# Ideal of nilpotent elements in non-commutative ring.

Let $R$ be a non-commutative ring such that every element is either invertible or nilpotent. I am trying to show that the set of nilpotent elements, denoted $I$, is a two sided ideal, but I am having problems with showing that the sum of two nilpotent elements is nilpotent.

Note that we cannot use the binomial theorem, since we are dealing with a noncommutative ring. Can someone provide me with a tip?

I proved that the product is nilpotent by noting that if $0\neq x\in I$ and $y\in R$ with $x^n=0$, assuming that $xy$ is invertible we get $x^{n-1} = x^{n-1}xy(xy)^{-1} = 0$, which by induction would lead to $x=0$, which is a contradiction.

Let $x\in R$ be nilpotent and $r\in R$ be arbitrary. If $xr$ is not nilpotent, it is invertible, so $xr(xr)^{-1}=1$ and therefore $x$ is right invertible. This is impossible, because from $x^n=0$ and $xy=1$ we get $x^{n-1}=0$, leading to a contradiction.

Similarly, $rx$ is nilpotent.

Suppose $x$ and $y$ are nilpotent, but $z=x+y$ is invertible. Then $$(x+y)z^{-1}=1$$ and so $$xz^{-1}=1-yz^{-1}$$ By what we have proved before, we know that $yz^{-1}$ is nilpotent. However, when $a\in R$ is nilpotent, $1-a$ is invertible; indeed, assuming $a^{n+1}=0$, we have $$(1-a)(1+a+\dots+a^{n})=1-a^{n+1}=1$$ and similarly on the other side. Therefore $xz^{-1}$ is invertible. Contradiction.

• Thank you for the response. I accept this answer over the other, because it does not assume knowledge on other subjects like the Jacobson radical.
– Marc
Feb 22, 2015 at 20:59
• @Marc You're welcome! Feb 22, 2015 at 21:00

Note: in case you only gave the question a quick glance instead of reading it, note that in this ring every element is either nilpotent or invertible and the OP proved that it follows that if $r$ is nilpotent and $x\in R$ then $xr$ is nilpotent. It follows by a similar argument that if $y$ is another element then $xry$ is nilpotent.

Consider the Jacobson radical of $R$, which is the ideal consisting of all elements $r$ such that for all $x,y\in R$ we have that $1+xry$ is invertible. You have already proven that if $r$ is nilpotent, then so is $xry$. Thus, at least in this ring, if $r$ is nilpotent then it is contained in the Jacobson radical because $1+a$ is invertible whenever $a$ is nilpotent. The Jacobson radical is a proper ideal and hence contains no invertible elements. Thus the Jacobson radical of $R$ consists exactly of the nilpotent elements since every element of $R$ is either nilpotent or invertible. Thus the set of nilpotent elements forms an ideal (namely the Jacobson radical).

• What about it? In principle the Jacobson radical could be properly contained in the set of nilpotent elements, so it is hard to see what you are driving at. Feb 22, 2015 at 18:44
• In general it can, but not in this ring. Feb 22, 2015 at 18:44
• @rschwieb I've edited my answer Feb 22, 2015 at 19:08
• Thank you for the response. I have accepted the other answer over this one, because the other one does not assume knowledge on other subjects, like the Jacobson radical, which makes it more accessible.
– Marc
Feb 22, 2015 at 20:59
• @Marc no problem. Feb 22, 2015 at 21:01