The set of all nilpotent elements is an ideal
Given that R is commutative ring with unity, I want show that set of all nilpotent elements is an ideal of R.
I know how to show ideal if set is given but here set is not given to me. Can anyone help me?
Solution 1:
The set is given: it is $$\{a \in R\mid a\text{ is nilpotent}\}.$$ Remember that $a\in R$ is nilpotent if and only if there exists $n\gt 0$ such that $a^n=0$.
Hints.
Is $0$ in the set?
If $a$ and $b$ are in the set, can you guarantee that a large enough power of $(a-b)$ is equal to $0$? Think binomial expansion.
If $a$ is in the set and $r\in R$, is $ra$ in the set? Here, too, commutativity will be key.
Added. For $2$: if $a^k=0$, then $a^r=0$ for all $r\geq k$; if $b^t=0$, then $b^s=0$ for all $s\geq t$. Can you find an $N$ such that each term in the binomial expansion of $(a-b)^N$ will have either $a^r$ with $r\geq k$ or $b^s$ with $s\geq t$?