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Matrix rings acting on right R-modules

Peter

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MHB Site Helper
Jun 22, 2012
2,918
In the book "Modules and Rings" by John Dauns he adopts the following notation in Exercises 1-5 (see attached)

"For additive groups A, B, C, D and [TEX] a \in A, b \in B, c \in C, d \in D [/TEX] we write

[TEX] [a,b; c,d] = \begin {vmatrix} a & b \\ c & d \end {vmatrix} [/TEX] ; [TEX] [A, B, C, D] = \begin {vmatrix} A & B \\ C & D \end {vmatrix} [/TEX] "


Exercise 1 (i), (ii) and (iii)

The matrix ring R acts on the right R-module M whose elements are row vectors:

Find (i) all submodules of M (ii) all right ideals of R, indicating which of these are ideals


[TEX] R = [ \mathbb{Z}_2, \mathbb{Z}_2; 0, \mathbb{Z}_2, ] , M = \mathbb{Z}_2 \oplus \mathbb{Z}_2, [/TEX]

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I am somewhat overwhelmed by this problem and, at the very least, need some help to get started

Peter

This has also been posted on MHF

Peter
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,492
You seem to have forgotten to join the attachment.

First, the notation $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is just a matrix, not a determinant. Next, $\begin{vmatrix} A & B \\ C & D \end{vmatrix}$ is a set consisting of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ where $a\in A$, …, $d\in D$.

So, if $M=\mathbb{Z}_2\oplus\mathbb{Z}_2$, then elements of $M$ have the form of a row vector $(x_1,x_2)$ where $x_i\in \mathbb{Z}_2$ for $i=1,2$. Such vectors are multiplied on the right by matrices from $[\mathbb{Z}_2,\mathbb{Z}_2;0,\mathbb{Z}_2]$. Such matrices have the form $\begin{vmatrix} a & b \\ 0 & d \end{vmatrix}$ where $a,b,d\in\mathbb{Z}_2$. Then $\{(0,x)\mid x\in\mathbb{Z}_2\}$ seems to be a submodule because it is a subgroup of $\mathbb{Z}_2\oplus\mathbb{Z}_2$ and multiplying such vectors by matrices produces vectors of the same shape. The trivial subgroup $\{(0,0)\}$ is also a submodule.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
You seem to have forgotten to join the attachment.

First, the notation $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is just a matrix, not a determinant. Next, $\begin{vmatrix} A & B \\ C & D \end{vmatrix}$ is a set consisting of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ where $a\in A$, …, $d\in D$.

So, if $M=\mathbb{Z}_2\oplus\mathbb{Z}_2$, then elements of $M$ have the form of a row vector $(x_1,x_2)$ where $x_i\in \mathbb{Z}_2$ for $i=1,2$. Such vectors are multiplied on the right by matrices from $[\mathbb{Z}_2,\mathbb{Z}_2;0,\mathbb{Z}_2]$. Such matrices have the form $\begin{vmatrix} a & b \\ 0 & d \end{vmatrix}$ where $a,b,d\in\mathbb{Z}_2$. Then $\{(0,x)\mid x\in\mathbb{Z}_2\}$ seems to be a submodule because it is a subgroup of $\mathbb{Z}_2\oplus\mathbb{Z}_2$ and multiplying such vectors by matrices produces vectors of the same shape. The trivial subgroup $\{(0,0)\}$ is also a submodule.
Thanks Evgeny, just working through your post and thinking about the exercie

By the way, sorry about the attachment - it is now attached.

Peter
 

Peter

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MHB Site Helper
Jun 22, 2012
2,918
Thanks Evgeny, just working through your post and thinking about the exercie

By the way, sorry about the attachment - it is now attached.

Peter
Can anyone help with part two of this problem - that is how does one go about finding all of the right ideals of R

Peter
 
Last edited:

Deveno

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MHB Math Scholar
Feb 15, 2012
1,967
One of the things I always do in situations like this, is set boundaries for "how bad the problem could be". Here, we are lucky, \(\displaystyle R\) is FINITE, it only has 8 elements. So, at worst, we are looking at 256 possible subsets. Furthermore, since an ideal is an additive subgroup, we only need to consider subsets of cardinalities 1,2 and 4. We have:

8 subsets of cardinality 1,
28 subsets of cardinality 2,
70 subsets of cardinality 4.

So we are, at worst, looking at 106 possible subsets, now.

There is clearly just one possible additive subgroup of order 1, the trivial subgroup (0 matrix), which is also clearly a right ideal. We can disregard the other 7 possible one-element subsets.

We only need consider 2-element subsets that include the identity, this gives us 7 to look at.

Similarly, we only need to consider 4-element subsets that also contain the identity, this gives us just 35 to look at.

So we are down to considering only 43 subsets, one of which is trivial, so really only 42.

Clearly, all 7 2-element subsets that we are considering form abelian groups of order 2.

A little reflection shows that all the possible 4-element abelian subgroups must be isomorphic to the Klein 4-group (every element of \(\displaystyle R\) is of order 2), and that we can do this by specifying any 2 non-identity elements. This means that having chosen TWO non-identity elements, our third choice is forced upon us.

Now there are 21 ways to pick 2 non-identity elements...BUT this leads to a 3-fold duplication (if we choose {a,b} and are forced to add c as well, then choosing {a,c} or {b,c} leads to the same subgroup). So we really have only 14 distinct additive subgroups to check out, 7 of order 2, and 7 of order 4. We could actually LIST these, if we had the patience.

But let's take the lazy road. Suppose:

\(\displaystyle A = \begin{pmatrix}a&b\\0&c \end{pmatrix}\) is of order 2. For this to be a right ideal, we require that:

\(\displaystyle AB = A\) or \(\displaystyle AB = 0\) for all \(\displaystyle B \in R\).

Writing:

\(\displaystyle B = \begin{pmatrix}x&y\\z&w \end{pmatrix}\)

\(\displaystyle AB = A\) leads to:

ax + bz = a
cz = 0
ay + bw = b
cw = c

If c is non-zero, this forces z = 0, which is not always the case. Similarly, if \(\displaystyle AB = 0\), we must also have z = 0 for non-zero c. This eliminates 4 of our 2-element subgroups, so we need only consider 3.

Now consider the element of \(\displaystyle R\):

\(\displaystyle B = \begin{pmatrix}0&1\\1&0 \end{pmatrix}\). This swaps the columns of \(\displaystyle A\), which means the only possible 2-element subgroup of \(\displaystyle R\) which might still be a right ideal is:

\(\displaystyle \{ \begin{pmatrix}0&0\\0&0 \end{pmatrix}, \begin{pmatrix}1&1\\0&0 \end{pmatrix} \}\)

However, it's easy to see that:

\(\displaystyle \begin{pmatrix}1&1\\0&0 \end{pmatrix}\begin{pmatrix}0&0\\0&1 \end{pmatrix} = \begin{pmatrix}0&1\\0&0 \end{pmatrix}\)

so there are NO 2-element right ideals. We're now down to checking just 7 subsets.

Using the "column-swapping" matrix we used above, we see that if:

\(\displaystyle \begin{pmatrix}a&b\\0&c \end{pmatrix} \in R\), then we must have:

\(\displaystyle \begin{pmatrix}b&a\\c&0 \end{pmatrix} \in R\).

Thus c = 0, which leaves just ONE of our 7 4-element subgroups as a possibility:

\(\displaystyle \{ \begin{pmatrix}0&0\\0&0 \end{pmatrix}, \begin{pmatrix}1&0\\0&0 \end{pmatrix}, \begin{pmatrix}0&1\\0&0 \end{pmatrix}, \begin{pmatrix}1&1\\0&0 \end{pmatrix} \}\)

I leave it to you to verify this is indeed a right ideal.
 

Peter

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MHB Site Helper
Jun 22, 2012
2,918
Deveno,

Thank you for an exceedingly helpful post ... this takes my study of ring and module theory on quite a way.

Thanks also for explaining your approach to the problem ... that provides a real learning experience for me and other members of Math Help Boards Linear and Abstract Algebra forum.

I am at my day job now, but will work through your post in detail this evening - Australian Eastern Time

Thank you again

Peter