Download e-book for kindle: Algèbre 1 [Lecture notes] by Laurent Berger

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Extra resources for Algèbre 1 [Lecture notes]

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En regardant les contenus, on voit que b · cont(P ) = a et donc finalement que P (X) = cont(P ) · P1 (X) · · · Pr (X). Ceci montre d’une part que A[X] est factoriel, et d’autre part qu’il n’y a pas d’autres irr´eductibles que ceux de A et les polynˆomes primitifs irr´eductibles dans K[X]. Enfin, il reste `a v´erifier l’unicit´e de la d´ecomposition. Si P (X) ∈ A[X] s’´ecrit P (X) = a1 · · · ar · P1 (X) · · · Ps (X), alors a1 · · · ar est une d´ecomposition de cont(P ) et est donc ˆ CHAPITRE 4. POLYNOMES ET CORPS FINIS 34 unique aux unit´es pr`es.

En fait, beaucoup des anneaux que l’on rencontre le sont. En voici un qui ne l’est pas : soit K un corps et A l’ensemble des suites (xn )n≥1 d’´el´ements de K (l’addition et la mutliplication ´etant terme a` terme). L’id´eal I des suites nulles a` partir d’un certain rang n’est alors pas de type fini. CHAPITRE 4 ˆ POLYNOMES ET CORPS FINIS Si A est un anneau, alors on note A[X] l’anneau des polynˆomes en X a` coefficients dans A. 1. Polynˆ omes et racines Si a ∈ A, alors on a un morphisme d’´evaluation P (X) → P (a) de A[X] dans A.

Soit {mi } une base de M et Ni = N ∩ (m1 , . . , mi ). Nous allons montrer par r´ecurrence sur i que Ni est libre de rang ≤ i. Comme N1 ⊂ (m1 ) A et que A est principal, N1 est de la forme (a1 m1 ) avec a1 ∈ A et il est donc libre de rang ≤ 1. Soit i ≥ 1 et I l’ensemble des a ∈ A tels qu’il existe x ∈ Ni+1 qui peut s’´ecrire x = b1 m1 + · · · + bi mi + ami+1 . C’est un id´eal de A et il est donc engendr´e par un ´el´ement ai+1 ∈ A. Si ai+1 = 0, alors Ni+1 = Ni et Ni+1 est bien libre de rang ≤ i + 1.

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Algèbre 1 [Lecture notes] by Laurent Berger

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