Commutative group schemes by F. Oort

By F. Oort

We limit ourselves to 2 features of the sphere of staff schemes, during which the consequences are particularly whole: commutative algebraic workforce schemes over an algebraically closed box (of attribute various from zero), and a duality conception challenge­ ing abelian schemes over a in the community noetherian prescheme. The prelim­ inaries for those issues are introduced jointly in bankruptcy I. SERRE defined houses of the class of commutative quasi-algebraic teams through introducing pro-algebraic teams. In char8teristic 0 the placement is obvious. In attribute diversified from 0 details on finite workforce schemee is required as a way to deal with crew schemes; this data are available in paintings of GABRIEL. within the moment bankruptcy those principles of SERRE and GABRIEL are prepare. additionally extension teams of hassle-free crew schemes are made up our minds. a guideline in a paper through MANIN gave crystallization to a fee11ng of symmetry relating subgroups of abelian kinds. within the 3rd bankruptcy we end up that the twin of an abelian scheme and the linear twin of a finite subgroup scheme are comparable in a really average manner. Afterwards we grew to become acutely aware precise case of this theorem used to be already recognized by means of CARTIER and BARSOTTI. purposes of this duality theorem are: the classical duality theorem ("duality hy­ pothesis", proved via CARTIER and by means of NISHI); calculation of Ext(~a,A), the place A is an abelian kind (result conjectured through SERRE); an evidence of the symmetry (due to MANIN) about the isogeny form of a proper staff connected to an abelian kind.

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Multiplying both members of the latter inequality by ca , we get (cy)(J, - ac(J,-1 (cy) ~ (1 - a) cet , y ~ 0. -1 = a, c= ( a )a::T a II we get 1 x=c= ( a )a::T · a here the equality occurs only when Thus, the function x et - ax, a > 1, a > 0, x ~ 0, 1 takes the least value in the point x = ( ~ ) i-a t equal to ex. a )a=-r- . The (1- a) ( a 3-0868 theorem is proved. J3 In particular, the function x 2 - ax (ex = 2) takes the 1 · t Iie point . Ieas t va1ue In x= ( Ta )~-1 a = 2' equal to 2 a ( a) 2-1 = - T.

Thus, the function y == cos x cos 2x takes the greatest value of 1 in the points x == 0, +2n, ±4n, . . The graph of the function y == cos x cos 2x is drawn in Fig. 3. y 1 Fig. ;) Problem 4. Find the least value of the function xa ax, + where a > 0, (X < 0, x ~ 0. Solution. Since (X < 0, then according to the inequality (12) (1 + z)a ~ 1 (Xz, + and the sign of equality holds only when z == O. Assuming 1 -t- z ~ y, z = y - 1, we get ya ~ 1 + (y - (X 1), 1J ~ 0, the sign of equality occurring only when y inequality it follows, that ya - (Xy ~ 1 - (X, (cy)a - Assuming a == -(Xca- l , X = = 1.

22) Proof. Suppose a~ + a~ b1 + b~ + + ... + a~ = A P, ... + b~ = Bq. Then the right member of the inequality (22) will be equal to (AP) 1 1 P (Bq) q = AB. : Since AP=a~+a~+ ... +a~= = A P dl + A P c~ + ... + A Pc~ = A P (c~ + c~ + ... + c~), then Cf +. c~ + ... + c~ = 1. In a similar way, it is checked that d1+~+ ... +~==1. AB (c! + ~t ) , cP a 2b2 -<'. AB d 1 q (-i- +-;) , (*) ·J From these inequalities it follows, that alb! + a2b2 + ... ( cl+c~+ ... (let us reca11 tha t 1 1 p -t- q = 1, c~ dl+~+ + c~ + ...

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