Let $X$ be a scheme and $\mathscr{L}$ be a line bundle on $X$. In a few proofs I have seen the scheme $$L = \mathscr{S}{\rm pec} \oplus_{n \in \mathbb{Z}} \mathscr{L}^{\otimes n} \to X$$ pop up. Does the scheme $L$ have a name? How should I be thinking about $L$?

-
I should add that I understand the obvious properties of $L$, e.g the pushforward of its structure sheaf is $\oplus_n \mathscr{L}^n$ – Daniel Barter Nov 24 '13 at 0:41
I think $L$ can be viewed as the total space of the line bundle $\mathscr{L}$, minus the image of the zero section. – Julian Rosen Nov 24 '13 at 1:47
What Jason said below, and Julian said above... except that it should be the total space of the dual line bundle (minus the zero section). – Marty Nov 24 '13 at 2:33
@Marty $\oplus_n \mathscr{L}^{\otimes n}$ is canonically isomorphic to $\oplus_n (\mathscr{L}^\vee)^n$ (by a map that negates degree), so we can identify the total space of a line bundle (minus zero section) with the total space of its dual (minus zero section). Concretely, a nowhere vanishing section $\gamma$ of $\mathscr{L}$ over some open $U\subset X$ determines a nowhere vanishing section $\tilde{\gamma}$ of $\mathscr{L}^\vee$ over $U$ by the condition $\tilde{\gamma}(\gamma)=1$. – Julian Rosen Nov 24 '13 at 2:57

Denote by $\mathcal{A}$ the $\mathbb{Z}$-graded sheaf of $\mathcal{O}_X$-algebras, $$\mathcal{A} = \oplus_{n\in \mathbb{Z}} \mathcal{L}^{\otimes n}.$$ There is a natural isomorphism of invertible $\mathcal{A}$-modules, $$\phi: \mathcal{A}\otimes_{\mathcal{O}_X}\mathcal{L} \to \mathcal{A}.$$ As a map of graded modules, this is homogeneous of degree $1$, i.e., the degree $0$ part, $\mathcal{A}_0\otimes_{\mathcal{O}_X}\mathcal{L} = \mathcal{L}$ gets mapped via the identity map to the degree $1$ part $\mathcal{A}_1=\mathcal{L}$. There is a unique extension of this map, $\phi_0=\text{Id}_{\mathcal{L}}$, to an isomorphism of $\mathcal{A}$-modules.

Denote by $\pi:L\to X$ the natural projection. Associated to this isomorphism $\phi$, there is an isomorphism of $\mathcal{O}_L$-modules, $\widetilde{\phi}:\pi^*\mathcal{L}\to \mathcal{O}_L$. This isomorphism is universal: for every $X$-scheme, $f:Y\to X$, and for every isomorphism $\psi:f^*\mathcal{L}\to \mathcal{O}_Y$ of $\mathcal{O}_Y$-modules, there is a unique morphism of $X$-schemes, $g:Y\to L$, such that $g^*\widetilde{\phi}$ equals $\psi$.

Although $\widetilde{\phi}$ is universal, of course it is not unique. In particular, on the product $\mathbb{G}_m\times L$, there is a modification of $\text{pr}_L^*\widetilde{\phi}$ that scales this by the universal invertible element in $\mathcal{O}_{\mathbb{G}_m}$. In particular, this modification of $\widetilde{\phi}$ gives rise to a morphism of $X$-schemes, $$m_L :\mathbb{G}_m\times L \to L.$$ Of course this is an action of $\mathbb{G}_m$ on $L$ over $X$. The corresponding $\mathbb{Z}$-grading of $\pi_*\mathcal{L}$ is the evident grading of $\mathcal{A}$. In particular, since $\mathcal{A}$ is isomorphic to $\mathcal{O}_X[t,t^{-1}]$ as graded sheaves of algebras locally on $X$, the action $m_L$ makes $L$ into a $\mathbb{G}_m$-torsor. This is the "$\mathbb{G}_m$-torsor naturally associated to $\mathcal{L}$".

Of course, as is always the case, about half of the mathematical word chooses the opposite convention. But that is no problem: whenever somebody talks about THE torsor associated to an invertible sheaf, just ask them for the universal property that their space satisfies, and then use that to calibrate with your own convention.

-

It is common to use the name "the $\mathbb{G}_m$-torsor associated to $\mathscr{L}$", and it is concisely written as $L = \underline{\operatorname{Isom}}_X(\mathscr{O}_X, \mathscr{L})$. However, an ambiguity under inversion means the $\mathbb{G}_m$-action is not completely determined by the scheme structure of $L$, so you might want to be careful about whether or not you have chosen a torsor structure.

-
That ambiguity is why I wrote things the way that I did. To calibrate conventions, ask your correspondent for his / her universal property of "the" torsor $L$. Then, with respect to that universal property, ask what corresponds to the multiplication map $\mathbb{G}_m\times L \to L$. In every case I have seen, this clears up everything. Unfortunately, in most cases, two correspondents with opposite conventions prefer to argue the merits of their own choice and belittle the opposing choice, rather than simply clarify which choice each has made. – Jason Starr Nov 24 '13 at 16:34