Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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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 N1 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).

Proof

technique · direct
1.1

By [L1] and [L2], the ideal Nil(R) is finitely generated. Choose generators a1,,ar, and for each i choose an integer ni1 with aini=0. Set N=n1++nr.

L1L2choose
2.1

Every element of Nil(R)N is a finite sum of monomials of total degree N in the generators a1,,ar. In each such monomial, some generator ai occurs at least ni times, so that monomial contains the factor aini=0 and therefore vanishes. Hence every monomial, and therefore every finite sum of them, is zero.

step 1.1algebra
3.1

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

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources