Does there exist a countably infinite extension ring of the rational numbers? This question came up as I was thinking about whether there could be a ring with the conventional addition and multiplication defined on it that contains all of $\mathbb{Q}$ and at least one other element and is countable. Intuitively, a set that is related to $\mathbb{Q}$ in the same way $\mathbb{Q}$ is related to $\mathbb{Z}$. I tried to attack the question using the fact that if a ring $B$ contains all of $A$ and at least one other element $c$, it must contain every element of the form $a_0 +a_1c +...  +a_nc^n$. i.e. every element of the polynomial ring $A[c]$. But unfortunately couldn't get anywhere that way. There is some related stuff on the internet. Notably, this and this (the second one seems to be evidence for the existence of such a set). The first link, however, refers to power sets which are different to 'polynomial sets' at least in the fact that a 'polynomial set' doesn't contain infinite polynomials (sequences of coefficients). Cantor's diagonal argument can't be applied for the same reason.
 A: Yes, $\Bbb Q[x]$, the ring of polynomials. If you want to have something "nicer", look at $\Bbb Q[\sqrt 2]$, or perhaps $\Bbb Q^{\text{alg}}$, the field of all algebraic numbers (which is a countable subfield of $\Bbb C$), or its intersection with $\Bbb R$.
All of these are countable. $\Bbb Q[x]$ is countable because the set of finite sequences from a countable set is itself countable, and $\Bbb Q[\sqrt 2]$ can be either seen as a quotient of $\Bbb Q[x]$ (so it is countable as the image of a countable set), or directly as $\{p+q\sqrt 2\mid p,q\in\Bbb Q\}\cong\Bbb Q\times\Bbb Q$. And finally, $\Bbb Q^{\text{alg}}$ is countable because each number is the root of a polynomial in $\Bbb Q$, and each polynomial has finitely many roots, so it is a countable union of finite sets, and therefore countable.
Similarly, you can consider $\Bbb Q(\pi)$, or even just adjoin any new symbol with some equation, $\langle\Bbb Q,x\mid x+x=0, x\neq 0\rangle$, so we add an element of additive order $2$, and the ring is no longer a field.
