4 added 215 characters in body

Just looking at stalks is not enough:

Suppose that $X$ is a nontrivial complex manifold. Let $i_x:x\to X$ denote the inclusion, and set $$\mathcal{F} =\bigoplus_{x\in X} i_{x*}\mathcal{O}_x$$ Notice that it is naturally an $\mathcal{O}_X$-module with $\mathcal{F}_x\cong \mathcal{O}_x$, and yet it is certainly not locally free.

(

Notes Rather than editing, I'll keep the original form of my answer in tact and add a few footnotes.

1. Of course, this $\mathcal{F}$ is not coherent.Our two answers are complementary.

2. (Re: UG's first comment.)

Perhaps it's not important, but since UG harbors doubts, here's I probably should have included the proof that $\mathcal{F}_x\cong \mathcal{O}_x$. Here it is. The left is the direct limit $$\varinjlim\bigoplus_{y\in U} \mathcal{O}_y$$ as $U$ shrinks to $x$. There is a projection $p$ to $\mathcal{O}_x$ which is surjective since it has a section. Suppose that $f=\sum f_y$ lies in the kernel of $p$. Shrink $U$ to avoid the support of $f$ (which excludes $x$). Then we see that the class of $f$ in the direct limit must be zero. (There is a reason I took the sum and not the product.)

3. (Re: Laurent's comment.) By $i_{x*}\mathcal{O}_x$, I meant the skyscraper sheaf associated to $\mathcal{O}_x$.

3 added 559 characters in body

Just looking at stalks is not enough:

Suppose that $X$ is a nontrivial complex manifold. Let $i_x:x\to X$ denote the inclusion, and set $$\mathcal{F} =\bigoplus_{x\in X} i_{x*}\mathcal{O}_x$$ Notice that it is naturally an $\mathcal{O}_X$-module with $\mathcal{F}_x\cong \mathcal{O}_x$, and yet it is certainly not locally free.

(Of course, this is not coherent. Our two answers are complementary.)

Perhaps it's not important, but since UG harbors doubts, here's the proof $\mathcal{F}_x\cong \mathcal{O}_x$. The left is the direct limit $$\varinjlim\bigoplus_{y\in U} \mathcal{O}_y$$ as $U$ shrinks to $x$. There is a projection $p$ to $\mathcal{O}_x$ which is surjective since it has a section. Suppose that $f=\sum f_y$ lies in the kernel of $p$. Shrink $U$ to avoid the support of $f$ (which excludes $x$). Then we see that the class of $f$ in the direct limit must be zero. (There is a reason I took the sum and not the product.)

2 added 69 characters in body

Just looking at stalks is not enough:

Suppose that $X$ is a nontrivial complex manifold. Let $i_x:x\to X$ denote the inclusion, and set $$\mathcal{F} =\bigoplus_{x\in X} i_{x*}\mathcal{O}_x$$ Notice that it is naturally an $\mathcal{O}_X$-module with $\mathcal{F}_x\cong \mathcal{O}_x$, and yet it is certainly not locally free.

(Of course, this is not coherent. Our two answers are complementary.)

1