GALOIS THEORY 89

EXAMPLE: Let f (x) = x

2

− x − 1. Then f(x) factors as

f (x) = (x − φ)(x − φ

) where φ = (1 +

√

5)/2 = 1.6180340 ...

and φ

= (1 −

√

5)/2 =−0.6180340 ... . So the roots of f (x) are

φ and φ

. By the way, φ is an interesting number called the

“golden ratio.” It is supposed to describe a particularly

pleasing proportion that occurs in nature. A recent book on

the subject is (Livio, 2002).

EXAMPLE: Let f (x) = 5(x − 1)(x + 2)(x − 3). Here we began

with f (x) already factored, so we can see that its roots are 1,

−2, and 3. We can multiply out f and see it explicitly as a

third-degree polynomial: f(x) = 5x

3

− 10x

2

− 25x + 30.

EXAMPLE: Consider f (x) = x

4

− 1. We expect four roots, but

to see them all we have to use the complex numbers. The

roots are 1, −1, i, and −i, and this corresponds to the

factorization f (x) = (x − 1)(x + 1)(x − i)(x + i).

The Field of Algebraic Numbers Q

alg

DEFINITION: A complex number is said to be algebraic if it

is the root of some Z-polynomial. The set of all algebraic

numbers is denoted by Q

alg

.

The set Q

alg

contains every integer n, because n is a root of

the equation x − n = 0. It also contains every fraction a/b, because

a/b is the solution of bx − a = 0. The set contains square roots of

every fraction a/b, because

a/b solves the equation bx

2

− a = 0.

(Notice in particular that this means Q

alg

contains i, the square

root of −1, because it is a solution of x

2

+ 1 = 0.) In fact, the set

Q

alg

contains nth roots of every fraction, because

n

a/b solves the

equation bx

n

− a = 0.

But there are many more complicated types of algebraic

numbers. This is partly because the sum, difference, product, and

quotient of any two algebraic numbers will also be algebraic.

Therefore, Q

alg

is a ﬁeld. This assertion is a difﬁcult theorem to

prove, which we will not do in this book. As an example, you can try

90 CHAPTER 8

the following exercise:

EXERCISE: Find a Z-polynomial that has a root

√

2 +

3

√

5.

SOLUTION: x

6

− 6x

4

− 10x

3

+ 12x

2

− 60x + 17. To ﬁnd this

solution, let x =

√

2 +

3

√

5. Then (x −

√

2)

3

= 5. Multiply out

the left-hand side, and you get x

3

− 3

√

2 x

2

+ 6x − 2

√

2 = 5.

Now rearrange this to be x

3

+ 6x − 5 = 3

√

2x

2

+ 2

√

2, square

both sides, and collect terms.

The exercise illustrates the following fact, which we will not

prove:

THEOREM 8.1: If α is an algebraic number that is a root of a

Z-polynomial of degree m, and β is an algebraic number that

is a root of a Z-polynomial of degree n, then α + β, α −β, αβ

and α/β (if β = 0) are algebraic numbers, each of which is a

root of some Z-polynomial of degree no more than mn.

More examples: The number π is not in Q

alg

, and this fact is very

difﬁcult to prove. The nineteenth-century mathematician Georg

Cantor proved that “almost all” real numbers are not in Q

alg

, but

it can be difﬁcult to decide whether a particular number is or is not

in Q

alg

.

One interesting question: Can every algebraic number be ob-

tained, starting from the integers, by the repeated operations of

addition, subtraction, multiplication, division, and

n

√

for various

n? The well-known answer is: No. For example, let w be a root of the

following Z-polynomial of degree 5: 2x

5

− 10x + 5. The Norwegian

mathematician Niels Abel proved that w could not be expressed

starting only with integers and using just addition, subtraction,

multiplication, division, and

n

√

.

There are lots of algebraic numbers, and they are connected in

many complicated ways. To start with, we have already stated the

fact that if you add, subtract, multiply or divide any two algebraic

numbers (except, of course, you cannot divide by 0), the answer is

again an algebraic number. It is also true that if you take a square

root, or cube root, or any other nth root of an algebraic number,

Start Free Trial

No credit card required