# Assignment 3, Problem 107

Let $R$ be a commutative ring with identity such that the identity map is the only ring automorphism of $R$. Prove that the set $N$ of all nilpotent elements of $R$ is an ideal of $R$

## Proposed solution (Avi Steiner)

Let $N$ be the set of nilpotent elements of $R$. Note that conjugation by a unit $u$ is an automorphism: $u(x+y)u^{-1}=uxu^{-1} + uyu^{-1}$, $uxyu^{-1}=uxu^{-1}uyu^{-1}$, and the map $x\mapsto uxu^{-1}$ is clearly a bijection (with inverse given by conjugation by $u^{-1}$). So, since there are no non-trivial automorphisms of $R$, all such conjugations must be trivial. Therefore, $R^\times$ commutes pointwise with all of $R$ and in particular is abelian.

Now, $1-a$ is a unit for any nilpotent $a$: If $a^n=0$, then

$(1+a+a^2+\cdots+a^{n-1})(1-a) = 1-a^n=1.$

So, letting $r\in R$, $a\in N$, we have

$r-ar=(1-a)r=r(1-a)=r-ra$,

meaning that $r$ and $a$ commute. Therefore, if $a^n=0$, we have that $(ra)^n=(ar)^n=a^nr^n=0$; i.e. $ra=ar\in N$. Finally, suppose $a^n=0$ and $b^m=0$. Since $a$ and $b$ commute, $(a+b)^{n+m}$ is a $\mathbb{Z}$-linear combination of terms of the form $a^{n+m-k}b^k$ for $0\leq k\leq n+m$. If $k\geq m$, then $b^m=0$, making the corresponding term vanish. If $k<m$, then $n+m-k>n$, so that $a^{n+m-k}=0$, again making the corresponding term vanish. Thus, $(a+b)^{n+m}=0$. Hence, $a+b\in N$. Thus, $N$ is an ideal.