MHB Every submodule has at least one set of generators?

[Math Amateur]

In Paul Bland's book: Rings and Their Modules, we read the following text at the start of Section 2.2 Free Modules:https://www.physicsforums.com/attachments/3388In the above text we read:

" ... ... Every submodule has at least one set of generators, namely the set $$N$$. ... "

Now, I know that this looks an extremely plausible statement ... but it worries me ... how would we prove it ...

Can someone please help?... ... reflecting ... ...Just thinking about what would be involved in the case where $$X$$ was finite ...

We would have $$X = \{ x_1, x_2, \ ... \ ... \ , x_n \} = N$$ ... ... where $$N$$ is taken/considered as a set ...so then,$$N = \sum \nolimits_X xR$$

= $$x_1R + x_2R + \ ... \ ... \ + x_nR$$

$$= \{ x_1r_1 + x_2r_2 + \ ... \ ... \ + x_nr_n \ | \ r_1, r_2, \ ... \ ... , r_n \in R \}$$Then, if for example we consider $$x_1 \in N$$, we have to have a sum as above such that:

$$x_1 = x_1r_1 + x_2r_2 + \ ... \ ... \ + x_nr_n $$ where $$r_1, r_2, \ ... \ ... , r_n \in R$$

Now in the above $$r_1 = 1$$ and $$r_2, \ ... \ ... , r_n = 0 $$

would be fine ... equation 'works'!

BUT ... what would we do in the case of a non-unital ring/module? - problem no $$r = 1$$!

Can someone please clarify this situation for me ... as well as indicating a proof for Bland's statement above ...Peter

 

[Euge]

The smallest submodule of $M$ containing $N$ is $N$ itself, so $N$ generates itself.

 

[Evgeny.Makarov]

The smallest submodule of $M$ containing $N$ is $N$ itself, so $N$ generates itself.

According to the book, the fact that if $N$ is generated by $X$, then $N$ is the intersection of all submodules that contain $X$ is a statement that can be proved. But the definition of what it means to be generated by $X$ is $N=\sum_{x\in X}xR$. So Peter's question is legitimate.

what would we do in the case of a non-unital ring/module? - problem no $$r = 1$$!

Wikipedia says:

"Authors who do not require rings to be unital omit condition 4 above in the definition of an R-module [which says that $x1_R=x$ where $1_R$ is $R$'s multiplicative identity], and so would call the structures defined above "unital left R-modules". In this article, consistent with the glossary of ring theory, all rings and modules are assumed to be unital."

Maybe Bland also assumes that in this context $R$ has a multiplicative identity?

 

[mathbalarka]

An $R$-module $N$ is said to be generated by a set $X \subseteq N$ if $N = \{x_1 r_1 + x_2 r_2 + \cdots + x_n r_n : x_i \in X, r_i \in R\}$. This in a sense means that $N$ is "spanned" over $R$ by the set $X$. Now $N$ is an $R$-module, so every element can trivially be written as an $N$-linear combination of elements from $R$.

It's just that definitions and rigor just goes above your head sometimes - when this happens, invoke the intuitive idea behind the veil of the definitions you are working with. It helps.

 

[Evgeny.Makarov]

An $R$-module $N$ is said to be generated by a set $X \subseteq N$ if $N = \{x_1 r_1 + x_2 r_2 + \cdots + x_n r_n : x_i \in X, r_i \in R\}$. This in a sense means that $N$ is "spanned" over $R$ by the set $X$. Now $N$ is an $R$-module, so every element can trivially be written as an $N$-linear combination of elements from $R$.

This is not obvious if $R$ does not have a multiplicative unit. This was the OP's question. Also, isn't it better to call $x_1 r_1 + x_2 r_2 + \cdots + x_n r_n$ a linear combination of elements from $N$ rather than $R$? When $R$ is a field and $N$ is a vector space, we don't refer to elements of $N$ as linear combinations of numbers (i.e., elements of $R$) with vector coefficients, but linear combinations of vectors with numerical coefficients.

 

[mathbalarka]

This is not obvious if $R$ does not have a multiplicative unit.

I don't care about rings without identity :p

Also, isn't it better to call $x_1 r_1 + x_2 r_2 + \cdots + x_n r_n$ a linear combination of elements from $N$ rather than $R$?

Indeed, it's a typo. I meant $R$-linear combination of elements from $N$.

 

[Euge]

Evgeny, I never consider Peter's questions as illegitimate. To cut things short, Bland's book assumes modules to be unital.

 

[Math Amateur]

Evgeny, I never consider Peter's questions as illegitimate. To cut things short, Bland's book assumes modules to be unital.

Thanks to Euge, Evgeny and Mathbalarka for a very helpful and interesting discussion ... ... it certainly cleared up the issue for me ...

Euge was certainly right with respect to Bland's assumption regarding rings. At the start of Chapter 1 Basic Properties of Rings and Modules, Bland defines a ring and indicates/implies that it does not have to have a multiplicative identity or unit, but then writes the following:View attachment 3389Thanks again to you all ... (my God, I am sounding like a Texan :) )

Peter