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| Autor |
 Allgemeine Gleichung 5ten Grades nicht auflösbar |
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LukasNiessen
Aktiv 
Dabei seit: 30.09.2019
Mitteilungen: 148
Herkunft: Nordrhein-Westfalen, Bonn, Weststadt
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Hallo,
mit der Galois-Theorie kann man leicht zeigen, dass die n-te allgemeine Gleichung für n größer 4 nicht auflösbar ist.
Ich habe aber mein Problem beim genauen Verständnis von was mit der n-ten allgemeinen Gleichung nun gemeint ist.
Die intuitivste "Erkenntnis" obiger Aussage wäre wohl: $aX^5 + bX^4+cX^3+dX^2+eX+f = 0$ hat keine allgemeine Lösungsmenge (durch Radikale), mit $a,b,c,d,e,f \in K$ mit K Körper.
---
Aber die n-te allgemeine Gleichung wird ganz anders definiert:
Sei k ein Körper und $L = Q(k(T_1,...,T_n))$ mit $T_1,...,T_n$ Variablen.
Dann ist der Körper der symmetrischen rationalen Funktionen K definiert als der Fixkörper $L^{S(n)}$ wobei $S(n)$ die symmetrische Gruppe sei.
Die Gleichung der Erweiterung $L/K$ ist:
$f(X) = \prod_{i=1}^{n} (X-T_i) \in k[T_1,...,T_n][X]$.
Man kann nun zeigen, dass $L/K$ galoissch mit Galois-Gruppe $S(n)$ ist und, dass L ein Zerfällungskörper von f über K ist.
---
Wendet man darauf die Theorie von Radikalerweiterungen an, folgt sofort, dass die allg. Gleichung n-ten Grades für n größer 4 nicht auflösbar ist, denn $S(n)$ ist es nicht.
Aber ich sehe nicht die Parallelen von dieser Definition der allg. Gleichung n-ten Grades und meinem anfangs erwähnten Polynom $aX^5 + bX^4+cX^3+dX^2+eX+f$.
Ich schätze $T_1,...,T_n$ sollen hier die allgemeinen Variablen sein (also quasi $a,...,f$).
Aber wieso entspricht dann $f(X) = \prod_{i=1}^{n} (X-T_i)$ dem Polynom $aX^5 + bX^4+cX^3+dX^2+eX+f$?
Wir erhalten doch bei $\prod_{i=1}^{n} (X-T_i)$ nur bestimmte Summen und Produkte von den $T_i$ als Koeffizienten vor X wenn man ausmultipliziert?
Und zwar ein elementarsymmetrisches Polynom aus K.
Außerdem haben wir noch einen alternierenden Faktor $(-1)^i$ davor.
Ich sehe also nicht, wie diese Definition von der allgemeinen n-ten Gleichung einen Schluss auf meine anfangs gennante intuitive "Erkenntnis" zulässt.
Ich hoffe man versteht, was ich meine.
Könnte vielleicht jemand helfen? Danke!
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄\(\endgroup\)
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 |      Beitrag No.1, eingetragen 2020-11-16
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Die allgemeine 'Gleichung' $n$-ten Grades über einem Körper $K$ ist ganz einfach das normierte Polynom $X_n + X_{n-1} T + \cdots + X_1 T^{n-1} + T^n$ über dem Funktionenkörper $K(X_1,\dotsc,X_n)$ mit der Variablen $T$. Nun wäre es hilfreich, eine Faktorisierung zu kennen. Wenn man den Ansatz $(T-Y_1) \dotsc (T-Y_n)$ macht, bekommt, dass $X_i$ bis auf den Faktor $(-1)^i$ das $i$-te elementarsymmetrische Polynom in $Y_1,\dotsc,Y_n$ ist (welches Grad $i$ hat). Nun sagt der Hauptsatz über symmetrische Polynome aus, dass $K(X_1,\dotsc,X_n) \to K(Y_1,\dotsc,Y_n)^{\Sigma_n}$ ein Isomorphismus ist. Wir können daher die allgemeine Gleichung auch als Polynom über $K(Y_1,\dotsc,Y_n)^{\Sigma_n}$ bzw. sogar $K(Y_1,\dotsc,Y_n)$ auffassen. Der Vorteil ist hier, dass man damit zum einen die Galoisgruppe leicht bestimmen kann (siehe etwa Bosch, Algebra), und zum anderen hat man über der Erweiterung $K(Y_1,\dotsc,Y_n)$ eine ganz konkrete Faktorisierung des Polynoms.
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LukasNiessen
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Danke!
Verstehe es richtig, dass dann der Körper K eigentlich garkeine Rolle spielt? Also ich meine welcher Körper K nun ist, ist doch eig. egal, oder?
Denn wenn wir die allgemeine Gleichung n-ten Grades jetzt so wie ich anfangs als $X5+aX4+bX3+cX2+dX+e$ interpretieren, dann betrachtet man $a,b,c,d,e$ ja nicht als Elemente von K sondern als Elemente von $K(X_1,...,X_5)$. Also quasi als "feste Variablen".
Und die Art wie wir "mit ihnen rechnen" können ist dann aber dennoch nicht ganz unabhängig von K. Bsplw. wäre für $K=\{0,1\}$ folgendes der Fall: $a + a = 2a = 0$. Nicht aber für $K = \IQ$.
Trotzdem bleibt aber das Resultat, dass die Gleichung nicht auflösbar ist für alle Körper K bestehen, richtig?
Also auch für zb so "einfache" Körper wie $\{0,1\}$.
Verstehe ich das richtig?
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄\(\endgroup\)
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2020-11-16 16:04 - LukasNiessen in Beitrag No. 2 schreibt:
Verstehe es richtig, dass dann der Körper K eigentlich garkeine Rolle spielt? Also ich meine welcher Körper K nun ist, ist doch eig. egal, oder?
Ja. Stelle dir hier $K$ als einen "initialen Koeffizientenkörper" vor. Die Koeffizienten der allgemeinen Gleichungen sind Variablen über $K$, nicht in $K$. Damit meine ich, dass die algebraisch unabhängig über $K$ sind. Gerade das zeichnet ihre Allgemeinheit aus. Sonst würde man eben nicht von einer allgemeinen, sondern einer speziellen Gleichung sprechen.
Denn wenn wir die allgemeine Gleichung n-ten Grades jetzt so wie ich anfangs als $X5+aX4+bX3+cX2+dX+e$ interpretieren, dann betrachtet man $a,b,c,d,e$ ja nicht als Elemente von K sondern als Elemente von $K(X_1,...,X_5)$. Also quasi als "feste Variablen".
Nicht als irgendwelche Elemente, sondern eben als diese Variablen $X_1,\dotsc,X_5$. Aber dann schreibe bitte nicht $X$ für die andere Variable.
Und die Art wie wir "mit ihnen rechnen" können ist dann aber dennoch nicht ganz unabhängig von K. Bsplw. wäre für $K=\{0,1\}$ folgendes der Fall: $a + a = 2a = 0$. Nicht aber für $K = \IQ$.
Ja gut, aber das ist einfach die allgemeine Konstruktion von $K(X_1,\dotsc,X_n)$. Dass manchmal $2 \cdot X_1 = 0$ gilt und manchmal nicht, hat nichts mit $X_1$, sondern mit $K$ zu tun.
Trotzdem bleibt aber das Resultat, dass die Gleichung nicht auflösbar ist für alle Körper K bestehen, richtig?
Ja.
Also auch für zb so "einfache" Körper wie $\{0,1\}$.
Das ist eine Menge, kein Körper. Du meinst etwas anderes.\(\endgroup\)
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