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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 $$L \oplus N(x) = K^2$$.
Can someone please critique my effort ... is it basically OK ...
Proof is as follows:
Let $$a \in L \oplus N(x)$$
Then $$ a = l_1 + n_1 $$
$$= \left(\begin{array}{cc}1\\0\end{array}\right) k_1 + \left(\begin{array}{cc}x\\1\end{array}\right) k_2 $$ where $$k_1, k_2 \in K$$
$$= \left(\begin{array}{cc}k_1\\0\end{array}\right) + \left(\begin{array}{cc}{k_2 x} \\k_2\end{array}\right)$$
$$= \left(\begin{array}{cc}{k_1 + k_2 x}\\k_2\end{array}\right) \in K^2 $$
Now let $$a \in K^2$$; that is $$a = \left(\begin{array}{cc}{k_1}\\k_2\end{array}\right)$$ for some $$ k_1, k_2 \in K $$
Now take $$k_1 = c_1 + k_2 x $$ where $$k_1, c_1, k_2$$ and $$x \in K$$. (This is permissible and possible since $$K$$ is a field; $$c_1$$, of course, may be negative)
Then $$a = \left(\begin{array}{cc}{c_1 + k_2 x }\\k_2\end{array}\right)$$
Therefore $$a = \left(\begin{array}{cc}{c_1 }\\0\end{array}\right) + \left(\begin{array}{cc}{ k_2 x }\\k_2\end{array}\right) $$
Therefore $$ a = \left(\begin{array}{cc}{1 }\\0\end{array}\right) c_1 + \left(\begin{array}{cc}{ x }\\1\end{array}\right) k_2 \in L \oplus N(x) $$
Thus we conclude that $$L \oplus N(x) = K^2$$.
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
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 $$L \oplus N(x) = K^2$$.
Can someone please critique my effort ... is it basically OK ...
Proof is as follows:
Let $$a \in L \oplus N(x)$$
Then $$ a = l_1 + n_1 $$
$$= \left(\begin{array}{cc}1\\0\end{array}\right) k_1 + \left(\begin{array}{cc}x\\1\end{array}\right) k_2 $$ where $$k_1, k_2 \in K$$
$$= \left(\begin{array}{cc}k_1\\0\end{array}\right) + \left(\begin{array}{cc}{k_2 x} \\k_2\end{array}\right)$$
$$= \left(\begin{array}{cc}{k_1 + k_2 x}\\k_2\end{array}\right) \in K^2 $$
Now let $$a \in K^2$$; that is $$a = \left(\begin{array}{cc}{k_1}\\k_2\end{array}\right)$$ for some $$ k_1, k_2 \in K $$
Now take $$k_1 = c_1 + k_2 x $$ where $$k_1, c_1, k_2$$ and $$x \in K$$. (This is permissible and possible since $$K$$ is a field; $$c_1$$, of course, may be negative)
Then $$a = \left(\begin{array}{cc}{c_1 + k_2 x }\\k_2\end{array}\right)$$
Therefore $$a = \left(\begin{array}{cc}{c_1 }\\0\end{array}\right) + \left(\begin{array}{cc}{ k_2 x }\\k_2\end{array}\right) $$
Therefore $$ a = \left(\begin{array}{cc}{1 }\\0\end{array}\right) c_1 + \left(\begin{array}{cc}{ x }\\1\end{array}\right) k_2 \in L \oplus N(x) $$
Thus we conclude that $$L \oplus N(x) = K^2$$.
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
Last edited: