Detailed proof of two claims by Mori

In this article, i will give a detail of a proof of (8.3) and (8.4) in [Mor], 1979.


(8.3) $O_S \otimes O_Z(S)\cong O_S(-1)$.


 Let $L \cong \mathbb{P} ^{n-2}$ be a hyperplane, then because $S$ is a section of $Z$ over $Y$, which means $\psi (S) \cong Y \cong \mathbb{P}^{n-1}$, so in $ \mathbb{P}^{n-1}$, we can easily find a curve not contains in a hyperplane, so we can find $C \cong \mathbb{P}^{1} \subset S$ be a straight line such that $\psi(C) \nsubseteq L$. Let $D = \psi^{-1}(L)$, and then $(C . D) = 1$ because straight line intersect hyperplane at only one point. Since $\phi(S) = R$ and $S \cap D \neq { \varnothing }$, then $R \in \phi(D)$, and $V = \phi^{-1} \phi(D) \supseteq S \cup D$ but $\phi(V - S) \cong \phi(D) - {R}$ and $\phi(D \cup S - S) \cong \phi(D) - {R}$ so $\phi^{-1} \phi(D) = S \cup D$, then we can write $\phi^{-1} \phi(D) = D + aS $ as divisors for some $a > 0$. By the projection formula, $(C, \phi^{-1}\phi(D)) = (\phi(C), \phi(D)) = 0$ because $\phi(C) = R $ then we can consider $\phi(C) = 0$ (by projection formula 3.17, [Deb]) . Thus $0 = 1 + a(S.C)$, whence $(S.C)= -1$, hence $S$ is not ample, thus $deg (O_Z(S) \times O_S) = -c$ $(c>0)$ and $O_Z(S) \otimes O_C \cong O_C(-1)$, thus $c = 1$ and $O_Z(S) \otimes O_S = O_S(-1)$.

(8.4) $\mathbb{P}^n \cong U$.

We have standard exact sequence:
$0 \longrightarrow O_Z(-S) \longrightarrow O_Z \longrightarrow O_S \longrightarrow 0$
taking tensor by $O_Z(S)$, we get:
$0 \longrightarrow O_Z \longrightarrow O_Z(S) \longrightarrow O_S \otimes O_Z(S) \longrightarrow 0$
Which will be :
$0 \longrightarrow O_Z \longrightarrow O_Z(S) \longrightarrow O_S(-1) \longrightarrow 0$
t
Taking the direct image of $\psi$, we arrive at a new exact sequence :
$0 \longrightarrow O_Y \longrightarrow \psi_{*}O_Z(S) \longrightarrow O_Y(-1) \longrightarrow 0$

($R^1 \psi_{*}O_Z = H^1(O_S)=0$, because $S \cong \mathbb{P}^{n-1}$). Because $Ext^{1}_{\mathbb{P}^{n-1}}(O(-1),O) = 0$ ($O(-1)$ is locally free so it will be "projective" in homological algera sense), and hence by theorem 7.31 in [Rot], $\psi_{*}O_Z(S) = O(-1) \oplus O$. Thus $Z \cong P_{\mathbb{P}^{n-1}}(O(-1) \oplus O$) ($Z \cong P_{\mathbb{P}^{n-1}}(\psi_{*}O_Z(S))$ as in previous part). By proposition 7.12 [II, Hart], section $S$ must equivalence to quotidien surjective $O(-1) \oplus O \longrightarrow L $, but $S$ is not ample so $L = O(-1)$. Now, we have to give a morphism from $Z$ to $\mathbb{P}^n$. Mori's idea, althrough is not quite natural, it still very good. The way to obtain that morphism is using linear system of an ample invertible sheaf over $Z$. Taking $\psi^{*}O_Y(1) \otimes O_Z(S)$, by projection formula: $\psi_{*}( \psi^{*}O_Y(1) \otimes O_Z(S)) \cong O_Y(1) \otimes \psi_{*}(O_Z(S)) \cong O_Y \oplus O_Y(1)$. Hence $\psi^{*}O_Y(1) \otimes O_Z(S)$ must has global section.and generated by global sections. Thus it gives a morphism $f$ from $Z$ to $\mathbb{P}^n$. Because $\psi^{*}O_Y(1) \otimes O_Z(S)$ doesn't has section on $S$ so $f(S)=K$ must be a point, hence induce a morphism $g$ from $Z|_{Z-S}$ to $\mathbb{P}^n - K$. Because $Z$ is projective space $\mathbb{P}^m$, hence we can give a morphism $h$ from $\mathbb{P}^n - K$ to $Z-S$ corresponds to invertible sheaf $f_{*}(\psi^{*}O_Y(1) \otimes O_Z(S))$, and $h$ is an inverse image of $g$, which means $Z-S \cong \mathbb{P}^n - K$. Since $\mathbb{P}^n$ is birational correspondence to $Z$ and $Z$  is birational correspondence to $U$, then $U$ is birational correspondence to $\mathbb{P}^n$. By Zariski's main theorem, so for a birational map $T$: $U \longrightarrow \mathbb{P}^n$, $T(U - R) \cong  \mathbb{P}^n - K$ which is bijective, $T$ is a isomorphism by a lemma below. Thus, we prove (8.4).

Lemma: Let $f: X \longrightarrow Y$ is a bijective, birational morphism over $Y$ normal and connected, then $f$ is isomorphism.

proof: (from mathoverflow)
Since $Y$ is normal, this is equivalent to assuming that $Y$ is irreducible.  Grothendieck's formulation of Zariski's Main Theorem is EGA $III_2$, Theorem 4.4.3, p. 136.  According to that theorem, there exists a factorization of $f$, $f=\overline{f}\circ i$ where $i:X\hookrightarrow\overline{X}$ is a dense open immersion, and where $\overline{f}:\overline{X}\to Y$ is a finite morphism.  Identify $X$ with the image of $i$.

Since $f$ is bijective on points, there exists a unique irreducible component $\overline{X}_0$ of $\overline{X}$ that dominates $Y$.  The restriction of $\overline{f}$ to $\overline{X}_0$ is a finite, surjective morphism that is generically one-to-one.  Since this morphism is birational, by the classical form of Zariski's Main Theorem, $\overline{f}$ restricts to an isomorphism from $\overline{X}_0$ to $Y$, cf. Mumford's, Red Book of Varieties and Schemes, p. 288. 

By way of contradiction, assume that $\overline{X}_0$ is a proper subset of $\overline{X}$.  Denote by $Z \subset \overline{X}$ the union of all irreducible components different from $\overline{X}_0$.  Since $X$ is a dense open in $\overline{X}$, also $X\cap Z$ is a dense open in $Z$.  Thus, also $(X\cap Z) \setminus (\overline{X}_0\cap Z)$ is a dense open subset of $Z$.  Denote by $W$ the image in $Y$ of this open subset.  Since $f$ is injective, the constructible subset $W$ is disjoint from the dense open $V=f(X\cap \overline{X}_0)$ of $Y$.  Thus, the closure of $W$ is disjoint from $V$.  Since $\overline{f}$ is finite, this closure equals the image of $Z$.  Thus, $X\cap \overline{X}_0$ is disjoint from $\overline{X}_0\cap Z$, i.e., $X\cap \overline{X}_0\cap Z$ is empty.  Thus, the open subsets $\overline{X}_0\setminus (\overline{X}_0\cap Z)$ and $Z\setminus (Z\cap \overline{X}_0)$ of $\overline{X}$ pullback to disjoint open subsets of $X$ that cover $X$.  Since $X$ is connected, this is a contradiction. 

Therefore, $\overline{X}_0$ equals all of $\overline{X}$.  Since $\overline{X}_0\to Y$ is an isomorphism, it follows that $f$ is an open immersion.  Since $f$ is also surjective, $f$ is an isomorphism.

(cf. https://mathoverflow.net/questions/264204/bijection-implies-isomorphism-for-algebraic-varieties)




[Mor] S.Mori, Projective Manifolds with Ample Tangent Bundles, Annals of mathematics, second series Vol. 110, No. 3 (Nov. 1979), pp. 593 - 606.
[Deb] O.Debarre, Introduction to Mori theory, cours de M2 2010-2011, université Paris Diderot.
[Rot] Joseph J.Rotman, An introduction to homological algebra, second ed., Springer.
[Hart] R.Hartshorne, Algebraic geometry, Springer.

Comments

  1. TTC28 September 2018 at 09:20

    Nice post. Can you explain what $O_S, O_Z$ and $U$ are?

    ReplyDelete
    Replies
    1. Bảo30 September 2018 at 12:18

      $O_S$ and $O_Z$ are structured sheaves associated with schemes $S$ and $Z$ respectedly. $U$ is $spec(\pi_{*}O_Z)$. Sorry for my late reply.

      Delete
    2. TTC4 October 2018 at 08:29

      Thanks. In fact I have no idea of what scheme is even though I learnt it when I was at Toulouse :D

      Delete

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