Noetherian not implies zero divisor in minimal prime

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

Related facts

 * Noetherian implies every element in minimal prime is zero divisor
 * Reduced Noetherian implies zero divisor in minimal prime
 * Reduced not implies zero divisor in minimal prime

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


 * $$R$$ is an affine ring over a field, hence is Noetherian.
 * The element $$y \in R$$ is a zero divisor, since $$xy = 0$$.
 * 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 $$k[x,y]/(x^2,xy,x) = k[x,y]/(x) \cong k[y]$$, 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.