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### Galois Theory

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Date: 06/27/97 at 13:26:36
From: Jack Moore

Does factorization of polynomials have anything in common with Galois
Theory? What are Galois groups?
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Date: 06/27/97 at 15:31:19
From: Doctor Rob

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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Associated Topics:
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