Submodules generated by A

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 ... "

(R is a ring with 1, M is a left module over R, A is a subset of M)

Can anyone formally and explicitly prove this statement?

Peter

Re: Submodules generated by A

Quote:

Originally Posted by

**Bernhard** 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 ... "

(R is a ring with 1, M is a left module over R, A is a subset of M)

Can anyone formally and explicitly prove this statement?

Peter

Just reflecting on my own question - it is easy to show that RA is a submodule of M, but what was worrying me was 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

We need to show that

Let

Now

So for

If then for since N is a submodule

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

So

Thus

So ... ... (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