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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 qchu.wordpress.com/2010/11/22/… . –  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

2 Answers 2

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).

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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 Lampret Jun 3 '12 at 21:07

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