Roland Dreier on Sat, 3 Oct 1998 13:11:11 -0500 (CDT) |
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sign error in factor over finite fields |
I have found what seems to be a bug in gp's factorization over extensions of prime fields. In the example that follows, the sign of the constant term of the linear term is incorrect. Roland GP/PARI CALCULATOR Version 2.0.11 (beta) i586 running linux (ix86 kernel) 32-bit version (readline enabled, extended help available) Copyright (C) 1989-1998 by C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier. Send bug reports, suggestions and patches to pari@math.u-bordeaux.fr Type ? for help. realprecision = 28 significant digits seriesprecision = 16 significant terms format = g0.28 parisize = 40000000, primelimit = 500000, buffersize = 30000 ? ff=ffinit(3,2,y) %1 = Mod(1, 3)*y^2 + Mod(1, 3) ? pol=Mod(Mod(1,3),ff)*x^3 - Mod(Mod(1,3)*y,ff) %2 = Mod(Mod(1, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x^3 + Mod(Mod(0, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x^2 + Mod(Mod(0, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x + Mod(Mod(2, 3)*y, Mod(1, 3)*y^2 + Mod(1, 3)) ? factor(pol) %3 = [Mod(Mod(1, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x + Mod(Mod(2, 3)*y, Mod(1, 3)*y^2 + Mod(1, 3)) 3] ? %[1,1]^3 %4 = Mod(Mod(1, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x^3 + Mod(Mod(1, 3)*y, Mod(1, 3)*y^2 + Mod(1, 3)) ? %-pol %5 = Mod(Mod(2, 3)*y, Mod(1, 3)*y^2 + Mod(1, 3)) ? factor(%4) %6 = [Mod(Mod(1, 3), Mod(1, 3)*y^2 + Mod(1, 3))*x + Mod(Mod(1, 3)*y, Mod(1, 3)*y^2 + Mod(1, 3)) 3]