Nilpotent = Ideal

**tttcomrader** · November 1st 2008, 03:54 PM

Show that nilpotent elements of a commutative ring R constitute an ideal of R.

Proof.

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

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

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

We need to show that $ar \in N$

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

That's it? Thanks.

**Jhevon** · November 1st 2008, 04:43 PM

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 $a \in R$ such that $a^n =0$ for some n>0.

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

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

We need to show that $ar \in N$

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

That's it? Thanks.

looking good so far to me.

another trivial thing you have to show though is that $N$ is closed under addition, that is, $x \pm y \in N$ for all $x,y \in N$

**ThePerfectHacker** · November 1st 2008, 04:43 PM

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 $a \in R$ such that $a^n =0$ for some n>0.

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

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

We need to show that $ar \in N$

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

That's it? Thanks.

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

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