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

[Math Amateur]

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.

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

Hi steenis ...

You write:

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


So we proceed as follows:

Assume that $$\phi$$ is not surjective ...

... then $$\text{ I am } \phi \ne N$$ ... indeed $$\text{ I am } \phi \subset N$$ ... that is a proper subset ...

Now define $$v \ : \ N \to N/ \text{ I am } \phi$$ where $$v(n) = n + \text{ I am } \phi$$ ... and hence $$v \ne 0 $$ ...But then ... since we have a non-zero mapping $$f = v$$ ... by the hypothesis of the converse problem we have that there exists a non-zero R-linear mapping $$h \ : \ M \to N$$ such that $$f \circ h = v \circ h \ne 0$$ ...

But ... all the R-linear mappings from $$M$$ to $$N$$ belong to $$\text{ Hom } (M, N)$$ ...

So ... that is ... $$\exists \ g^\star \ne 0$$ in $$\text{ Hom } (M, N)$$ such that $$v \circ g^\star \ne 0$$ ...But ... $$v \circ g^\star \ne 0 \Longrightarrow \exists \ x_{g^\star } \in M_{g^\star } = M$$ such that $$v( g^\star ( x_{g^\star } ) ) \ne 0$$ ...

Therefore $$g^\star (x_{g^\star}) \notin \text{ I am } \phi$$ ... ... ... ... ... (*)But $$\phi (( x_g) = \sum g ( x_g )$$ ... and this sum includes $$g^\star (x_{g^\star })$$ ... and this in turn implies $$g^\star ( x_{g^\star }) \in \text{ I am } \phi$$ ... but this contradicts (*) above ...Thus our assumption that $$\phi$$ is not surjective must be wrong ..

So we conclude that $$\phi$$ is surjective and hence M generates N ...
Is that correct ...?

Peter

 

[steenis]

I am sorry, I have no time today, I have other things to do, sorry.

 

[Math Amateur]

I am sorry, I have no time today, I have other things to do, sorry.

Sure ... fine steenis ... no worries ...

Talk to you when you're back ...

Hope all goes well ...

Peter

 

[steenis]

Well done, Peter, that is correct.
But I will rewrite it for you, because you have a tendency to make things too complicated.

There is an R-mao $h:M \to N$ such that $v \circ h \neq 0$.

So there is an $x \in M$ such that $v(h(x)) \neq 0$, this means $h(x) \notin \text{ I am } \phi$.

Take $((x_g)_ {\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_g$ such that $x_h = x$ and $x_g = 0$ for $g \neq h$.

Then $\phi((x_g)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } g(x_g) = h(x_h) = h(x) \in \text{ I am } \phi $, that is a contradiction.

So we conclude that $\phi$ is surjective and $\phi$ is a generator of M.

 

[Math Amateur]

Well done, Peter, that is correct.
But I will rewrite it for you, because you have a tendency to make things too complicated.

There is an R-mao $h:M \to N$ such that $v \circ h \neq 0$.

So there is an $x \in M$ such that $v(h(x)) \neq 0$, this means $h(x) \notin \text{ I am } \phi$.

Take $((x_g)_ {\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_g$ such that $x_h = x$ and $x_g = 0$ for $g \neq h$.

Then $\phi((x_g)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } g(x_g) = h(x_h) = h(x) \in \text{ I am } \phi $, that is a contradiction.

So we conclude that $\phi$ is surjective and $\phi$ is a generator of M.

Hi steenis ...Sorry to be late in replying but had to watch Australia play the Czech Republic in Austria ... Australia won 4-0 :) ... Australia's new coach is Bert van Marwijk from the Netherlands!

Thanks for improvements to proof ... I will work through them carefully tomorrow morning (Tasmanian time) ..

Thanks again ...

Peter

 

[steenis]

Of course, do not forget the important things ...(Smile)

 

[Math Amateur]

Of course, do not forget the important things ...(Smile)

Thanks again for all your help on this problem ...

Now going on to work on Bland Section 4.2 Noetherian and Artinian Modules ...

Am revising Propositions 4.2.3 and 4.24 ... and then moving on to other Propositions I have not worked on before ...

Also hoping before long to cover classical ring theory (Chapter 6) ...

Peter