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3. BEISPIELE FUR 41 z w } bildet −w z einen Teilring von M2 (C) und ihre Elemente heißen Quaternionen. Dass dies ein Teilring ist, sieht man leicht, bis auf die Abgeschlossenheit bzgl. ” · ”. 3. Die Menge H := {A ∈ M2 (C) ∃z, w ∈ C : A = z 1 w1 −w1 z1 · z 2 w2 −w2 z2 z 1 z 2 − w1 w2 z 1 w2 + w1 z 2 −z2 w1 − w2 z1 −w1 w2 + z1 z2 = ∈H Man hat die Norm N (A) := det(A) = |z|2 + |w|2 . Sei A ∈ H, A = 0. 2) adj(A) = z −w w z =⇒ A−1 = 1 adj(A) ∈ H. N (A) Folgerung: Die Quaternionen bilden einen Ring, in dem jedes Element = 0 ein Inverses hat.

Beweis. Wie in der Gruppentheorie ist injektiv ¨aquivalent zu ker ϕ = {0}. Sei also I = ker ϕ. 9 gilt entweder I = {0} oder I = K. Wegen ϕ(1) = 1 ist der zweite Fall ausgeschlossen und somit ker ϕ = {0} 38 KAPITEL 2. 11. (a) I R heißt Maximalideal :⇐⇒ I maximales Element von {I R I = R} bzgl der partiellen Ordnung ⊆. (b) I R heißt Primideal genau wenn I = R und wenn ab ∈ I =⇒ a ∈ I oder b ∈ I. 12. Sei R = I R. Dann gilt: (a) I Primideal ⇐⇒ R/I Integrit¨ atsbereich (b) I Maximalideal ⇐⇒ R/I K¨ orper Beweis.

H. 1. RINGE 35 Dadurch wird R2 zu einem Ring. Beachte, dass R2 kein Integrit¨atsbereich ist. Die Nullteiler haben die Form (x, 0) oder (0, y), denn (0, x) · (y, 0) = (0, 0). • Z/14Z ist auch kein Integrit¨atsbereich, weil 2 und 7 Nullteiler sind. 8. Bekanntlich ist Q der kleinste K¨orper, der Z enth¨alt. Das wollen wir verallgemeinern f¨ ur einen Integrit¨atsbereich. Wir wollen dazu den Quotientenk¨ orper konstruieren. Weil diese Konstruktion schon in der Linearen Algebra gemacht wurde, werden wir uns kurz halten.

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