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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
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