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xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1245
Herkunft: Bonn
 Beitrag No.1, eingetragen 2020-02-23 23:29    [Diesen Beitrag zitieren]
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Ich habe mittlerweile selbst eine Antwort gefunden und möchte diese hier der Vollständigkeit halber hinschreiben.

Es geht um den Beweis der folgenden Aussage:
Ein $s\in \Gamma(X_S,\c{O}_{X_S}([x_1]+\pts+[x_n]))$ finden, sodass
$$X_s\sube U=X_S-\Gamma_{x_1}\cup\pts\cup\Gamma_{x_n}$$ gilt.

Beweis:
Der Lesbarkeit halber schreibe ich $X$ und meine $X_S$.
Sei $D\colon=[x_1]+\pts+[x_n],\lineb\colon=\c{O}_X(D)$.
Sei $s\colon=s_D=1\in\Gamma(X,\lineb)$ der kanonische Schnitt.
Sei der Divisor $D$ durch ein Tupel $(U_i,f_i)_{i\in I}$ gegeben.
Es gilt $\Gamma_{x_1}\cup\pts\cup\Gamma_{x_n}=\o{Supp}(D)$.
Der Support von $D$ ist dann gegeben durch
$$\o{Supp}(D)=\set{x\in X}{(f_i)_x\not\in \c{O}_{X,x}^\tm,\tx{für ein}i\tx{sodass}x\in U_i}=\set{x\in X}{(f_i)_x\not\in \c{O}_{X,x}^\tm,\tx{für alle}i\tx{sodass}x\in U_i}$$.
Sei nun $x\not\in U$, also $x\in \o{Supp}(D)$ und sei $U_i,f_i$ ein Paar mit $x\in U_i$. Dann gilt $(f_i)_x\not\in \c{O}_{X,x}^\tm$.
Per Definition haben wir
$$\c{O}_{X}(D)\mid_{U_i}= \c{O}_{U_i}\pt f_i^{-1}$$ Das gibt einem die Trivialisierung
$$\c{O}_{X}(D)\mid_{U_i}\overset{\pt f_i}{\sto}\c{O}_{U_i}.$$ Diese schickt $s=1$ auf $f_i$.
Folglich gilt $s_x=(f_i)_x$.
Per Definition ist aber
$$X_s=\set{x\in X}{s_x\in \c{O}_{X,x}^\tm}$$.
Es gilt also $x\not\in X_s$, da wie oben geschildert $(f_i)_x\not\in\c{O}_{X,x}^\tm$ gilt.


Ich habe mal wieder eingesehen, wie einfach es doch sein kann, wenn man erstmal verstanden hat was das Problem eigentlich ist und wie gut ich doch darin bin, das Wesentliche zu übersehen :D.


 
\(\endgroup\)

xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1245
Herkunft: Bonn
 Themenstart: 2020-02-21 23:34    [Diesen Beitrag zitieren]
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Hallo zusammen.
Ich versuche mich gerade an der folgenden Aussage, und hänge an einer Stelle:
Seien wieder $X_S=S\tm_{\sp{k}}X$ wo $X$ eine glatte irreduzible Kurve ist und $\lineb',\lineb$ Geradenbündel.
Seien $x_1\cos x_n\in X(S)$ und sei wieder $U=X_S-\Gamma_{x_1}\cup\pts \cup \Gamma_{x_n}$ und $$\lineb\dashrightarrow \lineb'$$ eine meromorphe Abbildung von Geradenbündeln welche auf $X_S$ definiert seien, also ein Isomorphismus
$$\a\colon \lineb\mid_U \sto \lineb'\mid_U$$.
Dann gibt es ein $N$, groß genug, sodass sich $\a$ zu einem Morphismus
$$\lineb\to \lineb'(N[x_1]+\pts +N[x_n])$$ fortsetzt.

Hierbei steht $[x_i]$ für den (relativen) effektiven Cartier divisor der dem abgeschlossenen Unterschema $\Gamma_{x_i}\sube X_S$ entspricht.

Intuitiv ist das klar:
Da es nur endlich viele Punkte gibt, kann man einfach $N$ größer wählen als alle Polordnungen.

Mein Ansatz:
Ein Morphismus $\lineb\to \lineb'(N[x_1]+\pts+N[x_n])$
entspricht einem globalen Schnitt $\s\in \Gamma(\c{O}_X,\lineb^{-1}\ot_{X_S}\lineb'(N[x_1]+\pts N[x_n]))$.
So einen möchte ich jetzt bekommen indem ich
$\lineb'(N[x_1]+\pts+N[x_n])\cong \lineb'\ot_{\c{O}_{X_S}} (\c{O}_{X_S}([x_1]+\pts [x_n]))^{\ot N}$ benutze in Kombination mit der folgenden Aussage:[Theorem 7.22, Wedhorn]

Sei $\sc{M}$ ein Geradenbündel auf einem quasi-kompakten quasi-separablen Schema und sei $s$ ein Globaler Schnitt von $\sc{M}$.
Sei $\sc{F}$ eine quasi-kohärentes $\c{O}_X$-Modul.
Dann gibt es für jeden Schnitt $t'\in \Gamma(X_s,\sc{F})$ ein $N>0$ und einen Schnitt $t\in \Gamma(X,\sc{F}\ot \sc{M}^{\ot N})$
mit $t\mid_{X_s}=t'\ot s^{\ot N}$.

$X_s\colon=\set{x\in X}{s(x)\not=0 \tx{in}\lineb(x) }$

Für
$X=X_S, \sc{M}=\c{O}_{X_S}([x_1]+\pts+[x_n])$ und $\sc{F}=\lineb^{-1}\ot \lineb'$ gäbe es auf jeden Fall schonmal einen Globalen Schnitt von $\sc{M}$ da dieses Geradenbündel durch einen effektiven Cartier Divisor gegeben ist (zum Beispiel den kanonischen Schnitt). Um den Schnitt  $t'\in \Gamma(X_s,\sc{F})=\Gamma(X_s,\lineb^{-1}\ot\lineb')$ anzugeben müsste ich noch die Voraussetzung ins Spiel bringen, also die Meromorphe Abbildung.
Diese gibt schonmal einen Schnitt $\tau\in \Gamma(U,\lineb^{-1}\ot \lineb')$. Könnte man noch $X_s\sube U$ zeigen (i.e. ein $s$ finden, sodass das gilt), dann wäre man offensichtlich fertig.

Also die folgende Aussage:
Wenn $x\in \Gamma_{x_1}\cup\pts \cup \Gamma_{x_n}$($\not\in U$), dann
ist $s$ nicht invertierbar in $x$ ($\not\in X_s$), also $s=0$ in $\sc{M}(x)=\c{O}_{X_S}([x_1]+\pts [x_n])(x)$.

Es geht also intuitiv darum eine auf ganz $X_S$ definierte Funktion zu finden die entlang der Graphen von $x_1\cos x_n$ auch Pole der Ordnung höchstens $1$ haben dürfen, die auf ganz $\Gamma_{x_1}\cup\pts \cup \Gamma_{x_n}$ (punktweise) verschwindet.


Vielleicht hat jemand hier eine Idee wie man es machen könnte!?
Viele Grüße
XST


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