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Strukturen und Algebra » Körper und Galois-Theorie » Wieso ist das eine Galoiserweiterung?
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Universität/Hochschule J Wieso ist das eine Galoiserweiterung?
LukasNiessen
Aktiv Letzter Besuch: in der letzten Woche
Dabei seit: 30.09.2019
Mitteilungen: 132
Aus: Nordrhein-Westfalen, Bonn, Poppelsdorf
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Hallo,

Es seien $a,b \in \IC, a \neq b, a \neq -b$ Nullstellen von $(X^3-2)(X^2+3) \in \IQ[X]$ und $L=\IQ(a,b)$.
Man zeige, dass $L/\IQ$ galoisch ist, und bestimme die Galois-Gruppe sowie alle Zwischenkörper.

Natürlich ist die Erweiterung separabel da $\IQ$ perfekt ist.
Aber ich finde keinen guten Ansatz zu zeigen, dass sie auch normal ist.

Kann mir jemand helfen?

Danke!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
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Nuramon
Senior Letzter Besuch: in der letzten Woche
Dabei seit: 23.01.2008
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\(\begingroup\)\(\newcommand{\End}{\operatorname{End}} \newcommand{\id}{\operatorname{id}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\im}{\operatorname{im}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\d}{{\rm d}} \newcommand{\rg}{\operatorname{rg}} \newcommand{\spur}{\operatorname{spur}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\tr}{\operatorname{tr}}\)
Hallo,

es gibt im Wesentlichen zwei Fälle:
1. $a,b$ sind Nullstellen von $X^3-2$
2. $a$ ist Nullstelle von $X^3-2$ und $b$ von $X^2+3$

Im ersten Fall kannst du $X^3-2$ mit Vieta faktorisieren.
Im zweiten kannst du dir überlegen, dass $\IQ(b)$ eine primitive dritte Einheitswurzel enthält. Mit dieser und mit $a$ kannst du dann $X^3-2$ faktorisieren.
\(\endgroup\)


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LukasNiessen
Aktiv Letzter Besuch: in der letzten Woche
Dabei seit: 30.09.2019
Mitteilungen: 132
Aus: Nordrhein-Westfalen, Bonn, Poppelsdorf
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.2, vom Themenstarter, eingetragen 2020-09-09

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Hey,

danke für die Antwort!

Zu 1.:
Meinst du folgendes?
Sei also $X^3-2=(X-a)(X^2+pX+q)$. Mit dem Satz von Vieta folgt, dass für die Nullstellen $b, c$ von $(X^2+pX+q)$ gilt: $bc=q$ und daher gilt auch $c \in \IQ(a,b)$.

Zu 2.:
Leider weiß ich immer noch nicht ganz weiter. Einheitswurzeln kommen erst im nächsten Kapitel, ich habe mir aber mal die Definition angeschaut, doch ich sehe nicht, wieso $\IQ(b)$ eine dritte Einheitswurzel enthält.

Und ich sehe insbesondere auch nicht, wie wir eine solche Einheitswurzel zum faktorisieren von $X^3-2$ gebrauchen können, denn der konstante Teil des Polynoms ist ja 2 und nicht 1.

Danke!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne :-)
\(\endgroup\)


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Nuramon
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2020-09-09 11:39 - LukasNiessen in Beitrag No. 2 schreibt:
Zu 1.:
Meinst du folgendes?
Sei also $X^3-2=(X-a)(X^2+pX+q)$. Mit dem Satz von Vieta folgt, dass für die Nullstellen $b, c$ von $(X^2+pX+q)$ gilt: $bc=q$ und daher gilt auch $c \in \IQ(a,b)$.
Ja. Alternativ geht es auch so: Wenn $a,b,c$ die Nullstellen von $X^3-2$ sind, dann gilt $a+b+c=0$ (Koeffizient von $X^2$), also zerfällt $X^3-2 = (X-a)(X-b)(X+a+b)$ über $\IQ(a,b)$ in Linearfaktoren.


Zu 2.:
Leider weiß ich immer noch nicht ganz weiter. Einheitswurzeln kommen erst im nächsten Kapitel, ich habe mir aber mal die Definition angeschaut, doch ich sehe nicht, wieso $\IQ(b)$ eine dritte Einheitswurzel enthält.
Wie sieht so eine dritte Einheitswurzel denn aus? Schreib das mal explizit hin.


Und ich sehe insbesondere auch nicht, wie wir eine solche Einheitswurzel zum faktorisieren von $X^3-2$ gebrauchen können, denn der konstante Teil des Polynoms ist ja 2 und nicht 1.
Was sind denn die Nullstellen von $X^3-2$? Was fällt auf, wenn du zwei verschiedene Nullstellen miteinander vergleichst?
\(\endgroup\)


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LukasNiessen
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Perfekt! Ich sehe es nun auch, danke sehr.
Ich habe versucht die Aufgabe gänzlich zu lösen, also noch die Galois-Gruppe und die Zwischenkörper zu bestimmen.

Ich bin relativ weit gekommen, aber nicht ganz bis zum Schluss, und ich bin mir relativ unsicher ob das so überhaupt passt:

---------------------------------------------------

Im Falle 1, dass a,b Nullstellen von $X^2+3$ sieht man, dass die Erweiterung den Grad 2 hat und daher die Galois-Gruppe $S_2$ ist, daher gibt es auch keine (echten) Zwischenkörper.

Im Falle 2, dass a von $X^3-2$ und b von $X^2+3$ ist, sieht man (da dritte Einheitswurzel und b sie erzeugen), dass die Erweiterung den Grad 6 hat, daher $G = S_3$.

Im Falle 3, dass beide von $X^3-2$ sind:
Da $\IQ(a,b)$ stets Zerfällungskörper über dem Polynom ist und damit eindeutig ist, können wir a,b beliebig wählen. Etwa $a=\sqrt[3]{2}, b=(-1)^{\frac{2}{3}}\sqrt[3]{2}$. Dann gilt $[\IQ(a,b):\IQ]=[\IQ(a):\IQ]\cdot [\IQ(a,b):\IQ(a)] = 3\cdot 2 = 6$ und damit auch $G=S_6$.

Da wir in beiden Fällen die gleiche Galois-Gruppe sowie Körpererweiterung haben, ist es ausreichend einen Fall zu betrachten um alle Zwischenkörper zu bestimmen. Ich nehme Fall 3:

Eine echte Untergruppe von G hat entweder 2 oder 3 Elemente.
Wenn ich da nichts übersehe sind alle solche Untergruppen die folgenden:

Die Nullstellen um deren Permutationen es geht sind $\sqrt[3]{2} = a, (-1)^{\frac{2}{3}}\sqrt[3]{2} = b, -\sqrt[3]{-2} = c$

Ordnung 2:
$\{a \mapsto a, b \mapsto c, c \mapsto b \}$
$\{a \mapsto c, b \mapsto b, c \mapsto a \}$
$\{a \mapsto b, b \mapsto a, c \mapsto c \}$

Ordnung 3:
$\{a \mapsto b, b \mapsto c, c \mapsto a \}$

Die entsprechenden Fixkörper sind bei Ordnung 2:
$\IQ(a), \IQ(b), \IQ(c)$


Aber ich weiß nicht ganz, wie ich den Fixkörper bei der Untergruppe der Ordnung 3 bestimmen soll.

Und passt das so überhaupt? Es ist meine erste Aufgabe wo es darum geht die Zwischenkörper zu bestimmen :D

Danke!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
\(\endgroup\)


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Nuramon
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Dabei seit: 23.01.2008
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2020-09-11 16:51 - LukasNiessen in Beitrag No. 4 schreibt:
Im Falle 1, dass a,b Nullstellen von $X^2+3$ sieht man, dass die Erweiterung den Grad 2 hat und daher die Galois-Gruppe $S_2$ ist, daher gibt es auch keine (echten) Zwischenkörper.
Richtig, aber der Fall ist in der Aufgabenstellung durch die Bedingung $a\not= \pm b$ sowieso ausgeschlossen.


Im Falle 2, dass a von $X^3-2$ und b von $X^2+3$ ist, sieht man (da dritte Einheitswurzel und b sie erzeugen), dass die Erweiterung den Grad 6 hat, daher $G = S_3$.
Warum ist die Galoisgruppe $S_3$ und nicht z.B. $\IZ/6$?


Im Falle 3, dass beide von $X^3-2$ sind:
Da $\IQ(a,b)$ stets Zerfällungskörper über dem Polynom ist und damit eindeutig ist, können wir a,b beliebig wählen. Etwa $a=\sqrt[3]{2}, b=(-1)^{\frac{2}{3}}\sqrt[3]{2}$. Dann gilt $[\IQ(a,b):\IQ]=[\IQ(a):\IQ]\cdot [\IQ(a,b):\IQ(a)] = 3\cdot 2 = 6$ und damit auch $G=S_6$.
Die Notation $(-1)^\frac 23$ (und später auch $\sqrt[3]{-2}$) würde ich nicht verwenden. Besser wäre z.B. $\omega$ als primitive dritte Einheitswurzel (also als eine Lösung von $\omega^2+\omega+1 =0$) zu definieren und dann die Nullstellen von $X^3-2$ mit $\sqrt[3]2, \omega\sqrt[3]2, \omega^2\sqrt[3]2$ zu bezeichnen.


Da wir in beiden Fällen die gleiche Galois-Gruppe sowie Körpererweiterung haben, ist es ausreichend einen Fall zu betrachten um alle Zwischenkörper zu bestimmen. Ich nehme Fall 3:

Eine echte Untergruppe von G hat entweder 2 oder 3 Elemente.
Wenn ich da nichts übersehe sind alle solche Untergruppen die folgenden:

Die Nullstellen um deren Permutationen es geht sind $\sqrt[3]{2} = a, (-1)^{\frac{2}{3}}\sqrt[3]{2} = b, -\sqrt[3]{-2} = c$

Ordnung 2:
$\{a \mapsto a, b \mapsto c, c \mapsto b \}$
$\{a \mapsto c, b \mapsto b, c \mapsto a \}$
$\{a \mapsto b, b \mapsto a, c \mapsto c \}$

Ordnung 3:
$\{a \mapsto b, b \mapsto c, c \mapsto a \}$
Du könntest noch besser begründen, warum die Galoisgruppe gerade aus den Permutationen von $a,b,c$ besteht (das ist verwandt mit meiner Nachfrage zu Fall 2).


Die entsprechenden Fixkörper sind bei Ordnung 2:
$\IQ(a), \IQ(b), \IQ(c)$
Wie begründest du das?


Aber ich weiß nicht ganz, wie ich den Fixkörper bei der Untergruppe der Ordnung 3 bestimmen soll.
Um zu jeder Untergruppe direkt den entsprechenden Fixkörper zu bestimmen, müsstest du eigentlich eine $\IQ$-Basis von $\IQ(a,b)$ bestimmen, ausrechnen, wie die einzelnen Permutationen auf beliebige Linearkombinationen dieser Basis wirken und dann die Fixpunkte bestimmen.

Aber du kannst auch so argumentieren: Laut Hauptsatz der Galoistheorie gibt es 4 echte Zwischenkörper. Drei davon haben Erweiterungsgrad 3 und einer Grad 2. Es reicht also eigentlich aus, wenn du diese vier Zwischenkörper angeben kannst. Im Prinzip musst du also nur noch "erraten", was der Grad 2 Zwischenkörper ist und begründen, dass $\IQ(a),\IQ(b),\IQ(c)$ drei verschiedene Zwischenkörper vom Grad 3 sind.
Anschließend kannst du dir dann immer noch überlegen, welcher Zwischenkörper zu welcher Untergruppe korrespondiert.
\(\endgroup\)


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LukasNiessen
Aktiv Letzter Besuch: in der letzten Woche
Dabei seit: 30.09.2019
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Aus: Nordrhein-Westfalen, Bonn, Poppelsdorf
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2020-09-11 18:09 - Nuramon in Beitrag No. 5 schreibt:

Richtig, aber der Fall ist in der Aufgabenstellung durch die Bedingung $a\not= \pm b$ sowieso ausgeschlossen.
Stimmt


Warum ist die Galoisgruppe $S_3$ und nicht z.B. $\IZ/6$?
Weil $\IQ(a,b)$ der Zerfällungskörper des Polynoms ist. Wir haben einen Satz bewiesen, dass die Galois-Gruppe dann wine Untergruppe der Symmetrische Gruppe der Anzahl der Nullstellen des Polynoms ist, also hier $S_3$.


Wie begründest du das?
Da bin ich mir unsicher.
Wie könnte ich denn so eine Q-Basis bestimmen und sehen welche Linearkombinationen fix sind in diesem Beispiel?


und begründen, dass $\IQ(a),\IQ(b),\IQ(c)$ drei verschiedene Zwischenkörper vom Grad 3 sind.
Das würde ja daraus folgen, dass diese Körper tatsächlich die entsprechenden Fixkörper sind.


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
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Nuramon
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Stimmt, für die Grad 3 Zwischenkörper kannst du einfach so argumentieren: $\IQ(a),\IQ(b),\IQ(c)$ sind jeweils fix unter den entsprechenden Untergruppen und haben Grad 3, also sind dies tatsächlich die zugehörigen Fixkörper.

Um den Grad 2 Zwischenkörper zu finden, brauchst du eigentlich nur ein über $\IQ$ irreduzibles quadratisches Polynom zu finden, dass in $\IQ(a,b)$ zerfällt.
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2020-09-12 13:40 - Nuramon in Beitrag No. 7 schreibt:
Um den Grad 2 Zwischenkörper zu finden, brauchst du eigentlich nur ein über $\IQ$ irreduzibles quadratisches Polynom zu finden, dass in $\IQ(a,b)$ zerfällt.

Vielleicht übersehe ich irgendwas einfaches, aber ich sehe einfach nicht welches Polynom das sein könnte...


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
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Nuramon
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\(\begingroup\)\(\newcommand{\End}{\operatorname{End}} \newcommand{\id}{\operatorname{id}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\im}{\operatorname{im}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\d}{{\rm d}} \newcommand{\rg}{\operatorname{rg}} \newcommand{\spur}{\operatorname{spur}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\tr}{\operatorname{tr}}\)
Wie wäre es mit $X^2+3$?
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LukasNiessen
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Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.10, vom Themenstarter, eingetragen 2020-09-13

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Ja, stimmt! Ich habe nur $\sqrt[3]{-1}$ "isoliert" betrachtet und nicht die Darstellung als komplexe Zahl.

Also ist der entsprechende Zwischenkörper $E = \IQ(i \sqrt{3})$, denn die Erweiterung $E/\IQ$ ist vom Grad 2 und die Elemente von E bleiben unter allen Automorphismen der genannten Untergruppe der Ordnung 3 fix.

Passt das jetzt?

Vielen Dank!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
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Nuramon
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\(\begingroup\)\(\newcommand{\End}{\operatorname{End}} \newcommand{\id}{\operatorname{id}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\im}{\operatorname{im}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\d}{{\rm d}} \newcommand{\rg}{\operatorname{rg}} \newcommand{\spur}{\operatorname{spur}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\tr}{\operatorname{tr}}\)
Ja, passt.

Weil du danach gefragt hattest, wie man eine $\IQ$-Basis von $\IQ(a,b)$ bestimmt (sagen wir $a=\sqrt[3]2$ und $b=\omega a, c=\omega^2 a$):
Eine $\IQ$-Basis von $\IQ(a)$ ist $1,a,a^2$.
Eine $\IQ(a)$-Basis von $\IQ(a,b) = \IQ(a)(i\sqrt 3) = \IQ(a)(\omega)$ (mit $\omega^2+\omega+1 = 0$), ist $1,\omega$.
Folglich sind die paarweisen Produkte dieser Basisvektoren, also $1, a, a^2, \omega, \omega a, \omega a^2$ eine $\IQ$-Basis von $\IQ(a,b)$.

Betrachten wir jetzt die Untergruppe der Galoisgruppe von $\IQ(a,b)/\IQ$, die vom Automorphismus $\tau: a\mapsto b, b\mapsto c, c\mapsto a$ erzeugt wird.
Wegen $\omega = \frac ba$ gilt $\tau(\omega) = \frac cb$ und somit für beliebige $a_1,a_2,\ldots, a_6 \in \IQ$:
$$\begin{align*}
 \tau(a_1+a_2 a+ a_3 a^2 + a_4 \omega + a_5 \omega a+a_6 \omega a^2) &= a_1 + a_2b+ a_3b^2 + a_4 \frac cb + a_5 \frac cb b + a_6\frac cb b^2\\
&= a_1 + a_2 \omega a + a_3 \omega^2 a^2 + a_4 \omega+ a_5 \omega^2 a + a_6 a^2\\
&= a_1 + a_2 \omega a -a_3\omega a^2 - a_3 a^2 + a_4\omega -a_5\omega a - a_5 a + a_6a^2\\
&= a_1 -a_5 a + (a_6-a_3)a^2 + a_4\omega +(a_2-a_5)\omega a - a_3\omega a^2
\end{align*}$$ Durch Koeffizientenvergleich sieht man jetzt, dass die Fixpunkte von $\tau$ die Form $a_1 + a_4\omega$ haben müssen.
\(\endgroup\)


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LukasNiessen
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Perfekt, ich verstehe! Vielen Dank für deine Hilfe!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄



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