Subject: Re: [HM] radicals of higher degrees than degree p(x)
From: Abe Shenitzer (shenitze@pascal.math.yorku.ca)
Date: Wed Jan 05 2000 - 20:20:41 EST
I am not sure that I fully understand the question, but I hope that my
answer is of use.
Definition. Let f be a polynomial over a number field F. The equation
f(x)=0 is said to be solvable by radicals if all the roots of f can be
obtained from the elements of F by a finite sequence of rational
operations (addition, subtraction, multiplication, and division) and
extraction of n-th roots.
Example (of a quintic that is not solvable by radicals). The polynomial
f(x) = 2x^5 - 10x + 5
is irreducible over Q (the rationals) by the Eisenstein criterion. Let E
be the splitting field of f over Q. It can be shown that the Galois group
of E over Q is S_5. Hence our equation [all of whose roots are algebraic]
is not solvable by radicals.
Reference: Allan Clark, Elements of Abstract Algebra, Wadsworth, 1971.
See pp.130 and 143, respectively. (Almost any book on abstract algebra
contains this kind of information. I mention Clark because he is, to my
mind, a splendid writer.)
Regards,
Abe Shenitzer
On Wed, 5 Jan 2000, AJ Franco de Oliveira wrote:
>
> A p-g student asked why a root of a complex polynomial equation of
> degree n > 0, when algebraically expressible at all, cannot in an
> essential manner be expressed by an expression involving radicals
> of the form x^(1/m) with m > n. Anyone knows a theorem on this?
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