# Maps to projective space parameterized by (basepoint-free) linear series

In Vakil's book Foundations of Algebraic Geometry, he defines a linear series on a $$k$$-scheme $$X$$ as a vector space $$V$$, an invertible sheaf (a line bundle) $$X$$, and a linear map $$\lambda: V\to\Gamma(X, \mathscr{L})$$.

He doesn't define what it means for a linear series to be basepoint free, although he does say that it can be defined in a similar way as to what it means for a family of global sections of $$\mathscr{L}$$ to be basepoint-free (sections $$s_0, \ldots, s_n\in\Gamma(X,\mathscr{L})$$ are basepoint-free if they have no common zeros). I'm not sure what he means by this. How does the vanishing of sections related to linear maps?

Now, he claims that $$(n + 1)$$-dimensional linear series on a $$k$$-scheme $$X$$, with choice of basis, with base-point-free locus $$U$$ defines a morphism $$U\to \mathbb{P}^n_k$$. He says one can see this through the exercise that shows maps $$X\to \mathbb{P}^n_k$$ correspond to vector bundles $$\mathscr{L}$$ and $$(n+1)$$ global sections $$s_0, \ldots, s_n$$ of $$\mathscr{L}$$. While I've solved this exercise, I have no idea how this relates to linear series at all.

I'm hoping someone here can define what it means for a linear series to be basepoint free and what a basepoint-free locus is. I'd also like to understand how exactly $$(n + 1)$$-dimensional linear series on a $$k$$-scheme $$X$$, with choice of basis, with base-point-free locus $$U$$ defines a morphism $$U\to \mathbb{P}^n_k$$.

Is this the idea? Given such a series $$\lambda: V\to \Gamma(X,\mathscr{L})$$, say with chosen basis $$v_0, \ldots, v_n$$, we can take the sections $$s_i := \lambda(v_i)$$ and let $$U$$ be the complement of the vanishing locus of the $$s_i$$. Then, we get a line bundle $$\mathscr{L}_{\vert U}$$ on $$U$$ with sections $$s_{i\vert U}$$ that have no common zero. Then we just take the corresponding map $$U\to\mathbb{P}^n_k$$.

If this is the case, I'm still confused as he often refers to maps given by a linear series but without specifying that a basis was chosen. Wouldn't the choice of basis affect the map, then, as their images given different sections?

• @j.dmaths that's not an appropriate tag for this situation. Just because a problem mentions a concept doesn't mean a tag is necessary - would you add the addition tag to every question which has a $+$ in it? Commented Feb 14 at 14:25
• Ok , fair enough. I'll try to be more careful about tagging stuff correctly.
– J.D
Commented Feb 14 at 14:49

For any point $$x\in X$$, there's a restriction map $$\Gamma(X,L)\to L_x$$, so you can look at the composite map $$V\to L_x$$. If the image of $$V$$ under this map lies in $$\mathfrak{m}_xL_x$$, then $$x$$ is called a base point of $$V$$. The locus of $$x\in X$$ where $$x$$ is not a base point for $$V$$ is the base-point-free locus.
You're completely right about the connection with linear series: picking a basis $$v_0,\cdots,v_n$$ for $$V$$, then letting $$s_i=\lambda(v_i)$$ as you've done lets you apply what you know to get a map from the base-point-free locus of the $$s_i$$ to $$\Bbb P^n$$ (and this is the exact same as the base-point-free locus of the $$v_i$$ by construction).
To explain why we don't need to refer to the basis very much, two different choices of basis give the same map up to an automorphism of $$\Bbb P^n$$. So if you're focused on properties invariant under automorphisms of $$\Bbb P^n$$, you don't need to bother with the specific basis as much.