Noetherian implies every element in minimal prime is zero divisor

Statement
In a fact about::Noetherian ring, any element that lies in a fact about::minimal prime ideal must be a fact about::zero divisor.

Converse
The converse statement is true when the ring is reduced.

Generalizations
The assumption that the ring be Noetherian can be weakened to the assumption that every prime contains a minimal prime, and such that there are only finitely many minimal primes.

Facts used

 * 1) uses::Noetherian ring has finitely many minimal primes and every prime contains a minimal prime

Proof
Given: A Noetherian ring $$R$$

To prove: If $$x_1 \in R$$ lies in a minimal prime ideal of $$R$$, then $$x$$ is a zero divisor.

Proof: Since $$R$$ is a Noetherian ring, it has only finitely many minimal primes, say $$P_1, P_2, \ldots, P_n$$ (and without loss of generality, $$x_1 \in P_1$$. Moreover, any prime ideal contains a minimal prime ideal, so the nilradical of $$R$$ equals the intersection of the $$P_i$$s. In particular, the product of the $$P_i$$s lies inside the nilradical of $$R$$.

Let us next consider the product $$P_2 P_3 \ldots P_n$$. If this product is contained in $$P_1$$, then, one of the $$P_j$$s is contained in $$P_1$$, contradicting the assumption of minimality. Thus, $$P_2P_3\ldots P_n$$ is not contained in $$P_1$$, so we can find elements $$x_j \in P_j, 2 \le j \le n$$ such that $$x_2x_3 \ldots x_n \notin P_1$$.

Now consider the product $$x_1x_2x_3 \ldots x_n$$. This product lies in the nilradical, hence some power of it is zero. Thus, there exists a $$j$$ such that:

$$x_1^j(x_2x_3 \ldots x_n)^j = 0$$

However, the expression $$(x_2 x_3 \ldots x_n)^j$$ cannot be zero because $$x_2x_3 \ldots x_n \notin P_1$$, and therefore, is not nilpotent.

Thus, $$x_1^j$$ is a zero divisor, and hence, $$x_1$$ is a zero divisor (that's because the set of nonzerodivisors is a multiplicatively closed subset, so if $$x_1$$ were not a zero divisor, $$x_1^j$$ would also not be a zero divisor).