Noetherian modules

[Papapetros Vaggelis]

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 \(\displaystyle{b\,a=1}\) .

[Papapetros Vaggelis]

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,

\(\displaystyle{f(b\,a)=a\,b\,a=a=f(1)\implies b\,a=1}\).