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Extra resources for An easier solution of a Diophantine problem about triangles, in which those lines from the vertices which bisect the opposite sides may be expressed rationally

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39 § 5. Projektive Varietiiten und homogenes Spektrum Aufgaben: 1. G sei ein positiv graduierter Ring, wobei Go = K ein Karper mit unendlich vie len Elementen ist. Je n + 1 Elemente von G seien algebraisch abhiingig tiber K. a) Sind F" ... , F n + 1 E G homogene Elemente gleichen Grades und setzt man Fi := Fi - Ai Fn+ 1 (i = 1. • N - 1). b) Zu jedem endlich erzeugten homogenen Ideal leG gibt es homogene Elemente F" 00. , Fn). 2. c) 1st L ein beliebiger Erweiterungskarper von K, so ist jede projektive K-Varietiit V C IPn(L) Durchschnitt von n + 1 K-Hyperfliichen.

Das Spektrum eines Rings 6. 7. L/K sei eine K6rpererweiterung, wobei L algebraisch abgeschlossen ist. Flir einen Punkt x = (~I' ... DIn (L) sei I{)x : K [V] -+ L die Abbildung, die jeder Funktion fE K [V] den Funktionswert f(x) E L zuordnet. Die Abbildung x 1-+ I{)x ist eine Bijektion von V auf die Menge HomK (K [V], L) der K-Algebrahomomorphismen von K [V] in L. K sei ein beJiebiger K6rper, S die Menge aller Polynome aus K [XI. n(K) besitzen und I ein Ideal von K[XI' ... , Xn] mit Ins = 0. Dann besitzt I eine Nullstelle in ffi.

R2 a2 + P2) = r I r2 al a2 + rial P2 + r2 a2 PI + PI P2 E P () S im Widerspruch zu p () S = 0. p ist somit Primideal. Wendet man das Lemma auf S = {I} an, so ergibt sich die bekannte Tatsache, daB jedes Ideal I eines Rings R mit I :j: R in einem maximalen Ideal von R enthalten ist. Ferner hat man § 25 4. 5: Flir jedes Ideal I eines Rings R mit I Rad(I) = n '* R ist p. p:JI pE Spec (R) n Speziell ist p = Rad (0) die Menge aller nilpotenten Elemente von R. pE Spec(R) n Beweis: Geht man zu R/I liber, so erkennt man, da~ es gentigt, die zweite Aussage zu beweisen.

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An easier solution of a Diophantine problem about triangles, in which those lines from the vertices which bisect the opposite sides may be expressed rationally by Euler L.


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