Free module as a vector bundle over the spectrum

Question

I'm reading an article and I have a question about one of the steps.

Setting: k is a field, K is the function field of the projective line $P^1$ over $k$. Let $A=k[t,t^{-1}]$ be the ring of all the rational functions having poles only at $0,\infty$ . $V$ is a 2-dimensional vector space over $K$ and $M$ is a free module over $A$ containing a basis of $V$.

The author claims and I quote "The A-module M is free and can be regarded as a (trivial) rank 2 vector bundle on $Spec A = P^1 - \{0,\infty\}$".

I'm not familiar with vector bundles (I do know the definition), and I don't understand the duality between free modules over $A$ and vector bundles over $spec(A)$.

Any explanation is appriciated!

Thanks

Answer 1

$\newcommand{\Spec}{\operatorname{Spec}}$When we're talking about schemes, when you read "vector bundle" you should read "locally free sheaf". Here's what the latter term means precisely:

Let $X$ be a scheme and write $\mathcal O_U:=\mathcal O_X|_U$. We say an $\mathcal O_X$-module $\mathscr F$ is a locally free sheaf if for each $p\in X$, there is a set $I$ and an open neighborhood $U$ of $p$ such that $\mathscr F|_U\cong\mathcal O_U^{\oplus I}$ (note this is an isomorphism of $\mathcal O_U$-modules).

Usually we will be interested in when $I$ is finite and $\mathscr F$ has constant rank, so there is some integer $n$ such that $\mathscr F|_U\cong\mathcal O_U^{\oplus n}$ for our "trivial" neighborhoods $U$.

Now, here's a specific construction you should be aware of:

Let $A$ be a ring and $M$ an $A$-module. Let $X=\Spec A$ and $\mathcal O_X$ its structure sheaf. There is a natural $\mathcal O_X$-module $\widetilde M$ defined as follows: for $f\in A$, we let $\widetilde M(\Spec A_f)=M_f$, the localization of $M$ at $f$, which is an $A_f$-module. Since the open subsets $\Spec A_f$ form a base for $X$ as $f$ runs over all elements of $A$, this is enough to determine $\widetilde M$ as a sheaf on $X$, and you can double check any details to convince yourself that $\widetilde M$ is actually an $\mathcal O_X$-module.

Now you should practice unwinding definitions to convince yourself that the following is true:

The $\mathcal O_X$-module $\widetilde M$ is a locally free sheaf on $X=\Spec A$ if and only if $M$ is a locally free $A$-module, which means that there exists a set $\{f_1,\dots,f_n\}$ generating $A$ such that $M_{f_i}$ is a free $A_{f_i}$-module.

In particular, if $M$ is a free $A$-module, then our generating set above if just $\{f_1=1\}$, so $\widetilde M$ is actually a free $\mathcal O_X$-module, i.e. it is isomorphic to $\mathcal O_X^{\oplus 2}$.

This is why for a free $A$-module $M$, we can consider $M$ as a "trivial" locally free sheaf on $\Spec A$. As to why we identify "locally free sheaves" with "vector bundles", you can try to prove the following for yourself.

If $X$ is an $A$-scheme, then there is a one-to-one correspondence between rank $n$ locally free sheaves on $X$ and rank $n$ vector bundles on $X$, where a (rank $n$) vector bundle is an $A$-scheme $E$ along with a projection morphism $\pi:E\to X$ with the following property: for each $p\in X$ there is a neighborhood $U$ of $p$ and an isomorphism $\Phi:\pi^{-1}(U)\overset{\sim}\to U\times\Bbb A_A^n$ such that $\pi=\mathrm{pr}_2\circ\Phi$, where $\mathrm{pr}_2$ is the projection $\pi^{-1}(U)\times\Bbb A_k^n\to\Bbb A_A^n$.

If it's not immediately obvious, the "trivializable" neighborhoods $U$ in the above definition of vector bundle should correspond naturally to the "trivializable" neighborhoods $U$ where $\mathscr F|_U\cong\mathcal O_U^{\oplus n}$.