On Ring Theory

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
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On Ring Theory

#1

Post by Tsakanikas Nickos »

  1. Let \( \displaystyle R \) be a ring. If \( \displaystyle r \in R \) is nilpotent, then show that \( \displaystyle 1-r \) is invertible.
  2. Let \( \displaystyle R \) be a local ring. Show that the only idempotents in \( \displaystyle R \) are the trivial ones.
  3. Let \( \displaystyle R \) be a ring whose only idempotents are the trivial ones. Then all left-invertible and all right-invertible elements of \( \displaystyle R \) are invertible.
In all cases assume that \( R \) is an associative ring with unity!
Papapetros Vaggelis
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Re: On Ring Theory

#2

Post by Papapetros Vaggelis »

1. Since \(\displaystyle{r\in R}\) is nilpotent, there exists \(\displaystyle{n\in\mathbb{N}}\) such that \(\displaystyle{r^n=0}\) .

Then,

\(\displaystyle{\begin{aligned} (1-r)\,\sum_{k=0}^{n-1}r^k&=\sum_{k=0}^{n-1}(1-r)\,r^k\\&=\sum_{k=0}^{n-1}(r^k-r^{k+1})\\&=(1-r)+(r-r^2)+(r^2-r^3)+...+(r^{n-1}-r^n)\\&=1-r^n\\&=1\\&=\sum_{k=0}^{n-1}r^k\,(1-r)\end{aligned}}\)

So, \(\displaystyle{1-r}\) is invertible and \(\displaystyle{(1-r)^{-1}=\sum_{k=0}^{n-1}r^k}\) .

2. Let \(\displaystyle{r\in R}\) be an idempotent element of \(\displaystyle{R}\) , that is \(\displaystyle{r^2=r}\) .

SInce the ring is local, we have that \(\displaystyle{r}\) or \(\displaystyle{1-r}\) is invertible.

If \(\displaystyle{r}\) is invertible, then

\(\displaystyle{r^2=r\implies r^2\,r^{-1}=r\,r^{-1}\implies r=1}\)

If \(\displaystyle{1-r}\) is invertible, then

\(\displaystyle{r^2=r\implies r(1-r)=0\implies r(1-r)\,(1-r)^{-1}=0\implies r=0}\) .

In any case, \(\displaystyle{r\in\left\{0,1\right\}}\) and we are done.

3. Suppose that \(\displaystyle{r\in R}\) is a left-invertible element and \(\displaystyle{s\in R}\) is right invertible

element. Then, \(\displaystyle{x\,r=1}\) and \(\displaystyle{s\,y=1}\) for some \(\displaystyle{x\,,y\in R}\) .

\(\displaystyle{(r\,x)^2=r\,x\,r\,x=r\,x}\), which means that \(\displaystyle{r\,x}\) is an idempotent element of the ring,

so : \(\displaystyle{r\,x=0}\) or \(\displaystyle{r\,x=1}\).

If \(\displaystyle{r\,x=0}\), then, \(\displaystyle{r\,x=0\implies r\,x\,r=0\implies r=0}\), a contradiction.

Therefore, \(\displaystyle{r\,x=x\,r=1\implies r\in U(R)}\) .

Similarlly, \(\displaystyle{s\in U(R)}\) .
Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: On Ring Theory

#3

Post by Tsakanikas Nickos »

Papapetros Vaggelis wrote: If \(\displaystyle{1-r}\) is invertible, then
\(\displaystyle{r^2=r\implies r(1-r)=0\implies r(1-r)\,(1-r)^{-1}=0\implies r=0}\) .
Another way to prove that is to notice that \( \displaystyle 1-r \) is also an idempotent and then apply the method used in the first case.
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