Linear Algebra, Theory And Applications, 2012a

F.5. THE GALOIS GROUP 465
a rational function, a quotient of two polynomials which have coefficients in F. Furthermore,
if you do this, you can see right away that the resulting map form K to K is obviously an automorphism,
preserving the operations of multiplication and addition. Does it keep F fixed?
Of course. You don’t change the coefficients of the polynomials in the rational function
which are always in F. Thus every permutation of the roots determines an automorphism
of K. Now suppose σ is an automorphism of K. Does it determine a permutation of the
roots?
0=σ (p (r i )) = σ (p (σ (r i )))
and so σ (r i )isalsoaroot,sayr ij . Thus it is clear that each σ ∈ G (K, F) determines
a permutation of the roots. Since the roots are distinct, it follows that |G (K, F)| equals
the number of permutations of {r 1 , ··· ,r n } which is n! and that there is a one to one
correspondence between the permutations of the roots and G (K, F) . More will be done on
this later after discussing permutation groups.
This is a good time to make a very important observation about irreducible polynomials.
Lemma F.5.3 Suppose q (x) ≠ p (x) are both irreducible polynomials over a field F. Then
for K a field which contains all the roots of both polynomials, there is no root common to
both p (x) and q (x).
Proof: If l (x) is a monic polynomial which divides them both, then l (x) mustequal
1. Otherwise, it would equal p (x) andq (x) which would require these two to be equal.
Thus p (x) andq (x) are relatively prime and there exist polynomials a (x) ,b(x) having
coefficients in F such that
a (x) p (x)+b (x) q (x) =1
Now if p (x) andq (x) sharearootr, then(x − r) divides both sides of the above in K [x],
but this is impossible.
Now here is an important definition of a class of polynomials which yield equality in the
inequality of Theorem F.5.2.
Definition F.5.4 Let p (x) be a polynomial having coefficients in a field F. AlsoletK be a
splitting field. Then p (x) is separable if it is of the form
p (x) =
m∏
q i (x) k i
i=1
where each q i (x) is irreducible over F and each q i (x) has distinct roots in K. From the
above lemma, no two q i (x) share a root. Thus
has distinct roots in K.
p 1 (x) ≡
m∏
q i (x)
For example, consider the case where F = Q and the polynomial is of the form
(
x 2 +1 ) 2 (
x 2 − 2 ) 2
= x 8 − 2x 6 − 3x 4 +4x 2 +4
i=1
Then let K be the splitting field over Q, Q [ i, √ 2 ] .The polynomials x 2 + 1 and x 2 − 2are
irreducible over Q and each has distinct roots in K.
This is also a convenient time to show that G (K, F) forK a finite extension of F really
is a group. First, here is the definition.