| Themen-Übersicht |
xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1251
Herkunft: Augsburg
 |  Themenstart: 2019-11-03 20:19
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Hallo zusammen.
Sei $B$ ein Ring der Charakteristik $p$ und sei $\mathbb{G}_a\to \sp{B}$ die additive Gruppe über $B$.
Man kann dann den Endomorphismenring $\cEnd_B(\mathbb{G}_a)$ mit dem Ring der Frobeniuspolynome $B\{\tau\}$ identifizieren.
Das ist mir klar, bis auf ein kleines Detail, das ich mir nicht erschließen kann.
Der Beweis beruht auf der Tatsache, dass $B\{\tau\}$ kanonisch isomorph ist zu dem Ring der Additiven Polynome $B_{add}[x]$. Ein Polynom $f\in B[x]$ heißt in diesem Kontext additiv, falls $f(x+y)=f(x)+f(y)$ in $B[x,y]$ gilt und als Multiplikation nimmt man die Verkettung von Polynomen.
Da es einen natürlichen Isomorphismus
$$\cEnd_B(\A_B^1)\cong \End_{B-\o{Alg}}(B[x])\cong \abs{B[x]}$$ gibt muss man nur zeigen, dass ein Endomorphismus von Schemata $\varphi\colon \A_B^1\to \A_B^1$ genau dann ein Endomorphismus von Gruppenschemata $\varphi\colon \mathbb{G}_a\to\mathbb{G}_a$ ist, wenn das unter dem oben angegebenen Isomorphismus korrespondierende Polynom $f\in B[x]$ additiv ist.
Sei also $\varphi$ ein Endomorphismus der affinen Geraden und sei $f$ das korrespondierende Polynom.
Aus der Definition von $\varphi\tm \varphi$ folgt, dass man immer
$(1)$ $$\varphi\tm \varphi(x\ot 1)=f(x)\ot 1$$
und
$\varphi\tm \varphi (1\ot x)=1\ot f(x)$ hat.
Es ist auch klar, dass immer $f(x)\ot 1=f(x\ot 1)$ und $1\ot f(x)=f(1\ot x)$ gilt.
Die Bedingung, dass $\varphi$ ein Endomorphismus von Gruppenschemata ist ist äquivalnt zu
$(2)$ $$f(x)\ot 1+1\ot f(x)=\varphi\tm\varphi(x\ot 1 +1\ot x)$$.
Identifiziert man $B[x,y]\cong B[x]\ot_{B}B[x]$, dann bedeutet das, dass $f$ genau dann additiv ist, wenn $f(x\ot 1+1\ot x)=f(x\ot 1)+f(1\ot x)$ gilt.
Die einzige Aussage die es also zu zeigen gibt ist
$(3)$ $$f(x\ot 1+1\ot x)=\varphi\tm\varphi(x\ot 1+1\ot x)$$, denn dann ist $(2)$ offenbar äquivalent dazu, dass $f$ additiv ist.
Mir ist nicht klar, warum das gelten sollte.
In mindestens zwei Büchern die über dieses Thema berichten wird diese Stelle nicht begründet/näher ausgeführt und in einer Arbeit die mir vorliegt wird diese Gleichheit mit einem Ausrufezeichen versehen, welches nicht kommentiert wird. Außerdem schreibt der Autor die Gleichungen in einer falsch anmutenden Reihenfolge:
Die gelb markierten Gleichungen sind auf jeden Fall korrekt. Den Rest habe ich mit einem blauen Stift abgetrennt. Falls das Diagramm kommutiert, dann ist die Gleichung $F\tm F(T_1+T_2)=F(T_1)+F(T_2)$ korrekt, wenn $F$ additiv ist, dann gilt $F(T_1+T_2)=F(T_1)+F(T_2)$.
Ich würde mich sehr für eine Antwort interessieren.
Viele Grüße
\(\endgroup\)
|
Triceratops
Aktiv
Dabei seit: 28.04.2016
Mitteilungen: 5277
Herkunft: Berlin
 |  Beitrag No.1, eingetragen 2019-11-03 21:36
Bevor ich auf deinen Beweis eingehe, hier der Beweis, der mit der funktoriellen Definition von Gruppenschemata arbeitet.
$\mathbb{G}_a$ ist einfach der Vergissfunktor $\mathbf{CAlg}_B \to \mathbf{Ab}$, der die Multiplikation vergisst.
Ein Morphismus von Gruppenschemata ist eine natürliche Transformation der Funktoren.
Das heißt hier: Der zu einem Polynom $f \in B[x]$ gehörige Morphismus von Schemata $\mathbb{G}_a \to \mathbb{G}_a$ ist genau dann ein Morphismus von Gruppenschemata, wenn für alle (!) kommutativen $B$-Algebren $C$ die induzierte Abbildung $\mathbb{G}_a(C) \to \mathbb{G}_a(C)$, $c \mapsto f(c)$ ein Homomorphismus von abelschen Gruppen, also letztendlich additiv ist. Das bedeutet, es muss $f(c+c') = f(c)+f(c')$ für alle $c,c' \in C$ gelten. Das gilt insbesondere im universellen Fall $C = B[x,y]$ mit $c=x$ und $c'=y$. Umgekehrt reicht dieser universelle Fall auch schon aus, wie man leicht sieht.
Zu deinem Post nun:
Der Beweis mit der Grafik ist so zu verstehen: Die Kommutativität des ersten Diagramms ist zur Kommutativität des zweiten Diagramms äquivalent. Nun formt man diese äquvialent um:
Diagramm ist kommutativ
$\iff (F \otimes F) \circ \Gamma(\mu) = \Gamma(\mu) \circ F$
$\iff (F \otimes F)(\Gamma(\mu)(T)) = \Gamma(\mu)(F(T))$
$\iff (F \otimes F)(T_1 + T_2) = F(T_1 + T_2)$
$\iff F(T_1) + F(T_2) = F(T_1 + T_2)$
Und das war's schon.
|
xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1251
Herkunft: Augsburg
| Beitrag No.2, vom Themenstarter, eingetragen 2019-11-03 22:57
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2019-11-03 21:36 - Triceratops in Beitrag No. 1 schreibt:
Diagramm ist kommutativ
$\iff (F \otimes F) \circ \Gamma(\mu) = \Gamma(\mu) \circ F$
$\iff (F \otimes F)(\Gamma(\mu)(T)) = \Gamma(\mu)(F(T))$
$\iff (F \otimes F)(T_1 + T_2) = F(T_1 + T_2)$
$\iff F(T_1) + F(T_2) = F(T_1 + T_2)$
Und das war's schon. Hallo Triceratops.
Vielen Dank für die Antwort.
An einen funktoriellen Beweis habe ich noch gar nicht gedacht, da ich zuerst diesen Beweis verstehen wollte. Dennoch übertrifft der funktorielle Beweis meinen Beweis (mal wieder ;D).
Könntest du ausführen wieso
$(F \otimes F)(\Gamma(\mu)(T)) = \Gamma(\mu)(F(T))$
$\iff (F \otimes F)(T_1 + T_2) = F(T_1 + T_2)$ gilt?
Ich kann das nicht nachvollziehen.
Nehmen wir an $(F \otimes F)(\Gamma(\mu)(T)) = \Gamma(\mu)(F(T))$ gilt.
Die linke Seite ist gleich $(F\ot F)(T_1+T_2)$. Die Rechte Seite ist gleich $F(T_1)+F(T_2)$, denn es gilt $\Gamma(\mu)(F(T))=F(T)\ot 1+ F(T)\ot 1=F(T_1)+F(T_2)$. Warum sollte sie gleich $F(T_1+T_2)$ sein? Das war auch meine ursprüngliche Frage, nur dass ich eine andere Notation verwendet habe.
\(\endgroup\)
|
Triceratops
Aktiv
Dabei seit: 28.04.2016
Mitteilungen: 5277
Herkunft: Berlin
| Beitrag No.3, eingetragen 2019-11-04 07:35
Nein. Der Algebrahomomorphismus $\Gamma(\mu) : k[T] \to k[T_1,T_2]$ ist durch Fortsetzung von $T \mapsto T_1+T_2$ gegeben, also allgemein durch $f(T) \mapsto f(T_1+T_2)$. (Beachte, $f(T) \mapsto f(T_1)+f(T_2)$ gar kein Algebrahomomorphismus ist, schon $1 \mapsto 1$ ist falsch.)
|
xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1251
Herkunft: Augsburg
| Beitrag No.4, vom Themenstarter, eingetragen 2019-11-06 18:15
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2019-11-04 07:35 - Triceratops in Beitrag No. 3 schreibt:
Nein. Der Algebrahomomorphismus $\Gamma(\mu) : k[T] \to k[T_1,T_2]$ ist durch Fortsetzung von $T \mapsto T_1+T_2$ gegeben, also allgemein durch $f(T) \mapsto f(T_1+T_2)$. (Beachte, $f(T) \mapsto f(T_1)+f(T_2)$ gar kein Algebrahomomorphismus ist, schon $1 \mapsto 1$ ist falsch.) Oh Man! Stimmt! $\arr{k[T]}{F(T)}{\arr{k[T]}{\Gamma(\mu)}{k[T_1,T_2]}}$ ist ja durch $T\mapsto F(T)\mapsto F(T_1+F_2)$ gegeben. Wo war mein Kopf in dem Moment.. Ich bin froh, dass ich es jetzt verstanden habe, aber noch besser ist, dass der Beweis mit der funktoriellen Definition so einfach ist. Komischerweise wird es aber in allen Büchern zum Thema die ich habe mit dem "umständlichen" Beweis gemacht.
Vielen Dank\(\endgroup\)
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