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Comme l’anneau nul n’est pas un corps, A=I 6= 0 et I 6= A. Soit d’autre part un ide´al J de A contenant I. Si J 6= I, il existe ainsi a 2 J n I. Sa classe cl(a) 2 A=I est donc non nulle, donc inversible puisque A=I est un corps. Soit b 2 A tel que cl(a) cl(b) = 1. On a donc ab 1 2 I et comme a 2 J, 1 = ab (ab 1) 2 J. Par suite, J = A. Montrons re´ciproquement que si I 6= A et si tout ide´al de A contenant I est e´gal a` I ou A, alors A=I est un corps. De´ja`, A=I est non nul. 1. IDE´AUX PREMIERS, IDE´AUX MAXIMAUX 45 x 2 A=I est non nul, il existe a 2 A tel que x = cl(a), et l’on a a 62 I.

De´monstration. — a) Comme I 6= A, l’anneau A=I est non nul et posse`de donc un ide´al maximal. D’apre`s la proposition pre´ce´dente, celui-ci est de la forme m=I ou` m est un ide´al maximal de A contenant I. 1. IDE´AUX PREMIERS, IDE´AUX MAXIMAUX 47 b) Comme 0 62 S, l’anneau localise´ S 1 A n’est pas l’anneau nul. Soit m un ide´al maximal de S 1 A. Son image re´ciproque i 1 (m) dans A par l’homomorphisme canonique i : A ! S 1 A est alors un ide´al premier I de A disjoint de S. Un autre corollaire est la caracte´risation suivante des e´le´ments inversible d’un anneau.

On de´finit ensuite (a=s) + (b=t) = (at + bs)=st; (a=s) (b=t) = (ab=st): Ve´rifions d’abord que cette de´finition a un sens : si (a; s) que (at + bs; st) (a0 t + bs0 ; s0 t) (a0 ; s0 ), il faut montrer (a0 b; s0 t): et (ab; st) On a alors (at + bs)s0 t (a0 t + bs0 )st = t2 (as0 a0 s): Choisissons u 2 S tel que u(as0 a0 s) = 0 ; il en re´sulte que Ð u (at + bs)s0 t (a0 t + bs0 )st = 0 et donc (at + bs; st) (a0 t + bs0 ; s0 t). De meˆme, u(abs0 t a0 bst) = ubt(as0 a0 s) = 0 et donc (ab; st) (a0 b; st).

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