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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


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