Direct Sums of Copies of Modules - B&K - Exercise 2.1.6 (iii)

[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).

In Chapter2: Direct Sums and Short Exact Sequences we find Exercise 2.1.6 part (iii). I need some help to get started on this exercise.

Exercise 2.1.6 reads as follows:View attachment 3365
View attachment 3366
I am attempting to get started on part (iii) but need help.

In particular, could someone help me interpret \(\displaystyle M_k ( \text{End} (L) ) \) ... what entity is this? Looks like a matrix (or module?) multiplied by \(\displaystyle \text{End} (L) \) written as a matrix?

Are we to assume that as well as \(\displaystyle M\) being equal to \(\displaystyle L^{\Lambda}\) we also have that

\(\displaystyle M = M_1 \oplus M_2 \oplus M_3 \oplus \ ... \ ... \ \oplus M_k \oplus \ ... \ ... \ \oplus M_n\)

... ... and that \(\displaystyle M_k\) is the \(\displaystyle k\)th element of the direct sum written as a matrix?

Can someone please help clarify the notation and also get me started on the exercise.

Peter

***NOTE***

The exercise mentions idempotents as well as the full set of inclusions and projections. Thus, to ensure that MHB members can appreciate the context of the above post I am providing the relevant text on these topics from B&K.

The relevant text from B&K on standard inclusions and projects is as follows:View attachment 3367
https://www.physicsforums.com/attachments/3368
The relevant text from B&K on idempotents is as follows:
https://www.physicsforums.com/attachments/3369
https://www.physicsforums.com/attachments/3370

 

[Deveno]

To do (iii) I suggest tackling the $k = 2$ case, first, and applying the results of (i).

Can you think of a way to use induction on $k$ and the fact that:

$L^k \cong L^{k-1} \oplus L$?

(in other words, direct sums are associative "up to isomorphism").

The matrices detailed in this exercise are often called "block matrices", in which matrix "blocks" are treated as if they were matrix "entries" (this makes sense as long as the blocks are of "matching sizes" for the requisite multiplication).

 

[Math Amateur]

To do (iii) I suggest tackling the $k = 2$ case, first, and applying the results of (i).

Can you think of a way to use induction on $k$ and the fact that:

$L^k \cong L^{k-1} \oplus L$?

(in other words, direct sums are associative "up to isomorphism").

The matrices detailed in this exercise are often called "block matrices", in which matrix "blocks" are treated as if they were matrix "entries" (this makes sense as long as the blocks are of "matching sizes" for the requisite multiplication).

Thanks, I will try that ... BUT ... can you please clarify the meaning of \(\displaystyle M_k ( \text{End} (L) ) \) ... ...

Thanks again,

Peter

 

[Deveno]

Thanks, I will try that ... BUT ... can you please clarify the meaning of \(\displaystyle M_k ( \text{End} (L) ) \) ... ...

Thanks again,

Peter

I'm not familiar with your text, but it should be the same as $M_n(R)$ where instead of ring-elements of $R$, the matrix entries are endomorphisms of $L$. These, after all, DO form a ring.

 

[Math Amateur]

I'm not familiar with your text, but it should be the same as $M_n(R)$ where instead of ring-elements of $R$, the matrix entries are endomorphisms of $L$. These, after all, DO form a ring.

OK ... BUT ... just to confirm ... the answer to my question in the previous post, then ...

" ... ... Are we to assume that as well as \(\displaystyle M\) being equal to \(\displaystyle L^{\Lambda}\) we also have that

\(\displaystyle M = M_1 \oplus M_2 \oplus M_3 \oplus \ ... \ ... \ \oplus M_k \oplus \ ... \ ... \ \oplus M_n\)

... ... and that \(\displaystyle M_k\) is the \(\displaystyle k\)th element of the direct sum written as a matrix? ... ... "

Is yes ... ... ?

For some reason it just bothered me that both conditions were on M ... ... Peter

 

[Euge]

Hi Peter,

The set $M_k(\text{End } L)$ consists of all $k \times k$ matrices with entries in $\text{End } L$. Given $f \in \text{End}(L^k)$, define a matrix $(f_{ij}) \in M_k(\text{End } L)$ by setting $f_{ij} = \pi_i f\sigma_j$, for $1 \le i, j \le k$. This yields a map $T : \text{End}(L^k) \to M_k(\text{End } L)$, which you can show is an $R$-homomorphism. Conversely, given $f_{ij} \in M_k(\text{End } L)$, define an element $f\in \text{End}(L^k)$ by setting

\(\displaystyle f(x_1,\ldots, x_k) = \sum_{j = 1}^k (f_{1j}(x_j),\ldots, f_{kj}(x_j))\)

for all $(x_1,\ldots, x_k)\in L^k$. In other words,

\(\displaystyle f = \sum_{i, j = 1}^k \sigma_i f_{ij} \pi_j.\)

This determines a map $S : M_k(\text{End } L) \to \text{End}(L^k)$. For all $f \in \text{End}(L^k)$,

\(\displaystyle ST(f) = S((\pi_if\sigma_j)) = \sum_{i, j = 1}^k \sigma_i \pi_i f\sigma_j\pi_j = \sum_{j =1}^k f\sigma_j\pi_j = f\)

Thus $ST = id_{\text{End}(L^k)}$. For all $(f_{ij})\in M_k(\text{End } L)$,

\(\displaystyle TS(f_{ij}) = T\Bigl(\sum_{\ell, m = 1}^k \sigma_\ell f_{\ell m}\pi_m\Bigr) = \Bigl(\pi_i\Bigl(\sum_{\ell, m = 1}^k \sigma_\ell f_{\ell m}\pi_m\Bigr)\sigma_j\Bigr) \)

\(\displaystyle = \sum_{\ell,m = 1}^k (\pi_i\sigma_{\ell}f_{\ell m}\pi_m\sigma_j) = \sum_{\ell,m = 1}^k (\delta_{i\ell}f_{\ell m}\delta_{mj}) = (f_{ij})\)

Therefore, $TS = id_{M_k(\text{End } L)}$. This shows that $T$ is a one-to-one correspondence. Since $T$ is also an $R$-homomorphism, it is an isomorphism of $\text{End}(L^k)$ onto $M_k(\text{End } L)$.

I don't understand your second question. What is $L^\Lambda$ and what do you mean when you say "$M_k$ is the $k$th element of the direct sum written as a matrix"?

 

[Math Amateur]

Hi Peter,

The set $M_k(\text{End } L)$ consists of all $k \times k$ matrices with entries in $\text{End } L$. Given $f \in \text{End}(L^k)$, define a matrix $(f_{ij}) \in M_k(\text{End } L)$ by setting $f_{ij} = \pi_i f\sigma_j$, for $1 \le i, j \le k$. This yields a map $T : \text{End}(L^k) \to M_k(\text{End } L)$, which you can show is an $R$-homomorphism. Conversely, given $f_{ij} \in M_k(\text{End } L)$, define an element $f\in \text{End}(L^k)$ by setting

\(\displaystyle f(x_1,\ldots, x_k) = \sum_{j = 1}^k (f_{1j}(x_j),\ldots, f_{kj}(x_j))\)

for all $(x_1,\ldots, x_k)\in L^k$. In other words,

\(\displaystyle f = \sum_{i, j = 1}^k \sigma_i f_{ij} \pi_j.\)

This determines a map $S : M_k(\text{End } L) \to \text{End}(L^k)$. For all $f \in \text{End}(L^k)$,

\(\displaystyle ST(f) = S((\pi_if\sigma_j)) = \sum_{i, j = 1}^k \sigma_i \pi_i f\sigma_j\pi_j = \sum_{j =1}^k f\sigma_j\pi_j = f\)

Thus $ST = id_{\text{End}(L^k)}$. For all $(f_{ij})\in M_k(\text{End } L)$,

\(\displaystyle TS(f_{ij}) = T\Bigl(\sum_{\ell, m = 1}^k \sigma_\ell f_{\ell m}\pi_m\Bigr) = \Bigl(\pi_i\Bigl(\sum_{\ell, m = 1}^k \sigma_\ell f_{\ell m}\pi_m\Bigr)\sigma_j\Bigr) \)

\(\displaystyle = \sum_{\ell,m = 1}^k (\pi_i\sigma_{\ell}f_{\ell m}\pi_m\sigma_j) = \sum_{\ell,m = 1}^k (\delta_{i\ell}f_{\ell m}\delta_{mj}) = (f_{ij})\)

Therefore, $TS = id_{M_k(\text{End } L)}$. This shows that $T$ is a one-to-one correspondence. Since $T$ is also an $R$-homomorphism, it is an isomorphism of $\text{End}(L^k)$ onto $M_k(\text{End } L)$.

I don't understand your second question. What is $L^\Lambda$ and what do you mean when you say "$M_k$ is the $k$th element of the direct sum written as a matrix"?

Thanks Euge, I think you have answered my question ...

You write:

" ... ... I don't understand your second question. What is $L^\Lambda$ and what do you mean when you say "$M_k$ is the $k$th element of the direct sum written as a matrix"? ... ..."

The notation you ask about comes from B&K as follows:
https://www.physicsforums.com/attachments/3374
https://www.physicsforums.com/attachments/3375Just working through the details of your post now ...

Thanks again for your help ...

Peter

 

[Deveno]

OK ... BUT ... just to confirm ... the answer to my question in the previous post, then ...

" ... ... Are we to assume that as well as \(\displaystyle M\) being equal to \(\displaystyle L^{\Lambda}\) we also have that

\(\displaystyle M = M_1 \oplus M_2 \oplus M_3 \oplus \ ... \ ... \ \oplus M_k \oplus \ ... \ ... \ \oplus M_n\)

... ... and that \(\displaystyle M_k\) is the \(\displaystyle k\)th element of the direct sum written as a matrix? ... ... "

Is yes ... ... ?

For some reason it just bothered me that both conditions were on M ... ... Peter

I see no reason to suppose the $k$ in $M_k(\text{End}(L))$ refers to some summand of a direct sum. But, because of possible confusion like this, I prefer the notation:

$\text{Mat}_k(R)$ which is less ambiguous.