Does IM Form a Submodule of M?

[iamalexalright]

Homework Statement

Let M be a R-module and I is an ideal in R.
Let IM be the set of all finite sums of the form:
[tex]r_{1}v_{1} + ... + r_{n}v_{n}[/tex]
With [tex]r_{i} \in I[/tex] and [tex]v_{i} \in M[/tex]
Is IM a submodule of M?

Homework Equations

A submodule of an R-module M is a nonempty subset S of M that is an R-Module in its own right, under the operations obtained by restricting the operations of M to S.

The Attempt at a Solution

First I want to show that IM is a module itself.

I need to determine if IM is nonempty.
I is nonempty from givens and M is nonempty from givens so IM is nonempty.

Now we have two operations:
[tex]+^{IM}: IM x IM[/tex] and [tex]*^{IM}: R x IM[/tex]

Next I'll check if IM is an abelian group under addition:
Let [tex]a,b \in IM[/tex]
[tex]a + b =[/tex]
[tex]a_{1}v_{1} + ... + a_{n}v_{n} + b_{1}u_{1} + ... b_{n}u_{n} =[/tex]
[tex]b_{1}u_{1} + ... + b_{n}u_{n} + a_{1}v_{1} + ... a_{n}v_{n} =[/tex]
[tex]b + a[/tex]

Thus it is abelian.

Now I need to check if for all [tex]r,s \in R[/tex] and [tex]u,v \in IM[/tex]
these hold:

r(u + v) = ru + rv
(r + s)u = ru + su
(rs)u = r(su)
1u = u

These seem straight forward but I feel like I'm not understanding something (proof just seems wrong). Any help?

 

[micromass]

All these things are correct. And eventually you'll get there. But there's a much easier way to prove that something is a submodule!

Take M a R-module. And let [tex]N\subseteq M[/tex]. Then N is a submodule of M if and only if

• [tex]0\in N[/tex]
• [tex]\forall n,m\in N:~n+m\in N[/tex]
• [tex]\forall n\in N:~\forall r\in R:~r.n\in N[/tex]

So it suffices to show these 3 properties and you're done. You don't need to show commutativity and all that things! It's not wrong if you do, but it's superfluous...