# Generation of submodules

#### Peter

##### Well-known member
MHB Site Helper
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 [TEX] A \subseteq N [/TEX]

We need to show that [TEX] RA \subseteq N [/TEX]

Let [TEX] x \in RA [/TEX]

Now [TEX]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}^{+} [/TEX]

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

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

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

So [TEX] x \in N [/TEX]

Thus [TEX] x \in RA \Longrightarrow x \in N [/TEX]

So [TEX] A \subseteq N \Longrightarrow RA \subseteq N [/TEX] ... ... (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]

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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 $(\mathcal{N},\subseteq)$

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
One can argue by contradiction, here.

Suppose we have a submodule $$\displaystyle N$$ with:

$$\displaystyle A \subseteq N \subsetneq RA$$.

By your previous argument, since $$\displaystyle N$$ is a submodule containing $$\displaystyle A$$, we have:

$$\displaystyle RA \subseteq N \subsetneq RA$$ that is:

$$\displaystyle RA \neq RA$$, a contradiction. So no such $$\displaystyle N$$ can exist.