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 has
mainly featured the field of real or complex numbers but this procedure will show how to
obtain many other fields which could be used in most of what was presented earlier in
the book. Here is an important idea concerning equivalence relations which I hope is
familiar.
Definition 7.3.12Let S be a set. The symbol, ∼ is called an equivalence relation on S ifit satisfies thefollowing 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 7.3.13
[x]
denotes the set of all elements of S which are equivalent to x and
[x]
is called the equivalence class determined by x or just the equivalence class of x.
Also recall the notion of equivalence classes.
Theorem 7.3.14Let ∼ be an equivalence class defined on a set, S and let ℋ denote theset of equivalence classes. Then if
[x]
and
[y]
are two of these equivalence classes, eitherx ∼ y and
[x]
=
[y]
or it is not true that x ∼ y and
[x]
∩
[y]
= ∅.
Definition 7.3.15Let F be a field, for example the rational numbers, and denote by F
[x]
thepolynomials having coefficients in F. Suppose p
(x)
is a polynomial. Let a
(x)
∼ b
(x)
(a
(x)
issimilar to b
(x)
) when
a (x) − b (x) = k(x)p(x)
for some polynomial k
(x)
.
Proposition 7.3.16In 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 the polynomial p
(x)
in the above definition is
irreducible so I will begin assuming this.
Definition 7.3.17Let F be a field and let p
(x)
∈ F
[x]
be a monic irreducible polynomial ofdegree greater than 0. Thus there is no polynomial having coefficients in F which divides p
(x)
except for itselfand constants, and its leading coefficient is 1. For the similarity relation defined inDefinition 7.3.15, define the following operations on the equivalence classes.
[a(x)]
is anequivalence class means that it is the set of all polynomials which are similar to a
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
[q(x )]
∈ F
[x]
∕
(p(x))
where
[q(x)]
≠
[0]
.
Then q
(x)
is not a multiple of p
(x)
and so by the first part, q
(x)
,p
(x)
are relatively prime. Thus
there exist n
(x)
,m
(x)
such that
1 = n(x)q(x)+ m (x)p(x)
Hence
[1] = [1− n(x)p (x)] = [n(x)q(x)] = [n(x)][q(x)]
which shows that
[q(x)]
^{−1} =
[n (x)]
. Thus this is a field. The polynomial has a root in this field
because if
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 ]
∑ i ∑ i
i=0 ci[x] = i=0 cix = 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. ■
From the above theorem, it makes perfect sense to write b rather than
[b]
if b ∈ F. Then with
this convention,
[bϕ (x)] = [b][ϕ (x)] = b[ϕ(x)].
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 7.3.19The polynomials consisting of all polynomial multiples of p
(x)
, denotedby
(p(x))
is called an ideal. An ideal I is a subset of the commutative ring (Here the ringis F
[x ]
.) with unity consisting of all polynomials which is itself a ring and which has theproperty that whenever f
(x)
∈ F
[x]
, and g
(x)
∈ I, f
(x)
g
(x )
∈ I. In this case, you couldargue that
(p (x))
is an ideal and that the only ideal containing it is itself or the entire ringF
[x]
. This is called a maximal ideal.
Example 7.3.20The 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. Henceyou would have x^{2}− 2 =
(x+ a)
(x +b)
which requires that a + b = 0 so this factorizationis 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 form a + b
√2
.
In the above example,
[ ]
x2 +x
is not zero because it is not a multiple of x^{2}− 2. What is
[x2 + 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)]
=
[x2 + 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 7.3.21Let F ⊆ K be two fields. Then clearly K is also a vector space overF. Then also, 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 7.3.22Let 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=1 j=1
It follows that
{kjli}
is a spanning set. If
∑n ∑m
fjikjli = 0
i=1j=1
Then, since the l_{i} are independent, it follows that
m
∑ f k = 0
j=1 ji j
and since
{kj}
is independent, each f_{ji} = 0 for each j for a given arbitrary i. Therefore,
{kjli}
is
a basis. ■
Note that if p
(x)
were not irreducible, then you could find a field 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 7.3.23Let p
(x)
= x^{n} + a_{n−1}x^{n−1} +
⋅⋅⋅
+ a_{1}x + a_{0}be a polynomial with coefficients ina field of scalars F. There exists a larger field G and
{z1,⋅⋅⋅,zn}
contained in G, listed accordingto multiplicity, such that
n
p(x) = ∏ (x− zi)
i=1
This larger field is called a splittingfield. Furthermore,
[G : F] ≤ n!
Proof:From Proposition 7.3.18, 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
q1(x) = (x− z2)q2(x).
and so
p(x) = (x − z1)(x − z2)q2(x)
After n such extensions, you will have obtained the necessary field G.
Finally consider the claim about dimension. By Proposition 7.3.18, there is a larger field G_{1}
such that p
(x)
has a root a_{1} in G_{1} and
[G1 : F ]
≤ n. Then
p(x) = (x− a1)q (x)
Continue this way until the polynomial equals the product of linear factors. Then by Proposition
7.3.22 applied multiple times,