# 2024-09-03-12-46-25: Linear Systems

:::{.remark}
Recall that a linear system is defined by $L = \PP(V)$ where $V \subseteq H^0(F)$ is a finitely-generated vector space.
We define $\Base(L) = \intersect (s)$ for $s\in V$.
Recall that if $F$ is a line bundle then $F = \OO_X((s))$ for any effective $(s)$.
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:::{.theorem}
There is a bijection between maps $X \mapsvia{f} \PP^n$ with $f(X)$ not contained in a hyperplane and linear systems $L$ with $\Base(L) = \emptyset$.
One sets $F \da f^* \OO_{\PP^n}(1)$ in one direction.
In the other direction, writing $s = \sum c_i x_i$ with $x_0,\cdots, x_n$ local coordinates on $\PP^n$ and sets $D = (s)$.
:::

:::{.remark}
Recall that

- $f$ is injective $\iff L$ separates points.
- $f$ is an isomorphism $\iff$ additionally $L$ separates tangent directions.

Writing $\T_{X, p} = (\mfm_p/\mfm_p^2)\dual$, the latter condition means $T_p \to T_{f(p)}$ is injective, or equivalently $T\dual_{f(p)}\to T\dual_{p}$ surjective.
This is satisfied if the pullbacks $f^* y_1,\cdots, f^* y_n$ generate $\mfm_p/\mfm_p^2$, where $y_1,\cdots, y_n$ are local coordinates in $\PP^n$.
Note that a surjection on maximal ideals corresponds to a surjection on the completions of local rings $\hat{\OO_{f(p)}}\surjects \hat{\OO_p}$.
:::

:::{.remark}
For $X$ a smooth curve, these conditions simplify:

- $\Base(\abs{D}) = \emptyset$, or equivalently $h^0(D-p) = h^0(D)-1$.
    These numbers are equal iff every divisor passes through $p$.
    This is proved by Riemann-Roch.
- Separating points: $h^0(D-p-q) = h^0(D-p)-1$, so not all of the divisors that pass through $p$ also pass through $q$.
- Separating tangents: $h^0(D-2p) = h^0(D-p)-1$.

Note that on a surface, $\abs{D-p}$ is a just convenient notation for divisors that pass through $p$, since $D-p$ is not a divisor.
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:::{.theorem}
If $X$ is a smooth variety, then rational maps $X\rational \PP^n$ are in bijection with linear systems without base components, i.e. $\Base(L)$ does not contain a divisor.
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:::{.remark}
One can always remove the indeterminacy in codimension 1, but not necessarily in codimension 2.
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:::{.theorem}
Suppose $\phi: S\to S'$ is a map from a smooth surface to a projective surface.
Then there exists a diagram where $S_i = \Bl_{p_i} S_{i-1}$:

\[\begin{tikzcd}
	&&& {S_k} \\
	&& \dots \\
	& {S_1} \\
	{S_0 \da S} &&&&&& {S'}
	\arrow[from=1-4, to=2-3]
	\arrow["{f\qquad\text{(regular)}}", from=1-4, to=4-7]
	\arrow[from=2-3, to=3-2]
	\arrow[from=3-2, to=4-1]
	\arrow[dashed, from=4-1, to=4-7]
\end{tikzcd}\]
> [Link to Diagram](https://q.uiver.app/#q=WzAsNSxbMCwzLCJTXzAgXFxkYSBTIl0sWzEsMiwiU18xIl0sWzMsMCwiU19rIl0sWzYsMywiUyciXSxbMiwxLCJcXGRvdHMiXSxbMCwzLCIiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSwwXSxbMiw0XSxbNCwxXSxbMiwzLCJmXFxxcXVhZFxcdGV4dHsocmVndWxhcil9Il1d)


:::

:::{.remark}
Idea:

\includesvg{inkscape/2024-09-03_13-11.svg}

On the proof: let $D = D_0$ be a divisor in the linear system.
Resolve to get $D_1 = f_1^* D - m_1 E_1$ where $m_1 = \mult_p(L) \da \min \ts{\mult_p(D) \st D\in \L}$.
Check that $D_1^2 = D_0^2 - m_1^2$ and $D_i^2\geq 0$ for all $i$.
But the self-intersection can't decrease forever.
:::

:::{.remark}
Let $L = \PP^1$ in $\PP^2$ be the set of lines through zero.
Blowing up yields a fibration $\Bl_p \PP^2 \to \PP^1$ which is a ruled surface with fibers the exceptional curves.
This surface is $\FF_1$.
:::

:::{.theorem}
Let $f: S'\to S$ be a birational map between smooth surfaces where $f\inv$ is undefined at a point $p\in S$.
Let $\eps: \hat{S} \to S$ be the blowup at $p$, then $f$ factors through $\eps$.
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:::{.corollary}
Any birational map of smooth projective surfaces $S\to S'$ factors into a sequence of blowups and blowdowns.
:::

:::{.remark}
Blowup enough to make the rational map regular.
Either it's an isomorphism or undefined at a point.
In the latter case, blowup and apply the theorem.
Why this terminates: $f\inv(p)$ contains finitely many divisors.
:::

Toward proving the theorem:

:::{.lemma}
Let $f: S'\to S$ be birational with $g\da f\inv$ undefined at $p$.
Then $\exist C \subseteq S'$ a curve with $f(C) = p$.
:::

Why this implies the theorem:

:::{.proof}
Suppose $C\mapsto p$ and consider $\eps: \hat{S} = \Bl_p S\to S$.
Extend $f: S'\to S$ to $g = \eps\inv\circ f$.
If $g$ is a morphism, we're done.
If $g$ is undefined then there exists a curve in $\hat{S}$ which is contracted by $g\inv$.
Note that $g\inv$ is only undefined at finitely many points, and thus can't contract a curve.
So the exceptional point is contracted to a point $q$ in $S$.

> DZG: Missed parts here!

Conclusion: $\eps\inv(z)$ vanishes along $E$ with multiplicity $m \geq 2$.
Conclude that $f\inv(z)$ also vanishes along $C$ with multiplicity $\geq 2$, but this contradicts the construction of the blowup.
:::

:::{.remark}
Note that this proof does not use Castelnuovo's criterion.
:::