MHB Modules - Generating and Cogenerating Classes - Bland - Chapter 4, Section 4.1

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Chapter 4, Section 4.1 on generating and cogenerating classes and need help with the proof of Proposition 4.1.1.

Proposition 4.1.1 and its proof read as follows:

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I need some help with what seems a fairly intuitive step in the logic of the proof of $$(3) \Longrightarrow (4)$$ - see text above.In the proof of $$(3) \Longrightarrow (4)$$ Bland writes:

" ... ... Thus, by (3) there is a finite set $$F \subseteq \Delta$$ and an epimorphism $$ \phi \ : \ R^{ ( F ) } \longrightarrow M$$.

If we let

$$F = \{ 1,2, \ ... \ ... \ ,n \}$$

and if

$$\{ e_i \}_{i =1}^n$$

is the canonical basis for the free $$R$$-module $$R^{ ( n ) }$$, then the finite set

$$X = \{ \phi (e_i) \}_{i =1}^n$$

will generate $$M$$. ... ... "My question is the following:

Why, exactly, if $$\{ e_i \}_{i =1}^n$$ is the canonical basis for the free $$R$$-module $$R^{ ( n ) }$$, are we guaranteed that $$X$$ will generate $$M$$? [this does seem intuitive - but why EXACTLY! ]I would really appreciate some help with this issue.Peter***EDIT***

I now have a second question:

In the above text, Bland writes:

" ... ... Every R-module is the homomorphic image of a free R-module, ... ... "

So if that is true, then we have a set $$\Delta$$ and a homomorphism $$R^{ ( \Delta ) } \longrightarrow M$$ ... ...

... ... BUT ... ... Bland claims we have an epimorphism $$R^{ ( \Delta ) } \longrightarrow M$$ ...

How do we know that we not only have a homomorphism, but that we have an epimorphism?

Hep will be appreciated ...

 

[Euge]

Take $m \in M$. Since $\phi : R^{(n)} \to M$ is surjective, there exists an $x \in R^{(n)}$ such that $\phi(x) = m$. Since $\{e_i\}_{i = 1}^n$ is the canonical basis for $R^{(n)}$, there are unique $R$-scalars $r_1,\ldots, r_n$ such that $x = \sum_{i = 1}^n r_i e_i$. Since $\phi$ is an $R$-homomorphism,

$$m = \phi(x) = \sum_{i = 1}^n r_i \phi(e_i).$$

As $m$ is arbitrary, this shows that $\{\phi(e_i)\}_{i = 1}^n$ generates $M$.

 

[Math Amateur]

Take $m \in M$. Since $\phi : R^{(n)} \to M$ is surjective, there exists an $x \in R^{(n)}$ such that $\phi(x) = m$. Since $\{e_i\}_{i = 1}^n$ is the canonical basis for $R^{(n)}$, there are unique $R$-scalars $r_1,\ldots, r_n$ such that $x = \sum_{i = 1}^n r_i e_i$. Since $\phi$ is an $R$-homomorphism,

$$m = \phi(x) = \sum_{i = 1}^n r_i \phi(e_i).$$

As $m$ is arbitrary, this shows that $\{\phi(e_i)\}_{i = 1}^n$ generates $M$.

Thanks for your help, Euge ...

... much appreciated ...Can you please help with the second question ... ... as follows ... ...

In the above text, Bland writes:

" ... ... Every R-module is the homomorphic image of a free R-module, ... ... "

So if that is true, then we have a set Δ and a homomorphism $$R^{ ( \Delta ) } \longrightarrow M$$ ... ...

... ... BUT ... ... Bland claims we have an epimorphism $$R^{ ( \Delta ) } \longrightarrow M$$ ...

How do we know that we not only have a homomorphism, but that we have an epimorphism?

Hope you can help,

Peter
Peter

 

[Euge]

Since $M$ is the homomorphic image of a free $R$-module, $M = \phi(R^{(\Delta)})$ for some set $\Delta$ and some $R$-homomorphism $\phi$. This $\phi$ must then be an epimorphism from $R^{(\Delta)}$ onto $M$.

 

[Math Amateur]

Since $M$ is the homomorphic image of a free $R$-module, $M = \phi(R^{(\Delta)})$ for some set $\Delta$ and some $R$-homomorphism $\phi$. This $\phi$ must then be an epimorphism from $R^{(\Delta)}$ onto $M$.

Thanks Euge ... most helpful ... as usual ...

Peter