# prime ideals minimal over a zerodivisor

Let $$R$$ be a commutative ring with identity. If $$P$$ is a prime ideal of $$R$$ that is minimal over some zerodivisor of $$R$$, then must $$P$$ consist only of zerodivisors? I suspect not but I can't figure out how to construct a counterexample.

(The full context of the situation I am in is this. Suppose that $$P$$ is a prime ideal of $$R$$ such that $$PP^{-1} \neq P$$. Then there is an element $$a$$ of $$P$$ and an element $$b$$ of $$R-P$$ such that $$P = (aR :_R bR)$$. In this case, $$P$$ is necessarily minimal over $$a$$. My question is, if $$a$$ is a zerodivisor, can I conclude that $$P$$ consists only of zerodivisors?)

For the first question, consider $$R = \mathbb{Z}[X]/(X^3 - 1)$$, $$P = (x+ 1)$$ and $$a = (x + 1)(1 + x + x^2)$$ where $$x$$ denotes the image of $$X$$ in $$R$$.
In order to see that it provides us with a reduced one-dimensional Noetherian counter-example, note that $$R/P \simeq \mathbb{Z}/2\mathbb{Z}$$ and that $$(x - 1)a = 0$$. Hence $$P = (2, x -1)$$ is a maximal ideal of $$R$$ and $$a$$ is a zero-divisor. Any other prime ideal of $$R$$ containing $$a$$ has to contain $$Q = (1 + x + x^2)$$. As $$x + 1$$ doesn't divide $$1 + x + x^2$$ (since otherwise $$1 + x + x^2$$ would map to an even integer modulo $$(x - 1)$$), $$Q$$ is not a subset of $$P$$. Therefore $$P$$ is a minimal prime over $$a$$ which contains regular elements, namely $$2$$ and $$x + 1$$.
As the ideal $$P$$ is generated by the regular element $$x + 1$$, it is invertible, i.e., it satisfies $$PP^{-1} = R \neq P$$. It is easy to check that $$P = ((a) :_R (1 + x + x^2))$$, so that this counter-example satisfies also your extra requirements.
On the positive side, it is well-known that any minimal prime of a commutative ring with identity consists only of zero-divisors. Thus the answer is yes if $$a = 0$$.
Side note. In [1], D. Anderson and J. Pascual refers to the definition of Property (A): A commutative ring $$R$$ with identity enjoys property (A) if every faithful finitely generated ideal $$I$$ (i.e., $$I$$ is finitely generated and $$(0:_R I) = 0$$) is regular (i.e., $$I$$ contains at least one regular element). The authors remark that a ring in which $$0$$ has a primary decomposition satisfies Property $$(A)$$. I am not sure if this can be helpful in your context, but if your ring $$R$$ has Property $$(A)$$, then $$P$$ consists solely of zero-divisors if and only if $$P$$ is not faithful, which may be easier to check.