A counterexample from one of my other MO answers seems to work again. Let $k$ be an algebraically closed field. Let $R$ be the ring of functions $f: k^2 \to k$ such that there exists a polynomial $\overline{f} \in k[x,y]$ with
- $f(x,y) = \overline{f}(x,y)$ for all but finitely many $(x,y) \in k^2$ and
- $f(0,0) = \overline{f}(0,0)$.
$R$ is reduced: If $f(x,y) \neq 0$, then $f(x,y)^n \neq 0$ for all $n$. $\square$
Every element $f$ of $R$ is either a unit or a zero divisor:
Case 1: $f$ is nowhere zero. In this case, $\overline{f}$ must lie in $k^{\ast}$, as otherwise $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them. So $f^{-1}$ is equal to the nonzero constant $\overline{f}^{-1}$ at all but finitely many points, and $\overline{f}^{-1} \in R$. So, in this case, $f$ is a unit.
Case 2: $f(x_0,y_0)=0$. Without loss of generality, we may assume that $(x_0,y_0) \neq (0,0)$. This is because, if $f(0,0)=0$ then $\overline{f}(0,0)=0$, implying that $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them, so we can find some other $(x_0,y_0)$ at which $f$ also vanishes. Let $\delta(x,y)$ be $1$ if $(x,y) = (x_0, y_0)$ and $0$ otherwise. Then $f \delta=0$ and $\delta \neq 0$, showing that $f$ is a zero divisor. $\square$
The set of functions vanishing at $(0,0)$ is clearly a maximal ideal of $R$; which we will denote $(0,0)$. We claim that $R_{(0,0)} \cong k[x,y]_{(0,0)}$. Proof sketch: We claim that $f = \overline{f}$ in the localization. To see this, let $g$ vanish at the finitely many points where $f \neq \overline{f}$, but $g(0,0) \neq 0$. Then $fg=\overline{f} g$, and $g$ is invertible in the localization. This shows that $f s^{-1} = \overline{f} \overline{s}^{-1}$ for any $f$ and $s$. $\square$.
Clearly, $k[x,y]_{(0,0)}$ is not classical.
I think this construction can clearly be generalized to make classical rings which have any local ring of dimension $\geq 2$ as a localization.
