A theorem of extreme NBD

In this post, I will rewrite a prove for a proposition of extremal nbd from [Mor88].
First, there is some lemmas we need to know first.

Lemma 1:
For an extremal nbd $X \subset C \cong \mathbb{P}^{1}$:
$H^{1}(O_Z) = H^1(\omega_X \otimes O_Z) = 0$ for all $Z \subset X$ with $Z_{red}=C$.

Lemma 2:
Over a smooth projective curve, for any locally free module $ \xi$ of rank $r$, we have equation: $\chi (C, \xi) =$ deg (det $\xi$)+($1-g$)rk $\xi$.

Now we will come to the main theorem:
Theorem:  Let $X \subset C \cong \mathbb{P}^{1}$ is smooth extremal nbd. Then $O_C(K_X) \cong O_C(-1)$, $I_C/I_C^{2} \cong O_C \oplus O_C(1)$ and $|-K_X|$ has smooth member, where $I_C$ is an ideal sheaf of $C$.


 Proof of theorem. By the definition of extremal nbd, $K_X$ is not ample, hence $O_X(K_X) \otimes O_C$ must not be ample and deg  $O_X(K_X) \otimes O_C$ = deg $(K_X.C)$  $< 0$ (it cannot equal to $0$ because $K_X$ is not equivalence to $0$).
By Remainn-Roch theorem, $h^0(O_X(K_X) \otimes O_C) - h^1(O_X(K_X) \otimes O_C) = deg (O_X(K_X) \otimes O_C) + 1-g$.
Because $ H^1(\omega_X \otimes O_C) = 0$, hence $deg (O_X(K_X) \otimes O_C) = -1$, hence $(K_X.C) = -1$, thus $O_C(K_X) \cong O_C(-1)$.
Now we will prove next assertion. By a standard exact sequence in [Hart77], we have:
$ 0 \longrightarrow I_C/I_C^2 \longrightarrow \Omega_{X} \otimes O_C \longrightarrow \Omega_{C} \longrightarrow 0$.
By theorem 8.17 in [II, Hart77], $I_C/I_C^{2}$ and $\Omega_{C}$ is a locally free sheaf of rank 2, hence by a property of exterior power, we get:
$ \wedge^{2} (I_C/I_C^{2}) \otimes O_C(K_C) \cong O_C(K_X)$
$\Leftrightarrow \wedge^{2} (I_C/I_C^{2}) \cong O_C(K_X) \otimes O_C(-K_C)$
Hence, $deg \wedge^{2} (I_C/I_C^{2}) = (K_X.C) - deg O_C(-K_C)$,
But $O_C(-K_C) = \omega_C(-2)$ because $C \cong \mathbb{P}^1$, so $ deg \wedge^{2} (I_C/I_C^{2}) = -1+ 2 = 1$.
By exact sequence:
$0 \longrightarrow I_C/I_C^2 \otimes_{O_C} O_C(-1) \longrightarrow O_X/I_C^2 \otimes_{O_X} \omega_X \longrightarrow O_C(-1) \longrightarrow 0$,
because $H^1(O_X/I_C^2 \otimes \omega_X) =0$ by lemma 1 and $H^0(C(-1))=0$, then $H^1(I_C/I_C^2 \otimes O_C(-1)) = 0$. By lemma 2, $h^0(I_C/I_C^2 \otimes O_C(-1)) = deg ( \wedge^{2} (I_C/I_C^{2}) \otimes O_C(-1)) + 2 = 1$. Hence $I_C/I_C^2 \otimes O_C(-1) \cong O_C \oplus O_C(a)$, then $I_C/I_C^2 \cong O_C(1) \oplus O_C(a+1)$. Because $deg \wedge^{2} (I_C/I_C^{2}) =  1$, we have $a+1+1=1$, which means $a= - 1$ and thus $I_C/I_C^2 \cong O_C(1) \oplus O_C$.
For the last statement. Let $(D,x)$ be smooth divisor germ which meet $(C,x)$ tranversally at one point, then $(D,x)$ must extend to a divisor $D'$ in $X$. Hence $(D'.C) = (D.C) = 1 =(-K_X.C)$. By theorem 0.1.3 in [Mor88], the birational morphism between $X \supset f^{-1}(y) = C$ and $(Y, y)$ is the composition of a divisor contraction morphism and directed flip. By a property of flip in 0.4 [Mor88], $f': X' - C' \cong Y - y$ (1) and $p: X \rightarrow X'$ is birational divisor contraction morphism. For any two effective divisors $Z', T'$ over $X$ contracted to same divisor in $C$, it must be equivalence by statement (1). We know that $(D'.C) =(-K_X.C) = 1$, hence these two divisors must contracted to $O_C(1)$ and hence must be equivalence, hence $|-K_X|$ has a smooth member.

Reference:
[Mor88] S.Mori, Flip theorem and the existence of Minimal model program for 3-folds. Annals of mathematics, 1988.

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