Industrie des polynômes valeurs entières diviseurs nombres premiers

Description

Industrie des polynômes à valeurs entières : diviseurs, nombres premiers, « petit » théorème de Fermat, records I – L'USINE 2 II – LES DIVISEURS DES POLYNÔMES 3 Une bonne question 3 Quels diviseurs ? 3 On revient à P (x) = x 5 – x 4 Et la démonstration du théorème A ? 5 III – LE « PETIT » THÉORÈME DE FERMAT 6 En préambule : le triangle arithmétique de Pascal 6 Le « petit » théorème de Fermat 6 IV – LES POLYNÔMES QUI « DONNENT » DES NOMBRES PREMIERS 9 1772 : Euler 9 Comment fait-on pour trouver des records ? 9 Les « diviseurs premiers périodiques » 10 Le record de Ruby a-t-il été battu ? 11 1/12

• divisibilité par pa et par qb


• triangle arithmétique de pascal


• temps au temps


• polynôme


• formule


• boutons de réglage


• sacoche pleine de théorèmes

Sujets

• Polynôme
• Formule
• maths

Informations

• Publié par: pefav
• Nombre de lectures: 32
• Langue: English

Extrait

292 C.VOLL (APPENDIX BY A. BEAUVILLE)GAFA Appendix: Lineson Pfaffian Hypersurfaces by A. Beauville The aim of this appendix is to prove that a general pfaffian hypersurface n of degreer >2nŠ3 inPcontains no lines (Proposition 1). By a simple dimension count (see Corollary 4 below), it suffices to show that the variety of lines contained in the universal pfaffian hypersurface (that is, the hyper- surface of degenerate forms in the space of all skew-symmetric forms on a given vector space) has the expected dimension. We will deduce this from an explicit description of the pencils of degenerate skew-symmetric forms, which is the content of the proposition below. We work over an algebraically closed “eldk. We will need an elementary lemma: Lemma 4.Given a pencil of skew-symmetric forms on an-dimensional ! " n+1 vector space, there exists a subspace of dimensionwhich is isotropic 2 for all forms of the pencil. Proof. By induction onn, the casesn= 0 andn= 1 being trivial. Let ϕ+tψbe our pencil; we can assume thatϕis degenerate. LetDbe a line ⊥ contained in the kernel ofϕ, and letDbe its orthogonal with respect ⊥ ¯ toψ. Thenϕandψinduce skew-symmetric forms¯ϕandψonD /D; ! " nŠ1 by the induction hypothesis there exists a subspace of dimensionin 2 ¯ ⊥ ⊥ D /Dwhich is isotropic for¯ϕandψ. The pull-back of this subspace inD ! " n+1 has dimensionand is isotropic forϕandψ.  2 The following result must be well known, but I have not been able to “ nda reference: 1 Proposition 5.LetVbe a vector space of dimension2r, and(ϕ t ) ∈P a t pencil ofdegenerateskew-symmetric forms onV. There exists a subspace 1 L⊂Vof dimensionr+ 1which is isotropic forϕ t for allt∈P. Proof. Again we prove the proposition by induction onr, the caser= 1 be- ∗ ing trivial. The associated maps Φ t :V→Vform a pencil of singular linear maps. By a classical result in linear algebra (see [G, Chap.XII, Thm.4]),  ∗ there exist subspacesK∈VandL∈V, with dimK= dimL+ 1, such  that Φ t (K)⊂Lfor allt; equivalently, there exist subspacesKandLofV, with dimK+ dimL= 2r+ 1, which are orthogonal for eachϕ t . Replacing (K, L) by (K∩L, K+L) we may assumeK⊂L; the pencil (ϕ t ) restricted toLis singular onK, hence induces a pencil (¯ϕ t ) onL/K. Put dimK=p, so that dim(L/K) = 2r+ 1Š2p. By the above lemma there is a subspace