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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1The 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 $\mathbb{Z}_n^\mathbb{N}$. – Julien Apr 10 '13 at 02:52
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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 $2$.
Andreas Blass
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1What an interesting example. I had completely forgotten one could put such a structure on $\mathcal{P}(X)$. Thanks, +1. – Julien Apr 10 '13 at 03:01
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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.
Lubin
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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 $K : F_p = l^i$.