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Additional resources for Algebraic Geometry and Its Applications: Dedicated to Gilles Lachaud on His 60th Birthday (Series on Number Theory and Its Applications)
Suppose that the x1 -coordinate of P is not 0 (otherwise do the same proof with an other coordinate). 1 and because the xi xj (i, j = 1) terms in (x21 ) ◦ g −1 are 0 at g(P ) = (1 : 0 : 0). 2 and the form of f ◦ g −1 (x21 H(f ))(P ) = −(det g)−2 2(n − 1)2 r. e. P is a flex). The proof shows also that this method can fail if p divides 2(n−1). We then suggest the following strategy. Denote K a complete local field of characteristic 0, O its ring of integers, M its maximal ideal such that O/M ≃ k (O may be the ring of Witt vectors of k).
Orzech and M. Orzech. Plane algebraic curves, volume 61. , New-York, 1981. 21. C. Ritzenthaler. Point counting on genus 3 non hyperelliptic curves. In Algorithmic Number Theory Symposium - ANTS-VI, volume 3076 of LNCS, pages 379–394. Springer, 2004. 22. F. Abu Salem and K. Khuri-Makdisi. Fast Jacobian group operations for C3,4 curves over a large finite field. pdf. 23. J. Estrada Sarlabous, E. Reinaldo Barreiro and J. A. Pi˜ neiro Barcel´ o. On the Jacobian varieties of Picard curves: explicit addition law and algebraic structure.
Oyono, C. Ritzenthaler Table 5. ) compute u := u2D1 j1 = inv2 ·res22 , j2 = j13 , j3 = j1 v1 , j4 = j32 , j5 = j1 v0 , j6 = j3 (j4 +6j5 ); j7 = (v2 + v1 + v0 )(h3 + h2 + h1 ), j8 = (v2 − v1 + v0 )(h3 − h2 + h1 ), j9 = v2 h3 ; j10 = v0 h1 , j11 = (j7 + j8 )/2 − (j10 + j9 ), j12 = 3j3 + j2 j9 , j14 = j6 + j2 j11 ; j13 = 3(j5 + j4 ) − j2 + j2 ((((4v2 + 2v1 + v0 )(4h3 + 2h2 + h1 ) − j7 + j8 − j10 )/2 − 2(4j9 + j11 ))/3); u2 = j12 −u′2 , u1 = j13 −u′1 −u′2 u2 , u0 = −u′2 u1 +j14 −u′0 −u′1 (j12 −u′2 ); u = x3 + u2 x2 + u1 x + u0 total ∗3 ∗4 ∗3 ∗4 (M+I) (9M+SQ) j11 = v2 h2 , j12 = 3j3 , j13 = 3(j5 + j4 ) − j2 + j2 j11 ; j14 = j6 + j2 ((v2 + v1 )(h2 + h1 ) − (v1 h1 + j11 )); (5M) inv1 = c−1 10 , c14 d37 = −2t23 , d35 (M+I) (5M+SQ) j12 = 3j3 , j13 = 3(j5 + j4 ) − j2 , j14 = j6 + j2 (h1 v2 ); (2M) If h3 , h2 , h1 = 0 then replace ∗1 , ∗2 , ∗3 and ∗4 by inv1 = c−1 10 , c14 = inv1 , c15 = inv1 · c9 , c16 = 1; d41 = −2h0 t23 , d38 = 0, d39 = 0, d40 = 0, d42 = 0, d43 = 0, d44 = 0; (M+I) (M+SQ) j12 = 3j3 , j13 = 3(j5 + j4 ) − j2 , j14 = j6 ; Table 9.