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Körper und Galois-Theorie
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2020-12-07 18:54 - LineareAlgebruh im Themenstart schreibt:
Hallo.
Jede komplexe Zahl lässt sich darstellen als \(a+bi\) mit \(a,b \in \mathbb{R}\). Ich habe mich jetzt gefragt, ob sich dann auch jede reelle Zahl darstellen lässt als \( a+b \sqrt 2\), mit \(a,b \in \mathbb{Q}\)
Hallo.
Hier noch eine Alternative:
$\mathbb{C}$ ist ein zweidimensionaler $\mathbb{R}$-Vektorraum. Das meinst du ja mit der Aussage in der ersten Zeile. Jetzt fragst du dich, ob $\mathbb{R}$ ein zweidimensionaler $\mathbb{Q}$-Vektorraum ist und schlägst auch gleich eine konkrete Basis vor. Es reicht also festzustellen (finde drei linear unabhängige Elemente), dass aus $\lambda_1+\lambda_2\sqrt{2}+\lambda_3\sqrt{3}=0, \lambda_i\in \mathbb{Q}$ folgt $\lambda_1=\lambda_2=\lambda_3=0$. Wäre das nicht so, dann folgte(zum Beispiel im Fall, dass alle ungleich Null sind), indem du $-\lambda_1$ quadrierst, dass $\sqrt{6}=\sqrt{2}\sqrt{3}$ rational ist, was bekanntlich nicht stimmt. Die anderen Fälle gehen auch so ähnlich.
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Polynome
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Hi Lukas.
Sei $K$ ein Körper und $f$ ein nicht konstantes Polynom.
Dann stimmt die Menge der multiplen Wurzeln von $f$ in einem algebraischen Abschluss mit den gemeinsamen Wurzeln von $f$ und $f'$ überein.
Ist $f$ irreduzibel, dann kann $f$ nur dann multiple Wurzeln haben, wenn $f'$ identisch verschwindet.
Die Aufgabe würde ich so angehen:
Angenommen es gibt multiple Wurzeln,
dann gilt $f'=0$. Das ist äquivalent dazu, dass es ein Polynom $g\in K[X]$ mit $f(X)=g(X^p)$ gibt.
Wenn jetzt der Frobenius surjektiv ist, dann kannst du die $p$-ten Wurzeln der Koeffizienten von $g$ hernehmen und ein Polynom $G$ definieren (Wie in der Lösung). Dann hast du
$$f=G^p$$
im Widerspruch zur Irreduzibilität.
Grüße
XST\(\endgroup\)
Erfahrungsaustausch
Hi alle.
Weiß jemand wo man eine Kopie dieses Artikels bekommt:
"R.P. Steiner. "A theorem on the Syracuse problem". In: ed. by D. McCarthy and H. C. Williams. Congressus numerantium; 20. Proceedings of the 7th Manitoba Conference on Numerical Mathematics and Computation, September 29-October 1, 1977. Winnipeg: Utilitas Mathematica Pub., 1978, pp. 553-559."
Es ist ein Resultat im Zusammenhang mit der Collatz-Vermutung.
MfG
XST
Algebraische Geometrie
2020-06-27 17:16 - Slash in Beitrag No. 1 schreibt:
Ich weiß nicht, ob dieses Resultat passend ist für das was du vorhast.
Berg, L., and Meinardus, G., Functional equations connected with the Collatz problem, Results in Math. 25 (1994)
Hi Slash,
da ich ein absoluter Neuling auf diesem Gebiet bin und es auch ehrlichgesagt nur der Vorlesung zur Liebe mache ist das Resultat neu und interessant für mich. Vielen Dank.
Vielleicht melde ich mich hier nochmal, wenn ich Neuigkeiten habe.
MfG XST
Algebraische Geometrie
Hi zusammen.
Im Zuge einer Vorlesung die sich damit beschäftigt Arbeiten über die Collatz-Vermutung zu studieren kam mir die Frage, ob es bisher schon einen geometrischen Ansatz gibt, also mit Hilfe der Algebraischen Geometrie, oder ob das grundsätzlich eher aussichtslos ist.
Ich habe das hier gefunden, was ja sehr interessant ist, aber vielleicht mit dem ursprünglichen Problem nicht mehr viel zu tun hat.
Ich selbst hab das Gefühl, dass ein solcher Ansatz eher nicht funktionieren würde, ich kann mir auch nicht vorstellen wie man das Problem in die Algebraische Geometrie übersetzen könnte, aber im Forum gibt es glaube ich ein paar die sich schon mit dem Problem auseinandergesetzt haben.
Vielleicht seit ihr ja bei der Recherche nach Arbeiten schonmal auf etwas interessantes in diese Richtung gestoßen!?
Ich freue mich auch über Vorschläge zu interessanten Arbeiten über das Thema.
Es geht mir nicht darum das Problem zu lösen, was ich mit 100%ger Wahrscheinlichkeit nicht schaffen würde (:D), sondern darum, dass ich im Zuge der Vorlesung einen Vortrag über eine Arbeit vorbereiten soll. Da würde mich persönlich eben eine Arbeit mit geometrischen Mitteln sehr reizen, aber ich konnte nicht mehr finden als die oben zitierte.
MfG
XST
Ringe
Hi geeert,
schau mal hier .
Grüße
XST
Oder hast du diese Frage auf stackexchange gestellt? :D
Vielleicht steht in Eisenbud ja auch etwas dazu!?
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Für $1.241,85$ Euro ein bisschen Plastik kaufen ist mir zu teuer. Da spende ich das Geld lieber. Man kann diese Würfel auch mit einem Programm simulieren lassen, das ist kostenlos und reicht erstmal zum üben.
Ich würde als Mathegeschenk einen digitalen Stift empfehlen. Man spart Papier für Notizen und kann sich online besser über Mathe unterhalten.
Kostenpunkt etwa 30-50 Euro.
MfG\(\endgroup\)
Schwarzes Brett
@Tetsuya
Ob Serlo jetzt wirklich besser erklärt als Wikipedia sehe ich jetzt nicht sofort ein.
Fest steht, dass man bei Serlo aber schnell an Grenzen stößt. Wirklich viele Inhalte gibt es dort noch nicht, aber die Seite befindet sich ja auch noch im Aufbau.
Der Matheplanet ist hingegen einzigartig, hier gibt es sowohl interessante Artikel als auch eine Community die bei Fragen gezielt weiterhilft.
VG
XST
Schwarzes Brett
Analysis Eins:
"Stell Dir vor, hochwertige Bildung steht welt-weit allen Menschen komplett kostenlos zur Verfügung und wird von Kindern, SchülerInnen und Studierenden aktiv mitgestaltet."
Das gibt es doch schon lange in Form von Wikipedia.
Was genau ist also der Bewegungsgrund extra noch eine neue Platform aufzumachen?
Bisher sehe ich keinen Grund Serlo Wikipedia vorzuziehen:
Es gibt weniger Inhalte und man weiß auch nicht inwiefern die Inhalte auf Serlo korrekt sind (was man bei Wikipedia leider auch nicht immer weiß, aber dort schauen mehr Experten drauf würde ich mal vermuten).
Viele Grüße
XST
PS:Ich fühle mich hier auf dem MP wohl.
Relationen und Abbildungen
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2020-05-13 20:13 - helmetzer in Beitrag No. 4 schreibt:
2020-05-13 19:22 - xiao_shi_tou_ in Beitrag No. 3 schreibt:
Beachte auch, dass eine asymmetrische Relation nicht symmetrisch ist. Kann also eine symmetrische Relation asymmetrisch sein?
Auch wenn es an Haarspalterei grenzt: Bei einer leeren Relation sollte man das noch einmal überdenken.
Uups. Stimmt, so einfach ist es nicht.
Naja, dann muss man benützen, dass die Implikation $(a,b)\in R\implies (b,a)\not\in R$ immer gilt, da es keine $(a,b)\in R$ gibt.
Naja, Haarspalterei ist es ja nicht, wenn nur "wahr" und "nicht wahr" als Lösungen in Frage kommen und ich meine Bemerkung genau die falsche Lösung impliziert.
Gut, dass es dir aufgefallen ist.
VG XST\(\endgroup\)
Relationen und Abbildungen
Hi
Die leere Relation kann (auf einer nicht-leeren Menge wie den natürlichen Zahlen) nicht reflexiv sein, da es ja nicht einmal ein einziges Element gibt welches in Relation zu sich selbst steht, geschweige denn alle.
Beachte auch, dass eine asymmetrische Relation nicht symmetrisch ist. Kann also eine symmetrische Relation asymmetrisch sein?
Relationen und Abbildungen
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2020-05-09 13:52 - MePep in Beitrag No. 6 schreibt:
Moment, ist also \(\sim\) die Schnittmenge von \(\sim_{1} und \sim_{2}\) ?
Was sollte es sonst sein? Es gibt keine größere (also gröbere) Menge die sowohl in $R_{\sim_1}$ als auch in $R_{\sim_2}$ enthalten ist.
Du solltest noch nachschauen, dass die Schnittmenge eine Äquivalenzrelation definiert.
Und jetzt, wo du weißt wie die gröbste Verfeinerung definiert wird musst du auch nicht mehr den Umweg über die Teilmengen $R$ gehen, sondern kannst die Definition der gröbsten Verfeinerung direkt hinschreiben und alles nachrechnen.
Viel Erfolg\(\endgroup\)
Relationen und Abbildungen
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Hi.
Wenn $\sim$ eine Äquivalenzrelation auf einer Menge $X$ ist, dann kannst du eine Menge $R_{\sim}\sube X\tm X$ definieren durch
$(x,x')\in R_{\sim}$ genau dann wenn $x\sim x'$.
Es ist nicht schwer sich klarzumachen welche Eigenschaften diese Menge $R_\sim$ haben muss. Zum Beispiel muss sie die Diagonale enthalten, wegen Reflexivität.
Eine Äquivalenzrelation $\sim$ ist feiner als eine Äquivalenzrelation $\sim'$ genau dann wenn aus $x\sim x'$ immer $x\sim' x'$ folgt, also wenn $R_{\sim}\sube R_{\sim'}$ gilt.
Eine gemeinsame Verfeinerung von $\sim_1,\sim_2$ finden bedeutet also eine Menge $R\sube X\times X$ zu finden die sowohl in $R_{\sim_1}$ als auch in $R_{\sim_2}$ enthalten ist und eine Äquivalenzrelation definiert. Es ist klar, was es bedeutet die gröbste solche Menge zu finden. Welcher Kandidat kommt in Frage?\(\endgroup\)
Moduln
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Gegeben ist also eine kurze exakte Folge
$$0\to M\overset{f}{\to} N\overset{g}{\to} P\to 0$$ von $R$-Moduln.
Das ist schonmal die Voraussetzung.
Leider ist der Post abgeschnitten.
Was möchtest du zeigen/fragen?
MfG XST\(\endgroup\)
Holomorphie
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Sei $f(z)=u(z)+iv(z)$.
Dann ist $f$ genau dann komplex diffbar, wenn die Cauchy-Riemannschen Gleichungen
$$\partial_x u=\partial_y v, \partial_y u=-\partial_x v$$
gelten.
Die Bedingung in der Aufgabe
ist offenbar äquivalent zu
$$\partial_\theta u+i\partial_\theta v=i\partial_t u-\partial_t v,$$
also zu
$$\partial_\theta u=-\partial_t v, \partial_\theta v=\partial_t u.$$
Der Zusammenhang zwischen den $\partial_x u,\partial_y u,\partial_x v,\partial_y v$ und den $\partial_\theta u,\partial_\theta v,\partial_t u,\partial_t v$ ist durch die Kettenregel gegeben, zum Beispiel gilt:
$\partial_\theta u(x,y)=\partial_x u\partial_\theta x+\partial_y u\partial_\theta y.$
Viele Grüße
XST\(\endgroup\)
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Ist es dir überlassen die Definitionsbereiche zu wählen?
Beachte, dass man ganz einfache Beispiele von injektiven Funktionen mit $x\mapsto x^2$ bekommt, wenn man zum Beispiel als Definitionsbereich $\emptyset$ oder eine Einpunktmenge nimmt, denn dann gibt es gar nicht $2$ Punkte die auf ein und denselben Punkt abgebildet werden könnten 🙂.
Ansonsten kannst du so vorgehen wie oben beschrieben.\(\endgroup\)
Zahlentheorie
2020-04-04 16:25 - Nuramon in Beitrag No. 4 schreibt:
2020-04-04 16:09 - ziad38 in Beitrag No. 2 schreibt:
wooow Du hast Recht, warum habe ich nicht auch so gedacht obwohl ich 3 mal gelesen habe.
Ich habe mich auch schon mal über ein [1] am Ende einer Formel in einem Artikel gewundert, bis mir der Betreuer meiner Bachelorarbeit sagte, dass das eine Quellenangabe ist.
den beweis noch ?
Welche Fragen hast du zu dem Beweis?
[Die Antwort wurde nach Beitrag No.2 begonnen.] Naja, das ist ja noch halb so schlimm.
Ich hab Ewigkeiten versucht eine Aussage zu beweisen, die eigentlich als Definition gedacht war...Dennoch hat sich daraus jetzt etwas interessantes ergeben.
[Die Antwort wurde nach Beitrag No.11 begonnen.]
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2020-03-25 16:33 - Nuramon in Beitrag No. 14 schreibt:
2020-03-25 16:25 - xiao_shi_tou_ in Beitrag No. 13 schreibt:
Auf deinem Profil gibt es ein Latex Profil.
Da kannst du Abkuerzungen wie
copy \newcommand{\Q}{\mathbb{Q}}
hinschreiben. Aber dort keine Dollarzeichen verwenden.
Dann musst du nur noch
schreiben, wenn du $\Q$ haben willst.
Bei default gibt es auf dem MP schon \IQ, \IR, \IN, \IZ, \IC für $\IQ, \IR, \IN, \IZ, \IC$. Wenn man sich den einen extra Buchstaben leisten kann, dann braucht man muss man sich also dafr gar nicht extra Abkürzungen einrichten.
Gut zu wissen, wusst ich nicht👍😎.\(\endgroup\)
Teilbarkeit
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Hi.
In einem Ring nennt man multiplikative Inverse Einheiten .
Im Fall von $\Z$ sind das $\pm 1$.
Man bekommt im allgemeinen in einem sogenannten UFD (Ein Ring mit eindeutiger Primfaktorzerlegung) eine Zerlegung die eindeutig ist bis auf Multiplikation mit Einheiten.
Das kannst du alles nachlesen unter dem Stichwort UFD oder ZPE-Ring.
TeX-code kann man auch mit Dollarzeichen machen:TeX
copy $\mathbb{R}$
$\mathrm{ggT}$
$\mathbb{R}$
$\mathrm{ggT}$
[Die Antwort wurde nach Beitrag No.10 begonnen.]
PS: Auf deinem Profil gibt es ein Latex Profil.
Da kannst du Abkuerzungen wie
copy \newcommand{\Q}{\mathbb{Q}}
hinschreiben. Aber dort keine Dollarzeichen verwenden.
Dann musst du nur noch
schreiben, wenn du $\Q$ haben willst.
\(\endgroup\)
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2020-03-25 15:36 - OliverFuchs in Beitrag No. 5 schreibt:
Ja danke nochmal. Zum einen für die Willommensgrüße im Forum zum anderen
für die schnelle Hilfe. Das mit der Linearkombination von a und b,
hätte ich auch sehen können. Sorry
Aber das was der Kollege mit p/ggT(a^2,b^2) gesagt hat, da hat er recht.
Genau so habe ich es auch in meiner Version gemacht nur habe ich es
aus dem Grund nicht verwendet, weil ich mir nicht sicher war ob man
bei der Verwendung der Primzahl nicht Argumentte mit einbaut,
mit denen man auch die Primfaktorzerlegung zeigen kann. Da habe ich
zu wenig parat wie man diese zeigt und welch Argumente da eingehen.
Wenn ich mich aber recht aus der Vorlesung erinnere, so kann man die Primfaktorzerlegung in jedem Ring zeigen, in welchem auch die Divison mit
Rest gilt. Nur muss sie dann nicht eindeutig sein. Dazu, glaube ich wenigstens, benötigt man den Hauptsatz der Arithmetik (p/a1...an =>
p/ai 1<=i<=n). Wenn also bei der alternativen Beweisführung nicht beide
Dinge verwendet werden dann wäre der Beweis zulässig. Das muss
ich mir noch ansehen.
Danke nochmals für die Hilfe und den Willkommensgruß
lg Oliver
Aus algebraischer Sicht ist ja die PFZ in dem Ring $\Z$ auch nicht eindeutig, da man die Wahl zwischen $\pm p_i$ hat. Man legt sich nur traditionell auf $p_i>0$ fest.
Dass in jedem Ring mit Euklidischer Division es eine Primfaktorzerlegung gibt ist richtig.
Ich muss mich jetzt doch noch korrigieren
Man braucht die Primfaktorzerlegung schon. Tut mir leid für diese falsche Aussage.
[Die Antwort wurde nach Beitrag No.6 begonnen.] \(\endgroup\)