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### Algebraic Extensions

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Date: 06/28/97 at 07:18:19
From: BILL PHILLIPS
Subject: Algebraic extensions

What are algebraic extensions? Could you please give me some examples?

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Date: 06/30/97 at 13:12:53
From: Doctor Rob
Subject: Re: Algebraic extensions

Good question!

First you must start with a field, F. This is a set equipped with two
operations, addition and multiplication, which satisfy certain laws.
Among them are the commutative, associative, and distributive laws,
the existence of zero and one, and the existence of negatives and
inverses (except the inverse of zero).

Next you consider an algebraic equation in one indeterminate (or
unknown) whose coefficients lie in the field F. Another way to
describe this equation is a polynomial with coefficients in F set
equal to zero. Say it is called f(x) = 0. Let the degree of the
polynomial be n. For technical reasons, we want f(x) to be irreducible
over F, that is, to not be factorable into two polynomials of smaller
degree whose coefficients also lie in F.

Next take a to be a root of this equation, in a formal sense, that is,
f(a) = 0, but we don't have to worry about finding an expression for
a, just use the equation.

Next we consider a new set K, which is the set of all polynomials of
degree less than n in the variable a with coefficients in F. It turns
out that K is also a field, using the usual definition of polynomial
addition, and defining multiplication by multiplying polynomials
together in the usual way and then removing powers of a with exponents
at least n by using the relationship f(a) = 0.

This field K is called an algebraic extension of the field F, and is
often denoted K = F(a), which is read, "F with a adjoined." It is an
algebraic extension because a satisfies and algebraic equation,
f(a) = 0.  The opposite of an algebraic extension is a transcendental
extension, which is gotten by adjoining an element b which satisfies
no such equation. F(b) will then be the set of all quotients of
polynomials in the variable b with coefficients in F.

Examples:

1. Let the field F = R, the field of real numbers. Consider the
equation x^2 + 1 = 0. This is irreducible over R. Its root i
satisfies i^2 + 1 = 0, or i^2 = -1.  R(i) is the set of all
expressions of the form s + t*i.  Addition is defined by:

(s + t*i) + (u + v*i) = (s + u) + (t + v)*i

Multiplication is defined by:

(s + t*i)*(u + v*i) = s*u + (s*v + t*u)*i + (t*v)*i^2
= s*u + (s*v + t*u)*i + (t*v)*(-1)
= (s*u - t*v) + (s*v + t*u)*i

We are using the equation i^2 = -1. You can check that this set
forms a field by telling what zero and one are, and verifying the
existence of negatives and inverses (except for the inverse of
zero). This extension field is called the field of complex numbers,
and is denoted by C.

2. Let the field F = Q, the field of rational numbers. Consider the
equation x^3 - x - 1 = 0. This is irreducible over Q. Its root a
satisfies a^3 - a - 1 = 0, so a^3 = a + 1. F(a) is the set of all
expressions of the form r + s*a + t*a^2. Addition is defined by:

(r+s*a+t*a^2) + (u+v*a+w*a^2) = (r+u) + (s+v)*a + (t+w)*a^2

Multiplication is defined by:

(r + s*a + t*a^2)*(u + v*a + w*a^2)
= r*u + (r*v+s*u)*a + (r*w+s*v+t*u)*a^2 + (s*w+t*v)*a^3 + (t*w)*a^4
= r*u + (r*v+s*u)*a + (r*w+s*v+t*u)*a^2 + (s*w+t*v)*(a + 1) + (t*w)*
(a^2+a)
= (r*u+s*w+t*v) + (r*v+s*u+s*w+t*v+t*w)*a + (r*w+s*v+t*u+t*w)*a^2

Because we are using the equation a^3 = a + 1, and so
a^4 = a^2 + a. You can check that this set forms a field by telling
what zero and one are, and verifying the existence of negatives and
inverses (except for the inverse of zero).

Notice that we did not care exactly which root of f(x) = 0 the a we
chose happened to be.  It didn't really matter.

I hope this answers your question.  If not, write again, and we'll try
again.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
College Modern Algebra

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