Artinian Rings - Comments by P.M. Cohn, Ring Theory, page 66

[Math Amateur]

I am reading P.M. Cohn's book: Introduction to Ring Theory (Springer Undergraduate Mathematics Series) ... ...

I am currently focused on Section 2.3: Artinian Rings: The Semisimple Case

I need help with some comments made by Cohn in the introduction to Section 2.3 ...

The relevant comments by Cohn read as follows:https://www.physicsforums.com/attachments/4965In the above text, Cohn writes the following:

" ... ... It follows that every cyclic right \(\displaystyle R\)-module is Artinian, for any such module is of the form \(\displaystyle R/A\) for some right ideal \(\displaystyle A\). ... "


I do not understand how/why the above statement follows ... can someone help ... ?

In particular, why is any cyclic right \(\displaystyle R\)-module of the form \(\displaystyle R/A\) for some right ideal \(\displaystyle A\) ... and further, why does this fact make any cyclic right \(\displaystyle R\)-module Artinian ... ??

Hope someone can help ... ...

Peter

 

[Deveno]

Suppose $M$ is a cyclic right $R$-module. What this means is that if $m \in M$, then $m = xa$ for the generator $x$ of $M$, and some $a \in R$.

Our task now is to find some (right) ideal of $R$, to be the $A$ Cohn speaks of. To do so, it is first convenient to define a surjective right $R$-linear map: $R \to M$.

It is clear that the map $\phi: R \to M$ given by $a\phi = xa$ is surjective, so we verify $R$-linearity:

$(a + b)\phi = x(a + b) = xa + xb = a\phi + b\phi$

$(ar)\phi = x(ar) = (xa)r = (a\phi)r$, QED.

It follows that considering $R$ as a right $R$-module over itself, we have that $\text{ker }\phi$ (as the kernel of a right $R$-module homomorphism, that is, a (right) $R$-linear map) is an $R$-submodule of $R$, but the $R$-submodules of $R$ (considered as a right $R$-module) are *precisely* the right ideals of $R$.

Thus we may take $A = \text{ker }\phi$.

 

[Math Amateur]

Suppose $M$ is a cyclic right $R$-module. What this means is that if $m \in M$, then $m = xa$ for the generator $x$ of $M$, and some $a \in R$.

Our task now is to find some (right) ideal of $R$, to be the $A$ Cohn speaks of. To do so, it is first convenient to define a surjective right $R$-linear map: $R \to M$.

It is clear that the map $\phi: R \to M$ given by $a\phi = xa$ is surjective, so we verify $R$-linearity:

$(a + b)\phi = x(a + b) = xa + xb = a\phi + b\phi$

$(ar)\phi = x(ar) = (xa)r = (a\phi)r$, QED.

It follows that considering $R$ as a right $R$-module over itself, we have that $\text{ker }\phi$ (as the kernel of a right $R$-module homomorphism, that is, a (right) $R$-linear map) is an $R$-submodule of $R$, but the $R$-submodules of $R$ (considered as a right $R$-module) are *precisely* the right ideals of $R$.

Thus we may take $A = \text{ker }\phi$.

Hi Deveno,

Thanks for your help ... appreciate it ...

Just reflecting on your post ... ... thinking ... what does your logic demonstrate? ...Now ... it seems that what you have done implies that for a cyclic right \(\displaystyle R\)-module \(\displaystyle M\) we can define an epimorphism \(\displaystyle \phi\) such that

\(\displaystyle M \cong R/ \text{ker } \phi \)

by the First Isomorphism Theorem for Modules ... ... ... ... BUT ... ... I am still a bit perplexed ... where, in what you have said is it demonstrated or implied that \(\displaystyle M\) is Artinian ...

Perhaps it is obvious from what you have said ... but can you help further ...

Hope you can help ...

Peter

 

[Deveno]

$R$ is, by definition, a (right) Artinian ring. This means that any descending chain of right ideals stabilizes. Considered as a (right $R$-) MODULE, this means that any descending chain of submodules stabilizes.

If these submodules all contain the submodule (right ideal) $A$, they will *still* stabliize (because *any* means *any*, including descending chains of a certain type). But descending chains containing $A$ correspond to chains of $R/A$ (the correspondence theorem).

That is: $R$ (right) Artinian implies $R/A$ (right) Artinian, for any (right) ideal $A$.

But if $M$ is cyclic, $M \cong R/A$ for some right ideal $A$, so $M$ is an Artinian (right $R$-) module.

 

[Math Amateur]

$R$ is, by definition, a (right) Artinian ring. This means that any descending chain of right ideals stabilizes. Considered as a (right $R$-) MODULE, this means that any descending chain of submodules stabilizes.

If these submodules all contain the submodule (right ideal) $A$, they will *still* stabliize (because *any* means *any*, including descending chains of a certain type). But descending chains containing $A$ correspond to chains of $R/A$ (the correspondence theorem).

That is: $R$ (right) Artinian implies $R/A$ (right) Artinian, for any (right) ideal $A$.

But if $M$ is cyclic, $M \cong R/A$ for some right ideal $A$, so $M$ is an Artinian (right $R$-) module.

Thanks Deveno ... your post certainly clarifies the issue ...

Appreciate your help ...

Peter