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2008-05-12T16:27:38Z

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Vipul: New page: ==Statement== The set of nilpotent elements in a commutative unital ring is an ideal (this ideal is termed the nilradical). ==Proof== ''Commutativity'' is crucial to the pro...

2008-02-09T00:08:42Z

New page: ==Statement== The set of nilpotent elements in a commutative unital ring is an ideal (this ideal is termed the nilradical). ==Proof== ''Commutativity'' is crucial to the pro...

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==Statement==

The set of [[nilpotent element]]s in a [[commutative unital ring]] is an [[ideal]] (this ideal is termed the nilradical).

==Proof==

''Commutativity'' is crucial to the proof, as we shall see.

===Abelian group structure===

It is clear that <math>0</math> is nilpotent, and that if <math>x</math> is nilpotent, so is <math>-x</math>. We thus only need to show closure under addition. Suppose <math>x</math> and <math>y</math> are nilpotent with <math>x^m = y^n = 0</math>. Then consider:

<math>(x+y)^{m+n}</math>

We can expand this by the binomial theorem. We get a sum of monomials. For each monomial, either the power of <math>x</math> is at least <math>m</math> or the power of <math>y</math> is at least <math>n</math>. Thus, each of the monomials in the expansion is zero, and so the above expression simplifies to 0.

Commutativity is essentially to rewrite expressions like <math>xyxy</math> as a power of <math>x</math>, times a power of <math>y</math>.

===The ideal property===

We need to show that if <math>x</math> is nilpotent, and <math>a</math> is any ring element, then <math>ax</math> is nilpotent. Since there exists a <math>n</math> such that <math>x^n = 0</math>, we have <math>(ax)^n = a^nx^n = 0</math>. Here, we again use commutativity to move the <math>a</math>s past the <math>x</math>s.

Nilradical is an ideal - Revision history

Vipul: 1 revision

Vipul: New page: ==Statement== The set of nilpotent elements in a commutative unital ring is an ideal (this ideal is termed the nilradical). ==Proof== ''Commutativity'' is crucial to the pro...