ensemble des résultats de complexité algébriqué enoncés dans les sections précédentes Quand le résultat théorique n'est pas optimal, nous indiquons la nature du test qu'il faudrait effectuer pour vérifier, en pratique, si le résultat obtenu est optimal ou s'il contient une racine carrée non nécessaire. Notons qu'hormis le cas o` u la caractéristique de Segre est, et dans ce cas nous obtenons toujours le résultat optimal) ,
il existe des faisceaux ne contenant pas de quadriquessingulì eres De plus, pour les quadriques de rang maximal, la situation se complique nécessairement en augmentant la dimension de l'espace puisque dans P 3 (R), le choix des inerties possibles se limitè a, alors que dans les cas général, le nombre de cas possibles augmente. Nous devons donc nous poser deux questions, pour chaque espace P n (R), existe-t-il des quadriques de rang n + 1, pp.2-4 ,
il existe (u, v) et (s, t) dans P 1 (R) tels que ,
Posons enfin v = W , et nous avons bien trouvé, 1) deux points de P 1 (R) tels que (X, Y, Z, W ) = (ut, vs, us, vt) et XY = ZW . Supposons maintenant que X = 0. L'´ equation XY = ZW implique que Z = 0 ou W = 0. Supposons que Z = 0 (le cas W = 0 est identique par la symétrie de l'expression) Nous avons alors (ut, vs, us, vt) = (0, Y, 0, W ). v = 0 car sinon (ut, vs, us, vt) seraitégaìseraitégaì a (0, 0, 0, 0) ? P 3 (R) Nous pouvons donc poser v = 1, Du coup, issue.0 ,
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