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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?

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    $\begingroup$ @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? $\endgroup$
    – KReiser
    Commented Feb 14 at 14:25
  • $\begingroup$ Ok , fair enough. I'll try to be more careful about tagging stuff correctly. $\endgroup$
    – J.D
    Commented Feb 14 at 14:49

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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.

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