Noetherian modules
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- Community Team
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Noetherian modules
Let \(\displaystyle{M}\) be a left \(\displaystyle{R}\) - module and \(\displaystyle{f:M\to M}\) a
homomorphism.
1. If the module \(\displaystyle{M}\) is a \(\displaystyle{\rm{Noetherian}}\)- module and \(\displaystyle{f}\)
is onto \(\displaystyle{M}\), prove that \(\displaystyle{f}\) is isomorphism.
2. If \(\displaystyle{R}\) is a \(\displaystyle{\rm{Noetherian}}\) - ring and \(\displaystyle{a\,,b\in R}\)
such that \(\displaystyle{a\,b=1}\), then prove that \(\displaystyle{b\,a=1}\) .
homomorphism.
1. If the module \(\displaystyle{M}\) is a \(\displaystyle{\rm{Noetherian}}\)- module and \(\displaystyle{f}\)
is onto \(\displaystyle{M}\), prove that \(\displaystyle{f}\) is isomorphism.
2. If \(\displaystyle{R}\) is a \(\displaystyle{\rm{Noetherian}}\) - ring and \(\displaystyle{a\,,b\in R}\)
such that \(\displaystyle{a\,b=1}\), then prove that \(\displaystyle{b\,a=1}\) .
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- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Noetherian modules
Here is a solution :
1. It is sufficient to prove that \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\).
We have the increasing sequence of \(\displaystyle{R}\) - submodules of \(\displaystyle{M}\)
\(\displaystyle{\rm{Ker}(f)\subseteq \rm{Ker}(f^2)\subseteq ...}\).
Since \(\displaystyle{M}\) is a Noetherian module, we get
\(\displaystyle{\left(\exists\,n\in\mathbb{N}\right)\,\left(\forall\,k\in\mathbb{N}\right), k\geq n\implies\rm{Ker}(f^k)=\rm{Ker}(f^n)}\).
Also, since \(\displaystyle{f}\) is onto \(\displaystyle{M}\), we have that \(\displaystyle{f^n}\) is
onto \(\displaystyle{M}\)
Let \(\displaystyle{x\in\rm{Ker}(f)}\). Then, \(\displaystyle{x=f^n(y)}\) for some \(\displaystyle{y\in Y}\).
But, \(\displaystyle{0=f(x)=f^{n+1}(y)\implies y\in\rm{Ker}(f^{n+1})=\rm{Ker}(f^{n})}\) and then
\(\displaystyle{f^{n}(y)=0\implies x=0}\).
Therefore, \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\).
2. We deifne \(\displaystyle{f:R\to R\,,f(x)=a\,x}\).
For each \(\displaystyle{x\,,y\,,r\in R}\) holds
\(\displaystyle{f(x+r\,y)=(x+r\,y)\,a=x\,a+r\,(y\,a)=f(x)+r\,f(y)}\), which means that \(\displaystyle{f}\)
is \(\displaystyle{R}\) - homomorphism.
Also, if \(\displaystyle{y\in R}\), then \(\displaystyle{f(b\,y)=a\,b\,y=1\cdot y=y}\), so \(\displaystyle{f}\)
is onto \(\displaystyle{M}\).
Since \(\displaystyle{R}\) is a Noetherian \(\displaystyle{R}\) - module, the result 1.
gives us that \(\displaystyle{f}\) is an isomorphism. So,
\(\displaystyle{f(b\,a)=a\,b\,a=a=f(1)\implies b\,a=1}\).
1. It is sufficient to prove that \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\).
We have the increasing sequence of \(\displaystyle{R}\) - submodules of \(\displaystyle{M}\)
\(\displaystyle{\rm{Ker}(f)\subseteq \rm{Ker}(f^2)\subseteq ...}\).
Since \(\displaystyle{M}\) is a Noetherian module, we get
\(\displaystyle{\left(\exists\,n\in\mathbb{N}\right)\,\left(\forall\,k\in\mathbb{N}\right), k\geq n\implies\rm{Ker}(f^k)=\rm{Ker}(f^n)}\).
Also, since \(\displaystyle{f}\) is onto \(\displaystyle{M}\), we have that \(\displaystyle{f^n}\) is
onto \(\displaystyle{M}\)
Let \(\displaystyle{x\in\rm{Ker}(f)}\). Then, \(\displaystyle{x=f^n(y)}\) for some \(\displaystyle{y\in Y}\).
But, \(\displaystyle{0=f(x)=f^{n+1}(y)\implies y\in\rm{Ker}(f^{n+1})=\rm{Ker}(f^{n})}\) and then
\(\displaystyle{f^{n}(y)=0\implies x=0}\).
Therefore, \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\).
2. We deifne \(\displaystyle{f:R\to R\,,f(x)=a\,x}\).
For each \(\displaystyle{x\,,y\,,r\in R}\) holds
\(\displaystyle{f(x+r\,y)=(x+r\,y)\,a=x\,a+r\,(y\,a)=f(x)+r\,f(y)}\), which means that \(\displaystyle{f}\)
is \(\displaystyle{R}\) - homomorphism.
Also, if \(\displaystyle{y\in R}\), then \(\displaystyle{f(b\,y)=a\,b\,y=1\cdot y=y}\), so \(\displaystyle{f}\)
is onto \(\displaystyle{M}\).
Since \(\displaystyle{R}\) is a Noetherian \(\displaystyle{R}\) - module, the result 1.
gives us that \(\displaystyle{f}\) is an isomorphism. So,
\(\displaystyle{f(b\,a)=a\,b\,a=a=f(1)\implies b\,a=1}\).
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