New PDF release: Algebra I

By Walter Gubler

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Zu zeigen ist nun, dass g oder h ∈ R[x]∗ . 5) folgt, dass 1 = µ(f ) = µ(g) · µ(h). Insbesondere muss dann µ(g) = µ(h) = 1 gelten. Weil f irreduzibel in Q[x] ist, muss g oder h ∈ Q[x]∗ = Q\ {0} sein. A: g ∈ Q\ {0}. Aus g ∈ R[x] folgt g ∈ R\ {0}. Wegen µ(g) = 1 folgt g ∈ R∗ = R[x]∗ Damit folgt der erste Schritt. Beweis des Theorems. Weil Q[x] faktoriell ist, gilt f = f1 · . . · fr f¨ ur irreduzible Polynome fi ∈ Q[x]. Wir ersetzen fi durch R-Vielfache in R[x] mit Inhalt 1, dann gilt: f = µ(f ) · f1 · .

Wenn a | 1, dann heißt a eine Einheit von R. Nach Definition ist dies aquivalent dazu, dass a invertierbar ist bez¨ ¨ uglich der Multiplikation. Wie in der Gruppentheorie gesehen, ist R bzgl der Multiplikation ein Monoid und die Menge R∗ der Einheiten bildet eine Gruppe bzgl ” · ”. Wenn es f¨ ur a ∈ R ein c ∈ R\ {0} gibt, mit a · c = 0, dann heißt a Nullteiler in R. Man darf diesen Begriff nicht mit den Teilern von Null verwechseln! Jedes a ∈ R ist ein Teiler von Null in obigem Sinn, aber Nullteiler sind meist ganz spezielle Elemente in R.

3. Sei α ∈ K eine Nullstelle von f (x) ∈ K[x]. g(x) ∈ K[x] mit f (x) = (x − α)g(x). Das nennt man Abspalten der Nullstelle α. Beweis. h. h ∈ K Setzen wir die Nullstelle α in das obige Polynom ein, dann folgt h = 0. 4. Sei f (x) ∈ K[x]\ {0} vom Grad n. Dann hat f (x) h¨ ochstens n verschiedene Nullstellen in K. Beweis. Mit Induktion n folgt dies leicht durch Abspalten von Nullstellen. 5. Es gibt genau einen Ringhomomorphismus ϕ : Z −→ K. Es gilt ker(ϕ) = Zp f¨ ur genau eine Primzahl p oder f¨ ur p = 0.

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Algebra I by Walter Gubler

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New PDF release: Algebra I
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