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Let $K_1,\dotsc,K_n$ be finite fields and let $V$ be the variety of rings, generated by the $K_i$ (rings aren't necessarily unital). I want to figure out what $V$ looks like. By a theorem of Tarski, elements of $V$ are the quotients of subrings of direct products of (possibly infinite) families of the $K_i$. But what are these rings exactly?

One thing we can figure out are the possible characteristics of the elements of $V$. Since taking subrings and quotients decreases the characteristic and the characteristic of the product is the least common multiple of the characteristics of the factors, any ring in $V$ has a characteristic which is a squarefree integer, whose prime factors are among the characteristics of the $K_i$. On the other hand, not every such ring lives in $V$. For example, the multiplicative semigroup of any ring in $V$ must have finite exponent (since this is true for the products of the $K_i$), which means that things like polynomial rings over $K_i$ can't appear in $V$.

I tried looking at the simplest case where $V$ is generated by $\mathbb{Z}_2$, but I can't really picture what's going on. I have a feeling that in this case $V$ will be the class of Boolean rings, but I'm not even sure how to show this.

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$\mathbb{F}_2$ does indeed generate the Boolean rings. You should be able to piece together a proof from the material in… . – Qiaochu Yuan May 22 '12 at 18:20
The theorem you quote at the beginning is Birkhoff's HSP theorem; I don't know why you are attributing it to Tarski. – Arturo Magidin May 22 '12 at 18:51
@Magidin: I have always believed that Birkhoff proved that a class of algebras is a variety iff it is closed under $H$, $S$ and $P$. And, Tarski proved that the variety generated by a class $K$ of algebras is $HSP(K)$. Anybody can confirm this? – boumol May 22 '12 at 19:18
@boumol: Well, Burris and Sankappanavar seem to agree (Theorem 9.5, page 67, of their book. Though it seems to me that it is immediate that HSP(K) is a variety: $HH=H$, $PP=P$, and $SS=S$, a subalgebra of a homomorphic image is a homomorphic image of a subalgebra, so $SH\subseteq HS$, the product of homomorphic images is a homomorphic image of a product $PH\subseteq HP$, and a product of subalgebras is a subalgebra of products, so $PS\subseteq SP$. From this, it follows readily that $HSP(K)$ is closed under $H$, $S$, and $P$. – Arturo Magidin May 24 '12 at 3:46
@ArturoMagidin Indeed, I was following Burris & Sankappanavar's attribution to Tarski. I agree that the result isn't very complicated. – Miha Habič May 27 '12 at 19:14
up vote 4 down vote accepted

The following result should prove helpful, excerpted from Stanley Burris and John Lawrence, Term rewrite rules for finite fields (1991).

enter image description here enter image description here

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This was very useful. Thanks! – Miha Habič May 27 '12 at 19:13

Let $p$ be a prime. I claim that $V = \langle \mathbb{F}_p \rangle$ consists of those rings satisfying the identity $x^p =x$ for all elements $x$. For $p=2$ we recover boolean rings. [I have to admit that I only consider unital rings]

It is clear that every ring in $\langle \mathbb{F}_p \rangle$ satisfies the identity $x^p=x$. Now assume that $R$ is such a ring. By a Theorem of Herstein, $R$ is commutative. Clearly $x^2=0 \Rightarrow x=0$. This implies more generally, that $x^n=0 \Rightarrow x=0$ by an induction on $n$. Thus, $R$ is reduced, which means that the canonical map $R \to \prod_{\mathfrak{p}} R/\mathfrak{p}$, where the product ranges over all prime ideals of $R$, is injective. Now each $R/\mathfrak{p}$ is an integral domain satisfying the identity. One then sees that it has to be a field, actually of at most $p$ elements. Thus it has to be $\mathbb{F}_p$. This proves $R \in \langle \mathbb{F}_p \rangle$.

For $\langle \mathbb{F}_{p^d} \rangle$ similar arguments can be used, but it gets more complicated. Let me know if you are interested ...

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This result is very cool! You shoulkd probably add "...consists of those commutative rings satisfying...". Why does a field that satisfies $x^p\!=\!x$ contain at most $p$ elements? Exactly $p$ elements? Could you please (when you find time - no hurry) prove the characterization of $\langle \mathbb{F}_{p^d}\rangle$? What rings does the variety $\langle \mathbb{Z}_n\rangle$ contain? – Leon Jun 3 '12 at 21:07

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