I need to prove that a locally free sheaf of finite rank is finitely presented. Being homework, I'm not asking for a proof but rather for an explanation of why my counterexample is wrong. Here is the counterexample
Fix a field $\mathbb{F} $ and the ring $$A=\bigoplus_{i\in\mathbb{N}} \mathbb{F} $$
Consider $X=spec A$ and define a sheaf $\mathcal{F} $ to be $\mathbb{F} ^i$ when restricted to the ith copy of $spec \mathbb{F} $.
Now $\mathcal{F} $ is locally free of finite rank, but it is not even finitely generated as a sheaf. In fact if we have $$\mathcal{O} _X^n\longrightarrow \mathcal{F} \longrightarrow 0$$ Which is exact, it would be exact on the stalks, but if we choose a point $p$ in a copy $m>n$ and localize we would have $$\mathbb{F} ^n\longrightarrow \mathbb{F} ^m\longrightarrow 0$$ Which can't be exact.