Uncountable Ring with Finite Characteristic 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.
 A: The ring $(\mathbb{Z}/n\mathbb{Z})^\mathbb{N}$ is uncountable and has characteristic $n$.
A: 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.
A: 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.  
