MHB 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 $$M = M_1 \oplus M_2$$, an internal sum of right $$R$$-modules, and let $$\{ \sigma_1 , \sigma_2, \pi_1 , \pi_2 \}$$ be the corresponding set of inclusions and projections.

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

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

where $$\mu_{11}m_1 + \mu_{12}m_2$$ is in $$M_1$$

and

$$\mu_{21}m_1 + \mu_{22}m_2$$ is in $$M_2$$

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

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

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

where $$Hom(M_1, M_2)$$ is the set of all $$R$$-Module maps from $$M_1$$ to $$M_2$$.
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 $$M = M_1 \oplus M_2$$, an internal sum of right $$R$$-modules, and let $$\{ \sigma_1 , \sigma_2, \pi_1 , \pi_2 \}$$ be the corresponding set of inclusions and projections.

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

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

where $$\mu_{11}m_1 + \mu_{12}m_2$$ is in $$M_1$$

and

$$\mu_{21}m_1 + \mu_{22}m_2$$ is in $$M_2$$

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

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

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

where $$Hom(M_1, M_2)$$ is the set of all $$R$$-Module maps from $$M_1$$ to $$M_2$$.
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 $$m$$ mapped under $$\mu$$ gives us (neatly) one element from $$M_1$$ and one element from $$M_2$$?

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

Second clarification:

If we replaced $$(m_1 + m_2)$$ by $$(m_1, m_2)$$ 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

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

equal to

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

Is the justification for this that $$\pi_1$$ is a module homomorphism, $$\mu$$ 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 $$m$$ mapped under $$\mu$$ gives us (neatly) one element from $$M_1$$ and one element from $$M_2$$?

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

Second clarification:

If we replaced $$(m_1 + m_2)$$ by $$(m_1, m_2)$$ 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

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

equal to

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

Is the justification for this that $$\pi_1$$ is a module homomorphism, $$\mu$$ 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 $$M$$ as a 'column space' $$\begin{bmatrix} M_1 \\ M_2 \end{bmatrix}$$, show that $$\mu$$ can be represented as a matrix $$\begin{bmatrix}\mu_{11} & \mu_{12} \\ \mu_{21} & \mu_{22} \end{bmatrix}.$$"

If we take $$m = \begin{bmatrix} m_1 \\ m_2 \end{bmatrix} $$

where $$m_1 \in M_1$$ and $$m_2 \in M_2$$

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

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

$$= \begin{bmatrix} \mu_{11}m_1 + \mu_{12}m_2 \\ \mu_{21}m_1 + \mu_{22}m_2 \end{bmatrix}$$

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

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

Peter