Example of a finitely generated module with submodules that are not finitely generated

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 $\mathbb R[X_1,X_2,...,X_n,...]=\bigcup_{k\geq 1}\mathbb R[X_1,X_2,...,X_k] \;$ )

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.

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.