MHB 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:View attachment 6122In 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 ...

Peter

 

[GJA]

Hi Peter,

This is an interesting question

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 ...

Peter

in that the rich structure of each entry being a field is simultaneously helpful and problematic. Ultimately solving this problem (at least using the approach I found) relies on using the extra structure to your advantage when you need it and finding a way to temper it when you don't. Perhaps the best fact to know is, as I mentioned in your other recent post from Bland, that every nonzero element of a field is a unit (and so every field is Artinian). I played with this example for a bit, so that's the best I can do in terms of giving the intuition since it only became clear from the other side of the looking glass.

The following is an effort to provide a hint or two without simply giving you the answers:

For the right Artinian case try looking at what the multiplication of any right ideal

$$\begin{pmatrix} A & B \\ 0 & D \end{pmatrix}
$$

($A\subseteq\mathbb{Q},\, B,D\subseteq\mathbb{R}$) by an element of the ring would look like combined with the fact that every nonzero element of a field is a unit and see what you can deduce.

For the non-left-Artinian case since each entry is a field, we need to "de"-field something at some point if there is to be any hope of finding an example of a nonterminating decreasing chain of ideals in $R$. One place to start looking is to things of the form $\mathbb{Q}+\mathbb{Q}\sqrt{n}.$ This object is no longer a field (in some cases). This isn't all that's needed for the final answer to the problem (you will need to modify/play with this to get what you want), but it will hopefully help give you some direction to move in.

 

[Math Amateur]

Hi Peter,

This is an interesting question
in that the rich structure of each entry being a field is simultaneously helpful and problematic. Ultimately solving this problem (at least using the approach I found) relies on using the extra structure to your advantage when you need it and finding a way to temper it when you don't. Perhaps the best fact to know is, as I mentioned in your other recent post from Bland, that every nonzero element of a field is a unit (and so every field is Artinian). I played with this example for a bit, so that's the best I can do in terms of giving the intuition since it only became clear from the other side of the looking glass.

The following is an effort to provide a hint or two without simply giving you the answers:

For the right Artinian case try looking at what the multiplication of any right ideal

$$\begin{pmatrix} A & B \\ 0 & D \end{pmatrix}
$$

($A\subseteq\mathbb{Q},\, B,D\subseteq\mathbb{R}$) by an element of the ring would look like combined with the fact that every nonzero element of a field is a unit and see what you can deduce.

For the non-left-Artinian case since each entry is a field, we need to "de"-field something at some point if there is to be any hope of finding an example of a nonterminating decreasing chain of ideals in $R$. One place to start looking is to things of the form $\mathbb{Q}+\mathbb{Q}\sqrt{n}.$ This object is no longer a field (in some cases). This isn't all that's needed for the final answer to the problem (you will need to modify/play with this to get what you want), but it will hopefully help give you some direction to move in.

Thanks GJA ... appreciate the help ...

Still reflecting on what you have said ...

Peter