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 7.3.27The algebraic numbers A are thosenumbers which are in G and alsoroots of some polynomial p
(x )
having coefficients in F. Theminimal polynomial of a ∈ A isdefined to be the monic polynomial p
(x)
having smallest degree such that p
(a)
= 0.
The next theorem is on the uniqueness of the minimal polynomial.
Theorem 7.3.28Let a ∈ A. Then there exists a unique monic irreducible polynomial p
(x)
having coefficients in F such that p
(a)
= 0. This polynomial is the minimal 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 7.3.29For a an algebraicnumber, let deg
(a)
denote the degree of the minimalpolynomial of a.
Also, here is another definition.
Definition 7.3.30Let a_{1},
⋅⋅⋅
,a_{m}be in A. A polynomial in
{a1,⋅⋅⋅,am }
will be an expression ofthe form
∑ 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 thea_{k1⋅⋅⋅
kn}equal zero. The collection of such polynomials will be denoted by
F[a1,⋅⋅⋅,am].
Now notice that for a an algebraic number, F
[a]
is a vector space with field of scalars F.
Similarly, for
{a1,⋅⋅⋅,am}
algebraic numbers, F
[a1,⋅⋅⋅,am ]
is a vector space with field of scalars
F. The following fundamental proposition is important.
Proposition 7.3.31Let
{a1,⋅⋅⋅,am}
be algebraic numbers. Then
∏m
dim F[a1,⋅⋅⋅,am] ≤ deg (aj)
j=1
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 minimal polynomial of a be
p(x) = xn + an−1xn−1 + ⋅⋅⋅+ 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
{ }
1,a,a2,⋅⋅⋅,an−1
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
[a ,⋅⋅⋅,a ]
1 m
is obtained from all linear
combinations of products of
{ k1 k2 kn}
a1 ,a2 ,⋅⋅⋅,an
where the k_{i} are nonnegative integers. From the
first part, it suffices to consider only k_{j}≤ deg
(aj)
. Therefore, there exists a spanning set for
F
[a1,⋅⋅⋅,am ]
which has
m∏
deg(ai)
i=1
entries. By Theorem 7.2.4 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
dimF [a ,⋅⋅⋅,a ] ≡ p ≤ ∏ deg (a ) < ∞,
1 m j=1 j
it follows
1,g(a1,⋅⋅⋅,am),g(a1,⋅⋅⋅,am )2 ,⋅⋅⋅,g(a1,⋅⋅⋅,am)p
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)
be
as just mentioned. Then it has a minimal polynomial,
p (x ) = xq + aq−1xq−1 + ⋅⋅⋅+ a1x+ a0
where the a_{i}∈ F. Then a_{0}≠0 or else the polynomial would not be minimal. Therefore,
Now from this proposition, it is easy to obtain the following interesting result about the
algebraic numbers.
Theorem 7.3.32The algebraicnumbers A, those roots of polynomials in F
[x]
which arein G, are a field.
Proof: By definition, each a ∈ A has a minimal polynomial. Let a≠0 be an algebraic number
and let p
(x)
be its minimal polynomial. Then p
(x)
is of the form
xn + an−1xn−1 + ⋅⋅⋅+ a1x+ a0
where a_{0}≠0. Otherwise p(x) would not have minimal degree. Then plugging in a yields
( )
an−-1 +-an−1an−2 +-⋅⋅⋅+-a1-(−-1)
a a0 = 1.
and so a^{−1} =
(an−-1+an−1an−2+⋅⋅⋅+a1)(−1)
a0
∈ F
[a]
. By the proposition, every element of F
[a]
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 Problem 34
on Page 608 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
[5].
Proposition 7.3.33There exists γ suchthat F
[a1⋅⋅⋅an ]
= F
[γ]
.
Proof:To begin with, consider F
[α,β ]
. Let γ = α + λβ. Then by Proposition 7.3.31γ 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 minimal 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 − λβj)
and so it would be the case that α_{1} + λβ_{1}− λβ_{j} = α_{k} for some k. Hence
αk − α1
λ = 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 minimal polynomial of β with respect to the field F
[γ]
. Then this minimal
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.