# Can a ring of positive characteristic have infinite number of elements?

For curiosity: can a ring of positive characteristic ever have infinite number of distinct elements? (For example, in $\mathbb{Z}/7\mathbb{Z}$, there are really only seven elements.) We know that any field/ring of characterisitc zero must have infinite elements, but I am not sure what happens above.

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The Cartesian product $R\times S$ of two rings of characteristic $n$ has characteristic $n$. The same is true for any arbitrary product. So just take your favorite finite characteristic $n$ unital ring, which seems to be $\mathbb{Z}_n$, and then take the infinite product $\mathbb{Z}_n^\mathbb{N}$. – 1015 Apr 10 '13 at 2:52

Consider $\mathbf{Z}/p\mathbf{Z}[x]$.

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Besides @Brandon’s most economical example, any field (such as $\mathbb Z/p\mathbb Z$) has an algebraic closure, and an algebraically closed field can not be finite.

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Start with any infinite set $X$ and let $R$ be the set of all subsets of $X$. With the operations of symmetric difference (as addition) and intersection (as multiplication), $R$ is a ring of characteristic $2$.

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What an interesting example. I had completely forgotten one could put such a structure on $\mathcal{P}(X)$. Thanks, +1. – 1015 Apr 10 '13 at 3:01

The OP avers to have seen the rings $\mathbb{Z}/p\mathbb{Z}$ and wants to know whether there must be infinite rings of characteristic $p$. Although exhibiting an infinite ring of positive characteristic is easily done, it is not necessary to do so in order to affirmatively answer the question. Indeed, for each fixed prime $p$, the rings of characteristic $p$ are the class of models of a first order theory in the language of rings. We know that there exist arbitrarily large finite models. From this it follows purely model-theoretically that there must be infinite models. (This is Theorem 8 in these notes. Anyway this is like $2+2= 4$ to model theorists.) It goes like this: by the Compactness Theorem, the theory of rings of characteristic $p$ plus the family of statements "$R$ has cardinality at least $n$" for every integer $n$ is finitely satisfiable, so it must be satisfiable. Indeed, it follows from the Lowenheim-Skolem (see $\S$ 2.6 of loc. cit. for the statements...but not the proofs) that there are rings of characteristic $p$ of every infinite cardinality.

If you know algebra but not model theory, you will probably not be impressed by this application: it's easy enough to take an infinite product of copies of $\mathbb{Z}/p\mathbb{Z}$. But the principle is very general: for instance, it just as immediately implies that there are fields of characteristic $p$ and every infinite cardinality, which is slightly less trivial. And so on...

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Let $p$ be a prime, $l > 1$. The nested union $\cup_{i \in \mathbb{N}} K_i$ is yet another example; where $K_i$ is the unique field extension of $\mathbb{F}_p$ with $|K : F_p| = l^i$.

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