 # Let R be a commutative k -algebra, where k is a field of characteristic zero. Coul Brock Byrd 2022-07-10 Answered
Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero.
Could one please give an example of such $R$ which is also:
(i) Not affine (= infinitely generated as a $k$-algebra).
and
(ii) Not an integral domain (= has zero divisors).My first thought was $k\left[{x}_{1},{x}_{2},\dots \right]$, the polynomial ring over k in infinitely many variables, but unfortunately, it satisfies condition (i) only. It is not difficult to see that it is an integral domain: If $fg=0$ for some $f,g\in k\left[{x}_{1},{x}_{2},\dots \right]$, then there exists $M\in \mathbb{N}$ such that $f,g\in k\left[{x}_{1},\dots ,{x}_{M}\right]$, so if we think of $fg=0$ in $k\left[{x}_{1},\dots ,{x}_{M}\right]$, we get that $f=0$ or $g=0$, and we are done.
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For instance, $k\left[{X}_{n}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}n\in \mathbb{N}\right]/\left({X}_{n}^{2}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}n\in \mathbb{N}\right)$: the ring of polynomials in infinitely many variables quotiented by the ideal generated by the square of the variables.