## Lifting idempotents modulo nil ideals

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Definition 1. Let $I$ be a two-sided ideal of a ring $R.$ We say that an idempotent element $a \in R/I$ can be lifted if $a=e + I$ for some idempotent element $e \in R.$

Definition 2. A subset $S$ of a ring $R$ is called nil if every element of $S$ is nilpotent.

Problem. If $I$ is a nil ideal of $R,$ then every idempotent of $R/I$ can be lifted.

Solution. Suppose that $a=r+ I$ is an idempotent element of $R/I.$ Then $r-r^2 \in I$ and thus $(r-r^2)^n=0,$ for some integer $n \geq 1.$ Therefore

$r^n - r^{n+1} \sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1} = 0.$

So if we let

$s =\sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1},$

then $r^n=r^{n+1}s$ and $rs=sr.$ Now let $e=(rs)^n.$ Using the fact that $r^n=r^{n+1}s$ we will get $e^2=e.$ Also, since $r+ I = r^2+I,$ we have $r+I=r^k + I$ for all $k \geq 1.$ Therefore

$r+I=r^n+I=r^{n+1}s+I=(r^{n+1}+I)(s+I)=rs+I.$

Thus

$a=r+I=r^n + I=(r+I)^n=(rs+I)^n=(rs)^n+I=e+I. \Box$

Example. Let $R$ be a (left) Artinian ring and let $J$ be the Jacobson radical of $R.$ Since $J$ is nilpotent, and hence nil, every idempotent of $R/J$ can be lifted.

The power of $r$ in the first summation should be $i-1$.

But if $R$ has no multiplicative identity,
what is $r^0$ means?
Here is another solution,

http://goo.gl/zEK01L

http://goo.gl/FpncJD

• Yaghoub Sharifi says:

Yes, you’re right. It should be $i-1$. I just fixed it, Thank you!
In this blog all rings have $1$ unless otherwise specified. Thanks for the link though! 🙂