As an application of the theory of vector spaces, this section considers the problem of field extensions.
When you have a polynomial like x^{2}− 3 which has no rational roots, it turns out you can enlarge the field
of rational numbers to obtain a larger field such that this polynomial does have roots in this
larger field. I am going to discuss a systematic way to do this. It will turn out that for any
polynomial with coefficients in any field, there always exists a possibly larger field such that the
polynomial has roots in this larger field. This book mainly features the field of real or complex
numbers but this procedure will show how to obtain many other fields. The ideas used in this
development are the same as those used later in the material on linear transformations but slightly
easier.
Here is an important idea concerning equivalence relations which I hope is familiar. If not, see Section
1.3 on page 14.
Definition 3.4.1Let S be a set. The symbol, ∼ is called an equivalence relation on S if it satisfiesthefollowing axioms.
x ∼ x for all x ∈ S. (Reflexive)
If x ∼ y then y ∼ x. (Symmetric)
If x ∼ y and y ∼ z, then x ∼ z. (Transitive)
Definition 3.4.2
[x]
denotes the set of all elements of S which are equivalent to x and
[x]
is calledthe equivalence class determined by x or just the equivalence class of x.
Also recall the notion of equivalence classes.
Theorem 3.4.3Let ∼ be an equivalence classdefined on a set, S and let ℋ denote the set ofequivalence classes. Then if
[x]
and
[y]
are two of these equivalence classes, either x ∼ y and
[x]
=
[y]
or it is not true that x ∼ y and
[x]
∩
[y]
= ∅.
Definition 3.4.4Let F be a field, forexample the rational numbers, and denote by F
[x]
the polynomialshaving coefficients in F. Suppose p
(x)
is a polynomial. Let a
(x)
∼ b
(x)
(a
(x)
is similar to b
(x)
)when
a(x)− b(x) = k (x)p (x )
for some polynomial k
(x)
. Denote by
(p(x))
all polynomials of the form p
(x )
k
(x)
where k
(x)
is somepolynomial.
Proposition 3.4.5In the above definition, ∼ is an equivalence relation.
Proof: First of all, note that a
(x)
∼ a
(x )
because their difference equals 0p
(x)
. If a
(x)
∼ b
(x)
, then
a
(x)
− b
(x)
= k
(x)
p
(x)
for some k
(x)
. But then b
(x)
− a
(x)
= −k
(x )
p
(x)
and so b
(x)
∼ a
(x )
. Next
suppose a
(x )
∼ b
(x)
and b
(x)
∼ c
(x)
. Then a
(x)
−b
(x)
= k
(x)
p
(x)
for some polynomial k
(x)
and also
b
(x)
− c
(x)
= l
(x)
p
(x)
for some polynomial l
(x)
. Then
a(x)− c(x) = a(x)− b (x) +b (x) − c (x)
= k(x)p(x)+ l(x)p(x) = (l(x)+ k(x))p(x)
and so a
(x)
∼ c
(x)
and this shows the transitive law. ■
With this proposition, here is another definition which essentially describes the elements of the new
field. It will eventually be necessary to assume that the polynomial p
(x)
in the above definition is
irreducible so I will begin assuming this.
Definition 3.4.6Let F be a field and let p
(x)
∈ F
[x]
be a monic irreducible polynomial of degree greaterthan 0. Thus there is no polynomial having coefficients in F which divides p
(x)
except for itselfandconstants, and its leading coefficient is 1. For the similarity relation defined in Definition 3.4.4, define thefollowing operations on the equivalence classes.
[a(x)]
is an equivalence class meansthat it is the set of allpolynomials which are similar to a
This collection of equivalence classes is sometimes denoted by F
[x]
∕
(p(x))
. This is called a quotientspace.
Proposition 3.4.7In the situation of Definition 3.4.6where p
(x)
is a monic irreduciblepolynomial, the following are valid.
p
(x)
and q
(x)
are relatively prime for any q
(x)
∈ F
[x]
which is not a multiple of p
(x)
.
The definitions of addition and multiplication are well defined.
If a,b ∈ F and
[a]
=
[b]
, then a = b. Thus F can be considered a subset of F
[x]
∕
(p (x))
.
F
[x]
∕
(p(x))
is a field in which the polynomial p
(x )
has a root.
F
[x]
∕
(p(x))
is a vector space with field of scalars F and its dimension is m where m is thedegree of the irreducible polynomial p
(x)
.
Proof: First consider the claim about p
(x)
,q
(x)
being relatively prime. If ψ
(x)
is the greatest
common divisor, (the monic polynomial of largest degree which divides both) it follows ψ
(x)
is either equal
to p
(x)
or 1. If it is p
(x)
, then q
(x)
is a multiple of p
(x )
which does not happen. If it is 1, then by
definition, the two polynomials are relatively prime.
To show the operations are well defined, suppose
[a(x)] = [a′(x)],[b(x)] = [b′(x)]
It is necessary to show
[a(x)+ b(x)] = [a′(x)+ b′(x)]
[a(x)b(x)] = [a′(x)b′(x)]
Consider the second of the two.
a′(x)b′(x) − a (x)b(x)
= a′(x)b′(x) − a (x)b′(x)+ a (x)b′(x)− a(x)b(x)
′ ′ ′
= b (x)(a (x)− a (x))+ a (x)(b (x)− b(x))
Now by assumption
′
(a (x)− a (x))
is a multiple of p
(x)
as is
′
(b(x)− b(x))
, so the above is a
multiple of p
(x)
and by definition this shows
[a (x)b (x)]
=
[a′(x)b′(x)]
. The case for addition is
similar.
Now suppose
[a]
=
[b]
. This means a−b = k
(x)
p
(x)
for some polynomial k
(x)
. Then k
(x)
must equal
0 since otherwise the two polynomials a − b and k
(x)
p
(x)
could not be equal because they would have
different degree.
It is clear that the axioms of a field are satisfied except for the one which says that non zero elements of
the field have a multiplicative inverse. Let
. Thus this is a field. The polynomial has a root in this field because
if
m m −1
p(x) = x + am−1x + ⋅⋅⋅+ a1x + a0,
m m −1
[0] = [p(x)] = [x] +[am− 1][x] + ⋅⋅⋅+ [a1][x ]+ [a0]
Thus
[x]
is a root of this polynomial in the field F
[x]
∕
(p(x))
.
Consider the last claim. It is clear that F
[x]
∕
(p(x))
is a vector space with field of scalars F. Indeed, the
operations are defined such that for α,β ∈ F,
α[r(x)]+ β [b(x)] ≡ [αr(x)+ βb(x)]
It remains to consider the dimension of this vector space. Let f
(x)
∈ F
[x]
∕
(p(x))
. Thus
[f (x)]
is a typical
thing in F
[x ]
∕
(p (x ))
. Then from the division algorithm,
f (x) = p(x)q (x) +r (x )
where r
(x)
is either 0 or has degree less than the degree of p
(x)
. Thus
[r(x)] = [f (x)− p(x)q(x)] = [f (x)]
but clearly
[r(x)]
∈span
( )
[1],⋅⋅⋅,[x]m−1
. Thus span
( )
[1],⋅⋅⋅,[x]m −1
= F
[x ]
∕
(p (x))
. Then
{ m−1}
[1],⋅⋅⋅,[x]
is a basis if these vectors are linearly independent. Suppose then that
m∑−1 [m∑−1 ]
ci[x]i = cixi = 0
i=0 i=0
Then you would need to have p
(x)
∕∑_{i=0}^{m−1}c_{i}x^{i} which is impossible unless each c_{i} = 0 because p
(x)
has
degree m. ■
This shows how to enlarge a field to get a new one in which the polynomial has a root. By using a
succession of such enlargements, called field extensions, there will exist a field in which the given
polynomial can be factored into a product of polynomials having degree one. The field you obtain in this
process of enlarging in which the given polynomial factors in terms of linear factors is called a splitting
field.
Remark 3.4.8The polynomials consisting of all polynomial multiples of p
(x)
, denoted by
(p (x))
is called an ideal. An ideal I is a subset of the commutative ring (Here the ring is F
[x]
.) withunity consisting of all polynomials which is itself a ring and which has the property that wheneverf
(x )
∈ F
[x]
, and g
(x)
∈ I, f
(x)
g
(x)
∈ I. In this case, you could argue that
(p(x))
is an ideal andthat the only ideal containing it is itself or the entire ring F
[x]
. This is called amaximal ideal.
Example 3.4.9The polynomial x^{2}−2 is irreducible in ℚ
[x]
. This is because if x^{2}−2 = p
(x)
q
(x)
where p
(x)
,q
(x)
both have degree less than 2, then they both have degree 1. Hence you wouldhave x^{2}− 2 =
(x+ a)
(x + b)
which requires that a + b = 0 so this factorization is of the form
(x − a)
(x+ a)
and now you need to have a =
√-
2
∕∈
ℚ. Now ℚ
[x]
∕
( )
x2 − 2
is of the form a + b
[x]
where a,b ∈ ℚ and
[x]
^{2}− 2 = 0. Thus one can regard
[x]
as
√-
2
. ℚ
[x]
∕
( )
x2 − 2
is of the forma + b
√ -
2
.
In the above example,
[ 2 ]
x +x
is not zero because it is not a multiple of x^{2}− 2. What is
[ 2 ]
x + x
^{−1}?
You know that the two polynomials are relatively prime and so there exists n
(x )
,m
(x)
such
that
1 = n(x)(x2 − 2) +m (x)(x2 + x)
Thus
[m (x)]
=
[ 2 ]
x + x
^{−1}. How could you find these polynomials? First of all, it suffices to consider only
n
The above is an example of something general described in the following definition.
Definition 3.4.10Let F ⊆ K be two fields. Then clearly K is also a vector space over F. Thenalso, K is called a finite field extension of F if thedimension of this vector space,denoted by
[K : F]
is finite.
There are some easy things to observe about this.
Proposition 3.4.11Let F ⊆ K ⊆ L be fields. Then
[L : F ]
=
[L : K ]
[K : F ]
.
Proof:Let
{li}
_{i=1}^{n} be a basis for L over K and let
{kj}
_{j=1}^{m} be a basis of K over F. Then if l ∈ L,
there exist unique scalars x_{i} in K such that
∑n
l = xili
i=1
Now x_{i}∈ K so there exist f_{ji} such that
m∑
xi = fjikj
j=1
Then it follows that
∑n ∑m
l = fjikjli
i=1j=1
It follows that
{k l}
ji
is a spanning set. If
∑n m∑
fjikjli = 0
i=1 j=1
Then, since the l_{i} are independent, it follows that
m
∑
j=1fjikj = 0
and since
{kj}
is independent, each f_{ji} = 0 for each j for a given arbitrary i. Therefore,
{kjli}
is a basis.
■
You will see almost exactly the same argument in exhibiting a basis for ℒ
(V,W )
the linear
transformations mapping V to W.
Note that if p
(x)
were of degree n and not irreducible, then there still exists an extension G containing
a root of p
(x)
such that
[G : F]
≤ n. You could do this by working with an irreducible factor of
p
(x)
.
Theorem 3.4.12Let p
(x)
= x^{n} + a_{n−1}x^{n−1} +
⋅⋅⋅
+ a_{1}x + a_{0}be a polynomial with coefficients in a field ofscalars F. There exists a larger field G and
{z1,⋅⋅⋅,zn}
contained in G, listed according to multiplicity,such that
n∏
p(x) = (x − zi)
i=1
This larger field is called a splittingfield. Furthermore,
[G : F] ≤ n!
Proof:From Proposition 3.4.7, there exists a field F_{1} such that p
(x)
has a root, z_{1} (=
[x ]
if p is
irreducible.) Then by the Euclidean algorithm
p (x) = (x− z1)q1(x)+ r
where r ∈ F_{1}. Since p
(z1)
= 0, this requires r = 0. Now do the same for q_{1}
(x)
that was done for p
(x)
,
enlarging the field to F_{2} if necessary, such that in this new field