Showing every finite rank n vector bundle over the affine line is trivial of rank n

Question

In Ravi Vakil's algebraic geometry notes, exercise 13.2.C asks to show that every finite rank n vector bundle over $X = \mathbb{A}^1_k$ is actually free of rank n. The hint is to use the structure theorem for finitely generated modules over PIDs.

My question is how to show that the vector bundle being locally free implies it is globally finitely generated.

Let $\mathcal{F}$ be our vector bundle. If $\mathcal{F}(X)$ is finitely generated the structure theorem implies it's free, and clearly when we restrict down to one of our open sets over which is was free of rank n the ranks must agree.

But why is it finitely generated? Is it simply because if it were not I could construct a submodule of $\mathcal{F}(X)$ of rank greater than n? i.e. by choosing linearly independent elements.

Over affine varieties, all quasi-coherent sheaves are globally generated. — Mohan, Sep 08 '15 at 20:25
@Mohan What is a vector bundle on a scheme? Schemes aren’t manifolds. — user5826, Nov 11 '21 at 20:27
@user46372819. Vector bundles can be defined even over topological spaces, manifold is not necessary. — Mohan, Nov 11 '21 at 20:59
@Mohan In the affine line case, what is the field that the vector spaces are defined over? — user5826, Nov 11 '21 at 22:43
@Mohan Is it just $\mathbb R$? In the post, it says $\mathcal F$ is our vector bundle and then $\mathcal F(X)$ is finitely generated. But then what is $\mathcal F(X)$? — user5826, Nov 12 '21 at 17:13
@user46372819 $F(X)$ is the set of sections of the vector bundle. — Mohan, Nov 12 '21 at 23:21

score 2 · Answer 1 · answered Sep 09 '15 at 14:41

2

This comes down to showing that given a Noetherian ring A, an A-module M, and a finite open cover of SpecA by sets $D(f_i)$ along with the fact $M_{f_i}$ is Noetherian, then M is Noetherian, but this is a standard result.

answered Sep 09 '15 at 14:41

Garnet

986

Dear Garnet, could you tell me a reference? Thank you. – BenicioLima Apr 17 '17 at 16:25
Hi @BenicioLima, sorry for the late response. We really want to show that M is finitely generated, because then we're finitely generated over a Noetherian ring, hence Noetherian. The argument is similar to that given in 5.3.9 of Ravi Vakil's notes, albeit with a module so you'll need to be a bit careful. You can find them here: http://math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf
I'll try and look up a more direct reference once I have access to my textbooks tomorrow. I think Hartshorne might cover this.
– Garnet Apr 18 '17 at 17:39

Showing every finite rank n vector bundle over the affine line is trivial of rank n

1 Answers1