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:
r_{1}v_{1} + ... + r_{n}v_{n}
With r_{i} \in I and v_{i} \in M
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:
+^{IM}: IM x IM and *^{IM}: R x IM

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

Thus it is abelian.

Now I need to check if for all r,s \in R and u,v \in IM
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 N\subseteq M. Then N is a submodule of M if and only if

• 0\in N
• \forall n,m\in N:~n+m\in N
• \forall n\in N:~\forall r\in R:~r.n\in N

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...