Endomorphisms of Direct Sums - Berrick and Keating - Exercise 2.1.6 (i)

[Math Amateur]

Exercise 2.1.6 (i) of Berrick and Keating's book An Introduction to Rings and Modules reads as follows:Let \(\displaystyle M = M_1 \oplus M_2\), an internal sum of right \(\displaystyle R\)-modules, and let \(\displaystyle \{ \sigma_1 , \sigma_2, \pi_1 , \pi_2 \}\) be the corresponding set of inclusions and projections.

Given an endomorphism \(\displaystyle \mu\) of \(\displaystyle M\), define \(\displaystyle \mu_{ij} = \pi_i \mu \sigma_j\), an \(\displaystyle R\)-homomorphism from \(\displaystyle M_j\) to \(\displaystyle M_i, i,j = 1,2\).Show that for \(\displaystyle m = m_1 + m_2\), with \(\displaystyle m_1 \in M_1, m_2 \in M_2\) we have:

\(\displaystyle \mu(m) = ( \mu_{11}m_1 + \mu_{12}m_2) + ( \mu_{21}m_1 + \mu_{22}m_2)
\)

where \(\displaystyle \mu_{11}m_1 + \mu_{12}m_2\) is in \(\displaystyle M_1\)

and

\(\displaystyle \mu_{21}m_1 + \mu_{22}m_2\) is in \(\displaystyle M_2\)

Viewing \(\displaystyle M\) as a 'column space' \(\displaystyle \begin{bmatrix} M_1 \\ M_2 \end{bmatrix}\), show that \(\displaystyle \mu\) can be represented as a matrix \(\displaystyle \begin{bmatrix}\mu_{11} & \mu_{12} \\ \mu_{21} & \mu_{22} \end{bmatrix}.\)

Deduce that the ring of endomorphisms End(\(\displaystyle M\)) of \(\displaystyle M\) can be written as a ring of \(\displaystyle 2 \times 2\) matrices:

\(\displaystyle End(M) = \begin{bmatrix} End(M_1) & Hom(M_2, M_1) \\ Hom(M_1, M_2) & End(M_2) \end{bmatrix}\)

where \(\displaystyle Hom(M_1, M_2)\) is the set of all \(\displaystyle R\)-Module maps from \(\displaystyle M_1\) to \(\displaystyle M_2\).
Can someone please help me to get started on this exercise.

Help will be appreciated!

Peter

 

[Deveno]

Exercise 2.1.6 (i) of Berrick and Keating's book An Introduction to Rings and Modules reads as follows:Let \(\displaystyle M = M_1 \oplus M_2\), an internal sum of right \(\displaystyle R\)-modules, and let \(\displaystyle \{ \sigma_1 , \sigma_2, \pi_1 , \pi_2 \}\) be the corresponding set of inclusions and projections.

Given an endomorphism \(\displaystyle \mu\) of \(\displaystyle M\), define \(\displaystyle \mu_{ij} = \pi_i \mu \sigma_j\), an \(\displaystyle R\)-homomorphism from \(\displaystyle M_j\) to \(\displaystyle M_i, i,j = 1,2\).Show that for \(\displaystyle m = m_1 + m_2\), with \(\displaystyle m_1 \in M_1, m_2 \in M_2\) we have:

\(\displaystyle \mu(m) = ( \mu_{11}m_1 + \mu_{12}m_2) + ( \mu_{21}m_1 + \mu_{22}m_2)
\)

where \(\displaystyle \mu_{11}m_1 + \mu_{12}m_2\) is in \(\displaystyle M_1\)

and

\(\displaystyle \mu_{21}m_1 + \mu_{22}m_2\) is in \(\displaystyle M_2\)

Viewing \(\displaystyle M\) as a 'column space' \(\displaystyle \begin{bmatrix} M_1 \\ M_2 \end{bmatrix}\), show that \(\displaystyle \mu\) can be represented as a matrix \(\displaystyle \begin{bmatrix}\mu_{11} & \mu_{12} \\ \mu_{21} & \mu_{22} \end{bmatrix}.\)

Deduce that the ring of endomorphisms End(\(\displaystyle M\)) of \(\displaystyle M\) can be written as a ring of \(\displaystyle 2 \times 2\) matrices:

\(\displaystyle End(M) = \begin{bmatrix} End(M_1) & Hom(M_2, M_1) \\ Hom(M_1, M_2) & End(M_2) \end{bmatrix}\)

where \(\displaystyle Hom(M_1, M_2)\) is the set of all \(\displaystyle R\)-Module maps from \(\displaystyle M_1\) to \(\displaystyle M_2\).
Can someone please help me to get started on this exercise.

Help will be appreciated!

Peter

Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$.

Now $\mu_{11}(m_1) = \pi_1\mu\sigma_1(m_1) = \pi_1\mu(m_1 + 0)$

$\mu_{12}(m_2) = \pi_1\mu\sigma_2(m_2) = \pi_1\mu(0 + m_2)$

$\mu_{21}(m_1) = \pi_2\mu\sigma_1(m_1) = \pi_2\mu(m_1 + 0)$

$\mu_{22}(m_2) = \pi_2\mu\sigma_2(m_2) = \pi_2\mu(0+m_2)$.

Thus:

$(\mu_{11}(m_1) + \mu_{12}(m_2)) + (\mu_{21}(m_1) + \mu_{22}(m_2))$

$= (\pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2)) + (\pi_2\mu(m_1 + 0) + \pi_2\mu(0+m_2))$

$= (\pi_1\mu((m_1 + 0) + (0 + m_2)) + (\pi_2\mu((m_1 + 0) + (0 + m_2))$

$= (\pi_1\mu(m_1 + m_2)) + (\pi_2\mu(m_1 + m_2))$

$= (\pi_1\mu(m)) + (\pi_2\mu(m)) = \pi_1(n_1+n_2) + \pi_2(n_1 + n_2)$

$ = n_1 + n_2 = \mu(m)$.

Can you take it from here?

 

[Math Amateur]

Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$.

Now $\mu_{11}(m_1) = \pi_1\mu\sigma_1(m_1) = \pi_1\mu(m_1 + 0)$

$\mu_{12}(m_2) = \pi_1\mu\sigma_2(m_2) = \pi_1\mu(0 + m_2)$

$\mu_{21}(m_1) = \pi_2\mu\sigma_1(m_1) = \pi_2\mu(m_1 + 0)$

$\mu_{22}(m_2) = \pi_2\mu\sigma_2(m_2) = \pi_2\mu(0+m_2)$.

Thus:

$(\mu_{11}(m_1) + \mu_{12}(m_2)) + (\mu_{21}(m_1) + \mu_{22}(m_2))$

$= (\pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2)) + (\pi_2\mu(m_1 + 0) + \pi_2\mu(0+m_2))$

$= (\pi_1\mu((m_1 + 0) + (0 + m_2)) + (\pi_2\mu((m_1 + 0) + (0 + m_2))$

$= (\pi_1\mu(m_1 + m_2)) + (\pi_2\mu(m_1 + m_2))$

$= (\pi_1\mu(m)) + (\pi_2\mu(m)) = \pi_1(n_1+n_2) + \pi_2(n_1 + n_2)$

$ = n_1 + n_2 = \mu(m)$.

Can you take it from here?

Thanks for the start Deveno ... gives me confidence to continue ...

Will try to progress the solution a bit ...

Peter

 

[Math Amateur]

Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$.

Now $\mu_{11}(m_1) = \pi_1\mu\sigma_1(m_1) = \pi_1\mu(m_1 + 0)$

$\mu_{12}(m_2) = \pi_1\mu\sigma_2(m_2) = \pi_1\mu(0 + m_2)$

$\mu_{21}(m_1) = \pi_2\mu\sigma_1(m_1) = \pi_2\mu(m_1 + 0)$

$\mu_{22}(m_2) = \pi_2\mu\sigma_2(m_2) = \pi_2\mu(0+m_2)$.

Thus:

$(\mu_{11}(m_1) + \mu_{12}(m_2)) + (\mu_{21}(m_1) + \mu_{22}(m_2))$

$= (\pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2)) + (\pi_2\mu(m_1 + 0) + \pi_2\mu(0+m_2))$

$= (\pi_1\mu((m_1 + 0) + (0 + m_2)) + (\pi_2\mu((m_1 + 0) + (0 + m_2))$

$= (\pi_1\mu(m_1 + m_2)) + (\pi_2\mu(m_1 + m_2))$

$= (\pi_1\mu(m)) + (\pi_2\mu(m)) = \pi_1(n_1+n_2) + \pi_2(n_1 + n_2)$

$ = n_1 + n_2 = \mu(m)$.

Can you take it from here?

Hi Deveno ... just a minor clarification:

You write:

" ... ... Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$. ... ... "

How can we be sure that \(\displaystyle m\) mapped under \(\displaystyle \mu\) gives us (neatly) one element from \(\displaystyle M_1\) and one element from \(\displaystyle M_2\)?

Is it because \(\displaystyle \mu (m) = n\) for some \(\displaystyle n \in M\) and given the nature of the direct sum we have \(\displaystyle n = n_1 + n_2\) where \(\displaystyle n_1 \in M_1\) and \(\displaystyle n_2 \in M_2\)? Can you confirm that this is the reason?

Second clarification:

If we replaced \(\displaystyle (m_1 + m_2)\) by \(\displaystyle (m_1, m_2)\) in the proof/solution , would everything still hold? [I am thinking that given the nature of direct sums this would be OK.]

Peter

 

[Math Amateur]

Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$.

Now $\mu_{11}(m_1) = \pi_1\mu\sigma_1(m_1) = \pi_1\mu(m_1 + 0)$

$\mu_{12}(m_2) = \pi_1\mu\sigma_2(m_2) = \pi_1\mu(0 + m_2)$

$\mu_{21}(m_1) = \pi_2\mu\sigma_1(m_1) = \pi_2\mu(m_1 + 0)$

$\mu_{22}(m_2) = \pi_2\mu\sigma_2(m_2) = \pi_2\mu(0+m_2)$.

Thus:

$(\mu_{11}(m_1) + \mu_{12}(m_2)) + (\mu_{21}(m_1) + \mu_{22}(m_2))$

$= (\pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2)) + (\pi_2\mu(m_1 + 0) + \pi_2\mu(0+m_2))$

$= (\pi_1\mu((m_1 + 0) + (0 + m_2)) + (\pi_2\mu((m_1 + 0) + (0 + m_2))$

$= (\pi_1\mu(m_1 + m_2)) + (\pi_2\mu(m_1 + m_2))$

$= (\pi_1\mu(m)) + (\pi_2\mu(m)) = \pi_1(n_1+n_2) + \pi_2(n_1 + n_2)$

$ = n_1 + n_2 = \mu(m)$.

Can you take it from here?

Just another minor clarification, Deveno ... ...

In your working above, you put

\(\displaystyle \pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2) \)

equal to

\(\displaystyle \pi_1\mu((m_1 + 0) + (0 + m_2)) \)

Is the justification for this that \(\displaystyle \pi_1\) is a module homomorphism, \(\displaystyle \mu\) is also and so their composition is a module homomorphism?

Peter
Peter

 

[Deveno]

Hi Deveno ... just a minor clarification:

You write:

" ... ... Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$. ... ... "

How can we be sure that \(\displaystyle m\) mapped under \(\displaystyle \mu\) gives us (neatly) one element from \(\displaystyle M_1\) and one element from \(\displaystyle M_2\)?

Is it because \(\displaystyle \mu (m) = n\) for some \(\displaystyle n \in M\) and given the nature of the direct sum we have \(\displaystyle n = n_1 + n_2\) where \(\displaystyle n_1 \in M_1\) and \(\displaystyle n_2 \in M_2\)? Can you confirm that this is the reason?

Second clarification:

If we replaced \(\displaystyle (m_1 + m_2)\) by \(\displaystyle (m_1, m_2)\) in the proof/solution , would everything still hold? [I am thinking that given the nature of direct sums this would be OK.]

Peter

We have an INTERNAL direct sum so any element $m \in M$ has a UNIQUE expression as $m_1 + m_2$. This also holds for the element $\mu(m) \in M$.

So, yes, you are correct.

The "side by side" writing of $m_1 + m_2$ as $(m_1,m_2)$ is for an EXTERNAL direct product. We don't need to do that, here.

Just another minor clarification, Deveno ... ...

In your working above, you put

\(\displaystyle \pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2) \)

equal to

\(\displaystyle \pi_1\mu((m_1 + 0) + (0 + m_2)) \)

Is the justification for this that \(\displaystyle \pi_1\) is a module homomorphism, \(\displaystyle \mu\) is also and so their composition is a module homomorphism?

Peter
Peter

Yep.

 

[Math Amateur]

We have an INTERNAL direct sum so any element $m \in M$ has a UNIQUE expression as $m_1 + m_2$. This also holds for the element $\mu(m) \in M$.

So, yes, you are correct.

The "side by side" writing of $m_1 + m_2$ as $(m_1,m_2)$ is for an EXTERNAL direct product. We don't need to do that, here.

Yep.

Thanks Deveno ... hmmm ... missed the 'unique' bit ... thanks for the reminder ...

Peter

 

[Math Amateur]

Suppose $\mu(m) = \mu(m_1 + m_2) = n_1 + n_2$ with $n_1 \in M_1,n_2 \in M_2$.

Now $\mu_{11}(m_1) = \pi_1\mu\sigma_1(m_1) = \pi_1\mu(m_1 + 0)$

$\mu_{12}(m_2) = \pi_1\mu\sigma_2(m_2) = \pi_1\mu(0 + m_2)$

$\mu_{21}(m_1) = \pi_2\mu\sigma_1(m_1) = \pi_2\mu(m_1 + 0)$

$\mu_{22}(m_2) = \pi_2\mu\sigma_2(m_2) = \pi_2\mu(0+m_2)$.

Thus:

$(\mu_{11}(m_1) + \mu_{12}(m_2)) + (\mu_{21}(m_1) + \mu_{22}(m_2))$

$= (\pi_1\mu(m_1 + 0) + \pi_1\mu(0 + m_2)) + (\pi_2\mu(m_1 + 0) + \pi_2\mu(0+m_2))$

$= (\pi_1\mu((m_1 + 0) + (0 + m_2)) + (\pi_2\mu((m_1 + 0) + (0 + m_2))$

$= (\pi_1\mu(m_1 + m_2)) + (\pi_2\mu(m_1 + m_2))$

$= (\pi_1\mu(m)) + (\pi_2\mu(m)) = \pi_1(n_1+n_2) + \pi_2(n_1 + n_2)$

$ = n_1 + n_2 = \mu(m)$.

Can you take it from here?

You write:

" ... ... Can you take it from here? ... ..."

Well, the second part of the exercise looks pretty straightforward ... unless I am misinterpreting it ...

The second part of the exercise states:

"Viewing \(\displaystyle M\) as a 'column space' \(\displaystyle \begin{bmatrix} M_1 \\ M_2 \end{bmatrix}\), show that \(\displaystyle \mu\) can be represented as a matrix \(\displaystyle \begin{bmatrix}\mu_{11} & \mu_{12} \\ \mu_{21} & \mu_{22} \end{bmatrix}.\)"

If we take \(\displaystyle m = \begin{bmatrix} m_1 \\ m_2 \end{bmatrix} \)

where \(\displaystyle m_1 \in M_1\) and \(\displaystyle m_2 \in M_2\)

then we can represent \(\displaystyle \mu(m_1)\) as follows:

\(\displaystyle \mu (m_1) = \begin{bmatrix}\mu_{11} & \mu_{12} \\ \mu_{21} & \mu_{22} \end{bmatrix} \begin{bmatrix} m_1 \\ m_2 \end{bmatrix} \)

\(\displaystyle = \begin{bmatrix} \mu_{11}m_1 + \mu_{12}m_2 \\ \mu_{21}m_1 + \mu_{22}m_2 \end{bmatrix}\)

where \(\displaystyle \mu_{11}m_1 + \mu_{12}m_2 \in M_1\) and \(\displaystyle \mu_{21}m_1 + \mu_{22}m_2 \in M_2\)

Can you confirm that this is a satisfactory answer to this part of the exercise.

Peter