Each polynomial having coefficients in a field F has a splitting field. Consider the case of all
polynomials p
(x)
having coefficients in a field F ⊆ G and consider all roots which are also in G.
The theory of vector spaces is very useful in the study of these algebraic numbers. Here is a
definition.
Definition 3.4.16The algebraic numbers A are thosenumbers which are in G and also roots ofsome polynomial p
(x)
having coefficients in F. Theminimum polynomial^{1}of a ∈ A is defined to be the monic polynomial p
(x )
having smallest degree such that p
(a)
= 0.
The next theorem is on the uniqueness of the minimum polynomial.
Theorem 3.4.17Let a ∈ A. Then there exists a unique monic irreducible polynomial p
(x)
havingcoefficients in F such that p
(a)
= 0. This polynomial is the minimum polynomial.
Proof: Let p
(x)
be a monic polynomial having smallest degree such that p
(a)
= 0. Then p
(x )
is
irreducible because if not, there would exist a polynomial having smaller degree which has a as a root. Now
suppose q
(x)
is monic with smallest degree such that q
(a)
= 0. Then
q(x) = p(x)l(x)+ r(x)
where if r
(x )
≠0, then it has smaller degree than p
(x)
. But in this case, the equation implies r
(a)
= 0
which contradicts the choice of p
(x)
. Hence r
(x)
= 0 and so, since q
(x)
has smallest degree, l
(x)
= 1
showing that p
(x)
= q
(x)
. ■
Definition 3.4.18For a an algebraicnumber, let deg
(a)
denote the degree of the minimumpolynomial of a.
Also, here is another definition.
Definition 3.4.19Let a_{1},
⋅⋅⋅
,a_{m}be in A. A polynomial in
{a1,⋅⋅⋅,am}
will be an expression of theform
∑ k1 kn
ak1⋅⋅⋅kna1 ⋅⋅⋅an
k1⋅⋅⋅kn
where the a_{k1}
⋅⋅⋅
k_{n}are in F, each k_{j}is a nonnegative integer, and all but finitely many of the a_{k1⋅⋅⋅
kn}equalzero. The collection of such polynomials will be denoted by
F [a1,⋅⋅⋅,am ].
Now notice that for a an algebraic number, F
[a]
is a finite dimensional vector space with field of scalars
F. Similarly, for
{a ,⋅⋅⋅,a }
1 m
algebraic numbers, F
[a,⋅⋅⋅,a ]
1 m
is a finite dimensional vector space with
field of scalars F. The following fundamental proposition demonstrates this observation. This is a
remarkable result.
Proposition 3.4.20Let
{a1,⋅⋅⋅,am}
be algebraic numbers. Then
m
dim F [a ,⋅⋅⋅,a ] ≤ ∏ deg(a )
1 m j=1 j
and for an algebraic number a,
dim F[a] = deg(a)
Every element of F
[a1,⋅⋅⋅,am ]
is in A and F
[a1,⋅⋅⋅,am]
is a field.
Proof: Let the minimum polynomial of a be
n n−1
p (x) = x + an−1x + ⋅⋅⋅+ a1x +a0.
If q
(a)
∈ F
[a]
, then
q(x) = p(x)l(x)+ r(x)
where r
(x)
has degree less than the degree of p
(x)
if it is not zero. Hence q
(a)
= r
(a)
. Thus F
[a]
is
spanned by
{ 2 n−1}
1,a,a ,⋅⋅⋅,a
Since p
(x)
has smallest degree of all polynomials which have a as a root, the above set is also linearly
independent. This proves the second claim.
Now consider the first claim. By definition, F
[a1,⋅⋅⋅,am]
is obtained from all linear combinations of
products of
{ }
ak11,ak22,⋅⋅⋅,aknn
where the k_{i} are nonnegative integers. From the first part, it suffices to
consider only k_{j}≤ deg
(a)
j
. This is because a_{1}^{m} can be written as a linear combination of a_{1}^{k} for
k ≤ deg
(a )
1
. Therefore, there exists a spanning set for F
[a ,⋅⋅⋅,a ]
1 m
which has
∏m
deg(ai)
i=1
entries. By Theorem 3.2.9 this proves the first claim.
Finally consider the last claim. Let g
(a1,⋅⋅⋅,am)
be a polynomial in
{a1,⋅⋅⋅,am }
in F
[a1,⋅⋅⋅,am]
.
Since
∏m
dim F[a1,⋅⋅⋅,am] ≡ p ≤ deg(aj) < ∞,
j=1
it follows
2 p
1,g(a1,⋅⋅⋅,am ),g(a1,⋅⋅⋅,am) ,⋅⋅⋅,g (a1,⋅⋅⋅,am )
are dependent. It follows g
(a1,⋅⋅⋅,am)
is the root of some polynomial having coefficients in F. Thus
everything in F
[a1,⋅⋅⋅,am]
is algebraic. Why is F
[a1,⋅⋅⋅,am]
a field? Let g
(a1,⋅⋅⋅,am)
≠0 be in
F
[a1,⋅⋅⋅,am ]
. Then it has a minimum polynomial,
q q−1
p(x) = x + aq−1x + ⋅⋅⋅+ a1x+ a0
where the a_{i}∈ F. Then a_{0}≠0 or else the polynomial would not be minimum. You would have
is in A and
this shows that for every nonzero element of A, its inverse is also in A. What about products and sums of
things in A? Are they still in A? Yes. If a,b ∈ A, then both a + b and ab ∈ F
[a,b]
and from the proposition,
each element of F
[a,b]
is in A. ■
A typical example of what is of interest here is when the field F of scalars is ℚ, the rational numbers
and the field G is ℝ. However, you can certainly conceive of many other examples by considering the
integers mod a prime, for example (See Propositon 1.13.9 on Page 70 for example.) or any of the fields
which occur as field extensions in the above.
There is a very interesting thing about F
[a1 ⋅⋅⋅an]
in the case where F is infinite which
says that there exists a single algebraic γ such that F
[a1⋅⋅⋅an]
= F
[γ]
. In other words, every
field extension of this sort is a simple field extension. I found this fact in an early version of
[10].
Proposition 3.4.22There exists γ suchthat F
[a1⋅⋅⋅an ]
= F
[γ]
. Here each a_{i}is algebraic.
Proof:To begin with, consider F
[α,β ]
. Let γ = α + λβ. Then by Proposition 3.4.20γ is an algebraic
number and it is also clear
F[γ] ⊆ F [α,β]
I need to show the other inclusion. This will be done for a suitable choice of λ. To do this, it suffices to
verify that both α and β are in F
[γ]
.
Let the minimum polynomials of α and β be f
(x)
and g
(x )
respectively. Let the distinct roots
of f
(x)
and g
(x)
be
{α1,α2,⋅⋅⋅,αn}
and
{β1,β2,⋅⋅⋅,βm }
respectively. These roots are in
a field which contains splitting fields of both f
(x)
and g
(x)
. Let α = α_{1} and β = β_{1}. Now
define
h (x) ≡ f (α + λβ − λx ) ≡ f (γ − λx )
so that h
(β)
= f
(α )
= 0. It follows
(x − β)
divides both h
(x )
and g
(x)
. If
(x− η)
is a different linear
factor of both g
(x)
and h
(x)
then it must be
(x− βj)
for some β_{j} for some j > 1 because these are the
only factors of g
(x)
. Therefore, this would require
0 = h (βj) = f (α1 + λβ1 − λ βj)
and so it would be the case that α_{1} + λβ_{1}− λβ_{j} = α_{k} for some k. Hence
λ = αk −-α1
β1 − βj
Now there are finitely many quotients of the above form and if λ is chosen to not be any of them, then the
above cannot happen and so in this case, the only linear factor of both g
(x)
and h
(x)
will be
(x− β)
.
Choose such a λ.
Let ϕ
(x)
be the minimum polynomial of β with respect to the field F
[γ]
. Then this minimum
polynomial must divide both h
(x)
and g
(x)
because h
(β)
= g
(β)
= 0. However, the only factor these two
have in common is x − β and so ϕ
(x)
= x − β which requires β ∈ F
[γ]
. Now also α = γ − λβ
and so α ∈ F
[γ]
also. Therefore, both α,β ∈ F
[γ]
which forces F
[α,β]
⊆ F
[γ]
. This proves
the proposition in the case that n = 2. The general result follows right away by observing
that
F[a1⋅⋅⋅an] = F[a1⋅⋅⋅an−1][an]
and using induction. ■
When you have a field F, F
(a)
denotes the smallest field which contains both F and a. When a is
algebraic over F, it follows that F
(a)
= F
[a]
. The latter is easier to think about because it just involves
polynomials.