I was talking to this fellow mathematician, and told him how for my very personal taste, the research in algebraic geometry has sailed a bit too far ashore from the original question that motivated the subject: A classification for the solutions of polynomial equations. I told him that most of the heavy machinery in algebraic geometry can be given very good intuition - however, most courses neglect to do so.
Anyways, he was clearly amused and not quite in belief of my radical position that unconditional and solitary abstraction isn't the only way to do algebraic geometry. He teased me to then tell him what an *»ample«* line bundle was, without talking about tensor powers or commutative diagrams.
So, that's the reason for this post. I answered that ample line bundles are precisely those that induce finite morphisms to projective space, but I couldn't remember where I knew it from. I searched through all the major literature and could not find a reference
until someone on Mathoverflow helped me out. Since I complained so much about how noone ever gives intuition to these kinds of concepts in algebraic geometry, I decided I'd blog a bit about line bundles and how you should think about them.
How to think about Line Bundles
Let $\Bbb{K}$ be a field and $X$ a $\Bbb{K}$-variety. Let $\mathscr{L}$ be a line bundle on $X$. I think about $\mathscr{L}$ as a sheaf of functions on $X$, where for any open subset $U\subseteq X$, the function $f\in\mathscr{L}(U)$ can be "evaluated" at a $\Bbb{K}$-rational point $P\in U$ by declaring $f(P)$ to be the image of $f$ under the morphism
\[
\varepsilon_P: \mathscr{L}(U) \to \mathscr{L}_P \cong \mathcal{O}_{X,P} \twoheadrightarrow \mathcal{O}_{X,P}/\mathfrak m_P=\Bbb{K}.
\]
Of course, this depends on the local isomorphism $\mathscr{L}_P\cong\mathcal{O}_{X,P}$ that we choose, but if I choose the same local isomorphism for $f_0,\ldots,f_n\in\mathscr{L}(U)$, then $[f_0(P):\ldots:f_n(P)]\in\mathbb{P}^{n}$ is well-defined as a point in projective space, because choosing a different isomorphism means multiplying by some
$\alpha\in\mathcal{O}_{X,P}^\times$ which, after projecting, means multiplying (simultaneously) by some $\alpha\in\Bbb{K}^\times$. Of course, this is all void if the $f_i$ vanish simultaneously at $P$, because there is no origin in projective space. Let's assume that this does not happen for the moment, we will fix it soon.
We can also give a coordinate-free definition of the above morphism. I will replace $U$ by $X$ now, i.e. assume that $X$ is affine. Let $V\subseteq\mathscr{L}(X)$ be a finite-dimensional, linear $\Bbb{K}$-subspace of $\mathscr{L}(X)$. Then, we get a rational map $\phi_V:X\dashrightarrow\mathbb{P}(V^\ast)$ by defining $\phi_V(P) := [\varepsilon_P|_V]$,the projective equivalence class of the $\Bbb{K}$-linear form $\varepsilon_P|_V$. The set of points $P$ with $\varepsilon_P|_V=0$ is closed, so $\phi_V$ is defined on the complement.
How to think about Ample Line Bundles
Assume now that $X$ is a
complete variety, i.e. it is
proper over $\Bbb{K}$. This means that
it's almost projective, so maybe think of a projective variety. In this case, $\mathscr{L}(X)$ is of finite dimension as a $\Bbb{K}$-vector space. Let $\phi_\mathscr{L} :=\phi_{\mathscr{L}(X)}$. First of all, we say that $\mathscr{L}$ is
globally generated if $\phi_\mathscr{L}$ is a morphism and not just a rational map. In other words, for each point $P\in X$ there is some $f\in\mathscr{L}(X)$ with $f(P)\ne 0$. It is noteworthy that this does
not cover all cases, by a large degree. There are ample line bundles that have no global sections at all, which is something I didn't even know before researching this blagpost. See
this mathoverflow answer for an example. Anyway, globally generated line bundles is good for intuition, and that's what this post is about.
Under this assumption, the line bundle $\mathscr{L}$ is
ample if and only if $\phi_\mathscr{L}: X\to \mathbb{P}(\mathscr{L}(X)^\ast)$ is a
finite morphism. I am not going to prove it for you, but you can read this and more in Robert Lazarsfeld's book
»Positivity in Algebraic Geometry I«. Note that the statement is very hard to find in other graduate level books on algebraic geometry.
Now maybe you know the definition (or the result) that $\mathscr{L}$ is ample if some tensor power $\mathscr{E}=\mathscr{L}^{\otimes k}$ is
very ample. Something that
is in most books on algebraic geometry is the fact that the morphism $\phi_\mathscr{E}$ is an immersion if and only if $\mathscr{E}$ is very ample. In the case where $X$ is complete, this is equivalent to $\phi_\mathscr{E}$ being a
closed immersion. For this, you can refer to Hartshorne, Remark II.5.16.1.