|Peter-Lawrence . Montgomery on Sat, 5 Dec 1998 04:31:51 +0100 (MET)|
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|Re: bug in polredabs() (fwd)|
Gerhard Niklasch <email@example.com> writes > In response to > > Message-Id: <19981204193944.K14565@io.txc.com> > > Date: Fri, 4 Dec 1998 19:39:44 -0500 > > From: Igor Schein <firstname.lastname@example.org> > [...] > > So x^16+48 and x^16+3 generate the same number field, > > No, they don't define the same field. They define two distinct > fields which happen to have the same discriminant 2^48*3^15. > > (If they did, then the field would also contain a 16th root of > (-48)/(-3) = 16, or in other words, a 4th root of 2. But it doesn't > even contain a square root of 2 -- try to nffactor x^2-2 over the > result of nfinit(polredabs(y^16+3)). A field can contain a 16th root of 16 without a 4th root of 2. It might instead have a fourth root of -2 or a square root of +-1 +- sqrt(-1), I agree that the fields are not isomorphic. If we try to factor x^16 + 48 over Q[y] where y^16 + 3 = 0, we get only the four degree-4 factors of x^16 - 16 y^16. Peter Montgomery GP/PARI CALCULATOR Version 2.0.11 (beta) unknown 32-bit version (readline disabled, extended help not available) Copyright (C) 1989-1998 by C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier. Send bug reports, suggestions and patches to email@example.com Type ? for help. realprecision = 38 significant digits seriesprecision = 16 significant terms format = g0.38 parisize = 20000000, primelimit = 500000, buffersize = 30000 echo = 1 (on) ? factornf(x^16+48,y^16+3) %1 = [Mod(1, y^16 + 3)*x^4 + Mod(2*y^2, y^16 + 3)*x^2 + Mod(2*y^4, y^16 + 3) 1] [Mod(1, y^16 + 3)*x^4 + Mod(-2*y^2, y^16 + 3)*x^2 + Mod(2*y^4, y^16 + 3) 1] [Mod(1, y^16 + 3)*x^4 + Mod(-2*y^4, y^16 + 3) 1] [Mod(1, y^16 + 3)*x^4 + Mod(2*y^4, y^16 + 3) 1] ? quit Good bye!