uncountable ring with countable quotient ring?

Question

Insipired by this post: Countable number of subgroups $\implies $ countable group

I'm wondering if it is possible to have an uncountable commutative ring with identity $R$, such that $R/I$ is countable for some two-sided proper ideal $I$?

It seems to be a really basic problem, but I couldn't give a proof or a counterexample. Any help is really appreciated, thanks!

Answer 1

Let $R$ be any finite or countably infinite commutative ring except the one ring. Then $R[[x]]$, the ring of formal power series in the variable $x$ over $R$ is uncountable and $R[[x]]/(x)$ is isomorphic to $R$, so is finite or countably infinite.

Answer 2

I believe $R=\Bbb Z\times\Bbb C$ and $I=\{0\}\times\Bbb C$ works.

Answer 3

Yes. For instance, for any rings $A$ and $B$, the projection $A\times B\to B$ gives an isomorphism between $B$ and a quotient ring of $A\times B$. In particular, if $A$ is uncountable then $A\times B$ will be uncountable, but $B$ could be any ring at all. Or, for any ring $B$, you can form a polynomial ring $R$ over $B$ in uncountably many variables, which will be uncountable (assuming $B$ is nonzero) but has a quotient isomorphic to $B$ (just mod out the ideal generated by all the variables).

A more naturally occurring example is the ring $\mathbb{Z}_p$ of $p$-adic integers, which is uncountable, but the quotient $\mathbb{Z}_p/(p)\cong \mathbb{F}_p$ is finite.