Showing papers on "Generic polynomial published in 1991"
TL;DR: In this article, the regular version of the Inverse Galois Problem was reduced to finding one rational point on an infinite sequence of algebraic varieties, and it was shown that any finite group G is the Galois group of an extension L/P (x) with L regular over any PAC field P of characteristic zero.
Abstract: We reduce the regular version of the Inverse Galois Problem for any finite group G to finding one rational point on an infinite sequence of algebraic varieties. As a consequence, any finite group G is the Galois group of an extension L/P (x) with L regular over any PAC field P of characteristic zero. A special case of this implies that G is a Galois group over Fp(x) for almost all primes p.
182 citations
TL;DR: In this paper, the authors considered Abelian integrals associated with generic polynomials of a given degree n+1 and with polynomial 1-forms of degree
Abstract: The author considers Abelian integrals associated with generic polynomials of a given degree n+1 and with polynomial 1-forms of degree
43 citations
TL;DR: For any cyclic group G of odd order and any field F whose characteristic does not divide the order of G, it is possible to find a polynomial which parametrizes all extensions of F with group G as discussed by the authors.
Abstract: For any cyclic group G of odd order and any field F whose characteristic does not divide the order of G it is possible to find a polynomial which parametrizes all extensions of F with group G. In this paper such a polynomial is explicitly constructed for each such group G which is valid for all such fields F.
22 citations
TL;DR: In this article, the authors explicitly construct all of the quadratic extensions L of K having Galois group D 4, the Sylow subgroup of GL 2 (F 3 ) (resp. SL 2 ( F 3 ) or GL 2(F 3 )) over F, whenever such extensions exist.
6 citations
TL;DR: If there is no logarithmic derivative of a solution of small algebraic degree, then the solution z itself must be algebraic and the algebraicdegree of z can be bounded and allows a direct computation of the minimal polynomial Q(ϑ) of z.
Abstract: The known algorithms for computing a liouvillian solution of an ordinary homogeneous linear differential equation L(y) = 0 use the fact that, if there is a liouvillian solution, then there is a solution z whose logarithmic derivative z"/z is algebraic over the field of coefficients. Their result is a minimal polynomial for z"/z. In this paper we show that, if there is no logarithmic derivative of a solution of small algebraic degree, then the solution z itself must be algebraic and the algebraic degree of z can be bounded. This can be used to improve algorithms computing liouvillian solutions and allows a direct computation of the minimal polynomial Q(ϑ) of z. In order to improve the computation of the minimal polynomial Q(ϑ), we get a criterion, in terms of the differential Galois group, from which the sparsity of Q(ϑ) can be derived.
5 citations
TL;DR: The main result of as discussed by the authors is that if K is a global field and K solv is the maximal prosolvable extension of K, then the Frattini group of G(K) is trivial.
Abstract: LetK be a local field,T the maximal tamely ramified extension ofK, F the fixed field inK sof the Frattini subgroup ofG(K), andJ the compositum of all minimal Galois extensions ofK containingT. The main result of the paper is thatF=J. IfK is a global field andK solv is the maximal prosolvable extension ofK, then the Frattini group of\(\mathcal{G}\)% MathType!End!2!1!(K solv/K) is trivial.
4 citations
01 Jan 1991
TL;DR: In this article, it was shown that the polynomial X does not have controlled singularities so this case can indeed be discarded in the proof of Proposition 5, and a proof for the existence of identically zero polynomials is given.
Abstract: The proof of Proposition 5 of [3] is incomplete. With notation as in the paper, the possibility that the polynomial X~/q'P o + X~/~' P1 + X~/q" Pz in (11) could be identically zero was overlooked. We will sketch here a proof that in this case X does not have controlled singularities so this case can indeed be discarded in the proof of Proposition 5. 2 Let F = Z SIPS" be a generic polynomial of this form with degPi = )~, (so i=0 2 d = degF = 2q' + 1) with ~2 X~/q'Pi identically zero. Thus every common zero of Po /=0 and P1 is a zero of X~/q' P2 and, since we are in the generic case, is a zero of PE and gives a singular point of F = 0 with Jacobian ideal of multiplicity at least q' since c~F/~X i = P~'.