By Burkhard Külshammer
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Additional info for Algebra [Lecture notes]
Sample text
Satz: Sei R Integrit¨ atsbereich. Dann definiert man eine Aquivalenzrelation ∼ auf R × (R \ {0}) durch (a, b) ∼ (a′ , b′ ) :⇔ ab′ = a′ b. ¨ Bezeichnet man f¨ ur a, b ∈ R mit b = 0 die Aquivalenzklasse von (a, b) mit [a, b], so ist Q(R) := {[a, b] : a, b ∈ R, b = 0} ein K¨ orper mit [a, b] + [c, d] := [ad + bc, bd], [a, b][c, d] := [ac, bd]. Nullelement in Q(R) ist [0, 1], Einselement [1, 1], und es gilt: −[a, b] = [−a, b], [a, b]−1 = [b, a]. Die Abbildung ι : R → Q(R), a → [a, 1] ist ein unit¨ arer Monomorphismus, und f¨ ur a, b ∈ R mit b = 0 ist [a, b] = ι(a)ι(b)−1 .
Beweis: Offenbar ist f ein Homomorphismus und ker(f ) = I1 ∩ . . ∩ In . Homomorphiesatz: R/I1 ∩ . . ∩ In ∼ = f (R) ⊆ R/I1 × . . × R/In . Sei also Ii + Ij = R f¨ ur i = j. W¨ ahle aij ∈ Ii , bij ∈ Ij mit 1 = aij + bij (i = j). Dann ist sj := k=j akj ∈ Ii f¨ ur i = j und sj + Ij = k=j (1 − bkj ) + Ij = 1 + Ij . Sind r1 , . . , rn ∈ R n und ist r := j=1 rj sj , so ist n rj sj + Ii = ri si + Ii = ri + Ii r + Ii = f¨ ur i = 1, . . h. f (r) = (r1 + I1 , . . , rn + In ). Daher ist f surjektiv, und R/I1 ∩ .
Pn = s1 . . sk q1 . . ql mit irreduziblen Elementen r1 , . . , rm , s1 . . , sk in R und primitiven Polynomen p1 , . . , pn , q1 , . . , ql in R[X], die irreduzibel in K[X] sind. Da p1 . . pn und q1 . . 1 ein u ∈ R× mit r1 . . rm = s1 . . sk u−1 , p1 . . pn = q1 . . ql u. Da R faktoriell ist, muß k = m und ri zu si assoziiert f¨ ur i = 1, . . , k bei geeigneter Numerierung sein. Da K[X] faktoriell ist, muß l = n und pi zu qi f¨ ur i = 1, . . , l bei geeigneter Numerierung in K[X] assoziiert sein.