Let $R$ be a commutative ring with identity and $N$ be the set of all nilpotent elements of $R.$ Show that $N$ forms an ideal of $R$ and the quotient ring $R/N$ has no non-zero nilpotent elements. Is commutativity of the ring $R$ essential in the above result? Justify.
My solution is as follows:
If $R$ is a commutative ring and $N$ is the set of all nilpotent elements of $R$ then, $N=\{a\in R:a^n=0,$ for some integer $n\in\Bbb Z^+\}.$
Let $x,y\in N$ then, $\exists n_1,n_2\in \Bbb Z^+$ such that $x^{n_1}=y^{n_2}=0.$
Now, $(x+y)^{n_1+n_2}=x^{n_1+n_2}+\binom{n_1+n_2}{1}x^{n_1+n_2-1}y+\cdots +\binom{n_1+n_2}{n_1+n_2}y^{n_1+n_2}=0.$
This means, $x+y\in N.$
Again, $(-x)^{n_1}=(-1)^{n_1}x^{n_1}=0.$ This means, $-x\in N.$
So, $N$ is a subgroup of $R$ under addition.
Let $r\in R$ then, $rx,xr\in N$ as $(rx)^{n_1}=0=(xr)^{n_1}.$ So, $N$ is an ideal.
Now, we claim that, $R/N$ has no non zero nilpotent element. This is because, say, $w\in R/N$ is a nilpotent element in $R/N$. Now, $\exists r\in R$ such that $r+N=w.$ This means, $\exists n\in\Bbb Z^+$ such that $w^n=N=r^n+N.$ This means $r^n\in N.$ Now, this implies $r^{nk}=0,$ for some $k\in\Bbb Z^+.$ Hence, $r\in N$ as well. So, $w=r+N=N.$
But, $w$ is any arbitrary nilpotent element of $R/N$ and hence, $N$ is the only nilpotent element in $R/N.$ Thus, $R/N$ has no non-zero nilpotent element.
However, I feel that commutativity of the ring $R$ essential in the above result but I am unable to justify this claim. Any help regarding this will be highly appreciated.
The binomial theorem requires commutativity. And indeed we see that this part of the proof can fail in non-commutative rings, such as for instance the ring of $2\times 2$ integer matrices, where $$ \begin{bmatrix}0&1\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\1&0\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix} $$ is not nilpotent. So $N$ is not closed under addition, and can therefore not be an ideal (neither right, left, nor two-sided).