External Direct Sums and Direct Products .... Bland Problem 1, Section 2.1 ....

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with Problem 1(b) of Problem Set 2.1 ...

Problem 1(b) of Problem Set 2.1 reads as follows:
View attachment 8048I have had difficulty in formulating a rigorous and convincing proof of the statement in Problem 1(b) ... can someone please

(1) critique my attempt at a proof (see below)

(2) provide an alternate rigorous and convincing proof My attempt at a proof is as follows:

We need to demonstrate that \(\displaystyle \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha\) if and only if \(\displaystyle \Delta\) is a finite set ...Assume \(\displaystyle \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \)

The above equality would require all of the terms \(\displaystyle (x_\alpha)\) of \(\displaystyle \prod_\Delta M_\alpha\) to have a finite number of components or elements in each \(\displaystyle (x_\alpha)\) ... thus \(\displaystyle \Delta\) is a finite set ...
Assume \(\displaystyle \Delta\) is a finite set

... then \(\displaystyle \prod_\Delta M_\alpha\) has terms of the form \(\displaystyle (x_\alpha) = ( x_1, x_2, \ ... \ ... \ , x_n )\) for some \(\displaystyle n \in \mathbb{Z}\) ... ...

and

... \(\displaystyle \bigoplus_\Delta M_\alpha\) has the same terms given that each of the above terms \(\displaystyle (x_\alpha)\) has a finite number of components ...
Hope someone can indicate how to formulate a better proof ...

Peter

 

[steenis]

Assume \(\displaystyle \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \)
The above equality would require all of the terms \(\displaystyle (x_\alpha)\) of \(\displaystyle \prod_\Delta M_\alpha\) to have a finite number of components or elements in each \(\displaystyle (x_\alpha)\) ... thus \(\displaystyle \Delta\) is a finite set ...
Peter

I think this is not right

The above equality would require all of the terms \(\displaystyle (x_\alpha)\) of \(\displaystyle \prod_\Delta M_\alpha\) to have a finite number of components or elements in each \(\displaystyle (x_\alpha)\) that are not zero, I cannot conclude from this that $\Delta$ is a finite set.

Assume \(\displaystyle \Delta\) is a finite set
... then \(\displaystyle \prod_\Delta M_\alpha\) has terms of the form \(\displaystyle (x_\alpha) = ( x_1, x_2, \ ... \ ... \ , x_n )\) for some \(\displaystyle n \in \mathbb{Z}\) ... ...

and

... \(\displaystyle \bigoplus_\Delta M_\alpha\) has the same terms given that each of the above terms \(\displaystyle (x_\alpha)\) has a finite number of components ...
Peter

I think this is ok but I would write it down this way:
Assume \(\displaystyle \Delta\) is a finite set then \(\displaystyle \prod_\Delta M_\alpha\) has terms of the form \(\displaystyle (x_\alpha) = ( x_1, x_2, \ ... \ ... \ , x_n )\) for some \(\displaystyle n \in \mathbb{Z}\),
finitely many $x_\alpha$ can be non-zero, so $(x_\alpha) \in \bigoplus_\Delta M_\alpha$, so
$\prod_\Delta M_\alpha \subset \bigoplus_\Delta M_\alpha$
we already have
$\bigoplus_\Delta M_\alpha \subset \prod_\Delta M_\alpha$
and thus ...

 

[steenis]

Let's prove this the fancy way by using definition 2.1.6 on page 44

Let $\Delta$ be finite. Let $S=\prod_\Delta M_\alpha$.

Let $u_\alpha:M_\alpha \longrightarrow S:x \longmapsto (0 \cdots x \cdots 0) \in S=\prod_\Delta M_\alpha$ be the canonical injections, for all $\alpha \in \Delta$.

Let $\{f_\alpha : M_\alpha \longrightarrow N \}_\Delta$ be a family of R-maps.

Now define $\Phi:S=\prod_\Delta M_\alpha \longrightarrow N : (x_\alpha) \longmapsto \sum_\Delta f_\alpha (x_\alpha)$, this sum is a finite sum, so it exists.

Make a diagram, like the diagram in definition 2.1.6 !

Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$.

 

[Math Amateur]

Let's prove this the fancy way by using definition 2.1.6 on page 44

Let $\Delta$ be finite. Let $S=\prod_\Delta M_\alpha$.

Let $u_\alpha:M_\alpha \longrightarrow S:x \longmapsto (0 \cdots x \cdots 0) \in S=\prod_\Delta M_\alpha$ be the canonical injections, for all $\alpha \in \Delta$.

Let $\{f_\alpha : M_\alpha \longrightarrow N \}_\Delta$ be a family of R-maps.

Now define $\Phi:S=\prod_\Delta M_\alpha \longrightarrow N : (x_\alpha) \longmapsto \sum_\Delta f_\alpha (x_\alpha)$, this sum is a finite sum, so it exists.

Make a diagram, like the diagram in definition 2.1.6 !

Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$.

Well ... this is REALLY interesting and REALLY helpful ...

Peter

 

[Math Amateur]

I think this is not right

The above equality would require all of the terms \(\displaystyle (x_\alpha)\) of \(\displaystyle \prod_\Delta M_\alpha\) to have a finite number of components or elements in each \(\displaystyle (x_\alpha)\) that are not zero, I cannot conclude from this that $\Delta$ is a finite set.
I think this is ok but I would write it down this way:
Assume \(\displaystyle \Delta\) is a finite set then \(\displaystyle \prod_\Delta M_\alpha\) has terms of the form \(\displaystyle (x_\alpha) = ( x_1, x_2, \ ... \ ... \ , x_n )\) for some \(\displaystyle n \in \mathbb{Z}\),
finitely many $x_\alpha$ can be non-zero, so $(x_\alpha) \in \bigoplus_\Delta M_\alpha$, so
$\prod_\Delta M_\alpha \subset \bigoplus_\Delta M_\alpha$
we already have
$\bigoplus_\Delta M_\alpha \subset \prod_\Delta M_\alpha$
and thus ...

Hi steenis ... thanks again for your help ...

Can you give me some help/guidance on how to use Bland's definitions to show that \(\displaystyle \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \Longrightarrow\) \(\displaystyle \Delta\) is a finite set ...Peter

 

[Math Amateur]

Let's prove this the fancy way by using definition 2.1.6 on page 44

Let $\Delta$ be finite. Let $S=\prod_\Delta M_\alpha$.

Let $u_\alpha:M_\alpha \longrightarrow S:x \longmapsto (0 \cdots x \cdots 0) \in S=\prod_\Delta M_\alpha$ be the canonical injections, for all $\alpha \in \Delta$.

Let $\{f_\alpha : M_\alpha \longrightarrow N \}_\Delta$ be a family of R-maps.

Now define $\Phi:S=\prod_\Delta M_\alpha \longrightarrow N : (x_\alpha) \longmapsto \sum_\Delta f_\alpha (x_\alpha)$, this sum is a finite sum, so it exists.

Make a diagram, like the diagram in definition 2.1.6 !

Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$.

Hi steenis ... following your advice ... and assuming that \(\displaystyle \Delta\) is a finite set ... we want to show that:

$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.So ... first we show that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...Let \(\displaystyle (x_\alpha), (y_\alpha) \in \prod_\Delta M_\alpha\) and let \(\displaystyle r \in R\) Then \(\displaystyle \Phi ( (x_\alpha) + (y_\alpha) ) = \Phi ( (x_\alpha + y_\alpha) )\)\(\displaystyle = \sum_\Delta f_\alpha (x_\alpha + y_\alpha)\)\(\displaystyle = \sum_\Delta ( f_\alpha x_\alpha + f_\alpha y_\alpha )\) since \(\displaystyle f_\alpha\) is an \(\displaystyle R\)-linear mapping for all \(\displaystyle \alpha \in \Delta\) ...\(\displaystyle = \sum_\Delta f_\alpha x_\alpha + \sum_\Delta f_\alpha y_\alpha \)\(\displaystyle = \Phi ( ( x_\alpha ) ) + \Phi ( ( y_\alpha ) )\) ... ... ... ... ... (1)

We also have ... ...\(\displaystyle \Phi ( ( x_\alpha ) r ) = \Phi ( ( x_\alpha r ) )\) \(\displaystyle = \sum_\Delta f_\alpha (x_\alpha r )\)\(\displaystyle = \sum_\Delta (f_\alpha x_\alpha) r \) since \(\displaystyle f_\alpha\) is an \(\displaystyle R\)-linear mapping for all \(\displaystyle \alpha \in \Delta\) ...\(\displaystyle = [ \sum_\Delta (f_\alpha x_\alpha) ] r\) \(\displaystyle = [ \Phi ( ( x_\alpha ) ) ] r \) ... ... ... ... ... (2)
(1) (2) above \(\displaystyle \Longrightarrow\) \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear mapping ...Now we need to show that \(\displaystyle \Phi\) is well defined ... ... ?

... BUT ... what is involved in showing this ...

Can you help ...?

Peter

 

[steenis]

Hi steenis ... thanks again for your help ...
Can you give me some help/guidance on how to use Bland's definitions to show that
\(\displaystyle \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \Longrightarrow\) \(\displaystyle \Delta\) is a finite set ...
Peter

I see what you mean, but until now no I cannot, maybe someone else has a bright idea.
I keep looking.

$\Phi$ is well defined: take $(x_\alpha) = (y_\alpha)$, prove that $\Phi ((x_\alpha)) = \Phi((y_\alpha))$, use the fact that $f_\alpha$ is implicitly well-defined: $x=y$ then $f_\alpha(x)=f_\alpha(y)$.

 

[steenis]

Peter what do you think of this ?

We have to prove $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$ $\Rightarrow$ $\Delta$ is finite

So suppose $\Delta$ is infinte.

We may suppose $M_\alpha \neq 0$ for all $\alpha \in \Delta$.

Then for each $\alpha \in \Delta$ there is a nonzero $x_\alpha \in M_\alpha$ (Axiom of Choice).

Then, of course, $(x_\alpha) \in \prod_\Delta M_\alpha$ however $(x_\alpha) \notin \bigoplus_\Delta M_\alpha$.

And therefore $\bigoplus_\Delta M_\alpha \neq \prod_\Delta M_\alpha$

 

[Math Amateur]

Peter what do you think of this ?

We have to prove $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$ $\Rightarrow$ $\Delta$ is finite

So suppose $\Delta$ is infinte.

We may suppose $M_\alpha \neq 0$ for all $\alpha \in \Delta$.

Then for each $\alpha \in \Delta$ there is a nonzero $x_\alpha \in M_\alpha$ (Axiom of Choice).

Then, of course, $(x_\alpha) \in \prod_\Delta M_\alpha$ however $(x_\alpha) \notin \bigoplus_\Delta M_\alpha$.

And therefore $\bigoplus_\Delta M_\alpha \neq \prod_\Delta M_\alpha$

Hi steenis ...

I think the proof is valid ... EXCEPT ... I cannot see how we can justify making the assumption:

" ... ... suppose $M_\alpha \neq 0$ for all $\alpha \in \Delta$ ,,, ,,, "


... BUT ... since I cannot see how to prove the matter without making the assumption I think the best we can do is add the condition to the problem statement ...So ... our problem statement becomes

If \(\displaystyle \{M_\alpha \}_\Delta\) is a family of nontrivial modules ( that is \(\displaystyle M_\alpha \neq \{ 0 \}\) for all \(\displaystyle \alpha\) ) then

$\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$ $\Rightarrow$ $\Delta$ is finite... what do you think?

Peter

 

[steenis]

In fact you do the same as I did in post #8: leaving out the trivial $M_\alpha$, hoping that is possible without loss of generality.
Below another attempt.Let
$\Delta ' = \{\alpha \in \Delta \mbox{ } | \mbox{ } M_\alpha \neq 0 \}$
$\Delta '' = \{\alpha \in \Delta \mbox{ } | \mbox{ } M_\alpha = 0 \}$
then
$\Delta = \Delta ' + \Delta ''$

$\prod_\Delta M_\alpha = \prod_{\Delta '} M_\alpha \times \prod_{\Delta ''} M_\alpha$

$\bigoplus_\Delta M_\alpha = \bigoplus_{\Delta '} M_\alpha \oplus \bigoplus_{\Delta ''} M_\alpha$

Suppose $\Delta '$ is infinte. Then you can construct an element that is in $\prod_{\Delta '} M_\alpha$ but not in $\bigoplus_{\Delta '} M_\alpha$, see post #8.

Thus $\bigoplus_{\Delta '} M_\alpha \neq \prod_{\Delta '} M_\alpha$ but also $\bigoplus_\Delta M_\alpha \neq \prod_\Delta M_\alpha$.

Therefore $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha \Longrightarrow \Delta '$ is finite.

One cannot conclude that $\Delta$ is finite, because there may be an infinite number of trivial $M_\alpha$.

Maybe someone else can explain that we can suppose that all the $M_\alpha$ are non-trivial, without loss of generality.

 

[Math Amateur]

In fact you do the same as I did in post #8: leaving out the trivial $M_\alpha$, hoping that is possible without loss of generality.
Below another attempt.Let
$\Delta ' = \{\alpha \in \Delta \mbox{ } | \mbox{ } M_\alpha \neq 0 \}$
$\Delta '' = \{\alpha \in \Delta \mbox{ } | \mbox{ } M_\alpha = 0 \}$
then
$\Delta = \Delta ' + \Delta ''$

$\prod_\Delta M_\alpha = \prod_{\Delta '} M_\alpha \times \prod_{\Delta ''} M_\alpha$

$\bigoplus_\Delta M_\alpha = \bigoplus_{\Delta '} M_\alpha \oplus \bigoplus_{\Delta ''} M_\alpha$

Suppose $\Delta '$ is infinte. Then you can construct an element that is in $\prod_{\Delta '} M_\alpha$ but not in $\bigoplus_{\Delta '} M_\alpha$, see post #8.

Thus $\bigoplus_{\Delta '} M_\alpha \neq \prod_{\Delta '} M_\alpha$ but also $\bigoplus_\Delta M_\alpha \neq \prod_\Delta M_\alpha$.

Therefore $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha \Longrightarrow \Delta '$ is finite.

One cannot conclude that $\Delta$ is finite, because there may be an infinite number of trivial $M_\alpha$.

Maybe someone else can explain that we can suppose that all the $M_\alpha$ are non-trivial, without loss of generality.

Hi steenis ...

Sorry to be slow in answering ... have temporarily left my home state of Tasmania and am traveling through the mainland state of Victoria ...

Regarding your last post ... I agree with all your comments/statements ...

Peter

 

[Math Amateur]

Let's prove this the fancy way by using definition 2.1.6 on page 44

Let $\Delta$ be finite. Let $S=\prod_\Delta M_\alpha$.

Let $u_\alpha:M_\alpha \longrightarrow S:x \longmapsto (0 \cdots x \cdots 0) \in S=\prod_\Delta M_\alpha$ be the canonical injections, for all $\alpha \in \Delta$.

Let $\{f_\alpha : M_\alpha \longrightarrow N \}_\Delta$ be a family of R-maps.

Now define $\Phi:S=\prod_\Delta M_\alpha \longrightarrow N : (x_\alpha) \longmapsto \sum_\Delta f_\alpha (x_\alpha)$, this sum is a finite sum, so it exists.

Make a diagram, like the diagram in definition 2.1.6 !

Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$.

Hi steenis ... I have been reflecting on your proof using Proposition 2.1.6 ...

I must confess that I do not fully understand the underlying logic ..

So ... you write:

" ... ... Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$. ... ... "Can you explain how this process leads to the statement that

$\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$. Indeed ... how does definition 2.1.6 lead to $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$. Can you help?

Peter
*** EDIT ***

After a little more reflection, I now think that the logic underlying your approach using Definition 2.1.6 is as follows:

Let \(\displaystyle S = \prod_\Delta M_\alpha\) where \(\displaystyle \Delta\) is a finite set ... ...

... then ...

... prove that \(\displaystyle S \equiv\) external direct sum \(\displaystyle \equiv \bigoplus_\Delta M_\alpha\) ...

Thus ...

\(\displaystyle S = \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha\)
Is that a correct description of the logic underlying your approach ... ?

Peter

 

[Math Amateur]

Let's prove this the fancy way by using definition 2.1.6 on page 44

Let $\Delta$ be finite. Let $S=\prod_\Delta M_\alpha$.

Let $u_\alpha:M_\alpha \longrightarrow S:x \longmapsto (0 \cdots x \cdots 0) \in S=\prod_\Delta M_\alpha$ be the canonical injections, for all $\alpha \in \Delta$.

Let $\{f_\alpha : M_\alpha \longrightarrow N \}_\Delta$ be a family of R-maps.

Now define $\Phi:S=\prod_\Delta M_\alpha \longrightarrow N : (x_\alpha) \longmapsto \sum_\Delta f_\alpha (x_\alpha)$, this sum is a finite sum, so it exists.

Make a diagram, like the diagram in definition 2.1.6 !

Show that
$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique, let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$, show that $\Psi = \Phi$.

Definition 2.1.6 now says that: $\bigoplus_\Delta M_\alpha = \prod_\Delta M_\alpha$.

Hi steenis ...

Continuing the proof of ...

\(\displaystyle \Delta\) is a finite set \(\displaystyle \Longrightarrow \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \) ... ...Have already shown that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...Now show that \(\displaystyle \Phi\) is well defined ... Let \(\displaystyle (x_\alpha ) = ( y_\alpha )\)

Then ...

\(\displaystyle \Phi( ( x_\alpha ) )\)

\(\displaystyle = \sum_\Delta f_\alpha ( x_\alpha ) \)

\(\displaystyle = \sum_\Delta f_\alpha ( y_\alpha ) since x_\alpha = y _\alpha \Longrightarrow f_\alpha ( x_\alpha ) = f_\alpha ( y_\alpha )\)

\(\displaystyle = \Phi( ( y_\alpha ) ) \)Thus \(\displaystyle \Phi \) is well defined ...Is that correct?

Peter

 

[Math Amateur]

Hi steenis ...

Continuing the proof of ...

\(\displaystyle \Delta\) is a finite set \(\displaystyle \Longrightarrow \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \) ... ...Have already shown that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...Now show that \(\displaystyle \Phi\) is well defined ... Let \(\displaystyle (x_\alpha ) = ( y_\alpha )\)

Then ...

\(\displaystyle \Phi( ( x_\alpha ) )\)

\(\displaystyle = \sum_\Delta f_\alpha ( x_\alpha ) \)

\(\displaystyle = \sum_\Delta f_\alpha ( y_\alpha ) since x_\alpha = y _\alpha \Longrightarrow f_\alpha ( x_\alpha ) = f_\alpha ( y_\alpha )\)

\(\displaystyle = \Phi( ( y_\alpha ) ) \)Thus \(\displaystyle \Phi \) is well defined ...Is that correct?

Peter

Hi steenis ...

Continuing the proof of ...

\(\displaystyle \Delta\) is a finite set \(\displaystyle \Longrightarrow \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \) ... ...Have already shown that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...... and have shown that \(\displaystyle \Phi\) is well defined ...
Now show that \(\displaystyle \Phi \circ u_\alpha = f_\alpha\) for all \(\displaystyle \alpha\)
Let \(\displaystyle x_\alpha \in M_\alpha\) ...

... now ...

\(\displaystyle \Phi \circ u_\alpha ( x_\alpha )\)

\(\displaystyle = \Phi ( 0, 0, \ ... \ ... \ , x_\alpha, \ ... \ ... \ ,0 ) \)

\(\displaystyle = \sum_{\alpha} f_\alpha ( x_\alpha )\)

\(\displaystyle = f_1 (0) + f_2 (0) + \ ... \ ... \ , f_\alpha ( x_\alpha ) + \ ... \ ... \ , f_n (0) \) for some \(\displaystyle n \in \mathbb{N}\) ...

\(\displaystyle = 0 + 0 + \ ... \ ... \ , f_\alpha ( x_\alpha ) + \ ... \ ... \ , + 0 \) since \(\displaystyle f_\alpha\) is an \(\displaystyle R\)-linear map for all \(\displaystyle \alpha\)

\(\displaystyle = f_\alpha ( x_\alpha )\)

Is that correct?

Peter

 

[steenis]

Your posts #12, #13, and #14, all correct.
Have fun in Victoria.

 

[Math Amateur]

Your posts #12, #13, and #14, all correct.
Have fun in Victoria.

THanks Steenis ...

... very helpful to have your confirmation ...

Peter

 

[Math Amateur]

Hi steenis ...

Continuing the proof of ...

\(\displaystyle \Delta\) is a finite set \(\displaystyle \Longrightarrow \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \) ... ...Have already shown that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...... and have shown that \(\displaystyle \Phi\) is well defined ...
Now show that \(\displaystyle \Phi \circ u_\alpha = f_\alpha\) for all \(\displaystyle \alpha\)
Let \(\displaystyle x_\alpha \in M_\alpha\) ...

... now ...

\(\displaystyle \Phi \circ u_\alpha ( x_\alpha )\)

\(\displaystyle = \Phi ( 0, 0, \ ... \ ... \ , x_\alpha, \ ... \ ... \ ,0 ) \)

\(\displaystyle = \sum_{\alpha} f_\alpha ( x_\alpha )\)

\(\displaystyle = f_1 (0) + f_2 (0) + \ ... \ ... \ , f_\alpha ( x_\alpha ) + \ ... \ ... \ , f_n (0) \) for some \(\displaystyle n \in \mathbb{N}\) ...

\(\displaystyle = 0 + 0 + \ ... \ ... \ , f_\alpha ( x_\alpha ) + \ ... \ ... \ , + 0 \) since \(\displaystyle f_\alpha\) is an \(\displaystyle R\)-linear map for all \(\displaystyle \alpha\)

\(\displaystyle = f_\alpha ( x_\alpha )\)

Is that correct?

Peter

Hi steenis ...

Continuing the proof of ...

\(\displaystyle \Delta\) is a finite set \(\displaystyle \Longrightarrow \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha \) ... ...Have already shown that \(\displaystyle \Phi\) is an \(\displaystyle R\)-linear map ...... and have shown that \(\displaystyle \Phi\) is well defined ...
... and, further, have shown that \(\displaystyle \Phi \circ u_\alpha = f_\alpha\) for all \(\displaystyle \alpha\)

Now ... show that \(\displaystyle \Phi\) is unique ... ... ... let $\Psi:\prod_\Delta M_\alpha \longrightarrow N$ and $\Psi \circ u_\alpha = f_\alpha$,Then we have ...\(\displaystyle \Phi ( ( x_\alpha ) ) \)\(\displaystyle = \sum_\Delta f_\alpha ( x_\alpha )\) \(\displaystyle = \sum_\Delta \Psi \circ u_\alpha ( x_\alpha )\) \(\displaystyle = \Psi ( \sum_\Delta u_\alpha ( x_\alpha ) )\) \(\displaystyle = \Psi ( ( x_\alpha ) )\)Hence \(\displaystyle \Phi = \Psi\) ...
Now, we have shown that :$\bullet$ $\Phi$ is an R-map
$\bullet$ $\Phi$ is well defined
$\bullet$ $\Phi \circ u_\alpha = f_\alpha$, for all $\alpha \in \Delta$.
$\bullet$ $\Phi$ is unique... so that if \(\displaystyle \Delta\) is a finite set, then \(\displaystyle S = \prod_\Delta M_\alpha\) is the external direct sum of \(\displaystyle \{ M_\alpha \}\) ... ... ... that is \(\displaystyle S = \bigoplus_\Delta M_\alpha\) ... ...Thus we have shown that if \(\displaystyle \Delta\) is a finite set then \(\displaystyle S = \prod_\Delta M_\alpha = \bigoplus_\Delta M_\alpha\) ... ... Is that correct ...

Peter

 

[steenis]

It is correct Peter,