Re: Galois resolvent
- To: mathgroup at smc.vnet.net
- Subject: [mg96908] Re: [mg96892] Galois resolvent
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Fri, 27 Feb 2009 06:11:04 -0500 (EST)
- References: <200902261302.IAA26707@smc.vnet.net>
- Reply-to: drmajorbob at bigfoot.com
The link includes illegible items such as the question marks in R(t) = t^3 + 2B t^2 + (B^2 ? 4D)t ? C^2 = 0. Hence, I have no idea what you're asking. Bobby On Thu, 26 Feb 2009 07:02:29 -0600, Kent Holing <KHO at statoil.com> wrote: > Forthe quartic (*) x^4 + Bx^2 + Cx + D = 0 for integers B, C and D, > assume that as for the case C = 0 that all its roots are classically > contructible also for the case C /= 0. > > We can then show that the equation (*) is cyclic (i.e. its Galois group > = Z4) iff the splitting field of its Descartes resolvent is E = > Q[Sqrt[t0] /= Q for t0 the one and only integer roots t0 of the > resolvent. For details, see > http://mathforum.org/kb/thread.jspa?threadID=1903146. > > If the quartic (*) is cyclic, it should be possible using the above to > explicitly construct the so-called Galois resolvents of (*): The roots > x1, x2, x3 and x4 of the quartic (*) can be given by polynomials of r > with degree less or equal to 3 with rational coefficients for r an > arbitrarily root of the quartic. (I.e. the splitting field of the > quartic (*) when cyclic is Q[r] for r a root.) > > Can somebody, using Mathematica, explicitly determine these polynomial > representations of the roots of the quartic (*). The case C = 0 is easy. > But the case C /= 0 is indeed messy. > > Kent Holing, > NORWAY > -- DrMajorBob at bigfoot.com
- References:
- Galois resolvent
- From: Kent Holing <KHO@statoil.com>
- Galois resolvent