The next few sections have to do with fields and field extensions. There are many linear algebra
techniques which are used in this discussion and it seems to me to be very interesting.
However, this is definitely far removed from my own expertise so there may be some parts
of this which are not too good. I am following various algebra books in putting this
together.
Consider the notion of splitting fields. It is desired to show that any two are isomorphic,
meaning that there exists a one to one and onto mapping from one to the other which preserves all
the algebraic structure. To begin with, here is a theorem about extending homomorphisms.
[18]
Definition F.4.1Suppose F,Fare two fields and that f : F →Fis a homomorphism. Thismeans that
f (xy) = f (x) f (y), f (x + y) = f (x) + f (y)
An isomorphism is a homomorphismwhich is one to one and onto. A monomorphism is ahomomorphism which is one to one. An automorphism is an isomorphism of a single field.Sometimes people use the symbol ≃ to indicate something is an isomorphism. Then if p
(x)
∈ F
[x]
,say
∑n k
p(x) = akx ,
k=0
p
(x)
will be the polynomial inF
[x]
defined as
∑n
¯p(x) ≡ f (ak)xk.
k=0
Also consider f as a homomorphism of F
[x]
andF
[x]
in the obvious way.
f (p(x)) = ¯p (x)
It is clear that if f is an isomorphism of the two fields F,F, then it is also an isomorphism of thecommutative rings F
[x]
,F
[x]
meaning that it is one to one and onto and preserves the twooperations of addition and multiplication.
The following is a nice theorem which will be useful.
Theorem F.4.2Let F be a field and let r be algebraic over F. Let p
(x)
be the minimalpolynomial of r. Thus p
(r)
= 0 and p
(x)
is monic and no nonzero polynomial having coefficientsin F of smaller degree has r as a root. In particular, p
(x)
is irreducible over F. Then definef : F
[x]
→ F
[r]
, the polynomials in r by
( ∑m ) ∑m
f aixi ≡ airi
i=0 i=0
Then f is a homomorphism. Also, defining g : F
[x]
∕
(p (x))
by
g([q(x)]) ≡ f (q (x)) ≡ q (r)
it follows that g is an isomorphism from the field F
[x]
∕
(p(x))
to F
[r]
.
Proof:First of all, consider why f is a homomorphism. The preservation of sums is obvious.
Consider products.
( ) ( )
∑ ∑ ∑ ∑
f( aixi bjxj) = f ( aibjxi+j) = aibjri+j
i j i,j ij
( ) ( )
= ∑ airi∑ bjrj = f ∑ aixi f( ∑ bjxj)
i j i j
Thus it is clear that f is a homomorphism.
First consider why g is even well defined. If
[q(x)]
=
[q1 (x)]
, this means that
q1(x)− q(x) = p (x)l(x)
for some l
(x)
∈ F
[x]
. Therefore,
f (q (x)) = f (q(x))+ f (p(x)l(x))
1
= f (q(x))+ f (p(x))f (l(x))
≡ q(r)+ p(r)l(r) = q (r) = f (q(x))
Now from this, it is obvious that g is a homomorphism.
g([q(x)][q1(x)]) = g ([q(x)q1(x)]) = f (q(x)q1(x)) = q(r)q1(r)
g([q (x)])g([q1(x)]) ≡ q (r)q1(r)
Similarly, g preserves sums. Now why is g one to one? It suffices to show that if g
([q(x)])
= 0,
then
[q (x )]
= 0. Suppose then that
g([q(x)]) ≡ q(r) = 0
Then
q (x) = p(x)l(x)+ ρ(x)
where the degree of ρ
(x)
is less than the degree of p
(x)
or else ρ
(x)
= 0. If ρ
(x)
≠0, then it follows
that
ρ(r) = 0
and ρ
(x)
has smaller degree than that of p
(x)
which contradicts the definition of p
(x)
as the
minimal polynomial of r. Thus q
(x )
= p
(x)
l
(x)
and so
[q(x)]
= 0. Since p
(x)
is irreducible,
F
[x]
∕
(p(x))
is a field. It is clear that g is onto. Therefore, F
[r]
is a field also. (This was shown
earlier by different reasoning.) ■
Here is a diagram of what the following theorem says.
Extending f to g
f ¯
F →≃ F
p(x) ∈ F[x] →f p¯(x ) ∈ ¯F[x]
p(x) = ∑n a xk → ∑n f (a )xk = ¯p(x)
k=0 k k=0 k
p(r) = 0 g ¯p(¯r) = 0
F[r] →≃ ¯F[¯r]
r →g ¯r
The idea illustrated is the following question: For r algebraic over F and f an isomorphism of F
and F, when does there exist r algebraic over F and an isomorphism of F
[r]
and F
[¯r]
which
extends f? This is the content of the following theorem.
Theorem F.4.3Let f : F →Fbe an isomorphism and let r be algebraic over F withminimial polynomial p
(x)
. Then the following are equivalent.
There existsralgebraic overFsuch thatp
(¯r)
= 0 in which casep
(x)
is the minimalpolynomial ofr.
There exists g : F
[r]
→F
[¯r]
an isomorphism which extends f such that g
(r)
= r. Inthis case, there is only one such isomorphism.
Proof: 2.)⇒1.) Let g
(r)
= g
(¯r)
with g an isomorphism extending f,g
(r)
= g
(¯r)
. Then since
it is an isomorphism,
0 = g (p (r)) = p¯(g (r)) = p¯(¯r) (∗)
(∗)
Define β as β
([k(x)])
≡k
(¯r)
relative to this r≡ g
(r)
and let α : F
[x]
∕
(p(x))
→ F
[r]
be the
isomorphism mentioned in Theorem F.4.2 called g there, given by α
([k(x)])
≡ k
(r)
.
Thus
F[r] α← F [x]∕(p(x))→β ¯F[¯r]
Then if β is a well defined homomorphism, it follows that g must equal β ∘ α^{−1} because
so β is indeed well defined. It is clear from the definition that β is a homomorphism.
1.)⇒ 2.) Next suppose there exists r algebraic over F such that p
(¯r)
= 0. Why is p
(x)
the
minimal polynomial of r? Call it q
(x)
. There is no loss of generality because f is an isomorphism
so the minimal polynomial can be written this way. Then β
([q (x )])
≡q
(¯r)
= 0 = p
(¯r)
. Then
p
(x)
= q
(x)
m
(x )
+ R
(x)
where the degree of R
(x)
is less than the degree of q
(x )
or equal to
zero and so R
(¯r)
= 0 which is contrary to q
(x)
being minimal polynomial for r unless R
(x)
= 0.
Therefore, R
(x)
= 0. It follows that, since f is a isomorphism, we have p
(x)
= q
(x)
m
(x)
contrary
to p
(x)
being the minimal polynomial for r. Indeed, if the degree of q
(x)
is less than that of p
(x)
,
we have 0 = p
(r)
= q
(r)
m
(r)
and so one of q
(r)
,m
(r)
equals 0 contrary to p
(x)
having smallest
possible degree for sending r to 0. Thus the degree of q
(x)
is the same as the degree of
p
(x)
and since both are monic by definition, m
(x)
= 1. Hence p
(x)
= q
(x)
and so
p
(x)
= q
(x)
.
Now let α,β be defined as above. It was shown above that β is a well defined homomorphism.
It is also clear that β is onto. It only remains to verify that β is one to one and when this is done,
the isomorphism will be β ∘ α^{−1}. Suppose β
([k (x)])
≡k
(¯r)
= 0. Does it follow that
[k (x )]
= 0?
By assumption, p
(¯r)
= 0 and also,
¯k (x ) = ¯p(x)¯l(x)+ ¯ρ(x) (*)
(*)
where the degree of ρ
(x)
is less than the degree of p
(x)
which is the same as the degree of p
(x)
or
else it equals 0. It follows that ρ
(¯r)
= 0 and this is a contradiction because p
(x)
is the minimal
polynomial for r which was shown above. Hence k
(x )
= p
(x)
l
(x)
and since f is an
isomorphism, this says that k
(x)
= p
(x)
l
(x )
and so
[k(x)]
= 0. Hence β is indeed
one to one and so an example of g would be β ∘ α^{−1}. Also β ∘ α^{−1}
(r)
= β
([x ])
= r.■
What is the meaning of the above in simple terms? It says that the monomorphisms from F
[r]
to a field K containing F correspond to the roots of p
(x)
in K. That is, for each root of p
(x)
, there
is a monomorphism and for each monomorphism, there is a root. Also, for each root r of p
(x )
in
K, there is an isomorphism from F
[r]
to F
[¯r]
. Here p
(x)
is the minimal polynomial for
r.
Note that if p
(x)
is a monic irreducible polynomial, then it is the minimal polynomial for
each of its roots. Consider why this is. If r is a root of p
(x)
, then let q
(x)
be the minimal
polynomial for r. Then
p(x) = q (x)k (x)+ R(x)
where R
(x)
is 0 or else has smaller degree than q
(x)
. However, R
(r)
= 0 and this contradicts q
(x)
being the minimal polynomial of r. Hence q
(x)
divides p
(x)
or else k
(x)
= 1. The latter
possibility must be the case because p
(x)
is irreducible.
This is the situation which is about to be considered. It involves the splitting fields K,K of
p
(x)
,p
(x)
where η is an isomorphism of F and F as described above. See [18]. Here is a little
diagram which describes what this theorem says.
Definition F.4.4The symbol
[K : F]
where K is a field extension of F means the dimension ofthe vector space K with field of scalars F.
η
F →≃ ¯F
p(x) “ηp (x) = ¯p(x)” p¯(x)
ζi ¯
F[r1,⋅⋅⋅,rn] { →≃ F [¯r1,⋅⋅⋅,¯rn]
m ≤ [K : F]
i = 1,⋅⋅⋅,m, m = [K : F],¯ri ⁄= ¯rj
Theorem F.4.5Let η be an isomorphism from F toFand let K = F
[r1,⋅⋅⋅,rn]
,K =
F
[¯r1,⋅⋅⋅,¯rn]
be splitting fields of p
(x)
andp
(x)
respectively. Then there exist at most
[K : F]
isomorphisms ζ_{i} : K →Kwhich extend η. If
{¯r1,⋅⋅⋅,¯rn}
are distinct, then there exist exactly
[K : F]
isomorphisms of the above sort. In either case, the two splitting fields areisomorphicwith any of these ζ_{i}serving as an isomorphism.
Proof: Suppose
[K : F]
= 1. Say a basis for K is
{r}
. Then
{1,r}
is dependent and so there
exist a,b ∈ F, not both zero such that a + br = 0. Then it follows that r ∈ F and so in this case
F = K. Then the isomorphism which extends η is just η itself and there is exactly 1
isomorphism.
Next suppose
[K : F]
> 1. Then p
(x)
has an irreducible factor over F of degree larger than 1,
q
(x)
. If not, you would have
n n−1
p(x) = x + an−1x + ⋅⋅⋅+ an
and it would factor as
= (x − r1)⋅⋅⋅(x− rn)
with each r_{j}∈ F, so F = K contrary to
[K : F]
> 1.Without loss of generality, let the roots of q
(x)
in K be
{r1,⋅⋅⋅,rm }
. Thus
∏m ∏n
q(x) = (x− ri) , p(x) = (x− ri)
i=1 i=1
Now q
(x)
defined analogously to p
(x)
, also has degree at least 2. Furthermore, it divides
p
(x)
all of whose roots are in K. This is obvious because η is an isomorphism. You
have
l(x)q (x) = p(x) so ¯l(x)¯q(x) = ¯p(x).
Denote the roots of q
(x)
in K as
{¯r1,⋅⋅⋅,¯rm}
where they are counted according to
multiplicity.
Then from Theorem F.4.3, there exist k ≤ m one to one homomorphisms (monomorphisms) ζ_{i}
mapping F
[r1]
to K≡F
[¯r1,⋅⋅⋅,¯rn]
, one for each distinct root of q
(x )
in K. If the roots of p
(x)
are distinct, then this is sufficient to imply that the roots of q
(x )
are also distinct, and k = m, the
dimension of q
(x)
. Otherwise, maybe k < m. (It is conceivable that q
(x )
might have repeated
roots in K.) Then
[K : F] = [K : F[r1]][F [r1] : F]
and since the degree of q
(x)
> 1 and q
(x )
is irreducible, this shows that
[F[r1] : F]
= m > 1 and
so
[K : F[r1]] < [K : F]
Therefore, by induction, using Theorem F.4.3, each of these k ≤ m =
[F[r1] : F ]
one to one
homomorphisms extends to an isomorphism from K to K and for each of these ζ_{i}, there are no
more than
[K : F [r1]]
of these isomorphisms extending F. If the roots of p
(x)
are distinct, then
there are exactly m of these ζ_{i} and for each, there are
[K : F[r1]]
extensions. Therefore, if the
roots of p
(x)
are distinct, this has identified
[K : F [r1]]m = [K : F [r1]][F[r1] : F] = [K : F]
isomorphisms of K to K which agree with η on F. If the roots of p
(x)
are not distinct, then maybe
there are fewer than
[K : F]
extensions of η.
Is this all of them? Suppose ζ is such an isomorphism of K and K. Then consider its
restriction to F
[r1]
. By Theorem F.4.3, this restriction must coincide with one of the ζ_{i}
chosen earlier. Then by induction, ζ is one of the extensions of the ζ_{i} just mentioned.
■
Definition F.4.6Let K be a finite dimensional extension of a field F such that everyelement of K is algebraic over F, that is, each element of K is a root of some polynomialin F
[x]
. Then K is called a normal extensionif for every k ∈ K all roots of the minimalpolynomial of k are contained in K.
So what are some ways to tell that a field is a normal extension? It turns out that if K is a
splitting field of f
(x)
∈ F
[x ]
, then K is a normal extension. I found this in [18]. This is an
amazing result.
Proposition F.4.7Let K be a splitting field of f
(x)
∈ F
[x ]
. Then K is anormal extension.In fact, if L is an intermediate field between F and K, then L is also a normal extension ofF.
Proof:Let r ∈ K be a root of g
(x)
, an irreducible monic polynomial in F
[x]
. It
is required to show that every other root of g
(x)
is in K. Let the roots of g
(x)
in a
splitting field be
{r1 = r,r2,⋅⋅⋅,rm}
. Now g
(x)
is the minimal polynomial of r_{j} over F
because g