On Noetherian (Artinian) Modules

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

On Noetherian (Artinian) Modules

#1

Post by Tsakanikas Nickos »

Let \( \displaystyle R \) be a ring. Show that the following conditions are equivalent for a left \( \displaystyle R \)-module \( \displaystyle M \):
  1. \( \displaystyle M \) is a noetherian module.
  2. Every submodule of \( \displaystyle M \) is finitely generated.
  3. Every nonempty family of submodules of \( \displaystyle M \) contains a maximal element.
  4. Every submodule and every quotient module of \( \displaystyle M \) is a noetherian module.
Additionally, what is the dual version of these equivalences concerning artinian modules? State it and prove it.

Finally, show that
  1. any finite (direct) sum of artinian (noetherian) modules is artinian (noetherian) module as well.
  2. if an endomorphism of a noetherian (artinian) module is an epimorphism (monomorphism), then it is an isomorphism. (The converse is apparently true.)
  3. every left (right) semisimple ring is a left (right) artinian (and consequently noetherian) ring.
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: On Noetherian (Artinian) Modules

#2

Post by Papapetros Vaggelis »

\(\displaystyle{1)\implies 2)}\) : Suppose that \(\displaystyle{_{R}\,M}\) is a noetherian module.

Let \(\displaystyle{N}\) be a submodule of \(\displaystyle{_{R}\,M}\) . If \(\displaystyle{N=\left\{0\right\}}\), then the

submodule \(\displaystyle{N}\) is finitely generated. Let now \(\displaystyle{N\neq \left\{0\right\}}\). There is \(\displaystyle{x\in M}\)

such that \(\displaystyle{x_1\in N}\) and \(\displaystyle{x_1\neq 0}\) . If \(\displaystyle{R\,x_1=N}\), it's ok. If \(\displaystyle{R\,x_1\neq N}\), then

there is \(\displaystyle{x_2\in M}\) such that \(\displaystyle{x_2\in N}\) and \(\displaystyle{x_2\notin R\,x_1\,\,\,,R\,x_1+R\,x_2\subseteq N}\) .

If \(\displaystyle{R\,x_1+R\,x_2=N}\), then \(\displaystyle{N=<\left\{x_1,x_2\right\}>}\) .

On the other hand, if \(\displaystyle{R\,x_1+R\,x_2\neq N}\), then \(\displaystyle{x_3\in N\,,x_3\notin R\,x_1+R\,x_2}\)

for some \(\displaystyle{x_3\in M}\) . Following the same progress, we have an increasing sequence \(\displaystyle{\left(I_{n}\right)_{n\in\mathbb{N}}}\) , where

\(\displaystyle{I_{n}=\sum_{i=1}^{n}R\,x_{i}}\), of submodules of \(\displaystyle{_{R}\,M}\) . Since \(\displaystyle{_{R}\,M}\) is a noetherian module,

there is \(\displaystyle{k\in\mathbb{N}}\) such that \(\displaystyle{N=\sum_{i=1}^{k}I_{k}\implies N=<\left\{x_1,...,x_k\right\}>}\) .

\(\displaystyle{2)\implies 3)}\) : Let \(\displaystyle{S}\) be a nonempty family of submodules of \(\displaystyle{_{R}\,M}\) . Consider the

partial ordered set \(\displaystyle{\left(S,\subseteq\right)}\) and the result follows from \(\displaystyle{\rm{Zorn's}}\) lemma.

\(\displaystyle{3)\implies 1)}\) : Suppose that \(\displaystyle{\left(I_{n}\right)_{n\in\mathbb{N}}}\) is an increasing sequence of

submodules of \(\displaystyle{_{R}\,M}\). By setting \(\displaystyle{S=\left\{I_{n}:n\in\mathbb{N}\right\}}\) , we get that \(\displaystyle{S}\) is a

nonempty family of submodules of \(\displaystyle{_{R}\,M}\) and according to \(\displaystyle{3)}\) there is maximal element \(\displaystyle{I_{k}}\) .

If \(\displaystyle{n\geq k}\), then \(\displaystyle{I_{k}\subseteq I_{n}\implies I_{k}=I_{n}}\), so the left \(\displaystyle{R}\) - module \(\displaystyle{_{R}\,M}\)

is a noetherian module.

Therefore, \(\displaystyle{1)\iff 2)\iff 3)\iff 1)}\) . In order to complete this proof,

we'll prove that \(\displaystyle{1)\implies 4)}\) is true, cause the converse is apparently true.



\(\displaystyle{1)\implies 4)}\) : Suppose that \(\displaystyle{_{R}\,M}\) is a noetherian module. If \(\displaystyle{N}\) is a submodule of \(\displaystyle{_{R}\,M}\)

and \(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) is an increasing sequence of submodules of \(\displaystyle{N}\), then the sequence

\(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) is also an increasing sequence of the noetherian module \(\displaystyle{_{R}\,M}\), so :

there is \(\displaystyle{i\in\mathbb{N}}\) such that \(\displaystyle{K_{j}=K_{i}\,,\forall\,j\geq i}\) and thus \(\displaystyle{_{R}\,N}\) is a noetherian module too.

Now, let \(\displaystyle{N}\) be a submodule of \(\displaystyle{_{R}\,M}\) and let \(\displaystyle{_{R}\,\left(M/N\right)}\) be the quotient module.

Consider an increasing sequence \(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) of submodules of \(\displaystyle{_{R}\,\left(M/{N}\right)}\) .

Then, for each \(\displaystyle{n\in\mathbb{N}}\), there is a submodule \(\displaystyle{M_{n}}\) of \(\displaystyle{_{R}\,M}\) such that

\(\displaystyle{N\subseteq M_{n}\subseteq M}\) and \(\displaystyle{K_{n}=M_{n}/{N}}\) . Due to the fact that \(\displaystyle{K_{n}\subseteq K_{n+1}\,,\forall\,n\in\mathbb{N}}\),

we get \(\displaystyle{M_{n}\subseteq M_{n+1}\,,\forall\,n\in\mathbb{N}}\) and thus:

\(\displaystyle{\left(\exists\,i\in\mathbb{N}\right)\,\left(\forall\,j\in\mathbb{N}\right): j\geq i\implies M_{j}=M_{i}\iff K_{j}=K_{i}}\).

So, the module \(\displaystyle{_{R}\,\left(M/{N}\right)}\) is a noetherian module.

The dual version of these equivalences concerning artinian modules is :

For a left \(\displaystyle{R}\) - module \(\displaystyle{\left(M,+\right)}\) :

1. The module \(\displaystyle{_{R}\,M}\) is an artinian module.

2. Every quotient module is finitely cogenerated.

3. Every nonempty family of submodules of \(\displaystyle{_{R}\,M}\) contains a minimal element.
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