# Nilpotent = Ideal

• Nov 1st 2008, 03:54 PM
Nilpotent = Ideal
Show that nilpotent elements of a commutative ring R constitute an ideal of R.

Proof.

So nilpotent element is an element $\displaystyle a \in R$ such that $\displaystyle a^n =0$ for some n>0.

Denote the set of all nilpotent elements of R by N.

Pick $\displaystyle a \in N$ and $\displaystyle r \in R$

We need to show that $\displaystyle ar \in N$

$\displaystyle (ar)^n=a^nr^n=0$

That's it? Thanks.
• Nov 1st 2008, 04:43 PM
Jhevon
Quote:

Originally Posted by tttcomrader
Show that nilpotent elements of a commutative ring R constitute an ideal of R.

Proof.

So nilpotent element is an element $\displaystyle a \in R$ such that $\displaystyle a^n =0$ for some n>0.

Denote the set of all nilpotent elements of R by N.

Pick $\displaystyle a \in N$ and $\displaystyle r \in R$

We need to show that $\displaystyle ar \in N$

$\displaystyle (ar)^n=a^nr^n=0$

That's it? Thanks.

looking good so far to me.

another trivial thing you have to show though is that $\displaystyle N$ is closed under addition, that is, $\displaystyle x \pm y \in N$ for all $\displaystyle x,y \in N$
• Nov 1st 2008, 04:43 PM
ThePerfectHacker
Quote:

Originally Posted by tttcomrader
Show that nilpotent elements of a commutative ring R constitute an ideal of R.

Proof.

So nilpotent element is an element $\displaystyle a \in R$ such that $\displaystyle a^n =0$ for some n>0.

Denote the set of all nilpotent elements of R by N.

Pick $\displaystyle a \in N$ and $\displaystyle r \in R$

We need to show that $\displaystyle ar \in N$

$\displaystyle (ar)^n=a^nr^n=0$

That's it? Thanks.

Exactly. You need to show $\displaystyle rN \subseteq N$ and that it exactly what you shown.