Nilpotent Elements of Commutative Ring form Ideal

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Theorem

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.


The subset of nilpotent elements of $R$ form an ideal of $R$.


Proof

Let $N$ be the subset of nilpotent elements.

Because $0_R$ is nilpotent, $0_R \in N$ and so $N$ is non-empty.


Let $x \in N$ and $a \in R$.

We have:

\(\ds \exists n \in \Z_{>0}: \, \) \(\ds x^n\) \(=\) \(\ds 0_R\) Definition of Nilpotent Ring Element
\(\ds \leadsto \ \ \) \(\ds a^n \circ x^n\) \(=\) \(\ds 0_R\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds \paren {a \circ x}^n\) \(=\) \(\ds 0_R\) Power of Product of Commutative Elements in Semigroup
\(\ds \leadsto \ \ \) \(\ds a x\) \(\in\) \(\ds N\) Definition of $N$


Let $x, y \in N$.

Let $x^n = 0$ and $y^m = 0$.

By the Binomial Theorem, $\paren {x - y}^p = 0$ for $p \ge n + m - 1$.

Thus $x - y \in N$.


Thus from Test for Ideal, $N$ is an ideal.

$\blacksquare$


Also see


Sources

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