# Module and Ideal

## 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?

Related Calculus and Beyond Homework Help News on Phys.org
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...