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Algebre commutative (lecture notes) by Chambert-Loir A.

By Chambert-Loir A.

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Un ide´al I de A est dit maximal s’il est distinct de A et si les seuls ide´aux de A qui contiennent I sont I et A. On remarquera que cela signifie que I est un e´le´ment maximal de l’ensemble des ide´aux de A distincts de A pour la relation d’ordre donne´e par l’inclusion. Comme pour les ide´aux premiers, on peut donner une caracte´risation des ide´aux maximaux en termes de quotients. 7. — Un ide´al I d’un anneau A est maximal si et seulement si l’anneau quotient A=I est un corps. De´monstration.

Alors, pour tout P 2 S, PP0 2 T, d’ou` (PP0 )(a1 ; : : : ; an ) = 0 = P(a1 ; : : : ; an )P0 (a1 ; : : : ; an ) et donc P(a1 ; : : : ; an ) = 0, ainsi qu’il fallait de´montrer. 2. LE THE´ORE`ME DES ZE´ROS DE HILBERT 51 p p e) Comme pI I, on a Z ( I) Z (I). Re´ciproquement, soit (a1 ; : : : ; an ) 2 Z (I). Si P 2 I, soit m 1 tel que pPm 2 I. On a alors Pm (a1 ; : : : ; an ) = 0, d’ou` P(a1 ; : : : ; an ) = 0 et (a1 ; : : : ; an ) 2 Z ( I). 5 (pour les fe´rus de topologie). — La proposition pre´ce´dente peut s’interpre´ter en disant qu’il existe une topologie sur Cn dont les ferme´s sont les ensembles alge´briques.

Soit x un e´le´ment de A=I. On sait qu’il existe a 2 A tel que x = cl(a). Si a0 est un autre repre´sentant de x, donc tel que x = cl(a0 ), on a a0 a 2 I, donc, puisque I Ker f , f (a0 a) = 0 et par conse´quent, f (a) = f (a0 ). On peut ainsi poser f (x) = f (a) — le re´sultat est inde´pendant du repre´sentant a choisi. Il reste a` montrer que f est un homomorphisme d’anneaux. Comme cl(0A ) = 0A=I et cl(1A ) = 1A=I , on a bien f (0A=I ) = 0B et f (1A=I ) = 1B . 1. ANNEAUX QUOTIENTS 27 Il en re´sulte que f est un homomorphisme d’anneaux.

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