Definition of base-point-free sheaf

Question

Let $X$ be a scheme with an invertible sheaf $\mathcal{L}$. We call it base-point-free if for any $x\in X$ we can find an $s\in\Gamma(X; \mathcal{L})\equiv \mathcal{L}_X(X)$ such that $s(x)\neq 0$.

Question: What do we mean when we write $s(x)$? I know that this is supposed to emulate vector bundles, but abstractly, $s$ is an element of the module $\mathcal{L}_X(X)$. I can think of two ways to read this:

1: When $X=Proj S_\bullet$ for some graded polynomial ring $S_\bullet$, we can identify the line bundle's global sections with homogeneous polynomials of the appropriate degree. It then makes sense to speak of global sections with 'no common zeroes'.

2: We look at $s:X\rightarrow Spec(Sym\mathcal{L})$ as a morphism of sheaves and evaluate it at $x$.

Is there a cleaner way to understand what it means to evaluate a section $s(x)$ in the general setting, where $X$ is the Proj of any graded ring and $\mathcal{L}$ is just some abstract invertible sheaf?

Answer 1

$s(x)$ is the evaluation of $s$ under the homomorphism $\Gamma(X,\mathcal{L}) \to \mathcal{L}_x \cong \mathcal{O}_{X,x} \to k(x)$ from global sections to the residue field at $x$. The first map is the restriction map, the isomorphism comes from the fact that $\mathcal{L}$ is locally free of rank one, and the last map is the quotient by the unique maximal ideal. This is completely general for any line bundle on any locally ringed space.

Of course, as Peter LeFanu Lumsdaine points out in the comments, the identification $\mathcal{L}_x/\mathfrak{m}_x\mathcal{L}_x \cong k(x)$ is only as vector spaces over $k(x)$ and is very non-canonical - all you really get out of it is whether $s_x\in\mathfrak{m}_x\mathcal{L}_x$ or not. But this is exactly what you need to consider when you're talking about base-point-freeness, since you're trying to determine whether the natural morphism $\Gamma(X,\mathcal{L})\otimes \mathcal{O}_X \to \mathcal{L}$ is surjective or not, and surjectivity is equivalent to the image of $\Gamma(X,\mathcal{L})$ not being contained in $\mathfrak{m}_x\mathcal{L}_x$ under the restriction map.

Answer 2

Q: "Thanks, this makes much sense. Just one thing bothers me still: So is s(x)∈k(x)? But I believe the fibre of an invertible sheaf (over, say, the point x) is Lx/mX,xLx≅Lx⊗Lxk(x), which is in general a vector space over k(x) (where mX,x is the maximal ideal of OX,x). And would you not want s(x) to lie in the fibre over x? – usernam Commented 6 hours ago"

A: If $x\in X$ and $s\in \Gamma(X, L)$ we usually write $s(x) \in L(x):=L_x \otimes_{\mathcal{O}_{X,x}} \kappa(x)$ where $\kappa(x)$ is the residue field at $x$. In the case of an invertible sheaf this will be a one dimensional vector space over $\kappa(x)$. Note that $L(x)$ is not a ring, and for this reason we do not write $"L(x) \cong \kappa(x)"$ since the residue field is a ring.

An example: If $C:=\mathbb{P}^1$ is the projective line over a ring $k$ and if $L(d):=\mathcal{O}(d)$ with $d \geq 1$ and $U:=D(x_0)$ it follows

$$L(d)(U) \cong k[t]x_0^d$$

with $t:=x_1/x_0$ an independent variable over $k$. If $k$ is a field and $x:=(t-a)$ with $a\in k$ it follows the fiber of $L$ at $x$ is

$$L(d)(x) \cong k[t]/(t-a)x_0^d \cong kx_0^d$$

and $kx_0^d$ is a one dimensional vector space over $\kappa(x)$. You cannot multiply elements in $L(x)$ as you can in $\kappa(x)$.

Moreover: If you forget about the local basis elements $x_0^d, x_1^d$ you are not able to distinguish between $L(d)$ and $L(d_1)$ for differing $d \neq d_1$.