Exercise 1876Algebra › Rings and fieldsM1Added 3 Nov 2026

Nilpotent elements

#nilpotent#ring#quotient

Statement

Let $R$ be a commutative ring. An element $a\in R$ is nilpotent if $a^n=0$ for some $n\ge1$ . Show that if $a,b$ are nilpotent, then $a+b$ is nilpotent.

Hint

Use the binomial theorem: $(a+b)^{m+n}=\sum\binom{m+n}{k}a^kb^{m+n-k}$ . Each term has either $a^k$ with $k\ge m$ or $b^{m+n-k}$ with $m+n-k\ge n$ .

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Solution

Binomial expansion. Suppose $a^m=0$ , $b^n=0$ . In the sum $(a+b)^{m+n}=\sum_{k=0}^{m+n}\binom{m+n}{k}a^kb^{m+n-k}$ , each term has either $k\ge m$ (so $a^k=0$ ) or $m+n-k\ge n$ (so $b^{m+n-k}=0$ ).

Conclusion. Every term is zero, so $(a+b)^{m+n}=0$ : $a+b$ is nilpotent.

Remark. The set of nilpotent elements forms an ideal, called the nilradical of $R$ .

Nilpotent elements

Statement

Solution

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