Noetherian not implies zero divisor in minimal prime

Statement

It is possible to have a Noetherian ring with a zero divisor that is not contained in any minimal prime ideal.

Let $k$ be a field. Consider the ring $R = k [x, y] / (x^{2}, x y)$ . We have:

$R$ is an affine ring over a field, hence is Noetherian.
The element $y \in R$ is a zero divisor, since $x y = 0$ .
The only minimal prime in $R$ is the ideal generated by $x$ : Since $x$ is nilpotent, any prime ideal must contain $x$ , and hence must contain the ideal generated by $x$ . On the other hand, the ideal generated by $x$ is prime, because the quotient is isomorphic to $k [x, y] / (x^{2}, x y, x) = k [x, y] / (x) ≅ k [y]$ , which is an integral domain. Thus, $(x)$ is the unique minimal prime.
$y$ is not in the ideal generated by $x$ : This is clear from the above remarks: the quotient $R / (x)$ can naturally be identified with a polynomial ring in $y$ , hence $y$ is nonzero in the quotient, and not in $(x)$ .

Thus, we have a Noetherian ring with a zero divisor outside all minimal prime ideals.