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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



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


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


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


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_is. In particular, the product of the P_is 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_js 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).