Start on Bland Problem 1, Problem Set 4.1: Generating & Cogenerating Modules

[steenis]

Very good.

Mention that (2) follows from prop.2.1.5

Mention that $g_\alpha(y_\alpha) \in \text{ I am } g_\alpha$

therefore $x = g((y_\alpha)) = \Sigma_\Delta g_\alpha(y_\alpha) \in \Sigma_\Delta \text{ I am } g_\alpha$Very good that you changed the polluted notation to avoid confusion.

 

[Math Amateur]

Very good.

Mention that (2) follows from prop.2.1.5

Mention that $g_\alpha(y_\alpha) \in \text{ I am } g_\alpha$

therefore $x = g((y_\alpha)) = \Sigma_\Delta g_\alpha(y_\alpha) \in \Sigma_\Delta \text{ I am } g_\alpha$Very good that you changed the polluted notation to avoid confusion.

Thanks for those points of improvement ...

The previously polluted notation ... :) ... was a helpful lesson for me ...

Thanks for all your help ... I learned a lot ...

Peter

 

[steenis]

Ok, now prove the converse direction.

I have to go now, I have other things to do.
I will check your posts this evening or tomorrow, dutch time.

 

[Math Amateur]

Ok, now prove the converse direction.

I have to go now, I have other things to do.
I will check your posts this evening or tomorrow, dutch time.

Oh Lord ... forgot the converse ... !

Had gone onto something else ... see new problem ...

But will return to converse...

Peter

 

[Math Amateur]

Let $$M$$ and $$N$$ be R-modules ...

We want to show that:

$$M$$ generates $$N \Longleftrightarrow \exists$$ a subset $$H \subseteq \text{ Hom }_R (M, N)$$ such that $$N = \sum_H \text{Im } f$$In this post we show that

$$M$$ generates $$N \Longrightarrow \exists$$ a subset $$H \subseteq \text{ Hom }_R (M, N)$$ such that $$N = \sum_H \text{Im } f$$So we assume that $$M$$ generates $$N$$ ...

Now ... $$M$$ generates $$N$$

$$\Longrightarrow \exists$$ a set $$\Delta$$ and an epimorphism $$g \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N$$ ... ... by definition 4.1.2 ...

Now, for each $$\alpha \in \Delta$$ define $$g_\alpha = g \circ i_\alpha$$ where $$i_\alpha \ : \ M \rightarrow \bigoplus_\Delta M $$ ... that is, $$i_\alpha$$ is the canonical inclusion/injection ... ...

Now, since $$g$$ and $$i_\alpha$$ are homomorphisms, it follows that each $$g_\alpha \in \text{ Hom }_R (M, N)$$ ... ...Define $$H = \{ g_alpha \mid \alpha \in \Delta \}$$

We now claim that $$N = \sum_H \text{Im } g_\alpha$$ where $$g_\alpha \ : \ M \rightarrow N$$ is as defined above ...

But, we have that $$\sum_H \text{Im } g_\alpha \subseteq N$$ since each $$g_\alpha$$ is a mapping from $$M$$ to $$N$$ ...

So we need to show $$N \subseteq \sum_H \text{Im } g_\alpha$$ So ... let $$x \in N$$ ...

Now, $$g$$ is an epimorphism from $$M^{ ( \Delta ) } = \bigoplus_\Delta M to N$$ ...

... so ... $$\exists$$ some $$( y_\alpha ) \in M^{ ( \Delta ) } = \bigoplus_\Delta M$$ such that $$g( ( y_\alpha ) ) = x$$ ... ... ... (1)

But $$g( ( y_\alpha ) ) = \sum_\Delta g_\alpha ( y_\alpha )$$ ... ... ... (2) (1)(2) $$\Longrightarrow g_\alpha ( y_\alpha ) = x$$

Therefore $$x \in \sum_H \text{Im } g_\alpha $$Is that correct?

(Note: most unsure of last step ... )

Peter

Let $$M$$ and $$N$$ be R-modules ...

We want to show that:

$$M$$ generates $$N \Longleftrightarrow \exists$$ a subset $$H \subseteq \text{ Hom }_R (M, N)$$ such that $$N = \sum_H \text{Im } g$$In this post we show that

$$\exists$$ a subset $$H \subseteq \text{ Hom }_R (M, N)$$ such that $$N = \sum_H \text{Im } g \Longrightarrow M$$ generates $$N$$ So ... assume that $$\exists$$ a subset $$H \subseteq \text{ Hom }_R (M, N)$$ such that $$N = \sum_H \text{Im } g $$ ...

Let $$H = \{ h_\alpha \mid \alpha \in \Delta \}$$

Define $$f \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N$$ where $$f = \sum_H g = \sum_\Delta h_\alpha$$ ...

But ... given this we know that then $$f(( x_\alpha )) = \sum_\Delta h_\alpha ( x_\alpha )$$ ...Now ,,, we claim that $$f$$ is an epimorphism ...To demonstrate this let $$x \in N$$ ...

Then $$x \in \sum_H \text{Im } g = \sum_\Delta \text{Im } h_\alpha$$ ...

Therefore $$\exists \ y \in M^{ ( \Delta ) }$$ such that $$f(y) = \sum_\Delta h_\alpha (y) = x $$ ...

So $$f$$ is an epimorphism ... and hence $$M$$ generates $$N$$ ...

Is that correct?

Peter

 

[steenis]

Very good, Peter, I knew you can do it !A very small comment, the last lines are a little bit confusing and not quite correct.
I would write that something like this

Let $y \in N = \sum_\Delta \text{ I am } h_\alpha$.

Then there are $x_\alpha \in M$ for all $\alpha \in \Delta$, such that $y = \sum_\Delta h_\alpha(x_\alpha)$.

Let $x = (x_\alpha)_\Delta$, then $x \in M^{(\Delta)}$ and $f(x) =f((x_\alpha)) = \sum_\Delta h_\alpha(x_\alpha)$ = y.

Therefore $f$ is an epimorpism.

 

[Math Amateur]

Very good, Peter, I knew you can do it !A very small comment, the last lines are a little bit confusing and not quite correct.
I would write that something like this

Let $y \in N = \sum_\Delta \text{ I am } h_\alpha$.

Then there are $x_\alpha \in M$ for all $\alpha \in \Delta$, such that $y = \sum_\Delta h_\alpha(x_\alpha)$.

Let $x = (x_\alpha)_\Delta$, then $x \in M^{(\Delta)}$ and $f(x) =f((x_\alpha)) = \sum_\Delta h_\alpha(x_\alpha)$ = y.

Therefore $f$ is an epimorpism.

Thanks for all the help, Steenis ... thanks to you I understand a lot more than when I started working on the problem ...

Regarding the last few lines of the above proof I thought I'd write out explicitly what is going for the case $$\Delta = \{ 1,2, \ ... \ ... \ , \ n \}$$

We have $$f \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N$$ where $$f = \sum_H g = \sum_\Delta h_\alpha$$ ... and $$h_\alpha \ : \ M \rightarrow N$$ ... ...

... and so ... $$f(( x_\alpha )) = \sum_\Delta h_\alpha ( x_\alpha )$$ ...
Now to show or illustrate explicitly in the case of \Delta = \{ 1,2, \ ... \ ... \ , \ n \} how f is an epimorphism ... ( ... not meant to be a proof ... just an illustration ...)Assume $$y_1 \in \text{Im } h_1 \Longrightarrow \exists \ x_1$$ such that $$h_1 ( x_1 ) = y_1$$ ...

and ... $$y_2 \in \text{Im } h_2 \Longrightarrow \exists \ x_2$$ such that $$h_2 ( x_2 ) = y_2$$

... ... ... ... ...

... ... ... ... ...

and ... $$y_n \in \text{Im } h_n \Longrightarrow \exists \ x_n$$ such that $$h_n ( x_n ) = y_n $$ Then ... $$\exists \ y \in N$$ such that $$y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n$$ ... ... since $$N$$ is a module ...

and then $$y \in \text{Im } h_1 + \in \text{Im } h_2 + \in \text{Im } h_3 + \ ... \ ... \ + \text{Im } h_n$$Now $$y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n \Longrightarrow y = h_1(x_1) + h_2(x_2) + h_3(x_3) + \ ... \ ... \ + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) $$

That is $$y = f(x)$$ where $$x = ( x_1, x_2, x_3, \ ... \ ... \ , x_n ) = (x_\alpha)_\Delta$$ ...

So $$f$$ is an epimorphism ...Is that basically correct ...

Peter

 

[steenis]

It is basically correct, but change:

Then ... $$\exists \ y \in N$$ such that $$y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n$$ ... ... since $$N$$ is a module ...

and then $$y \in \text{Im } h_1 + \in \text{Im } h_2 + \in \text{Im } h_3 + \ ... \ ... \ + \text{Im } h_n$$

Now $$y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n \Longrightarrow y = h_1(x_1) + h_2(x_2) + h_3(x_3) + \ ... \ ... \ + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) $$

into:
Define $y = y_1 + y_2 + \cdots + y_n$ then $y \in N$, because N is a module.

And $y = h_1(x_1) + h_2(x_2) + \cdots + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) \in \Sigma_\Delta \text{ I am } h_\alpha = N$