Relationship between R and a submodule of a R-module

Question

Consider a commutative ring with unity, call it $\mathbf{R}$. Suppose that $\mathbf{S}$ is a finitely generated $\mathbf{R}$-module and has a submodule $\mathbf{T}$. Then I was wondering if it holds that $\mathbf{R} \subseteq \mathbf{T}$ or even $\mathbf{T} \subseteq \mathbf{R}$. Then I considered the commutative ring $\mathbb{Z}$ with the ideal $2\mathbb{Z}$. And I set

$\mathbf{S}=\mathbb{Z}(\sqrt{2})$

$\mathbf{T}=2\mathbb{Z}(\sqrt{2})$

Then clearly neither of $\mathbf{R} \subseteq \mathbf{T}$ or $\mathbf{T} \subseteq \mathbf{R}$ holds. Is this a valid counterexample; i.e., is $\mathbf{T}$ valid submodule (I guess so since multiplying $a+b\sqrt{2}$ in $\mathbf{S}$ and $2c+2d\sqrt{2}$ gives always a number in $\mathbf{T}$, where $a,b,c,d \in \mathbb{Z}$)

Answer 1

Sorry, don't have enough Rep points to comment, but I agree with Ferra, this is a nonsense question, as by definition, an $R$-module $A$ is defined to be a function from $R\times A \to A$, so it makes no sense to ask if $R\subset T$.

As an example, remember any abelian group is automatically a $\mathbb{Z}$-module, hence if $G$ is some abelian group, and $g\in G$, then 3$g$ = $g+g+g$, so I can't see how you can make the connection that $\mathbb{Z}\subset G$.

Now let $G$ be the group of polynomial functions, and let $S\subset G$ be all the polynomials with no constant term. Then this submodule is all polynomials, and their are no constants, so.....

I'm going to bed here, but just to be completely explicit, let your group be $\mathbb{Z}[X]$ under addition, and $S$ be as above, then as a $2\mathbb{Z}$-submodule, you can see S isn't contained in... and vice versa.

I agree with Ferra Stucky. I certainly had fun thinking about it too, but as Ferra said, apriori and all. Pedals and wheels are required for a bike to move, but this doesn't mean they are the same thing.

Answer 2

In general, $T$ need not contain (a copy of) the $R$-module $R$, nor does $T$ have to be contained in (a copy of) the $R$ module $R$.

You could take, for example, the $\Bbb Z$ module $M=\Bbb Z/4\Bbb Z$. It does not contain a copy of $\Bbb Z$ because, well, it is finite, and couldn't possibly contain the infinite set $\Bbb Z$.

Secondly $M$ can't be isomorphic to a submodule of $\Bbb Z$ because all the submodules of $\Bbb Z$ look like $n\Bbb Z$ for some integer $n$, and these are all infinite except for the zero ideal.

Given a finitely generated $R$ module $M$ generated by $n$ elements, the best you can say in general is that $M$ is isomorphic to a quotient of the free module $R^n$.