# Locally free sheaf of rank n on $A^1_k$ is trivial of rank n

The question is how to prove the title, this is actually exercise 13.2.C. from Vakil's notes.

The hint is to use the structure theorem for f.g. module over PID. Since to be quasi-coherent is a local property and locally free sheaf is quasi-coherent, we know the sheaf on $A^1_k$ is given by some $k[x]$-module M, and note that $k[x]$ is a PID.

So I guess we first need to prove M is finitely generated. Is it true that being a finitely generated module is a local property? Also, how should I prove M is actually free? Is being free module also a local property? At least for module over PID? It seems to me that this is what the hint about. Could someone help? Thanks!

• Cover the base with basic affine sets $D(f_i)$. Choose elements of $M$ which generate $M_{f_i}=k[x]^n_{f_i}$. Show that for any global section, equivalently element $s$ of $M$, we can restrict to the affine cover, write it using our elements restricted to that part, and then glue our elements together. Then uniqueness shows that this must also be $s$. That shows $M$ is finitely generated.... – Eoin Jun 3 '16 at 6:07
• (Cont'd) Then $M$ is finitely generated, and locally free, hence projective and $M$ is over a PID so projective implies free. – Eoin Jun 3 '16 at 6:08

## 2 Answers

Locally free sheaves of rank $n$ on $\mathbb A^1_k$ correspond one-one to finitely generated projective modules of rank $n$ over the ring $k[x]$.
In this correspondance the sheaf $\mathcal E$ on $\mathbb A^1_k$ is sent to the module $\Gamma(\mathbb A^1_k,\mathcal E)$ and trivial locally free sheaves are sent to free modules.
This correspondence is a particular case of an equivalence of categories proved by Serre in 1955 in his ground-breaking article FAC, Chapitre II, 50, Proposition 4, page 242].
Since a finitely generated module over a PID is projective if and only it is free ( Dummit-Foote, Section 12.1, exercise 21) you get your result.
[That exercise is very easy: since a projective module over a domain has no torsion (since it is a sub-module of a free module) the classification of finitely generated modules over a PID given in theorem 6 of the same section immediately solves the exercice]

• You don't need finite generation to show that $M$ torsion free implies flat over a domain. We will use the ideal criterion for flatness: $M$ is flat if and only if $Tor(A/I,M) = 0$ for every ideal $I \subseteq M$. If $A$ is a PID and $M$ is torsion free, then $I = (a)$ for some $a \in A$ and the exactness of $0 \to M \stackrel{a\cdot}{\to} M \to M/aM \to 0$ gives the result. – Ben Lim Jun 3 '16 at 7:51
• Side note: We haven't interacted in many years! I used to be BenjaLim on this site! – Ben Lim Jun 3 '16 at 7:52
• Dear @Ben, I didn't mention flatness in my answer, which I tried to keep as elementary as possible. And I am very happy to interact with you again: I hope it will happen often ! – Georges Elencwajg Jun 3 '16 at 8:00

Eoin's answer (in the comments) can be further shortened by noting that locally free sheaves of finite rank over a noetherian ring is coherent, so $M$ being finitely generated comes directly. After that use the fact that $M$ is projective (as it is locally free of constant rank) and since your ring is a PID it is in fact free.

The much more difficult problem is whether the same conclusion holds for $\mathbb A^n_k$. The answer is yes. This is known as the Quillen–Suslin theorem.

Let me ask two related questions which I haven't really tried to answer at the moment (maybe I will also ask them separately)

a)Over an UFD, all locally free sheaves of rank $1$ is trivial (the converse is also true, with noetherian and normal assumption). Give an example of a non trivial locally free sheaf of rank $>1$.

b)Is it possible to characterize rings for which locally free sheaves of rank $\le n$ is trival but there is a nontrivial locally free sheaf of rank $n+1$?