# If $R$ is a commutative ring, the nilpotents form necessarily an ideal of $R$? [duplicate]

This is an algebra question from an exam a few years ago:

Let $R$ be a ring, and let $N = \{a \in R: a^n = 0 \text{ for some } n \in \mathbb{N}, (n \text{ depends on } a) \}$. Prove or disprove: If $R$ is commutative, then $N$ is an ideal of $R$?

So, clearly $N$ is nonempty and it is easy to show that it is closed under multiplication. However, for showing that $N$ is closed under subtraction, I believe this is not possible for we do have that $R$ is commutative, however, we don't have that $R$ is of characteristic $p$ and so the freshman's dream is just that...a dream. So, I would conclude that $N$ is not an ideal. Am I right about this?

I don't really understand the solution given in: If $x,y$ are nilpotent and commute, $x+y$ is nilpotent. and also isn't it not necessarily true that $a^m + b^n = (a+b)^{m+n}$?

## marked as duplicate by user26857, user147263, Macavity, Claude Leibovici, Jyrki LahtonenJun 29 '15 at 5:57

• If $a^n=0$ and $b^m=0$, what can you say about $(a-b)^{n+m}$? – WillO Jun 28 '15 at 17:06
• This really is a duplicate. I don't know why people are voting to reopen. – Matt Samuel Jul 16 '15 at 4:15

If $a\in N$ and $r\in R$, then $(ar)^n=a^nr^n=0$ by commutativity of $R$, so $ar\in N$.
Next suppose that $a,b\in N$. Then there are $n,m$ for which $a^n=b^m=0$. Since the ring is commutative, we may apply the Binomial theorem to obtain$$(a+b)^{n+m}=\sum_{i=0}^{n+m}\binom{n+m}{i}a^ib^{n+m-i}.$$
Now observe that either $i\geq n$ or $n+m-i\geq m$ (if neither were true, then $i+(n+m-i)<n+m$, a contradiction). Hence each term vanishes. Consequently $(a+b)^{n+m}=0$ so $a+b\in N$. Thus $N$ is an ideal.
By the way, $N$ is called the nilradical of $R$.
As stated in a comment, expand out $(a-b)^{N}$ and verify that for each term, either the power on $a$ is at least $N/2$ or the power on $b$ is at least $N/2$. Thus if you have some common $n$ such that $a^n = b^n = 0$, just set $N = 2n$.