I Example from Bland - Right Artinian but not Left Artinian ...

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand Example 6 on page 109 ... ...

Example 6 reads as follows:

In the above example Bland asserts that the matrix ring$\begin{pmatrix} \mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R} \end{pmatrix}$

is right Artinian but not left Artinian ...Can someone please help me to prove this assertion ...My thoughts on how to do this are limited ... but include reasoning from the fact that the entries in the matrix ring are all fields and thus the only ideals are $\{ 0 \}$ and the whole ring(field) ... and so the chains of such ideals should terminate ... but this seems to imply that the matrix ring is both left and right Artinian ...

Hope someone can help ...Peter

 

[fresh_42]

My thoughts on how to do this are limited ... but include reasoning from the fact that the entries in the matrix ring are all fields and thus the only ideals are $\{ 0 \}$ and the whole ring(field) ... and so the chains of such ideals should terminate ... but this seems to imply that the matrix ring is both left and right Artinian ...

Let me tell you, how I tackled the problem.

Firstly, what is right Artinian, and what left Artinian?
Artinian means to satisfy the descending chain condition (DCC), i.e. $M \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_n$ must become stable, that is $S_n=S_{n+1}$ for some $n$.
So the task is:

1. Find such a chain, that does not satisfy the DCC if considered as left modules.
2. Show that all chains satisfy the DCC if considered as right modules.

Since $R = \begin{bmatrix}\mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R}\end{bmatrix}$ our modules are of the form $M=\begin{bmatrix} U & V \\ 0 & W \end{bmatrix}$ with $U\subseteq \mathbb{Q}$ and $V,W \subseteq \mathbb{R}$.

Next I multiplied $R\cdot M$ and $M \cdot R$ to find out the difference.

We have three "submodules" $U,V,W$ to choose freely in the case of $M$ being a left module, since one counterexample already does the job. Keep it easy! This means, choose the zero module $\{0\}$ where possible, i.e. where there is no difference between right and left, and concentrate on the rest. As in the example of left Noetherian and right Noetherian, multiples of $2^n$ or $2^{-n}$ could be helpful.
Now 1. can be solved.

Remains 2. Here we have to deal with arbitrary $U,V,W$ and the multiplication $M\cdot R$ from the right hopefully already proves that any chain $M \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_n$ stabilizes pretty fast, e.g. by showing $S_1 \cdot R = M$.

$S_1 \cdot R = M$ means $S_1 \cdot R \supseteq M$ has to be shown, since $S_1 \cdot R \subseteq S_1 \subseteq M$ is clear by the definition of right modules.

Edit: I find it easier to reserve the term ideal for two sided ideals and talk of left and right modules instead of left and right ideals, but this is a spleen. You may substitute the word module by ideal in the above. I'm used to say $R$ is Artinian (left, right or both), iff it is as $R$-(left, right or both) module of itself.

 

[Math Amateur]

Let me tell you, how I tackled the problem.

Firstly, what is right Artinian, and what left Artinian?
Artinian means to satisfy the descending chain condition (DCC), i.e. $M \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_n$ must become stable, that is $S_n=S_{n+1}$ for some $n$.
So the task is:

1. Find such a chain, that does not satisfy the DCC if considered as left modules.
2. Show that all chains satisfy the DCC if considered as right modules.

Since $R = \begin{bmatrix}\mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R}\end{bmatrix}$ our modules are of the form $M=\begin{bmatrix} U & V \\ 0 & W \end{bmatrix}$ with $U\subseteq \mathbb{Q}$ and $V,W \subseteq \mathbb{R}$.

Next I multiplied $R\cdot M$ and $M \cdot R$ to find out the difference.

We have three "submodules" $U,V,W$ to choose freely in the case of $M$ being a left module, since one counterexample already does the job. Keep it easy! This means, choose the zero module $\{0\}$ where possible, i.e. where there is no difference between right and left, and concentrate on the rest. As in the example of left Noetherian and right Noetherian, multiples of $2^n$ or $2^{-n}$ could be helpful.
Now 1. can be solved.

Remains 2. Here we have to deal with arbitrary $U,V,W$ and the multiplication $M\cdot R$ from the right hopefully already proves that any chain $M \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_n$ stabilizes pretty fast, e.g. by showing $S_1 \cdot R = M$.

$S_1 \cdot R = M$ means $S_1 \cdot R \supseteq M$ has to be shown, since $S_1 \cdot R \subseteq S_1 \subseteq M$ is clear by the definition of right modules.

Edit: I find it easier to reserve the term ideal for two sided ideals and talk of left and right modules instead of left and right ideals, but this is a spleen. You may substitute the word module by ideal in the above. I'm used to say $R$ is Artinian (left, right or both), iff it is as $R$-(left, right or both) module of itself.

Thanks fresh_42 ... your help is much appreciated...

Working through your post and reflecting ...

Peter