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Galois Theory

Date: 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 

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/   
Associated Topics:
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