MHB Matrix rings acting on right R-modules - Dauns - Exercises 1-5 no 2

[Math Amateur]

In Exercise 2 of Exercises 1-5 of John Dauns' book "Modules and Rings" we are given

R = \begin{pmatrix} \mathbb{Z}_4 & \mathbb{Z}_4 \\ \mathbb{Z}_4 & \mathbb{Z}_4 \end{pmatrix} and M = \overline{2} \mathbb{Z}_4 \times \overline{2} \mathbb{Z}_4

where the matrix ring R acts on a right R-module whose elements are row vectors.

Find all submodules of M

Helped by Evgeny's post (See Math Help Boards) I commenced this problem as follows:

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Typical elements of R are R = \begin{pmatrix} r_1 & r_2 \\r_3 & r_4 \end{pmatrix}

where r_1, r_2, r_3, r_4 \in \mathbb{Z}_4 = \{ \overline{0}, \overline{1}, \overline{2}, \overline{3} \}

Typical elements of M are (x,y) where x, y \in \overline{2} Z_4

Now the elements of \overline{2} Z_4 = \overline{2} \times \{ \overline{0}, \overline{1}, \overline{2}, \overline{3} \} = \{ \overline{0}, \overline{2}, \overline{4}, \overline{6} \} = \{ \overline{0}, \overline{2}, \overline{0}, \overline{2} \} = \{ \overline{0}, \overline{2} \}

Now to find submodules! (Approach is by trial and error - but surely there is a better way!)

Consider a set of the form N_1 = \{ (x, 0) | \in \overline{2} |mathbb{Z}_4 - that is x \in \{ 0, 2 \}

Let r \in R and then test the action of R on M i.e. N_1 \times R \rightarrow N_1 - that is test if n_1r |in N_1

Now (x, 0) \begin{pmatrix} r_1 & r_2 \\r_3 & r_4 \end{pmatrix} = (r_1x, r_2x )

But now a problem I hope someone can help with!

How do we (rigorously) evaluate r_1x and r_2x and hence check whether (r_1x, r_2x) is of the form (x, 0) [certainly does not look like it but formally and rigorously ...?]

An example of my thinking here

If r_1 = \overline{3} and x = \overline{2} then (roughly speaking!) r_1 x = \overline{3} \overline{2} = \overline{6} = \overline{2}

In the above I am assuming that in \overline{2} \mathbb{Z}_4 that that \overline{0}, \overline{4}, \overline{8}, ... = \overline{0}

and that

that \overline{2}, \overline{6}, \overline{10}, ... = \overline{2}

but I am not sure what I am doing here!

Can someone please clarify this situation?

Further, can someone please comment on my overall approach to the Exercise - I am not at all sure regarding how to check for submodules and certainly lack a systematic approach ...

Be grateful for some help ...

Peter

 

[Evgeny.Makarov]

Where are all algebraists on this forum? Yes, algebra and logic (my specialty), somehow along with number theory, constitute a single "specialty" for the purpose of a Ph.D. thesis in Russia, but this is a relict from a former era.

Consider a set of the form N_1 = \{ (x, 0)\mid x \in \overline{2} \mathbb{Z}_4\} - that is x \in \{ 0, 2 \}

Let r \in R and then test the action of R on M i.e. N_1 \times R \rightarrow N_1 - that is test if n_1r |in N_1

Now (x, 0) \begin{pmatrix} r_1 & r_2 \\r_3 & r_4 \end{pmatrix} = (r_1x, r_2x )

If r_1 = \overline{3} and x = \overline{2} then (roughly speaking!) r_1 x = \overline{3} \overline{2} = \overline{6} = \overline{2}

I agree with your approach. If $r_2=3$ and $x=2$, then the second component of the resulting pair is $r_2x=2$. But we assumed that the submodule was $\{(0,0),(2,0)\}$. Therefore, it is not a submodule. Similarly, $\{(0,0),(0,2)\}$ is not a submodule, either. The only submodules I see are the trivial one and $M$ itself.

 

[Math Amateur]

Where are all algebraists on this forum? Yes, algebra and logic (my specialty), somehow along with number theory, constitute a single "specialty" for the purpose of a Ph.D. thesis in Russia, but this is a relict from a former era.
I agree with your approach. If $r_2=3$ and $x=2$, then the second component of the resulting pair is $r_2x=2$. But we assumed that the submodule was $\{(0,0),(2,0)\}$. Therefore, it is not a submodule. Similarly, $\{(0,0),(0,2)\}$ is not a submodule, either. The only submodules I see are the trivial one and $M$ itself.

Thanks Evgeny, really appreciate your help.Peter

 

[Deveno]

The first thing that pops out at me is that [math]M[/math] is finite, it only has 4 elements:

(0,0), (0,2), (2,0) and (2,2).

Such a finite set has only 16 possible subsets, and we can eliminate 8 of these straight-away because they do not include the additive identity (0,0).

Now, suppose we have a submodule [math]N[/math] containing (0,2). Taking:

[math]a = \begin{pmatrix}0&1\\1&0 \end{pmatrix} \in R[/math],

we find that [math](0,2)a = (2,0) \in N[/math] and thus:

[math](2,2) = (0,2) + (2,0) \in N[/math], hence [math]N = M[/math].

Similar reasoning shows that if a submodule contains (2,0) it likewise is all of [math]M[/math].

Finally, if a submodule contains (2,2), then taking:

[math]a' = \begin{pmatrix}0&0\\0&1 \end{pmatrix} \in R[/math],

we find (0,2) is in this submodule and thus (by above) it is all of [math]M[/math] as well.

So there are NO submodules (which are, after all, subsets) of cardinality 2, and also (by Lagrange, since submodules must also be abelian subgroups), there can be no submodules of cardinality 3.

This leaves just the trivial submodule, and all of [math]M[/math], and nothing remains to be shown, as these are trivially submodules.

 

[Math Amateur]

The first thing that pops out at me is that [math]M[/math] is finite, it only has 4 elements:

(0,0), (0,2), (2,0) and (2,2).

Such a finite set has only 16 possible subsets, and we can eliminate 8 of these straight-away because they do not include the additive identity (0,0).

Now, suppose we have a submodule [math]N[/math] containing (0,2). Taking:

[math]a = \begin{pmatrix}0&1\\1&0 \end{pmatrix} \in R[/math],

we find that [math](0,2)a = (2,0) \in N[/math] and thus:

[math](2,2) = (0,2) + (2,0) \in N[/math], hence [math]N = M[/math].

Similar reasoning shows that if a submodule contains (2,0) it likewise is all of [math]M[/math].

Finally, if a submodule contains (2,2), then taking:

[math]a' = \begin{pmatrix}0&0\\0&1 \end{pmatrix} \in R[/math],

we find (0,2) is in this submodule and thus (by above) it is all of [math]M[/math] as well.

So there are NO submodules (which are, after all, subsets) of cardinality 2, and also (by Lagrange, since submodules must also be abelian subgroups), there can be no submodules of cardinality 3.

This leaves just the trivial submodule, and all of [math]M[/math], and nothing remains to be shown, as these are trivially submodules.

Thanks Deveno ... most helpful ...

Peter