GAP command
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From: Chip Eastham on 10 May 2010 13:13 On May 10, 10:04 am, Gerry <gerry...(a)gmail.com> wrote: > On May 8, 12:36 am, Chip Eastham <hardm...(a)gmail.com> wrote: > > > > > On May 7, 5:43 pm, galoisgroupquestion > > > <galoisgroupquest...(a)hotmail.com> wrote: > > > > 07.05.2010 10:35, galoisgroupquestion kirjoitti: > > > > > Hi, I need to verify some of my Galois group > > > > exercises using GAP. > > > > > > How do I find the Galois group of the polynomial > > > > > x^4 + 2x^2 + x +3 over Q using GAP? > > > > > > Can I also find the splitting field of it using > > > > GAP? > > > > > > How do I also find the Galois group of the same > > > > polynomial over GF(7) rather than Q? > > > > > > I googled and tried it, but it was not successful > > > > so far. > > > > > > Any help will be appreciated. > > > > > > Thanks. > > > > > Listen some good reggae (for example Dr Mooch: > > > > Virginia) and > > > > think the problem anew. > > > > > Rike > > > > Don't reply if you have no intention to help. > > > It is a waste of time to read your stupid reply indeed. > > > I like Rike's suggestion very much and appreciate > > the time taken to reply. > > > However I will suggest that while the polynomial > > x^4 + 2x^2 + x + 3 is irreducible over Q, it is > > not irreducible over Z/7Z. I believe the GAP > > command you want requires an irreducible polynomial > > as input. See: > > >http://www.gap-system.org/Manuals/doc/htm/ref/CHAP056.htm > > > regards, chip- Hide quoted text - > > > - Show quoted text - > > Use > > p1:=X^4+2*X^2+X+3; > g1:=Galois(p1); > > gives "S4" > > The roots are : > > x1= 1/2*sqrt(r)-1/2*sqrt(-4-r-2/sqrt(r)) > x2= 1/2*sqrt(r)+1/2*sqrt(-4-r-2/sqrt(r)) > x3=-1/2*sqrt(r)-1/2*sqrt(-4-r+2/sqrt(r)) > x4=-1/2*sqrt(r)+1/2*sqrt(-4-r+2/sqrt(r)) > > and > > s = 1/2(-389+3*I*sqrt(11631)) > r = -4/3 + 40/(3*s^(1/3))+1/3*s^(1/3) > > So s is a root of the quadratic : > > x^2+389*x+64000 > > and r is a root of the cubic : > > x^3+4*x^2-8*x-1 > > i don't know the answer to yout other questions. Well, it's not really my question, but notice that with your notation: p1:=X^4+2*X^2+X+3; if we were to work over Z/7Z, then at X = 1 we'd have p1 = 0. Using synthetic division over Z/7Z: p1 = (X-1)(X^3 + X^2 + 3*X + 4) If the cubic factor were reducible over Z/7Z, it would have a linear factor, i.e. a root in Z/7Z. But these are quickly ruled out, esp. if we check +/- pairs together by computing the odd and even terms of the cubic. Now the splitting field of p1 over Z/7Z must be the splitting field of its cubic factor, and the only possibilities are that the splitting field is isomorphic to GF(7^3) or to GF(7^6), depending on whether adjoining one root is sufficient to split the cubic. Syntax for constructing an algebraic extension of a field is described here: http://www.gap-system.org/Manuals//doc/htm/ref/CHAP065.htm and moreover once the field extension is in hand the GaloisGroup command may be applied. regards, chip |