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]
```