a galois related problem
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From: Kent Holing on 11 Feb 2010 20:11 Can something general be said about the following: For an irreducible equation with rational coefficents of degree n and both real and complex roots, what is the maximum number of real roots if the Galois group is Dn (2n elements)? Note that if r is a real root of such an equation then all other real roots of the equation are in Q[r]. Example for n = 6: Here we can show that the equation cannot have 4 real roots if G = D6, so the maximum number of real roots is 2. Example for n = 5: Here the equation cannot have 3 real roots if G = D6 (if so, G = S6). (It is wellknown that if n = p a prime, then an irreducible equation with rational coefficients having exactly p-2 real roots has the Galois group Sp.) So, the maximum number of real roots is 1. Kent Holing
From: victor_meldrew_666 on 12 Feb 2010 09:48 On 12 Feb, 11:11, Kent Holing <K...(a)statoil.com> wrote: > Can something general be said about the following: > For an irreducible equation with rational coefficents of degree n and both real and complex roots, > what is the maximum number of real roots if the Galois group is Dn (2n elements)? I presume when you say "complex roots" you actually mean "non-real roots". Let's think about the action of complex conjugation. This induces a permutation of the roots which has the cycle structure 1^r 2^s where r > 0 and s > 0 by your hypothesis. This is an element of D_n lying in the point stabilizer. Thus it must have cycle structure 1^1 2^(n-1)/2 if n is odd and 1^2 2^(n-2)/2 if n is even. So the answer is 1 real root if n is odd 2 real roots if n is even.
From: Kent Holing on 12 Feb 2010 04:58 Can you (anybody) provide a septic with G = D7 and 3 real roots?
From: Chip Eastham on 12 Feb 2010 15:32 On Feb 12, 2:58 pm, Kent Holing <K...(a)statoil.com> wrote: > Can you (anybody) provide a septic with G = D7 and 3 real roots? Let's review the argument already advance. The Galois group is transitive on the roots of an irreducible polynomial (you asked for coefficients to be rational, and for reasons we need not go into here, they irreducible polynomial may without loss of generality be taken to be monic with integer coefficients). Now complex automorphism will be an element of this Galois group, and if there are 3 real roots, complex automorphism "fixes" just those three roots (and transposes the other two pairs). Now consider the symmetries D7 of the regular septagon. If any three distinct vertices are fixed, so are the rest of them (e.g. there can be no rotation or reflection). So the case of exactly three real roots cannot arise if the Galois group is D7. regards, chip
From: Timothy Murphy on 12 Feb 2010 19:25
Chip Eastham wrote: > Let's review the argument already advance. > > The Galois group is transitive on the roots of an irreducible > polynomial (you asked for coefficients to be rational, and for > reasons we need not go into here, they irreducible polynomial > may without loss of generality be taken to be monic with integer > coefficients). > > Now complex automorphism will be an element of this Galois group, > and if there are 3 real roots, complex automorphism "fixes" just > those three roots (and transposes the other two pairs). > > Now consider the symmetries D7 of the regular septagon. If any > three distinct vertices are fixed, so are the rest of them > (e.g. there can be no rotation or reflection). So the case of > exactly three real roots cannot arise if the Galois group is D7. The action of D_n on a regular n-gon gives an embedding of D_n in S_n. But does every embedding of D_n in S_n arise in this way? -- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |