Linear Algebra, Theory And Applications, 2012a

F.8. CONDITIONS FOR SEPARABILITY 473
p ′ (x) cannot be zero in this case because
p ′ (r n )=
n−1
∏
j=1
(r n − r j ) ≠0
because it is the product of nonzero elements of K. Similarly no term in the sum for p ′ (x)
can equal zero because
∏
(r i − r j ) ≠0.
j≠i
Then if q (x) is a monic polynomial of degree larger than 1 which divides p (x), then the
roots of q (x) inK are a subset of {r 1 , ··· ,r n }. Without loss of generality, suppose these
roots of q (x) are{r 1 , ··· ,r k } ,k≤ n − 1, since q (x) divides p ′ (x) which has degree at most
n − 1. Then q (x) = ∏ k
i=1 (x − r i) but this fails to divide p ′ (x) as polynomials in K [x] and
so q (x) fails to divide p ′ (x) as polynomials in F [x] either. Therefore, q (x) =1andsothe
two are relatively prime.
The following lemma says that the usual calculus result holds in case you are looking at
polynomials with coefficients in a field of characteristic 0.
Lemma F.8.3 Suppose that F has characteristic 0. Then if f ′ (x) =0, it follows that f (x)
is a constant.
Proof: Suppose
f (x) =a n x n + a n−1 x n−1 + ···+ a 1 x + a 0
Then take the derivative n − 1 times to find that a n multiplied by a positive integer ma n
equals 0. Therefore, a n = 0 because, by assumption ma n ≠0ifa n ≠0. Now repeat the
argument with
f 1 (x) =a n−1 x n−1 + ···+ a 1 x + a 0
and continue this way to find that f (x) =a 0 ∈ F.
Now here is a major result which applies to fields of characteristic 0.
Theorem F.8.4 If F is a field of characteristic 0, then every polynomial p (x) , having
coefficients in F is separable.
Proof: It is required to show that the irreducible factors of p (x) havedistinctrootsin
K a splitting field for p (x). So let q (x) be an irreducible monic polynomial. If l (x) isa
monic polynomial of positive degree which divides both q (x) andq ′ (x) , then since q (x) is
irreducible, it must be the case that l (x) =q (x) which forces q (x) to divide q ′ (x) . However,
thedegreeofq ′ (x) is less than the degree of q (x) so this is impossible. Hence l (x) =1and
so q ′ (x) andq (x) are relatively prime which implies that q (x) has distinct roots.
It follows that the above theory all holds for any field of characteristic 0. For example,
if the field is Q then everything holds.
Proposition F.8.5 If a field F has characteristic p, then p is a prime.
Proof: First note that if n · 1=0, if and only if for all a ≠0,n· a = 0 also. This just
follows from the distributive law and the definition of what is meant by n · 1, meaning that
you add 1 to itself n times. Suppose then that there are positive integers, each larger than
1 n, m such that nm · 1=0. Then grouping the terms in the sum associated with nm · 1,
it follows that n (m · 1) = 0. If the characteristic of the field is nm, this is a contradiction
because then m · 1 ≠ 0 but n times it is, implying that n