Non-zero idempotent element is not nilpotent

Question

I have a problem that I need to solve but I have trouble in solving the following question.

Question is;

Let $a \in R$ be a nonzero idempotent. Show that $a$ is not nilpotent. ($R$ is a ring)

I will appreciate your help.

Thanks in advance.

Answer 1

Since $a$ is an idempotent element then $$a^2=a$$ hence we have $$\forall n\in \Bbb N,\qquad a^n=a\ne0$$ hence $a$ isn't nilpotent.

Answer 2

Assume that $a$ is idempotent and nilpotent. Assume that $a \neq 0$. So there exists $k>0$ such that $a^k=0=a$ absurd. The last equality holds by the idempotency of $a$ plus the associativity of multiplication in a ring.

Here we prove that is a zero divisor (it is a common exercise).

By hypothesis $$a^2=a$$ so $$a^2-a=0=a(a-1)$$ if $a \neq 1$. (The case $a=1$ is clear.)

Answer 3

Let $R$ be a ring, let $a \in R$ be idempotent, i.e. $a^2 = a$, and nilpotent, i.e. $a^m = 0$ for some positive integer $m$. Then $a = 0$.

Proof: $a^2 = a \Rightarrow a^m = a$ for all positive integers $m \ge 2$, by a simple induction: $a^k = a \Rightarrow a^{k + 1} = aa^k = a^2 = a$. ($a^2 = a$ being the base case.) Thus if $a^m = 0$, $a = 0$. The case $m = 1$ yields $a = 0$ even more trivially. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!