Free module finite generated with a submodule not finite generated. Anyone knowns an example of a R-module finite genereted with a submodule not finite generated?
I find the following example: Taking the set of function $f:[0,1]\rightarrow\mathbb{R}$ seen as module of it self. This is finite generated. If we take the subset of functions $f$ such that $f(x)=0$  for all $x\in[0,1]$ except a finite numbers of points then we will get a submodule wich is not finite generated. Why?
This is a different example of the others I saw. 
Thanks a lot!!!
 A: Let $R$ be the ring of function $f:[0,1]\to\mathbb R$. Let $M$ be the $R$-module of functions $f$ such that $f(x)=0$ for all $x\in[0,1]$ except for a finite numbers of points.
Assume on contrary that $M$ is finitely generated with generators $g_1,\ldots,g_n$.
For each $a\in [0,1]$ let
$\chi_a(x)=
\begin{cases}
1&x=a,\\
0&x\neq a.
\end{cases}
$
The functions $\chi_a,a\in[0,1]$ clearly generates $M$.
Thus each $g_i$ is a linear combination of the $\chi_a$.
Then there exists $a_1,\ldots,a_m\in[0,1]$ such that each $g_i$ is linear combination of $\chi_{a_1},\ldots,\chi_{a_m}$.
Consequenlty $\chi_{a_1},\ldots,\chi_{a_m}$ generates $M$.
Let $a\in[0,1]$ distinct from each $a_j$.
Since $\chi_a$ is not a linear combination of $\chi_{a_j},j$ this lead to a contradiction.
A: Another example is $$A = \lbrace f: \mathbb{R} \to \mathbb{R} \rbrace$$ as module over itself, and the submodule generated by the functions $f$ with compact support, which can't be finitely generated.
A: Let $R=k[x_1,x_2,\ldots,x_n,\ldots ]$. Now $R$ as a module over itself is finitely generated, but clearly $I=(x_1,x_2,\ldots,x_n,\ldots )$ is not finitely generated.
