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Galois resolvent

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96892] Galois resolvent
  • From: Kent Holing <KHO at statoil.com>
  • Date: Thu, 26 Feb 2009 08:02:29 -0500 (EST)

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


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