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.
[20]
Definition 10.4.1Suppose F,Fare two fields and that f : F →Fis a homomorphism. This meansthat
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 two operations ofaddition and multiplication.
The following is a nice theorem which will be useful.
Theorem 10.4.2Let F be a field and let r be algebraic over F. Let p
(x)
be the minimum polynomial of r.Thus p
(r)
= 0 and p
(x)
is monic and no nonzero polynomial having coefficients in F of smaller degree hasr as a root. In particular, p
(x)
is irreducible over F. Then define f : F
[x]
→ F
[r]
, the polynomials in rby
( 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 (∑ axi∑ b xj) = f( ∑ a bxi+j) = ∑ a bri+j
i i j j i,j ij ij i j
( ) ( )
∑ i∑ j ∑ i ∑ j
= air bjr = f aix f ( bjx )
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 (q1 (x)) = f (q (x))+ f (p (x )l(x))
= 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 minimum
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
≃f ¯
p(x)∑ ∈n F [x] → ∑n ¯p(x) ∈ F [x]
p(x) = k=0 akxk → k=0 f (ak)xk = ¯p (x)
p (r) = 0 ¯p(¯r) = 0
F [r] g→ ¯F [¯r]
≃g
r → ¯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 10.4.3Let f : F →Fbe an isomorphism and let r be algebraic over F with minimialpolynomial p
(x)
. Then the following are equivalent.
There existsralgebraic overFsuch thatp
(¯r)
= 0 in which casep
(x)
is the minimumpolynomial ofr.
There exists g : F
[r]
→F
[¯r]
an isomorphism which extends f such that g
(r)
= r. In this 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 10.4.2 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
This is because g is a homomorphism which takes r to r. It only remains to verify that β is well
defined.
Why is β well defined? Suppose
[k(x)]
=
[k′(x )]
so that k
(x)
− k^{′}
(x)
= l
(x)
p
(x)
. Then since f is a
homomorphism, it follows from ∗ that
¯ ¯′ ¯ ¯ ¯′ ¯
k(x)− k (x) = l(x)¯p(x) ⇒ k (¯r)− k(¯r) = l(¯r)¯p(¯r) = 0
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 minimum
polynomial of r? Call it q
(x)
. There is no loss of generality because f is an isomorphism so
the minimum 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 minimum 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 minimum 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 minimum 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 minimum polynomial for
r.
Note that if p
(x)
is a monic irreducible polynomial, then it is the minimum polynomial for each of its
roots. Consider why this is. If r is a root of p
(x)
, then let q
(x)
be the minimum 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 minimum 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 [20]. Here is a little diagram which describes
what this theorem says.
Definition 10.4.4The symbol
[K : F ]
where K is a field extension of F means the dimension of the vectorspace K with field of scalars F.
F →η ¯F
≃
p(x) “ηp (x )ζ =i ¯p(x)” ¯p (x)
F[r1,⋅⋅⋅,rn] { →≃ ¯F [¯r1,⋅⋅⋅,¯rn]
m ≤ [K : F ]
i = 1,⋅⋅⋅,m,
m = [K : F ],¯ri ⁄= ¯rj
Theorem 10.4.5Let η be an isomorphism from F toFand let K = F
[r1,⋅⋅⋅,rn]
,K = F
[r¯1,⋅⋅⋅,r¯n]
be splitting fields of p
(x)
andp
(x)
respectively. Then there exist at most
[K : F]
isomorphismsζ_{i} : K →Kwhich extend η. If
{¯r1,⋅⋅⋅,¯rn}
are distinct, then there exist exactly
[K : F]
isomorphismsof the above sort. In either case, the two splitting fields areisomorphic with any of these ζ_{i}servingas 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
p(x) = xn + an−1xn−1 + ⋅⋅⋅+ 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
{r,⋅⋅⋅,r }
1 m
. 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) sol(x)¯q(x) = p¯(x ).
Denote the roots of q
(x)
in K as
{r¯1,⋅⋅⋅,¯rm }
where they are counted according to multiplicity.
Then from Theorem 10.4.3, 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 10.4.3, 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
[r ]
1
. By Theorem 10.4.3, 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 10.4.6Let K be a finite dimensional extension of a field F such that every element ofK is algebraic over F, that is, each element of K is a root of some polynomial in F
[x]
. Then K iscalled a normal extensionif for every k ∈ K all roots of the minimum polynomial of k are containedin 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 [20]. This is an amazing
result.
Proposition 10.4.7Let K be a splitting field of f
(x)
∈ F
[x]
. Then K is anormal extension. Infact, if L is an intermediate field between F and K, then L is also a normal extension of F.
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