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  Galois-Gruppe bestimmen |
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LukasNiessen
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Dabei seit: 30.09.2019
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Herkunft: Nordrhein-Westfalen, Bonn, Weststadt
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Hallo!
Es geht nochmal darum eine Galois-Gruppe zu bestimmen, und zwar folgenden Polynoms:
$X^8 - 2 \in \IQ[X]$.
---
Der Zerfällungskörper ist $\IQ(\sqrt[8]{2}, i)$ und die Erweiterung hat den Grad 16.
Es gilt:
$\text{Gal}(\IQ(\sqrt[8]{2}, i) / \IQ) \subset S_8$.
---
Ich komme hier aber nicht weiter, wie kann ich nun die 16 Automorphismen der Galois-Gruppe bestimmen?
Danke!
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄\(\endgroup\)
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ollie3
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Dabei seit: 21.02.2016
Mitteilungen: 62
 |      Beitrag No.1, eingetragen 2020-10-23
|
Hallo,
ich glaube nicht, das das so richtig
ist. Müsste nicht die prim.8. Einheitswuzel
mit im zerfällunskörper sein?
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ollie3
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Dabei seit: 21.02.2016
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| Beitrag No.2, eingetragen 2020-10-23
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Sorry,
das war ein Denkfehler, der zerfällungskörper
war doch richtig...
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Triceratops
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Dabei seit: 28.04.2016
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 |  Beitrag No.3, eingetragen 2020-10-23
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Triceratops
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Dabei seit: 28.04.2016
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| Beitrag No.4, eingetragen 2020-10-23
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@ollie3: Das stimmt schon, aber die primitive 8. Einheitswurzel ist $$\sqrt{i} = \frac{1+i}{\sqrt{2}} = \frac{1+i}{\sqrt[8]{2}^4},$$sodass man genauso gut $i$ als Erzeuger nehmen kann.
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LukasNiessen
Aktiv 
Dabei seit: 30.09.2019
Mitteilungen: 148
Herkunft: Nordrhein-Westfalen, Bonn, Weststadt
| Beitrag No.5, vom Themenstarter, eingetragen 2020-10-26
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Perfekt, danke euch!
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
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LukasNiessen hat die Antworten auf ihre/seine Frage gesehen.
LukasNiessen hat selbst das Ok-Häkchen gesetzt. |  [Neues Thema]  [Druckversion] |
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