Linear Algebra, Theory And Applications, 2012a

472 FIELDS AND FIELD EXTENSIONS
the rule for differentiation which we all learned in calculus. Thus the derivative is defined
as follows.
(
an x n + a n−1 x n−1 + ···+ a 1 x + a 0
) ′
≡ na n x n−1 + a n−1 (n − 1) x n−2 + ···+ a 1
This kind of formal manipulation is what most students do anyway, never thinking about
where it comes from. Here na n means to add a n to itself n times. With this definition, it
is clear that the usual rules such as the product rule hold. This discussion follows [17].
Definition F.8.1 A field has characteristic 0 if na ≠0for all n ∈ N and a ≠0. Otherwise
a field F has characteristic p if p · 1=0for p · 1 defined as 1 added to itself p times and p
is the smallest positive integer for which this takes place.
Note that with this definition, some of the terms of the derivative of a polynomial could
vanish in the case that the field has characteristic p. I will go ahead and write them anyway.
For example, if the field has characteristic p, then
(x p − a) ′ =0
because formally it equals p · 1x p−1 =0x p−1 , the 1 being the 1 in the field.
Note that the field Z p does not have characteristic 0 because p·1 = 0. Thus not all fields
have characteristic 0.
How can you tell if a polynomial has no repeated roots? This is the content of the next
theorem.
Theorem F.8.2 Let p (x) be a monic polynomial having coefficients in a field F, andletK
beafieldinwhichp (x) factors
p (x) =
n∏
(x − r i ) , r i ∈ K.
i=1
Then the r i are distinct if and only if p (x) and p ′ (x) are relatively prime over F.
Proof: Suppose first that p ′ (x) andp (x) are relatively prime over F. Since they are not
both zero, there exists polynomials a (x) ,b(x) having coefficients in F such that
a (x) p (x)+b (x) p ′ (x) =1
Now suppose p (x) has a repeated root r. TheninK [x] ,
p (x) =(x − r) 2 g (x)
and so p ′ (x) =2(x − r) g (x)+(x − r) 2 g ′ (x). Then in K [x] ,
(
)
a (x)(x − r) 2 g (x)+b (x) 2(x − r) g (x)+(x − r) 2 g ′ (x) =1
Then letting x = r, it follows that 0 = 1. Hence p (x) has no repeated roots.
Next suppose there are no repeated roots of p (x). Then
p ′ (x) =
n∑ ∏
(x − r j )
i=1 j≠i