By J. Rafael Sendra
The vital challenge thought of during this creation for graduate scholars is the selection of rational parametrizability of an algebraic curve and, within the confident case, the computation of an excellent rational parametrization. This quantities to settling on the genus of a curve: its whole singularity constitution, computing common issues of the curve in small coordinate fields, and developing linear structures of curves with prescribed intersection multiplicities. The ebook discusses a variety of optimality standards for rational parametrizations of algebraic curves.
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Extra resources for Rational Algebraic Curves: A Computer Algebra Approach (Algorithms and Computation in Mathematics)
Sample text
Pm . In this situation, C1 , C2 ∈ H(d, P1 + · · · + Pm ), and therefore H(d, P1 + · · · + Pm ) has positive dimension. But, if d ≥ 3, then d(d + 3) − d2 ≤ 0. 2 So the actual dimension and the expected dimension µ do not agree. We illustrate this reasoning by a specific example. 46 2 Plane Algebraic Curves 4 y2 –4 –2 2 x 4 –2 –4 Fig. 6. 61. Let us consider the projective cubics C1 and C2 defined by the polynomials z 2 x − y 3 + 3yz 2, and z 2 y − x3 + 3xz 2 , respectively (see Fig. 6). The intersection points of the two cubics are the real points √ √ 5−1 − 5−1 : :1 , P2 = P1 = (0 : 0 : 1), 2 2 √ √ √ √ 5−1 − 5−1 5+1 − 5+1 P3 = : :1 , P4 = : :1 , 2 2 2 2 √ √ 5+1 − 5+1 P5 = : :1 , P6 = (2 : 2 : 1), 2 2 √ √ P7 = (−2 : −2 : 1), P8 = (− 2 : 2 : 1), √ √ P9 = ( 2 : − 2 : 1).
That is, one might expect the dimension of H(d, r1 P1 + · · · + rm Pm ) to be m µ = max −1, ri (ri + 1) d(d + 3) − 2 2 i=1 . 1) However, this number is only a lower bound, since these constraints may be dependent. 59. Let P1 , . . , Pm ∈ P2 (K), and r1 , . . , rm ∈ N. Then, for every d ∈ N, m ri Pi )) ≥ dim(H(d, i=1 d(d + 3) − 2 m i=1 ri (ri + 1) . 2 44 2 Plane Algebraic Curves m Proof. First, we observe that the theorem holds if d(d + 3) < i=1 ri (ri + 1). So let us assume that d(d+3) ≥ m i=1 ri (ri +1).
64, in order 50 2 Plane Algebraic Curves to prove the result for P1 + · · · + Pi we just have to show that there exists a point Q such that linear equation introduced by Q is linearly independent from the linear equations generated by the divisor P1 + · · · + Pi−1 . 64, we get rank(A(P1 , . . , Pi )) = max{rank(A(Q1 , . . , Qi )) | Q1 × · · · × Qi ∈ (P2 (K))i } = rank(A(P1 , . . , Pi−1 , Q)) = i. Now, take C ∈ H(d, P1 +· · ·+Pi−1 ) and Q ∈ P2 (K)\C; observe that H(d, P1 + · · · + Pi−1 ) = ∅.