Uncountable Ring with Finite Characteristic

Question

Any good examples of these? Countable is easy, and uncountable is easy if I don't care about the proof being constructive but I really want something I can get a solid grip on. So nothing requiring choice.

Answer 1

The ring $(\mathbb{Z}/n\mathbb{Z})^\mathbb{N}$ is uncountable and has characteristic $n$.

Answer 2

Boolean algebras can be formed from the subsets of any set and have characteristic $2$.

So take the subsets of the natural numbers, which are uncountable. Addition is disjoint union and multiplication is intersection. The additive identity is the empty set, and the multiplicative identity is $\mathbb N$.

And you can do the same with subsets of the real numbers.

Answer 3

Fix a positive integer $n$.

Let $X$ be a nonempty set, and let $R_X = (\mathbb{Z}/n\mathbb{Z})[\{t_x\} \mid x \in X]$ be the polynomial ring over $\mathbb{Z}/n\mathbb{Z}$ in a set of indeterminates indexed by $X$. Then

$\# R_X = \max (\# X, \aleph_0)$.

Thus for every infinite cardinal $\kappa$ there is a ring $R$ of cardinality $\kappa$ and characteristic $n$. On the other hand, let $m \mid n$. Then

$\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$

is a ring of characteristic $n$ and order $mn$. Since any ring $R$ of characteristic $n$ admits $\mathbb{Z}/n\mathbb{Z}$ as a subring, if $R$ is finite then its order must be divisible by $n$, i.e., of the form $mn$ for some $m \in \mathbb{Z}^+$. Thus we have determined all possible cardinalities of rings of characteristic $n$.

Notice that the infinite case was actually easier than the finite case. There is a general principle here. Having characteristic $n$ is expressible as a sentence in the (countable) language of rings. Moreover $\mathbb{Z}/n\mathbb{Z}[t]$ is an infinite ring of characteristic $n$. It then follows from the Lowenheim-Skolem Theorem that there are characteristic $n$ rings of all infinite cardinalities. However this result does not say anything about the cardinalities of finite models.