Linear Algebra, Theory And Applications, 2012a

F.8. CONDITIONS FOR SEPARABILITY 471
θ is a homomorphism. Why is θ onto? This follows right away from Theorem F.4.5. Note
that K is the splitting field of p (x) overL j since L j ⊇ F. Also if σ ∈ G (L j , F) soitisan
automorphism of L j , then, since it fixes F, p (x) =¯p (x) inthattheorem. Thusσ extends to
ζ, an automorphism of K. Thus θζ = σ. Why is θ one to one? If θ [σ] =θ [α] , this means
σ = α on L j . Thus σα −1 is the identity on L j . Hence σα −1 ∈ G (K, L j ) which is what it
means for [σ] =[α].
There is an immediate application to a description of the normal closure of an algebraic
extension F [a 1 ,a 2 , ··· ,a m ] . To begin with, recall the following definition.
Definition F.7.2 When you have F [a 1 , ··· ,a m ] with each a i algebraic so that F [a 1 , ··· ,a m ]
is a field, you could consider
m∏
f (x) ≡ f i (x)
i=1
where f i (x) is the minimal polynomial of a i .ThenifK is a splitting field for f (x) , this K
is called the normal closure. It is at least as large as F [a 1 , ··· ,a m ] and it has the advantage
of being a normal extension.
Let G (K, F) ={η 1 ,η 2 , ··· ,η m } . The conjugate fields are the fields
η j (F [a 1 , ··· ,a m ])
Thus each of these fields is isomorphic to any other and they are all contained in K. LetK ′
denote the smallest field contained in K which contains all of these conjugate fields. Note
that if k ∈ F [a 1 , ··· ,a m ]sothatη i (k) is in one of these conjugate fields, then η j η i (k) is
also in a conjugate field because η j η i is one of the automorphisms of G (K, F). Let
S = { k ∈ K ′ : η j (k) ∈ K ′ each j } .
Then from what was just shown, each conjugate field is in S. Suppose k ∈ S. What about
k −1 ?
η j (k) η j
(
k
−1 ) = η j
(
kk
−1 ) = η j (1) = 1
and so ( η j (k) ) −1 ( )
= ηj k
−1
. Now ( η j (k) ) −1
∈ K
′
because K ′ is a field. Therefore,
(
η ) j k
−1
∈ K ′ . Thus S is closed with respect to taking inverses. It is also closed with
respect to products. Thus it is clear that S is a field which contains each conjugate field.
However, K ′ was defined as the smallest field which contains the conjugate fields. Therefore,
S = K ′ and so this shows that each η j maps K ′ to itself while fixing F. Thus G (K, F) ⊆
G (K ′ , F) . However, since K ′ ⊆ K, it follows that also G (K ′ , F) ⊆ G (K, F) . Therefore,
G (K ′ , F) =G (K, F) , and by the one to one correspondence between the intermediate fields
and the Galois groups, it follows that K ′ = K. This proves the following lemma.
Lemma F.7.3 Let K denote the normal extension of F [a 1 , ··· ,a m ] with each a i algebraic
so that F [a 1 , ··· ,a m ] is a field. Thus K is the splitting field of the product of the minimal
polynomials of the a i . Then K is also the smallest field containing the conjugate fields
η j (F [a 1 , ··· ,a m ]) for {η 1 ,η 2 , ··· ,η m } = G (K, F).
F.8 Conditions For Separability
So when is it that a polynomial having coefficients in a field F is separable? It turns out
that this is always the case for fields which are enough like the rational numbers. It involves
considering the derivative of a polynomial. In doing this, there will be no analysis used, just