Galois TheoryDate: 06/27/97 at 13:26:36 From: Jack Moore Subject: Something about Galois Does factorization of polynomials have anything in common with Galois Theory? What are Galois groups? Date: 06/27/97 at 15:31:19 From: Doctor Rob Subject: Re: Something about Galois Yes! :-) That is an excellent question. If a polynomial is irreducible, that is, if it cannot be factored, then a root of it adjoined to the rational numbers forms an algebraic number field. By "adjoined," we mean that we are considering the set of all rational functions with denominator not divisible by that polynomial, with the root substituted for the independent variable. Perhaps an example would help. Consider the equation x^3 - 2 = 0. The polynomial on the lefthand side of the equation is irreducible as a polynomial over the rational numbers Q. Let a be the cube root of 2, one of the roots of this cubic equation. Then if we look at the set of all rational functions in indeterminate y over Q, that is, quotients of polynomials with coefficients in Q, and we substitute for y the root a, we will get a large set of real numbers. Of course we have to throw out any which have denominator zero after the substitution, but those are just those such that y^3 - 2 divides the denominator. Some of the numbers have more than one representation in this form, so we pick just one special one. It turns out that every number like this can be represented uniquely in the form c(2)*a^2 + c(1)*a + c(0), where the c(i)'s are rational numbers. This set forms what is called a field: it has addition, subtraction, zero, multiplication, division (except by zero), and one, which satisfy all the usual laws that rational numbers (or real numbers, or complex numbers) do: Commutative, Associative, Distributive, etc. This field is denoted Q(a), and is called an algebraic number field. If the polynomial is not irreducible, the resulting object is not a field, because division is not defined for some nonzero denominators. Another example is the extension of the real numbers R by adjoining a root of the polynomial x^2 + 1. That root is called "i", and the resulting field is the field of complex numbers, called C. Then C = R(i). What is Galois theory? It is a study of field extensions like the ones I just described. What are Galois groups? They are groups of automorphisms of field extensions which leave the ground field fixed. An automorphism is a one-to-one and onto homomorphism from the extension field to itself. A homomorphism is a function which has the property that f(a+b) = f(a)+f(b) and f(a*b) = f(a)*f(b). An example of an automorphism of C which leaves R fixed is complex conjugation: s + i*t |--> s - i*t, where s and t are real. It turns out that the Galois group of C over R is a group of two elements, namely complex conjugation and the identity map. One thing every element of the Galois group does is to permute the roots of the polynomial among themselves. In order to do this properly, the extension field must contain *all* the roots of the polynomial. Note that Q(a) contains only real numbers, so the two complex cube roots of 2 are not in this field. They are a*(-1+i*Sqrt[3])/2 and a*(-1-i*Sqrt[3])/2. We have to make the field larger to have the Galois group properly defined, by adjoining w = (-1+i*Sqrt[3])/2 to Q(a), to get Q(a)(w) = Q(a,w). Now the Galois group of this field over Q is a noncommutative group of order 6, isomorphic to the symmetric group on three letters, S3. Note that finding these roots is a problem in factorization of the polynomial into linear factors, but with coefficients not in Q. Sometimes it happens that a polynomial which is irreducible over R will be factorable over C. For example, this is the case with: x^2 + 2*x + 2 = (x + 1 - i)*(x + 1 + i) Thus factorization of polynomials and field extensions are further related. I hope this helped. If you have more questions, write again. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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