# 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!!!

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.
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.
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.
• take any $n$ generators, then choose an $x_i$ that is not among them. – user114539 Jun 18 '14 at 16:59