Indecomposable modules - example from Berrick and Keating

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

At present I am focussed on Chapter 2: Direct Sums and Short Exact Sequences.

Example 2.1.2 (i) on pages 38-39 reads as follows:https://www.physicsforums.com/attachments/2957
View attachment 2958In the above text B&K write:

" ... ... Clearly K is indecomposable as a module ... "

Question 1:

Can someone please explain exactly why this is the case? How can we demonstrate rigorously that K is indecomposable?

Question 2:

I have attempted to show that \(\displaystyle L \oplus N(x) = K^2\).

Can someone please critique my effort ... is it basically OK ...

Proof is as follows:

Let \(\displaystyle a \in L \oplus N(x)\)

Then \(\displaystyle a = l_1 + n_1 \)

\(\displaystyle = \left(\begin{array}{cc}1\\0\end{array}\right) k_1 + \left(\begin{array}{cc}x\\1\end{array}\right) k_2 \) where \(\displaystyle k_1, k_2 \in K\)

\(\displaystyle = \left(\begin{array}{cc}k_1\\0\end{array}\right) + \left(\begin{array}{cc}{k_2 x} \\k_2\end{array}\right)\)

\(\displaystyle = \left(\begin{array}{cc}{k_1 + k_2 x}\\k_2\end{array}\right) \in K^2 \)

Now let \(\displaystyle a \in K^2\); that is \(\displaystyle a = \left(\begin{array}{cc}{k_1}\\k_2\end{array}\right)\) for some \(\displaystyle k_1, k_2 \in K \)

Now take \(\displaystyle k_1 = c_1 + k_2 x \) where \(\displaystyle k_1, c_1, k_2\) and \(\displaystyle x \in K\). (This is permissible and possible since \(\displaystyle K\) is a field; \(\displaystyle c_1\), of course, may be negative)

Then \(\displaystyle a = \left(\begin{array}{cc}{c_1 + k_2 x }\\k_2\end{array}\right)\)

Therefore \(\displaystyle a = \left(\begin{array}{cc}{c_1 }\\0\end{array}\right) + \left(\begin{array}{cc}{ k_2 x }\\k_2\end{array}\right) \)

Therefore \(\displaystyle a = \left(\begin{array}{cc}{1 }\\0\end{array}\right) c_1 + \left(\begin{array}{cc}{ x }\\1\end{array}\right) k_2 \in L \oplus N(x) \)

Thus we conclude that \(\displaystyle L \oplus N(x) = K^2\).

Can someone please confirm that this is OK ... or alternatively amend/critique the argument ...

Hope someone can help.

Peter

Notes: B&K definitions and notation

B&K's definition of the internal direct sum is as follows:

View attachment 2959
View attachment 2960

B&K then point out that the definition of internal direct sum can be restated as follows:

https://www.physicsforums.com/attachments/2961

Finally, just before the example above, B&K define decomposable module, complement and indecomposable module as follows:

https://www.physicsforums.com/attachments/2962

 

[Deveno]

Personally, I would write:

Let $c = k_1 - k_2x$, since $k_1,k_2,x$ are given.

You should recognize this as essentially the same example I gave in another thread, and expresses the same idea: direct sum decompositions are not unique.

So what one looks for, is "what invariants" can we find in direct sum decompositions. In this case, we have "essentially two kinds" of direct sum decompositons, corresponding to the following partitions of 2 (the dimension of $\mathcal{K}^2$):

2 = 2 + 0 <---this doesn't represent a "true" decomposition, one factor is trivial
2 = 1 + 1

Either one summand is a trivial subspace, or we have two subspaces of dimension one.

The condition (DS2) in this limited example, means that:

$\begin{pmatrix}k_1\\k_2 \end{pmatrix} \in L \cap N(x) \implies k_1 = k_2 = 0$.

That we must have $k_2 = 0$ is easy to see from membership in $L$. But if:

$\begin{pmatrix}k_1\\0 \end{pmatrix} = \begin{pmatrix}cx\\c\end{pmatrix}$

it is evident that $c = 0$, so $k_1 = 0x = 0$.

I point this out because you have shown that $\mathcal{K}^2 = L + N(x)$, and not that the sum is direct (it is not merely sufficient to show we can form any element of $\mathcal{K}^2$ as a sum of an element in $L$ and an element of $N(x)$, we must show this sum is uniquely determined).

Other than that, what you have is OK.

To answer your first question:

Suppose that $\mathcal{K} = M \oplus N$. Assume $M$ is non-trivial (non-zero), so we have $m \neq 0 \in M$.

Now $m \in \mathcal{K}$ since $M$ is a sub-module of $\mathcal{K}$. Since $\mathcal{K}$ is a field, we have:

$\dfrac{1}{m} \in \mathcal{K}$, and thus $\dfrac{1}{m}\cdot m = 1_{\mathcal{K}} \in M$ (since $M$ is a $\mathcal{K}$-module).

It thus follows that for any $k \in \mathcal{K}$, we have $k = k\cdot 1_{\mathcal{K}} \in M$.

Hence $M = \mathcal{K}$.

Since our sum is direct: $M \cap N = \{0_{\mathcal{K}}\} = \mathcal{K} \cap N = N$, so that $N$ is the 0-submodule.

 

[Math Amateur]

Personally, I would write:

Let $c = k_1 - k_2x$, since $k_1,k_2,x$ are given.

You should recognize this as essentially the same example I gave in another thread, and expresses the same idea: direct sum decompositions are not unique.

So what one looks for, is "what invariants" can we find in direct sum decompositions. In this case, we have "essentially two kinds" of direct sum decompositons, corresponding to the following partitions of 2 (the dimension of $\mathcal{K}^2$):

2 = 2 + 0 <---this doesn't represent a "true" decomposition, one factor is trivial
2 = 1 + 1

Either one summand is a trivial subspace, or we have two subspaces of dimension one.

The condition (DS2) in this limited example, means that:

$\begin{pmatrix}k_1\\k_2 \end{pmatrix} \in L \cap N(x) \implies k_1 = k_2 = 0$.

That we must have $k_2 = 0$ is easy to see from membership in $L$. But if:

$\begin{pmatrix}k_1\\0 \end{pmatrix} = \begin{pmatrix}cx\\c\end{pmatrix}$

it is evident that $c = 0$, so $k_1 = 0x = 0$.

I point this out because you have shown that $\mathcal{K}^2 = L + N(x)$, and not that the sum is direct (it is not merely sufficient to show we can form any element of $\mathcal{K}^2$ as a sum of an element in $L$ and an element of $N(x)$, we must show this sum is uniquely determined).

Other than that, what you have is OK.

To answer your first question:

Suppose that $\mathcal{K} = M \oplus N$. Assume $M$ is non-trivial (non-zero), so we have $m \neq 0 \in M$.

Now $m \in \mathcal{K}$ since $M$ is a sub-module of $\mathcal{K}$. Since $\mathcal{K}$ is a field, we have:

$\dfrac{1}{m} \in \mathcal{K}$, and thus $\dfrac{1}{m}\cdot m = 1_{\mathcal{K}} \in M$ (since $M$ is a $\mathcal{K}$-module).

It thus follows that for any $k \in \mathcal{K}$, we have $k = k\cdot 1_{\mathcal{K}} \in M$.

Hence $M = \mathcal{K}$.

Since our sum is direct: $M \cap N = \{0_{\mathcal{K}}\} = \mathcal{K} \cap N = N$, so that $N$ is the 0-submodule.

Thanks Deveno ... very helpful ...

You write:

" ... ... You should recognize this as essentially the same example I gave in another thread, and expresses the same idea: direct sum decompositions are not unique. ... ..."

Indeed ... how myopic of me! ... yes, see that now ... it is a great example as it does show clearly that direct sums are not unique! ...

Now just a clarification:

You write:

" ... ... The condition (DS2) in this limited example, means that:

$\begin{pmatrix}k_1\\k_2 \end{pmatrix} \in L \cap N(x) \implies k_1 = k_2 = 0$.

That we must have $k_2 = 0$ is easy to see from membership in $L$. But if:

$\begin{pmatrix}k_1\\0 \end{pmatrix} = \begin{pmatrix}cx\\c\end{pmatrix}$

it is evident that $c = 0$, so $k_1 = 0x = 0$.

I point this out because you have shown that $\mathcal{K}^2 = L + N(x)$, and not that the sum is direct (it is not merely sufficient to show we can form any element of $\mathcal{K}^2$ as a sum of an element in $L$ and an element of $N(x)$, we must show this sum is uniquely determined). .. ... "

... well yes ... really missed a very important point there!

BUT ... you write as if ...

" ... ...
$\begin{pmatrix}k_1\\k_2 \end{pmatrix} \in L \cap N(x) \implies k_1 = k_2 = 0$ means that the sum of an element in L and an element in N(x) is uniquely determined ... but why exactly does this follow ...?

Can you help?

Peter

 

[Deveno]

Ok, suppose we have for a vector space:

$V = \langle U,W\rangle = U+W = \{u+w: u\in U,w \in W\}$.

If $u+w = u'+w'$ with $u \neq u'$ or $w \neq w'$ (so the sum is not unique), we have:

$u - u' = w' - w$.

Since the LHS in in $U$, and the RHS is in $W$, and at least one side is not 0 (so both sides must be), we have:

$u - u' = w - w' \in U \cap W \neq \{0\}$.

On the other hand, if $U \cap W = \{0\}$, then for any two sums for which:

$u + w = u' + w'$, we have $u - u' = w - w' = 0$, so that $u = u'$ and $w = w'$.

I think this was addressed in another thread.

 

[Math Amateur]

Ok, suppose we have for a vector space:

$V = \langle U,W\rangle = U+W = \{u+w: u\in U,w \in W\}$.

If $u+w = u'+w'$ with $u \neq u'$ or $w \neq w'$ (so the sum is not unique), we have:

$u - u' = w' - w$.

Since the LHS in in $U$, and the RHS is in $W$, and at least one side is not 0 (so both sides must be), we have:

$u - u' = w - w' \in U \cap W \neq \{0\}$.

On the other hand, if $U \cap W = \{0\}$, then for any two sums for which:

$u + w = u' + w'$, we have $u - u' = w - w' = 0$, so that $u = u'$ and $w = w'$.

I think this was addressed in another thread.

Thanks Deveno ... your help has been extensive and is much appreciated ... just now working through all posts again! ... thanks to you and Euge, I now have a much clearer picture of direct sums and products ... but do need to review your posts again ...

Thanks so much ...

Peter