Take the sheaf $\mathbf{H}_{\mathbb{C}}$ of holomorphic functions on $\mathbb{C}$. In a book I am reading they define the subsheaf $I$ by setting the stalks to be
$I_{z}:=\begin{cases} 0\quad \text{if } z=0\\ \mathbf{H}_{\mathbb{C},z} \quad \text{else} \end{cases}$
and claim that this sheaf is not locally finitely generated as $\mathbf{H}_{\mathbb{C}}$-module.
Could anyone explain to me why this is true?
Let $U$ some neighbouhood of $0$. I will have to show that there is no surjective morphism of sheaves
$(\mathbf{H}_{\mathbb{C}}^{n})_{\vert U}\rightarrow I_{\vert U}$.
(Clearly, the only point around which finding a suitable surjective map could possibly fail is $0$.)
We can actually show that there is no non-zero morphism $f:\mathbf{H}_{\mathbb{C}}|_U\to I|_U$ if $U$ is connected. This will prove what you want. To show this, note that $s=f(1)$ (where 1 is the constant unit section of $\mathbf{H}_{\mathbb{C}}$) is a section of $I_z$ hence a section of $\mathbf{H}_{\mathbb{C}}$ since $I$ is a subsheaf of $\mathbf{H}_{\mathbb{C}}$. We have $s_0=0$ since $I_0=0$ hence $s|_V=0$ in some neighborhood $V$ of $0$. By the principle of isolated zeros, $s=0$. It follows that $f=0$.