets of $\mathrm{Spec}(R)$, which in turn correspond to idempotent elements. Since $\sigma Y = Y$ and $Y$ is clopen, the image of $Y$ in $X/C_2$ is clopen, and we are done.

Now let us treat $u \chi_{Z} + u^2 \chi_{\sigma Z}$. Since $Z$ is open, we may write $Z$ as a union of non-empty sets of the form $\{f \in X : f(r_1)=i_1,\dotsc,f(r_n)=i_n\}$ for some $r_k \in R$ and $i_k \in \mathbb{F}_4$. Since $Z \cap \sigma Z = \emptyset$, not all $i_k$ can be contained in $\mathbb{F}_2$. Since $f(r)=u^2$ is equivalent to $f(r^2)=u$, we may therefore assume that $i_1=u$. If $i_k=0$, rewrite the relation $f(r_k)=0$ as $f(r_1-r_k)=u$. If $i_k=1$, replace $r_k$ by $r_k-1$ and reduce to $i_k=0$. If $i_k=u^2$, replace $r_k$ by $r_k^2$ and reduce to $i_k=u$. Hence, each set may be written as $\{f \in X : f(r_1)=\dotsc=f(r_k)=u\}.$ We claim that this already equals $\{f \in X : f(r)=u\}$ for some $r \in \langle r_1,\dotsc,r_k \rangle$. By induction, it suffices to assume $k=2$. Then, it suffices to find a polynomial $p \in \mathbb{F}_2[x,y]$ such that for $a,b \in \mathbb{F}_4$ we have $p(a,b)=u$ if and only if $a=b=u$. A possible choice is $p := x (1 - (x-y)^3).$ In fact, if $a \neq b$ in $\mathbb{F}_4$, then $p(a,b)=0$, and otherwise $p(a,b)=a$. This implies the desired property.

We have proven that $Z$ is a union of sets of the form $\{f \in X : f(r)=u\}$ for various $r \in R$. Since $Z$ is assumed to be closed and therefore compact, finitely many $r$ suffice. We claim that there is some $r \in R$ such that $Z = \{f \in X : f(r)=u\}$. By induction, we may assume that $Z = \{f \in X : f(r_1)=u\} \cup \{f \in X : f(r_2)=u\}$. Since $Z \cap \sigma Z = \emptyset$, we see that $f(r_1)=u \Rightarrow f(r_2)^2 \neq u$ for all $f \in X$, and likewise $f(r_2)=u \Rightarrow f(r_1)^2 \neq u$. Therefore, similar to the technique before, it suffices to find a polynomial $p \in \mathbb{F}_2[x,y]$ such that for $a,b \in \mathbb{F}_4$ with $(a,b) \notin \{(u,u^2),(u^2,u)\}$ we have $p(a,b)=u$ if and only if ($a=u$ or $b=u$). We define $p_1:=x^2+x$, $p_2:=(x^2+x+1)(y^2+y)$ and finally $p := x p_1 + y p_2$. As polynomial functions on $\mathbb{F}_4 \times \mathbb{F}_4$, we have $p_1 = \chi_{\{u,u^2\} \times \{0,1,u,u^2\}}$, $p_2 = \chi_{\{0,1\} \times \{u,u^2\}}$. It follows that $p = x \chi_{\{u,u^2\} \times \{0,1,u,u^2\}} + y \chi_{\{0,1\} \times \{u,u^2\}}.$ Hence, $p(a,b)=u \Leftrightarrow a=u \vee b=u$ except for $p(u^2,u)=u^2$. Since we don't have to care for $(u^2,u)$, that's enough.

We have proven that $Z = \{f \in X : f(r)=u\}$ for some $r \in R$. It follows that $\sigma Z = \{f \in X : f(r)=u^2\}$. Thus, if we substract $(f \mapsto f(r))$ from $u \chi_{Z} + u^2 \chi_{\sigma Z}$, we obtain a function which has only values in $\{0,1\}$. We already know that such a function lies in the image of the counit, hence $u \chi_{Z} + u^2 \chi_{\sigma Z}$ lies in the image, too. $\square$

Thanks to Jyrki Lahtonen for pointing out the polynomials in the proof above. In some sense, we don't really have to write these polynomials down since their existence is guaranteed by the finite case of the claim which is easy to deal with since every finite $4$-ring is a direct product of copies of $\mathbb{F}_2$ or $\mathbb{F}_4$.

A Pierce sheaf (or its corresponding ring of sections) where all the stalks are either the two- or the four-element field. The base is the maximal ideal space, and this partitions into those corresponding to maximal ideal of index two, and the rest. It's been a long time since I've thought about Pierce sheaves, but the ones that are of index two are probably closed in the space. So the H$^1$ space is likely connected to this problem.

Edit. Here is a stab at showing the rings are classifiable, and fairly elementary. Perhaps I made a mistake somewhere. [I did not have Pierce's AMS Memoir handy.] The claim is that the classification boils down to just the pair consisting of the maximal ideal space $X$, and the closed subset $Y$ corresponding to maximal ideals of index two.

Let $X$ be a boolean space, that is, a zero-dimensional (totally disconnected) compact space; examples include Cantor sets), and let $Y$ be a closed subset. Let $R = C(X,F_4)$ (continuous functions with values in the (discrete) four-element field; this consists precisely of things of the form $\chi_U + u \chi_V + v\chi_W$, where $U,V,W$ are pairwise disjoint (possibly empty) clopen subsets of $X$, $\chi_U$ etc represent their indicator functions, and $u,v$ are the elements of $F_4\setminus F_2$. Let $Y$ be a closed subset of $X$, and set $$S\equiv S(X,Y) = \{f \in R \mid f(y)\in F_2 { \rm \ for\ all }\ y \in Y \},$$ where we regard $F_2$ as a subset of $F_4$ in the usual way. Then $S$ certainly satisfies $a^4 = a$ for all $a \in S$. Moreover, the pair $(X,Y)$ is a complete invariant for $S$ constructed in this manner.

Now given an arbitrary ring $T$ satisfying $a^4 = a$ for all $a \in T$, by ancient results, this is commutative, von Neumann regular, and its maximal ideal space, $X$, is a boolean space. Modulo each $x \in X$, the quotient is either $F_4$ or $F_2$. The set $Y$, consisting of points $x\in X$ such that $a^2-a \in x$ for all $a \in T$, is closed, and we have an embedding $T \subseteq S(X,Y)$ (from the Pierce sheaf representation of $T$ as a ring of continuous sections). We want to show equality.

The map $\alpha:a \mapsto a^2$ (the Frobenius map in this context) is a ring endomorphism of $T$ (fixing all the idempotents, corresponding to clopen subsets of $X$), whose image is a Boolean algebra, and its maximal ideal space is the same, $X$. By the usual representation theorem of Boolean algebras, we obtain a subring $T_0 = C(X,F_2)$ of $T$, and $\alpha(R) = T_0$.

Select $t \in T$; if $t^2 = t$, then $t \in T_0$ already. Otherwise, the set $Z$ consisting of the points where $z(t^2 - t) = 0$ is clopen and contains $Y$. For $s \in S(X,Y)$, let $F$ be the set of points in $X$ that kill $s^2 - s$, so that $F$ is clopen and contains $E$. Then $s|F$ is idempotent, so we can write $s|F = \chi_G$ where $G$ is a clopen subset of $F$. Let $E$ be the projection of $F$; then we have $Es = \chi_G$, so $Es \in T_0$.

Now consider $(1-E)s$; this vanishes on $F$ and is not zero on $F^c$, moreover, on the complement, it can only take the two values, $u,v \in F_4 \setminus F_2$. The set of points in $F^c$ where $z(s) = u$ is a clopen set (again) in $X$, call it $H_u$, and its complement in $F^c$ is also clopen and we have $s = \chi_G + u \chi_{H_u} + v\chi_{H_v}$.

It suffices to show that $u \chi_{H_u}$ belongs to $T$. Given $z \in H_u$, there exists $t_z$ such that $z(t_z) = u$; hence there exists a clopen neighbourhood of $z$, $V_z$, such that $y(t_z) = u$ for all $y \in V_z$. The set of these $V_z$ form an open covering of $H_u$, and since the space is totally disconnected and compact, we can find a finite disjoint set of clopen neighbourhoods $V_i$ covering $H$, and corresponding $t_i = \chi_{V_i}t_i$ in $T$ with $t_i|V_i = u\chi_{V_i}$ (the constant function $u$ on $V_i$) and $\cup V_i = H_u$. Then $\sum t_i = u \chi_{u}$, so the latter belongs to $T$. Hence $S(X,Y) = T$.

The conclusion is that $T$ is just $S(X,Y)$, and so the pair $(X,Y)$ is a complete invariant.

Edit #2. There is another way to deal with this, via inverse limits. Any finitely generated subring of $T$ is actually finite, and thus semisimple (in the strong sense); hence isomorphic to $C(E, F_2) \times C(D, F_4)$, where $D$ and $E$ are finite sets. As $T$ is the direct limit of its finitely generated subrings, we obtain $X$ as an inverse limit of the finite sets $D \dot\cup E$ (they should be subscripted to indicate their dependence on the finite set of generators), and the inverse limit of the $E$s is an inverse limit of clopen sets inside $X$, hence a closed subset, and the details should be routine.

• Thanks, but that seems to me what I already know. I am interested in a more explicit classification (or even better, an equivalence of categories). – Martin Brandenburg Sep 27 '14 at 1:10
• In his book "Modules Over Commutative Regular Rings", Pierce studies as an example rings satisfying $a^n=a$ and their corresponding structure sheaves. But does the book also offer a classification? – Martin Brandenburg Sep 27 '14 at 19:13
• I've just seen your edit. Thank you, I will look at it! – Martin Brandenburg Sep 28 '14 at 20:34
• I think you mean that $T_0$ is the fixed ring of $\alpha$, not $\alpha(R)$ (the latter is $R$ since $\alpha$ is an involution). – Laurent Moret-Bailly Sep 29 '14 at 18:34
• Yes, I should have said fixed ring, not the image; the important thing is that all the idempotents of $R$ belong to $T$, which is obvious anyway. – David Handelman Sep 29 '14 at 23:44