MHB Is RA the Smallest Submodule of M Containing A?

[Math Amateur]

On page 351 Dummit and Foote make the following statement:

"It is easy to see using the Submodules' Criterion that for any subset A of M, RA is indeed a submodule of M and is the smallest submodule of M which contains A ... "

I am not sure how to (formally and explicitly) prove this statement.

However, reflecting on the above, it is easy to show that RA is a submodule of M, but what is worrying me is the formal proof that RA is the smallest submodule of M that contains A.

However following the statement:

"It is easy to see using the Submodules' Criterion that for any subset A of M, RA is indeed a submodule of M and is the smallest submodule of M which contains A ... " Dummit and Foote write:

"i.e. any submodule of M which contains A also contains RA"

[I still find it perplexing that this actually shows that RA is the smallest submodule of M which contains A but anyway ... this is, I think, not hard to prove ...}

So to show that any submodule of M which contains A also contains RA

Let N be a submodule of M such that A \subseteq N

We need to show that RA \subseteq N

Let x \in RA

Now RA = \{ r_1a_1 + r_2a_2 + ... ... + r_ma_m \ | \ r_1, r_2, ... ... , r_m \in R, \ a_1, a_2, ... ... , a_m \in A, m \in \mathbb{Z}^{+}

So x = r_1a_1 + r_2a_2 + ... ... + r_ma_m for r_1, r_2, ... ... , r_m \in R, \ a_1, a_2, ... ... , a_m \in A

If A \subseteq N then r_ia_i \in N for 1 \le i \le n since N is a submodule

and then the addition of these elements, visually, r_1a_1 + r_2a_2 + ... ... + r_ma_m also is in N (if two elements belong to a submodule then so does the element that is formed by their addition)

So x \in N

Thus x \in RA \Longrightarrow x \in N

So A \subseteq N \Longrightarrow RA \subseteq N ... ... (1)

However, I am still not completely sure how to formally show that RA is the smallest submodule of M that contains A i.e. how does the implication (1) demonstrate this - can someone help by showing this explicitly and formally?

Peter

[This has also been posted on MHF]

 

[Fernando Revilla]

However, I am still not completely sure how to formally show that RA is the smallest submodule of M that contains A i.e. how does the implication (1) demonstrate this - can someone help by showing this explicitly and formally?

Consider the set
$$\mathcal{N}=\{N\subseteq M:N\text{ submodule of }M\text{ and }A\subseteq N \}$$
Then, $\subseteq$ is an order relation on $\mathcal{N}$, $RA\in \mathcal{N}$ and $RA\subseteq N$ for all $N\in\mathcal{N}.$ This means that $RA$ is the smallest element related to $(\mathcal{N},\subseteq)$

 

[Deveno]

One can argue by contradiction, here.

Suppose we have a submodule [math]N[/math] with:

[math]A \subseteq N \subsetneq RA[/math].

By your previous argument, since [math]N[/math] is a submodule containing [math]A[/math], we have:

[math]RA \subseteq N \subsetneq RA[/math] that is:

[math]RA \neq RA[/math], a contradiction. So no such [math]N[/math] can exist.