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The nilradical of a Noetherian ring is nilpotent

Statement

Let $R$ be a Noetherian commutative ring. Then its nilradical $Nil (R)$ is a nilpotent ideal: there exists an integer $N \geq 1$ such that $Nil (R)^{N} = (0)$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ .

[L1]

The nilradical is the ideal of nilpotent elements (The nilradical and reduced rings).

[L2]

In a Noetherian commutative ring, every ideal is finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

Proof

technique · direct

1.1L1L2choose

By [L1] and [L2], the ideal $Nil (R)$ is finitely generated. Choose generators $a_{1}, \dots, a_{r}$ , and for each $i$ choose an integer $n_{i} \geq 1$ with $a_{i}^{n_{i}} = 0$ . Set $N = n_{1} + \dots + n_{r}$ .

2.1step 1.1algebra

Every element of $Nil (R)^{N}$ is a finite sum of monomials of total degree $N$ in the generators $a_{1}, \dots, a_{r}$ . In each such monomial, some generator $a_{i}$ occurs at least $n_{i}$ times, so that monomial contains the factor $a_{i}^{n_{i}} = 0$ and therefore vanishes. Hence every monomial, and therefore every finite sum of them, is zero.

3.1step 2.1∎

Thus $Nil (R)^{N} = (0)$ for the integer $N$ chosen in step 1.1.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Hochster, Introduction to Commutative Algebra, Math 614 notes (2020) (standard reference, not scraped)
The Stacks Project, Section 10.32: Rings and modules with finiteness conditions (standard reference, not scraped)