Modules of Finite Length - Cohn, page 61

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks Deveno ... just reading your post carefully and studying and reflecting on what you have written ...

Peter

 

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks for the help, Deveno ... but ... just a simple clarification ...

You write:

" ... ... Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules). ... ... "
Can you please explain exactly why $$N$$ is simple (i.e. why 1-dimensional subspaces of a vector space are simple $F$-modules)?
Peter

 

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks Deveno ...

Now you write:

" ... ... Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$? ... ... "I need some help with this exercise ... especially with the issue of showing that modules are simple (that is that they contain no proper non-trivial submodules) ... ...

Thoughts so far are as follows:

We are asked:

" ... ... Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$ ... ... "

so then we have that

$$ U = \langle v + N\rangle $$

$$= \{ f ( v + N ) \ | \ f \in F \} $$

$$= \{ f v + N ) \ | \ f \in F \} $$
... and then we are asked to show that:

$$\{ 0 \} = N/N \subset U \subset M/N $$

is a composition series for $$M/N$$So we need to show that the composition factors $$C_i / C_{i - 1}$$ are simple for

$$C_1 \subset C_2 \subset C_3 $$ $$= \{ 0 \} = N/N \subset U \subset M/N$$That is, we have to show that:

(1) $$(U)/(N/N) = U/\{ 0 \} = U$$ is simple

and

(2) $$(M/N) / U = (M/N)/ (L/N) $$
BUT ... for (2) above we have that

$$(M/N)/ (L/N) \cong M/L$$

so, essentially, we have to show $$M/L$$ is simple
But ... how to show $$U$$ and $$M/L$$ are simple ? ... ...

Can you please help?Peter

***EDIT***

You asked about the dimension of $$L$$ ... ...

Since $$N \subset L \subset M$$

we have that:

$$ \text{ dim } N \lt \text{ dim } L \lt \text{ dim } M$$

BUT ... $$\text{ dim } N = 1$$ and $$\text{ dim } M = 3$$

So ... $$\text{ dim } L = 2$$

 

[Deveno]

Let's answer the question:

1) Is a one-dimensional subspace $U$ of a vector space $V$ over a field $F$, a simple $F$-module?

Suppose we have, for the sake of argument:

$\{0_V\} \subset W \subseteq U$.

Since $U$ is one-dimensional, any basis for $U$ is of the form $\{u_1\}$.

This means, given any $u \in U$, there is some $\alpha \in F$ with $u = \alpha u_1$, and moreover, $\alpha$ is unique.

For $\{0_V\} \subset W$ (that is, for this containment to be PROPER), we must have $w \neq 0_V \in W$.

Since $W \subseteq U$, we have $w \in U$. So $w = \beta u_1$.

Since $u_1$ is a basis element, $u_1 \neq 0_V$. If $\beta = 0_F$, we have $w = 0$, contradicting our choice of $w$.

Since $F$ is a field, $\beta^{-1} \in F$, and we have, for any $u \in U$:

$u = \alpha u_1 = [\alpha(\beta^{-1}\beta)]u_1 = (\alpha\beta^{-1})(\beta u_1) = (\alpha\beta^{-1})w \in W$.

This shows that ANY non-zero subspace of $U$ is $U$ itself, that is: $U$ is simple, as it has no non-trivial proper subspaces.

Now if $M$ has dimension 3, and $L$ has dimension 2, $M/L$ has dimension 1. From the above, $M/L$ is simple. This means that:

$N/N \subset L/N \subset M/N$

is a composition series for $M/N$, since:

$(M/N)/(L/N) \cong M/L$ (if $M/L$ had a non-trivial proper subspace, so would $(M/N)/(L/N)$).

 

[Math Amateur]

Let's answer the question:

1) Is a one-dimensional subspace $U$ of a vector space $V$ over a field $F$, a simple $F$-module?

Suppose we have, for the sake of argument:

$\{0_V\} \subset W \subseteq U$.

Since $U$ is one-dimensional, any basis for $U$ is of the form $\{u_1\}$.

This means, given any $u \in U$, there is some $\alpha \in F$ with $u = \alpha u_1$, and moreover, $\alpha$ is unique.

For $\{0_V\} \subset W$ (that is, for this containment to be PROPER), we must have $w \neq 0_V \in W$.

Since $W \subseteq U$, we have $w \in U$. So $w = \beta u_1$.

Since $u_1$ is a basis element, $u_1 \neq 0_V$. If $\beta = 0_F$, we have $w = 0$, contradicting our choice of $w$.

Since $F$ is a field, $\beta^{-1} \in F$, and we have, for any $u \in U$:

$u = \alpha u_1 = [\alpha(\beta^{-1}\beta)]u_1 = (\alpha\beta^{-1})(\beta u_1) = (\alpha\beta^{-1})w \in W$.

This shows that ANY non-zero subspace of $U$ is $U$ itself, that is: $U$ is simple, as it has no non-trivial proper subspaces.

Now if $M$ has dimension 3, and $L$ has dimension 2, $M/L$ has dimension 1. From the above, $M/L$ is simple. This means that:

$N/N \subset L/N \subset M/N$

is a composition series for $M/N$, since:

$(M/N)/(L/N) \cong M/L$ (if $M/L$ had a non-trivial proper subspace, so would $(M/N)/(L/N)$).

Thanks for your help Deveno ... just working carefully through your post now ...

Peter