Example of a finitely generated module with submodules that are not finitely generated I'm looking for an example of a finitely generated module with submodules that are not finitely generated.
I've found a similar question dealing with group (i.e. an example of a finitely generated group with subgroups that are not finitely generated). But I can't figure out whether that question do help to this one.
And I actually want to find a more "module-like" example rather than an example derived from a 'strange' group.
Can you please help? Thank you!
 A: Regarding the "hint" above. Let $f \in I$. Then $f$ has no constant term and has only finitely many of the variables $X_i$. Now suppose that $f_1,\dots,f_k \in I$ generate $I$ over $A$, in other words, suppose that $I$ is finitely generated over $A$. Let $n$ be a large enough natural number so that none of the $f_i$ contains the variable $X_n$. Now $X_n \in I$ therefore $X_n= p_1f_1 + \ldots + p_kf_k$. In this expression, set $X_1= \ldots = X_{n-1}=0$ and $X_n=X_{n+1}=\ldots = 1$. One gets that $0=1$, a contradiction.
A: The ideal $I=\langle X_1,X_2,...,X_n,... \rangle \subset \mathbb R[X_1,X_2,...,X_n,...]=A$ can be seen as a submodule of the free $A$-module of dimension one $A=A^1$, and that module is not finitely generated. Do you see why?
(Hint: even in a polynomial ring with infinitely many indeterminates, each polynomial involves only finitely many variables. In other words $\mathbb R[X_1,X_2,...,X_n,...]=\bigcup_{k\geq 1}\mathbb R[X_1,X_2,...,X_k] \;$  )
A: Here's a fairly simple example (of a non-Noetherian ring): the ring $R$ of polynomials in one indeterminate $X$ having rational coefficients but with an integer constant term. Its ideal of elements with zero constant term is not finitely generated as an $R$-module.
A: I was stuck with the problem as well and I found many examples but they were quite vague (in the sense that there is not a clear reasoning). Now let me restate an example and make it as clear as crystal!
A ring $R$ is a $R$-module itself. Its submodules are just its ideals. Let us consider $R=\mathbb R[x_1,x_2, ..., x_n, ...]$ and $I=<x_1,x_2,..,x_n,...,>$.
$R$ is finitely generated by $1_{\mathbb R}$ (as an element in $R$). (Yes! I have stated the generator clearly!)
We claim that $I$ is not finitely generated (by elements in $R$).
Proof:
We first consider the case that $I$ can be generated by a single generator. First we note that, any non-zero constant $c\in \mathbb R$ can not generate $I$ because $1_{\mathbb R} \cdot c=c\notin I.$ Next, we note that for all $j\geq 1$, $x_j$ cannot generate $I$ because there is no $r\in R$ such that $r\cdot x_j =x_k$ for $k\neq j$. Similar reasoning works for all polynomials with degree $1$, by noting that a polynomial has only finitely many terms but we have infinitely many $x_i$'s in $I$. The case that the generator (as a polynomial) has degree $\ge 2$ is even simpler: if the leading term is $a_n x_n^s$ for some $s\ge 2$, then $x_n (\in I)$ can not be generated. So, $I$ cannot be generated by a single generator.
The case of finitely many generators is similar. The essence is that, we have infinitely many $x_i$'s but the generating set is finite. Together with a similar reasoning as above, we can prove the claim.
