Locally free sheaf on a connected ringed space is free

Question

Hartshorne says if a locally free sheaf on a connected ringed space has a constant rank. Can I deduce further that the sheaf is actually free (globally) on the ringed space? If not is there any counter example?

Answer 1

Given the ring $\Bbb Z_6$ of integers modulo $6$, and the $\Bbb Z_6$-module $\Bbb Z_2$ (with module structure given by the standard quotient map $\Bbb Z_6\to\Bbb Z_2$ and regular multiplication), let $X = \operatorname{Spec}\Bbb Z_6$ be our ringed space and the induced $\mathscr O_X$-module $\Gamma = \widetilde{\Bbb Z_2}$ be our sheaf. I claim that it is locally free, but not free.

$X$ has two points, $(2)$ and $(3)$, and its distinguished open sets are: $$ X = D_1 = D_5\\ \{(3)\} = D_2 = D_4\\ \{(2)\} = D_3\\ \varnothing = D_0 $$ We have $$ \begin{array}{|c|c|c|} \hline U&\mathscr O_X(U)&\Gamma(U)\\ \hline D_1&\Bbb Z_6&\Bbb Z_2\\ D_2&\Bbb Z_3&0\\ D_3&\Bbb Z_2&\Bbb Z_2\\ \hline \end{array} $$ $D_2$ and $D_3$ covers all of $X$, and on both of those, $\Gamma$ is free. However, $\Gamma$ is not a free sheaf globally.

Answer 2

No, you may not deduce that a locally free sheaf is free. Here is a simple counterexample.

Consider the line bundle $\mathcal{O}_{\Bbb P^n}(-1)$ on $\Bbb P^n$ (which is connected). Then $\Gamma(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(-1))=0$, so any morphism from a free sheaf $\mathcal{O}_{\Bbb P^n}^{\oplus I}\to\mathcal{O}_{\Bbb P^n}(-1)$ is zero. In particular, $\mathcal{O}_{\Bbb P^n}^{\oplus I}\not\cong\mathcal{O}_{\Bbb P^n}(-1)$, so it isn't a free sheaf, but $\mathcal{O}_{\Bbb P^n}(-1)$ is still locally free.