9.5. A FEW GENERALIZATIONS 239

Note that the field Zp does not have characteristic 0 because p · 1 = 0. Thus not allfields have characteristic 0.

How can you tell if a polynomial has no repeated roots? This is the content of the nexttheorem.

Theorem 9.5.5 Let p(x) be a monic polynomial having coefficients in a field F, and let Kbe a field in which p(x) factors

p(x) =n

∏i=1

(x− ri) , ri ∈K.

Then the ri are distinct if and only if p(x) and p′ (x) are relatively prime over F.

Proof: Suppose first that p′ (x) and p(x) are relatively prime over F. Since they are notboth 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. Then inK [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) = ∑

ni=1 ∏ j ̸=i (x− r j).

p′ (x) cannot be zero in this case because p′ (rn) = ∏n−1j=1 (rn− r j) ̸= 0 because it is the

product of nonzero elements of K. Similarly no term in the sum for p′ (x) can equal zerobecause ∏ j ̸=i (ri− r j) ̸= 0. Then if q(x) is a monic polynomial of degree larger than 1which divides p(x), then the roots of q(x) inK are a subset of {r1, · · · ,rn}. Without loss ofgenerality, suppose these roots of q(x) are {r1, · · · ,rk} , k≤ n−1, since q(x) divides p′ (x)which has degree at most n− 1. Then q(x) = ∏

ki=1 (x− ri) 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) = 1 and so the two are relatively prime. ■

The following lemma says that the usual calculus result holds in case you are lookingat polynomials with coefficients in a field of characteristic 0.

Lemma 9.5.6 Suppose that F has characteristic 0. Then if f ′ (x) = 0, it follows that f (x)is a constant.

Proof: Supposef (x) = anxn +an−1xn−1 + · · ·+a1x+a0

Then0xn +0xn−1 + · · ·+0x+0 = nanxn−1 +an−1 (n−1)xn−2 + · · ·+a1

Therefore, each coefficient on the right is 0. Since the field has characteristic 0 it followsthat each ak = 0 for k ≥ 1. Thus f (x) = a0 ∈ F. ■

If F has characteristic p as in Zp for p prime, this is not true. Indeed, xp− 1 is notconstant but has derivative equal to 0.

Now here is a major result which applies to fields of characteristic 0.