- Let \( \displaystyle R \) be a ring. If \( \displaystyle r \in R \) is nilpotent, then show that \( \displaystyle 1-r \) is invertible.
- Let \( \displaystyle R \) be a local ring. Show that the only idempotents in \( \displaystyle R \) are the trivial ones.
- 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.
On Ring Theory
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On Ring Theory
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Re: On Ring Theory
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, \(\displaystyle{s\in U(R)}\) .
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, \(\displaystyle{s\in U(R)}\) .
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Re: On Ring Theory
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.Papapetros Vaggelis wrote: If \(\displaystyle{1-r}\) is invertible, then
\(\displaystyle{r^2=r\implies r(1-r)=0\implies r(1-r)\,(1-r)^{-1}=0\implies r=0}\) .
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