- #1

iamalexalright

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## 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?