INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

22 MezzettiClaim. ψ d is an isomorphism.Assuming the claim, we are able to compute dim ker φ d = 2 ( d3), therefore( ) d + 1dim Im φ d = 33( d− 23)which coincides with the dimension of I h (X) d previously computed. This provesthat φ d is surjective for all d and concludes the proof of the Proposition.Proof of the Claim. Let (G 0 , G 1 , G 2 ) belong to kerφ d . This means that thefollowing matrix N with entries in K[x 0 , x 1 , . . ., x 3 ] is degenerate:N :=⎛⎝ G ⎞0 G 1 G 2x 0 x 1 x 2⎠x 1 x 2 x 3Therefore, the rows of N are linearly dependent over the quotient field of thepolynomial ring K(x 0 , . . ., x 3 ). Since the last two rows are independent, thereexist reduced rational functions a 1a 0, b 1b 0∈ K(x 0 , x 1 , x 2 , x 3 ), such thatand similarlyG 0 = a 1a 0x 0 + b 1b 0x 1 = a 1b 0 x 0 + a 0 b 1 x 1a 0 b 0G 1 = a 1b 0 x 1 + a 0 b 1 x 2a 0 b 0, G 2 = a 1b 0 x 2 + a 0 b 1 x 3a 0 b 0The G i ’s are polynomials, therefore the denominator a 0 b 0 divides the numeratorin each of the three expressions on the right hand side. Moreover, if p is a primefactor of a 0 , then p divides the three products b 0 x 0 , b 0 x 1 , b 0 x 2 , hence p divides b 0 .We can repeat the reasoning for a prime divisor of b 0 , so obtaining that a 0 = b 0(up to invertible constants). We get:G 0 = a 1x 0 + b 1 x 1b 0, G 1 = a 1x 1 + b 1 x 2b 0, G 2 = a 1x 2 + b 1 x 3b 0,therefore b 0 divides the numeratorsc 0 := a 1 x 0 + b 1 x 1 , c 1 := a 1 x 1 + b 1 x 2 , c 2 := a 1 x 2 + b 1 x 3 .Hence b 0 divides also x 1 c 0 − x 0 c 1 = b 1 (x 2 1 − x 0x 1 ) = −b 1 F 2 , and similarly x 2 c 0 −x 0 c 2 = b 1 F 1 , x 2 c 1 − x 1 c 2 = −b 1 F 0 . But F 0 , F 1 , F 2 are irreducible and coprime, sowe conclude that b 0 | b 1 . But b 0 and b 1 are coprime, so finally we get b 0 = a 0 = 1.□