By Burkhard Külshammer

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Xn bn = ax1 y1 + . . + axn yn = a(x1 y1 + . . h. a x1 b1 + . . + xn bn . 43 44 KAPITEL 12. 3 F¨ ur Elemente a, b eines Integrit¨ atsbereiches R sind ¨ aquivalent: (1) a b und b a. (2) ∃u ∈ U (R) : b = ua. (3) (a) = (b). Definition: In diesem Fall nennt man a und b assoziiert. 1. (1)⇒(2): Seien a, b ∈ R mit a b und b a. Dann existieren x, y ∈ R mit b = ax, a = by. h. a(1 − xy) = 0. Im Fall a = 0 ist b = ax = 0x = 0, also b = a1, und im Fall a = 0 ist xy = 1, also x ∈ U (R). (2)⇒(3): Ist u ∈ U (R) mit b = au, so ist (b) = bR = auR ⊆ aR = (a).

An . 5. ⇐: Sind r1 , . . , rn ∈ R mit r1 a1 +. +rn an = 1, so ist jeder gemeinsame Teiler von a1 , . . , an auch Teiler von 1, also eine Einheit in R. 7 Ein Integrit¨ atsbereich R heißt euklidischer Ring, wenn eine Abbildung H : R \ {0} → IN0 mit folgenden Eigenschaften existiert: (i) F¨ ur a, b ∈ R mit b = 0 existieren q, r ∈ R mit a = bq + r, wobei r = 0 oder H(r) < H(b). (ii) F¨ ur a, b ∈ R mit ab = 0 ist H(a) ≤ H(ab). Bemerkung: In manchen B¨ uchern wird (ii) nicht verlangt. Man nennt H(a) die H¨ ohe von a ∈ R \ {0}.

Iii) u ∈ U (R) ∧ a ∈ R ⇒ u a. (iv) a ∈ R ∧ a 1 ⇒ a ∈ U (R). (v) a, b, c ∈ R, a b, b c ⇒ a c. (vi) a, b1 , . . , bn , x1 , . . , xn ∈ R ∧ a b1 ∧ . . ∧ a bn ⇒ a x1 b1 + . . + xn bn . Beweis: (i) a = a1, 0 = a0. h. a = 0. (iii) a = u · u−1 a. (iv) a 1 ⇒ ∃b ∈ R : ab = 1 ⇒ a ∈ U (R). (v) Sind a, b, c ∈ R mit a b und b c, so existieren x, y ∈ R mit b = ax, c = by. Folglich ist c = by = a · xy, also a c. (vi) Seien a, b1 , . . , bn , x1 , . . , xn ∈ R mit a b1 , . . , a bn . Dann existieren y1 , .

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Algebra [Lecture notes] by Burkhard Külshammer


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