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An element of a ring $R$ is nilpotent if $a^n=0$ for some $n \ge 1$.

How do I show that additive inverse of $a$ , $-a$ is also nilpotent?

The ring is commutative but may not have a unit element.

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2 Answers 2

up vote 3 down vote accepted

Using the distributive property, $ab+(-a)b=(a+(-a))b=0\cdot b=0$. Therefore, $$ (-a)b=-(ab)\tag{1a} $$ Also, $ab+a(-b)=a(b+(-b))=a\cdot0=0$. Therefore, $$ a(-b)=-(ab)\tag{1b} $$ Furthermore, since $a+(-a)=0$, we have $$ -(-a)=a\tag{2} $$ Using $(1)$ and $(2)$, it is easy to show by induction that $$ (-a)^k=\left\{\begin{array}{}a^k&\text{if }k\text{ is even}\\-(a^k)&\text{if }k\text{ is odd}\end{array}\right.\tag{3} $$ The fact that $a$ is nilpotent and $(3)$ shows that $-a$ is nilpotent.

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I am simply filling in the details of Kb100's suggestion and including $-(-a)=a$, which is needed to show $(3)$. –  robjohn Nov 13 '11 at 18:01

Try first proving that $a(-b)=(-a)b=-(ab)$ and then (by induction) that $(-a)^n$ is $a^n$ if $n$ is even and $-(a^n)$ if $n$ is odd.

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