Algebra I by Walter Gubler

By Walter Gubler

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Sei L/K eine K¨ orpererweiterung und sei S ⊂ L. Dann gibt es einen kleinsten Teilring von L, der K und S umfasst. Die Existenz folgt daraus, dass es sicher einen Teilring gibt, der K und S umfasst, und zwar L. Weiter ist der Durchschnitt einer beliebigen Familie von Teilringen wieder ein Teilring und somit ist K[S] := R R⊇S∪K der kleinste Teilring, der S und K enth¨alt. Dabei l¨auft R im Index u ¨ber alle Teilringe von L, die K und S enthalten. 10. Sei K[(xs )s∈S ] der Polynomring in den Variablen (xs )s∈S (a) Dann gibt es genau einen Ringhomomorphismus ϕ : K[(xs )s∈S ] −→ L mit ϕ(α) = α∀α ∈ K und mit ϕ(xs ) = s∀s ∈ S (b) K[S] = bild(ϕ) = {p(s1 , .

Folgt sofort. 9. Jeder Teilk¨ orper von K hat dieselbe Charakteristik wie K. 5. 2 K¨ orpererweiterungen Bis zum Schluss der Vorlesung werden wir K¨orpererweiterungen untersuchen. Die zentralen Resultate werden in der Galoistheorie gemacht. In diesem Abschnitt werden wir die Grundlagen bereitstellen. Wie immer sei K ein K¨ orper. ¨ KAPITEL 3. 1. Eine K¨ orpererweiterung L von K ist ein K¨orper L ⊇ K so, dass Addition und Multiplikation von L auf K mit derjenigen von K u ¨bereinstimmt. Dann ist K ein Teilk¨orper (oder Unterk¨orper) von L und L heißt Oberk¨ orper.

L K-linear unabh¨angige Elemente aus L und seien γ1 , . . , γm L-linear unabh¨angige Elemente aus M . In einem ersten Schritt zeigen wir, dass (βi γj )1≤i≤l,1≤j≤m K-linear unabh¨ angige Elemente aus M sind. Seien λij ∈ K mit λij βi γj = 0. Zu zeigen ist λij = 0∀i, j. 1≤i≤l,1≤j≤m m l 0= m λij βi γj = j=1 i=1 l γj j=1 m λij βi = i=1 m ( j=1 j=1 λij βi )γj ∈K ∈L L l Weil γ1 , . . , γm L-linear unabh¨angig ist, folgt λij βi = 0∀j = 1, . . , m. i=1 Weil die λij ∈ K und weil β1 , . . , βl K-linear unabh¨angig ist, folgt damit λij = 0∀i = 1, .

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