# 2024-10-22-12-46-33

:::{.remark}
Recall Castelnuovo's rationality criterion: let $S$ be a smooth projective surface, then $S$ is rational iff $q\da h^1(\OO_X) = h^0(\Omega_X) = 0$ and $P_2 \da h^0(2K_S) = 0$.
As a corollary, $S$ is rational iff $S$ is *unirational*, i.e. there is a dominant map $\PP^2 \rational S$ of some degree $d$.
We discussed minimal surfaces, for which there are two cases:

- $S$ is ruled, so $S \birational C\times \PP^1$ for some curve $C$.
    This implies that a generic point is contained in a rational curve.
- $S$ is not ruled.
    In this case, there are few rational curves.

We showed that if $S$ is not ruled, there exists a unique minimal model in every birational equivalence class.
That is, if $f: S'\birational S$ with $S, S'$ minimal surfaces, then $f$ is an isomorphism.
If $S$ is ruled, minimal models are not unique -- blow up any point in a fiber of $S\to C$ to transform one $(0)$-curve to a union of two $(-1)$-curves, and blowdown the other curve to get another ruled surface.
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:::{.theorem}
If $S$ is a rational minimal surface, then $S\cong \PP^2$ or $\FF_n$ for some $n$.
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:::{.theorem}
If $S$ is ruled and not rational, then $S\cong \PP_C(\mce)$ is a geometrically ruled surface where $C$ is a curve of genus $g\geq 1$.
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:::{.remark}
Important trick: suppose $C$ is an effective irreducible curve where $\abs{K_S + C} = \emptyset$.
Suppose also that $\chi(\OO_S) = 1$, noting that $\chi(\OO_S) = 1-q+p_2$ which vanishes for rational surfaces.
Then $C\cong \PP^1$.

Recall that $\chi(\OO(D)) = \chi(\OO_S) + {1\over 2}D(D-K_S)$.
Applying this to $D = K+C$ yields $\chi(K+C) = 1 + {1\over 2}(K+C)C = p_a(C)$.
On the other hand, $\chi(K+C) = h^0(K+C) - h^1(K+C) +h^0(-C)$.
Here $h^0(K+C) = h^0(-C) = 0$, so $\chi(K+C)\leq 0$, and thus $p_a(C)\leq 0$, forcing $p_a(C) = 0$ and $C\cong \PP^1$.

Recall that if $\nu: \tilde C\to C$ is the normalization, then $p_a(C) \geq p_a(\tilde C) = g(\tilde C)$.

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:::{.corollary}
Suppose $\chi(\OO_S) = 1$ and $C = \sum a_i C_i$ is an effective curve, and suppose $\abs{K+C} = \emptyset$.
Then each $C_i \cong \PP^1$ since $\abs{K+C_i} = \emptyset$.
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:::{.remark}
Consider $S = \PP^2$; the key is to look at minimal rational curves on $S$.
Note that this generalizes to higher dimensions, viz. Mori theory.
For $\PP^2$, these minimal curves will be lines.
If $C$ is a line, then $KC = -3 < 0$ and $\abs{K+C} = \abs{-3H + H} = \emptyset$.

For $S = \FF_n$, note that this is a $\PP^1$ fibration over $\PP^1$ and the minimal rational curves are its fibers.
Letting $C$ be a fiber, we have $C^2 = 0$ and by the genus formula, $KC = -2$.
Moreover $\abs{K+C} = \emptyset$, using the following observation:
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:::{.remark}
Suppose $A$ is an irreducible effective curve with $A^2 \geq 0$ with $D.A < 0$.
Then $D$ is not effective and $\abs{D} = \emptyset$.
This follows from $D = nA + \sum a_i B_i$ with $AB_i \geq 0$ and $nA^2\geq 0$.
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:::{.remark}
One can then check that if $K+C$ is effective, then its restriction to a fiber is still effective.
However, $K_{\PP^1} = \OO_{\PP^1}(-2)$ is not effective.
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:::{.remark}
We now prove that if $S$ is rational and minimal, then $S\cong \PP^2, \FF_n$.
Consider the set of irreducible curves $\ts{C \cong \PP^1 \subseteq S \st C^2 \geq 0}\neq \emptyset$.
This is nonempty by a previous result on existence of special curves on such surfaces.
Consider the subset of such curves with minimal square $C^2 = m$.
Fix a hyperplane section $H$, so $C.H = \deg(C)$, and consider the further subset with minimal $C.H$.
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:::{.proposition}
Step 1: in $\abs{C}$, every curve is irreducible and isomorphic to $\PP^1$.
Thus in this minimal covering set, the curves do not split into multiple components.
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:::{.proof}
Suppose $D \in \abs{C}$ and write $D = \sum a_i C_i$, then all $C_i \cong \PP^1$.
Then $\abs{K+D} = \abs{K+C} = \emptyset$.
This follows from the "useful observation", since $(K+C)C = 2p_a(C) - 2 = -2$.
Note that $H.C_i < H.C$.

> We'll return to this proof later.

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:::{.proposition}
Step 2: $\dim \abs{C} \leq 2$.
Note that if one considers the divisors in $\abs{C}$ passing through a fixed point $p$, the dimension either stays the same or decreases by one.
This is a $\codim \leq 1$ condition.
Those passing through $p$ with multiplicity $\geq 2$ is a $\codim \leq 3$ condition.
Given $\OO_S\surjects \OO_S/\mfm^2$, we have $H^0(\OO(C)) \to H^0(\OO(C)\tensor \OO_S/\mfm^2)$ and $\OO_S/\mfm^2$ has dimension 3, so the kernel has dimension at most 3.
Supposing $\dim \abs{C} \geq 3$, there exists a curve $D\in \abs{C}$ passing through $p$ with multiplicity $\geq 2$, which is thus a singular curve -- but this contradicts step 1.
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:::{.proposition}
Step 3: there is a SES $0\to \OO_S \to \OO_S(C) \to \OO_C(C)\to 0$ gotten by twisting $\OO_S(-C)\injects \OO_S \surjects \OO_C$ by $C$.
Note that $\OO_C(C) = \OO_{\PP^1}(m)$.
Taking the LES, note that $H^1(\OO_S)=0$ by rationality, so we have
\[
0\to H^0(\OO_S) = \CC \to H^0(\OO_S(C))\to H^0(\OO_{\PP^1}(m))\to 0
.\]

We claim $\abs{C}$ has no base points.
Note that $h^0(\OO_S(C)) = m$, and by the LES, $\dim \abs{C} = m+1$ since $h^0(\OO_{\PP^1}(m)) = m+1$.
Since $m+1\leq 2$, we have $m\leq 1$ and thus $m=0, 1$.
If $m=0$, then $\dim \abs{C} = 1$ yielding a morphism $S\to \PP^1$ with $C$ a fiber, making $S$ ruled by a previous theorem.
Since $S$ is minimal, this yields $\FF_n$, since it is a geometrically ruled surface over $\PP^1$.
If $m=1$, $\dim \abs{C} = 2$ and this yields a morphism $S\to \PP^2$.
Since the degree is $C^2 = 1$, this is an isomorphism.
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:::{.remark}
Next time: finish step 1, understand the minimal surfaces which are rational and not ruled.
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