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Fearless Symmetry by Robert Gross, Avner Ash

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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
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,

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