MHB Modules Generated by Sets of Submodules .... .... Bland Problem 2, Problem Set 4.1 .... ....

[Math Amateur]

Good, take your time

As I see it ... we are given a family of R-modules $\{M_\alpha \mid \alpha \in \Delta\}$ ... then we can take any R-module N and any family of R-linear mappings $\{\phi_\alpha \mid M_\alpha \to N\}$ and then be assured that there exists a uniqe R-linear mapping $\phi:\bigoplus_\Delta M_\alpha \to N$ satisfying

$\phi \circ i_\alpha = \phi_\alpha$, where

$i_\alpha : M_\alpha \to \bigoplus_\Delta M_\alpha$ are the canonical injections ...

I believe that is the correct logic for constructing $$\phi$$ ...Now to repeat the process with $$\Delta = \text{ Hom } ( M, N )$$ it seems to me we just replace $$\Delta$$ with $$\text{ Hom } ( M, N )$$ ... this is where my understanding may be a bit shallow since replacing the index set doesn't seem to me to make much difference to the proof ... so I may be missing something of significance ...

Peter

 

[steenis]

That is a correct interpretation. But now you are applying this proposition, sot you have to fill in the "unknowns", as you well know.

$\text{Hom }(M, N)$ is a very special index-set. Furthermore, if you follow the procedure you see that the R-maps $\phi_\alpha$ are not defined. So you have to define them.

 

[Math Amateur]

That is a correct interpretation. But now you are applying this proposition, sot you have to fill in the "unknowns", as you well know.

$\text{Hom }(M, N)$ is a very special index-set. Furthermore, if you follow the procedure you see that the R-maps $\phi_\alpha$ are not defined. So you have to define them.

When you defined the $$\phi_\alpha$$ as $$\phi_\alpha = s$$ where $$s \in \text{ Hom } ( M, N ) $$ ... ... was that a sufficiently precise definition ... or do we need to define the mapping between elements of the domain and image sets ...

I understand that ..." ... $\text{Hom }(M, N)$ is a very special index-set. ... " but how it affects the logic of the proof apart form specifying the index I don't see ...

Still thinking ...

Peter

 

[steenis]

No, because we use $\text{Hom }(M, N)$ as an index set.

If you have a theorem and you want to apply this theorem what do you do ?

 

[Math Amateur]

No, because we use $\text{Hom }(M, N)$ as an index set.

If you have a theorem and you want to apply this theorem what do you do ?

You make sure that the conditions in the hypothesis are met ,,,

 

[steenis]

So do that, apply prop.2.1.5 to our situation

 

[Math Amateur]

So do that, apply prop.2.1.5 to our situation

I suspect that will require some forward looking and creative definitions ... but I'll try ..

I am still very unsure how using $$\text{ Hom } ( M, N )$$ as an index affect things or enters the proof in any significant way ...

Must go to sleep now ... it is after midnight ...

Will continue in the morning ...

Peter*** EDIT ***

Do you feel able to give any further hints on $$N/ \text{ I am } \phi $$

 

[steenis]

Suppose $\phi$ is not surjective, what can you say about $\text{ I am } \phi$ and $N/\text{ I am }\phi$ ?

 

[Math Amateur]

Suppose $\phi$ is not surjective, what can you say about $\text{ I am } \phi$ and $N/\text{ I am }\phi$ ?

If \phi is not surjective then $$\text{ I am } \phi \ne N$$ and indeed $$\text{ I am } \phi \subset N$$ ...

But $N/\text{ I am }\phi$ ... ... what can we say ...?

PeterEDIT ... are you thinking of Correspondence Theorem ..?

 

[steenis]

If $\text{ I am }\phi \neq N$ then $N/\text{ I am }\phi \neq 0$. What can you say about the canonical projection $v:N \to N/\text{ I am }\phi$ ?

 

[Math Amateur]

If $\text{ I am }\phi \neq N$ then $N/\text{ I am }\phi \neq 0$. What can you say about the canonical projection $v:N \to N/\text{ I am }\phi$ ?

You can say that $$v \ne 0$$ ... that is v is not equal to the zero map ...

 

[steenis]

what next ?

 

[Math Amateur]

what next ?

Not sure what else follows ...

But it is nearly 2am ...

Can you give me a hint to continue with tomorrow morning ,,.

Falling asleep ...

Peter

 

[steenis]

Apply the hypothesis ...

 

[steenis]

In mathematics, an index set is a set whose members label (or index) members of another set. For instance, if the elements of a set $A$ may be indexed or labeled by means of a set $J$, then $J$ is an index set. The indexing consists of a surjective function from $J$ onto $A$ and the indexed collection is typically called an (indexed) family, often written as $(A_j)_{j\in J}$, or $\{x_j \in A\}_{j\in J}$. [See: Munkres – Topology (2000) p.36]
As far as I know, every set can be used as an index-set, as long as it is a set.

 

[Math Amateur]

Apply the hypothesis ...

In an attempt to apply the hypothesis in a way that takes account of the hints you have given ... in the hope that something emerges ... like a contradiction or a chance to apply a contrapositive ... Put $$f = v$$ where $$v \ : \ N \to N / \text{ I am } \phi$$ is such that $$v(x) = \overline{x} + \text{ I am } \phi$$ ...Also put $$h = \phi_\alpha$$ ...Then we have the situation in Figure 2 below ...https://www.physicsforums.com/attachments/8164

BUT ... the situation with the definitions I've made is perplexing ... if $$\phi$$ is not an epimorphism ... then $$f = v \ne 0$$ as required by the converse hypothesis ... and I suspect (but not completely sure... ) that $$f \circ h = v \circ \phi_\alpha$$ is also not equal to zero ... but then the conditions of the converse hypothesis are fulfilled ... assuming that (contrary to the implications of the problem statement) $$\phi$$ is NOT an epimorphism ... ? ... something wrong ... possibly $$f \circ h = v \circ \phi_\alpha$$ is NOT non-zero ... Can you clarify ...

Are you able to hint or show me how to make a better set of definitions that still take account of your hints regarding $$\text{ I am } \phi$$ and $$N / \text{ I am } \phi$$ ...Peter
***EDIT***

Attempt to investigate whether $$f \circ h \ne 0$$We have ...

$$f \circ h ( x_\alpha ) = v \circ \phi_\alpha ( x_\alpha )$$$$= v \circ \phi \circ i_\alpha ( x_\alpha ) $$$$= v \circ \phi ( 0 ... ... x_\alpha ... ... 0 ) $$ $$= v ( \sum_{ \text{ Hom } ( M, N ) } ( x_\beta )$$ where $$x_\beta = 0$$ when $$\beta \ne \alpha$$ and $$x_\beta = x_\alpha$$ when $$\beta = \alpha$$$$= v \circ \phi_\alpha ( x_\alpha )$$ since $$\phi_\alpha (0) = 0$$ $$= \overline{ \phi_\alpha ( x_\alpha ) } + \text{ I am } \phi $$
... ... hmmm ... don't seem to be getting anywhere ... is $$\overline{ \phi_\alpha ( x_\alpha ) } + \text{ I am } \phi \ne 0$$ ... ?Peter

 

[steenis]

What is $\phi_\alpha$ ? As I told you in post #18, we cannot go further without knowing $\phi_\alpha$

In the posts #14, #18, #20, #23, #28, and #32, I insisted that you define $\phi_\alpha$. As long as you refuse to do so, I cannot help you, and I will stop here.
Furthermore, I told you several times what $\phi_\alpha$ is, clearly you refuse to accept that.
So come back if you are willing to define $\phi_\alpha$.

 

[Math Amateur]

What is $\phi_\alpha$ ? As I told you in post #18, we cannot go further without knowing $\phi_\alpha$

In the posts #14, #18, #20, #23, #28, and #32, I insisted that you define $\phi_\alpha$. As long as you refuse to do so, I cannot help you, and I will stop here.
Furthermore, I told you several times what $\phi_\alpha$ is, clearly you refuse to accept that.
So come back if you are willing to define $\phi_\alpha$.

Sorry steenis ... I see that you did in fact suggest a definition for $$\phi_\alpha$$ ... my apologies for forgetting this ...

You suggested defining $$\phi_g = g$$ where $$g \in \text{ Hom } ( M, N )$$ ...

so that we can construct a unique R-map ...

$$\phi \ : \bigoplus_{ \text{ Hom } ( M, N ) } M_g \to N$$ satisfying

$$\phi \circ i_g = \phi_g = g $$

where $$i_g \ : \ M_g \to \bigoplus_{ \text{ Hom } ( M, N ) } M_g$$ are the canonical injections ...The constructed R-map $\phi$ is then defined on $\bigoplus_{\text{Hom }(M, N)} M_g$ in this way:

$\phi((x_g)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } g(x_g)$ for $((x_g)_{\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_g$

but ... (my apologies) ... I have an issue ... ... the family of mappings $$\{ \phi_g \ : \ M_g \to N \}$$ are already R-linear maps or R-module homomorphisms (see Proposition 2,1.5 ) ... so how does putting $$\phi_g = s$$ where $$g \in \text{ Hom } ( M, N )$$ ... that is declaring $$\phi_g$$ a member of $$\text{ Hom } ( M, N )$$ ... that is saying that $$\phi_g$$ is a homomorphism ... how does that advance our knowledge of the specific nature of the $$\phi_g$$ ... they were already homomorphisms by Proposition 2,1,5 ...Can you clarify ...

Maybe I am missing something regarding the index set being $$\text{ Hom } ( M, N )$$ ... ... Peter

 

[steenis]

Define the unknown $x$ to be the known $3$ then: $x = 3$

You define the unknown $\phi_f$ to be the known $f$ then : $\phi_f = f$, is perfectly allright

If you define $\phi_f = f$ for $f \in \text{Hom }(M, N)$ then your set $\{\phi_f = f | f \in \text{Hom }(M, N)\}$ becomes $\text{Hom }(M, N)$:

$\{\phi_f = f | f \in \text{Hom }(M, N)\} = \text{Hom }(M, N)$

Don't use s as an index-variable, we talking about R-maps so use f, g, h, or k or so.

 

[Math Amateur]

Define the unknown $x$ to be the known $3$ then: $x = 3$

You define the unknown $\phi_f$ to be the known $f$ then : $\phi_f = f$, is perfectly allright

If you define $\phi_f = f$ for $f \in \text{Hom }(M, N)$ then your set $\{\phi_f = f | f \in \text{Hom }(M, N)\}$ becomes $\text{Hom }(M, N)$:

$\{\phi_f = f | f \in \text{Hom }(M, N)\} = \text{Hom }(M, N)$

Don't use s as an index-variable, we talking about R-maps so use f, g, h, or k or so.

Changed s to g ... ... see post #48 ...

Do you have any comment about my choices for h, f and N' in post #46 ... given that now we would have $$h = \phi_g = g$$ ... ...

Peter

 

[steenis]

Sorry I overlooked the g, I thought you were using s
No comment on h and f in #46

Like you said in post #49, we want to declare that unknown $\phi_g$ to be a member of $\text{Hom }(M, N)$, that is what we want to do, you said it yourself. In that case $\phi_g$ is defined.
To make the assignment easy, we define $\phi_g = g \in \text{Hom }(M, N)$

 

[Math Amateur]

Sorry I overlooked the g, I thought you were using s
No comment on h and f in #46

Like you said in post #49, we want to declare that unknown $\phi_g$ to be a member of $\text{Hom }(M, N)$, that is what we want to do, you said it yourself. In that case $\phi_g$ is defined.
To make the assignment easy, we define $\phi_g = g \in \text{Hom }(M, N)$

I was using s ... but changed it to g based on your remarks ..

 

[steenis]

Hi Peter,

Do you want to continue this exercise ? If you want we can start from the beginning to get you on track.

Do not give up.

 

[Math Amateur]

Hi Peter,

Do you want to continue this exercise ? If you want we can start from the beginning to get you on track.

Do not give up.

Hi steenis ... yes of course ...

... good idea ... I value your help and assistance ...

Peter

 

[steenis]

Good, you are not angry at me.
We will make a fresh atart.
Give me some time.
I,ll be back very soon.

 

[Math Amateur]

Good, you are not angry at me.
We will make a fresh atart.
Give me some time.
I,ll be back very soon.

OK ... great ...

Thanks steenis ...

Peter

 

[steenis]

Bland - Rings and Their Modules (2011)
Exercise 2 of Problem Set 4.1

$\Leftarrow)$ $M$ and $N$ are right $R$-modules.
For each nonzero R-linear mapping $f:N \to N’$ (for any $R$-module $N'$) there is an R-linear mapping $h:M \to N$ such that $fh \neq 0$. Prove that $M$ generates $N$.

According to definition 4.1.2. we have to find an epimorpism $\phi:M^{(\Delta)} \to N$ for some index-set $\Delta$.

$M^{(\Delta)} = \bigoplus_\Delta M_\alpha$, where $M_\alpha = M$, for all $\alpha \in \Delta$.

I suggested to take $\Delta = \text{Hom }(M,N)$ as our index-set. Why? Our hypothesis says that we can get an $R$-map $h:M \to N$ such that $fh \neq 0$, for every nonzero R-map $f:N \to N’$, that map $h$ has to come from somewhere. Sometimes you have to except something and see at the end why.

Every set can be taken as an index-set, see post #45. At this moment I could not find more information on index-sets, maybe you can find something.

We have to find an $R$-map $\phi:\bigoplus_\Delta M_\alpha \to N$. I suggested to use proposition 2.1.5. to construct $\phi$. I think it is the only tool we have to do that. (After we have found $\phi$, we will prove that it is surjective)Proposition 2.1.5
If $\{M_\alpha | \alpha \in \Delta\}$ is a family of R-modules, then the direct sum $(\bigoplus_\Delta M_\alpha, i_\alpha)$,
has the property that for every $R$-module $N$ and every family $\{\phi_\alpha:M_\alpha \to N | \alpha \in \Delta\}$
there is a unique R-map $\phi:\bigoplus_\Delta M_\alpha \to N$
such that $\phi \circ i_\alpha = \phi_\alpha$, for all $\alpha \in \Delta$,
where the $i_\alpha$ are the canonical injections $i_\alpha:M_\alpha \to \bigoplus_\Delta M_\alpha$ for $\alpha \in \Delta$,

The $R$-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:

$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$Now I want you to write down the premises or conditions of this theorem and translate them into the premises in our case. For instance: the $\Delta$ in the theorem, becomes $\text{Hom }(M,N)$ in our case, and so on. (see post #35). Use a better index-variable than $\alpha$ ...
Good luck.

 

[Math Amateur]

Bland - Rings and Their Modules (2011)
Exercise 2 of Problem Set 4.1

$\Leftarrow)$ $M$ and $N$ are right $R$-modules.
For each nonzero R-linear mapping $f:N \to N’$ (for each $R$-module $N'$) there is an R-linear mapping $h:M \to N$ such that $fh \neq 0$. Prove that $M$ generates $N$.

According to definition 4.1.2. we have to find an epimorpism $\phi:M^{(\Delta)} \to N$ for some index-set $\Delta$.

$M^{(\Delta)} = \bigoplus_\Delta M_\alpha$, where $M_\alpha = M$, for all $\alpha \in \Delta$.

I suggested to take $\Delta = \text{Hom }(M,N)$ as our index-set. Why? Our hypothesis says that we can get an $R$-map $h:M \to N$ such that $fh \neq 0$, for every nonzero R-map $f:N \to N’$, that map $h$ has to come from somewhere. Sometimes you have to except something and see at the end why.

Every set can be taken as an index-set, see post #45. At this moment I could not find more information on index-sets, maybe you can find something.

We have to find an $R$-map $\phi:\bigoplus_\Delta M_\alpha \to N$. I suggested to use proposition 2.1.5. to construct $\phi$. I think it is the only tool we have to do that. (After we have found $\phi$, we will prove that it is surjective)Proposition 2.1.5
If $\{M_\alpha | \alpha \in \Delta\}$ is a family of R-modules, then the direct sum $(\bigoplus_\Delta M_\alpha, i_\alpha)$,
has the property that for every $R$-module $N$ and every family $\{\phi_\alpha:M_\alpha \to N | \alpha \in \Delta\}$
there is a unique R-map $\phi:\bigoplus_\Delta M_\alpha \to N$
such that $\phi \circ i_\alpha = \phi_\alpha$, for all $\alpha \in \Delta$,
where the $i_\alpha$ are the canonical injections $i_\alpha:M_\alpha \to \bigoplus_\Delta M_\alpha$ for $\alpha \in \Delta$,

The $R$-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:

$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$Now I want you to write down the premises or conditions of this theorem and translate them into the premises in our case. For instance: the $\Delta$ in the theorem, becomes $\text{Hom }(M,N)$ in our case, and so on. (see post #35). Use a better index-variable than $\alpha$ ...
Good luck.

Hi steenis ...

Will now try to write down the premises or conditions of this theorem and translate them into the premises in our case. We set up the premises of Proposition 2,1,5 if we accept the existence of a family of modules $\{M_\alpha \mid \alpha \in \Delta\}$ and define or specify (in some way) an R-module N and a family of R-maps $\{\phi_\alpha:M_\alpha \to N | \alpha \in \Delta\}$ ... ... For our problem we define $$\Delta = \text{ Hom } (M, N)$$ ...

... and the family of modules is $$\{ M_g \mid g \in \text{ Hom } (M, N) \} $$

The $$M_\alpha = M_g = M$$ for all $$g \in \text{ Hom } (M, N)$$ ...

We further specify $$\phi_\alpha = \phi_g = g \in \text{ Hom } (M, N)$$ ...

... (that is the premises) ...
... then ...... there is a unique R-map $$\phi \ : \ \bigoplus_{\text{ Hom } (M, N)} M_g \to N$$

such that ...

$$\phi \circ i_g = \phi_g = g$$ for all $$g \in \text{ Hom } (M, N)$$ ...

where the $$i_g$$ are the canonical injections $$i_g \ : \ M_g \to \bigoplus_{\text{ Hom } (M, N)} M_g$$ for all $$g \in \text{ Hom } (M, N)$$ ...then the R-map $$\phi$$ is defined on $$\bigoplus_{\text{ Hom } (M, N)} M_g$$ as follows:$$\phi ( ( x_g)_{\text{ Hom } (M, N)} ) = \sum_{\text{ Hom } (M, N)} \phi_g ( x_g )$$ for $$( x_g)_{\text{ Hom } (M, N)} \in \bigoplus_{\text{ Hom } (M, N)} M_g $$... and then ... by the hypothesis of the problem we are given an R-module $$N'$$ and an R-map $$f \ : \ N \to N$$' and an R-map $$h \ : \ M \to N$$ such that $$f \circ h \ne 0$$ ...
Hope the above is basically correct ...

Peter

 

[steenis]

Very good.

Notice that $\{\phi_g = g : M_g = M \to N\} = \text{Hom }(M, N)$, so you translate $\{\phi_\alpha:M_\alpha \to N | \alpha \in \Delta\}$ in the theory to
$\text{Hom }(M, N) $ in our case. If you replace $\phi_g$ by $g$ in your last formula, it will even more clear.
It then becomes:
$\phi((x_g)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } g(x_g)$ for $((x_g)_ {\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_g$

$g(x_g)$: notice that g acts on the $g$-th coordinate of $((x_g)_ {\text{Hom }(M, N)})$

The hypothesis does not give a $N'$, no we can apply the hypothesis using any appropriate $N'$, (I made an edit in post #57), we use that in the foloowing

Now prove that $\phi$ is surjective. Suppose $\phi$ is not surjective and work to a contradaction.

- - - Updated - - -

I have made an edit in #57, see #59.

 

[steenis]

You should now have this diagram:

\begin{tikzpicture}[>=stealth]
\node (A) at (1,0) {$M_g= M$};
\node (B) at (1,-5) {$\bigoplus_{Hom (M, N)} M_g$};
\node (C) at (6,-5) {$N$};
\draw[->] (A) -- node

{$i_g$} (B);
\draw[->] (A) -- node

{$g$} (C);
\draw[->] (B) -- node [below] {$\phi$} (C);
\end{tikzpicture}

I have made an edit in #57, see #59.​