nonNoetherian ring of countable cardinality

Question

I recently asked myself if I could find a nonNoetherian ring (commutative w/ one) of countable cardinality.

I could not.

My wealth of nonNoetherian rings is small and usually relies on taking $k[x_{1}, ....]$ modulo something, a nonfinite direct product of rings, or some other very large object.

I was thinking I could take my countable ring, write down the nonzero non-one elements as $x_{1}, ...$ and think of it as a quotient of $k[x_{1}, ....]$. The kernel of this map is very large and you can find a representative for every coset that is a sum of a bunch of monomials, but there is absolutely no reason (for me) to think some a finite polynomial ring over k surjects onto this quotient.

Any ideas would be appreciated.

Answer 1

You have the ring of all algebraic integers (the integral closure of $\mathbf Z$ in $\overline{\mathbf Q}$).

Also the ring of polynomials $\mathbf Z[x_1, x_2, \dots,x_n,\dots]$.

The ring of integer-valued polynomials (the subring of $\mathbf Q[x]$ which take integer values on integers).

Answer 2

You could just use the field of two elements $F_2$ and look at $\oplus_{i=1}^\infty F_2$ with an identity adjoined (it is simply the subring of $\prod_{i=1}^\infty F_2$ generated by the identity and $\oplus_{i=1}^\infty F_2$.

Answer 3

$A[(T_i)_{i\in I}]$ for any non-zero countable ring $A$ and any strictly countable set $I$.