Noetherian implies every element in minimal prime is zero divisor

This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness

Statement

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

Converse

The converse statement is true when the ring is reduced. Further information: reduced Noetherian implies zero divisor in minimal prime

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. 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, $P_{2}P_{3}\ldots P_{n}$ is not contained in $P_{1}$, so we can find elements $x_{j}\in P_{j},2\leq j\leq n$ such that $x_{2}x_{3}\ldots x_{n}\notin P_{1}$.

Now consider the product $x_{1}x_{2}x_{3}\ldots x_{n}$. This product lies in the nilradical, hence some power of it is zero. Thus, there exists a $j$ such that:

$x_{1}^{j}(x_{2}x_{3}\ldots x_{n})^{j}=0$

However, the expression $(x_{2}x_{3}\ldots x_{n})^{j}$ cannot be zero because $x_{2}x_{3}\ldots x_{n}\notin P_{1}$, 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).