Theorie algebrique des nombres. Deuxieme et troisieme cycles by Pierre Samuel
By Pierre Samuel
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1019906418 ) and the computer took too much time for Γ(6, 6, 1) . 5. The solution for the case p = 2 uses the concept of hypercube representation of a polynomial. in ti11 · · · tinn . ··· f (t1 , . . , tn ) = i 1 +···+i n =d The sum here runs over the lattice points in the hyperplane i1 + · · · + in = d of the n dimensional cube 0 ≤ iν ≤ d , ν = 1, . . , n . Note that the number of lattice points in this cube is (d + 1)n , growing exponentially in n for ﬁxed d . There is another way of writing the same polynomial, namely f (t1 , .
Then h(P ) = − deg(Z) min ordZ (fj ), j Z where Z ranges over all prime divisors and the degree is with respect to a ﬁxed ample class. In particular, the height of a rational function f ∈ K(X)× is h(f ) = h((1 : f )) = − deg(Z) min(0, ordZ (f )). Z Thus h(f ) = 0 if and only if f has no poles. By h(f ) = h(f −1 ) , this is equivalent to div(f ) = 0 . If X is normal, a function without poles is regular (R. 3A), hence constant on the irreducible components of XK . 15). 6. 1. jn tj11 · · · tjnn = f (t1 , .
The constants cp (d, e) and kp (d, e) are related by cp (d, e) = d+e kp (d, e). d Proof: Let f (t1 , . . , tn ) be a homogeneous polynomial of degree d and let F be the symmetrical step function on [0, 1)d given by F (x1 , . . , xd ) = nd/p 1 ∂d f d! ∂ti1 · · · ∂tid for i1n−1 ≤ x1 < in1 , . . , idn−1 ≤ xd < ind . Also, let g(t1 , . . , tn ) be a homogeneous polynomial of degree e and deﬁne G in the same way as F . e! (d + e)! F (xK )G(xL ) (K,L)∈sh(d,e) . p The rest of the proof is an approximation argument.