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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?

Thank you in advance.

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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Marc Feb 22 '15 at 20:59
  • $\begingroup$ @Marc You're welcome! $\endgroup$ – egreg Feb 22 '15 at 21:00
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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).

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  • $\begingroup$ 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. $\endgroup$ – rschwieb Feb 22 '15 at 18:44
  • $\begingroup$ In general it can, but not in this ring. $\endgroup$ – Matt Samuel Feb 22 '15 at 18:44
  • $\begingroup$ @rschwieb I've edited my answer $\endgroup$ – Matt Samuel Feb 22 '15 at 19:08
  • $\begingroup$ 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. $\endgroup$ – Marc Feb 22 '15 at 20:59
  • $\begingroup$ @Marc no problem. $\endgroup$ – Matt Samuel Feb 22 '15 at 21:01

protected by user26857 Aug 12 '15 at 16:43

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