3.1. Multiple roots.

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Let f, g ∈ F[X]. Even when f and g have no common factor in F[X], you might expect

that they could acquire a common factor in Ω[X] for some Ω ⊃ F. In fact, this doesn’t

happen—gcd’s don’t change when the field is extended.

Proposition 3.1. Let f and g be polynomials in F[X], and let Ω ⊃ F. If r(X) is the

gcd of f and g computed in F[X], then it is also the gcd of f and g in Ω[X]. In particular,

if f and g are monic and irreducible and f _= g, then they do not have a common root in

any extension field of F.

Proof. Let rF (X) and rΩ(X) be the greatest common divisors of f and g in F[X] and

Ω[X] respectively. Certainly rF (X)|rΩ(X) in Ω[X]. The Euclidean algorithm shows that

there are polynomials a and b in F[X] such that

a(X)f(X) + b(X)g(X) = rF (X).

Since rΩ(X) divides f and g in Ω[X], it divides the left-hand side of the equation, and

therefore also the right. Hence rΩ = rF .

For the second statement, note that the hypotheses imply that gcd(f, g) = 1 (in F[X]).

Hence they can’t have a common factor X − α in any extension field.

The proposition allows us to write gcd(f, g), without reference to a field.

Let f ∈ F[X], and let f(X) =

_

(X − αi)mi , αi distinct, be a splitting of f over some

large field Ω ⊃ F. We then say that αi is a root of multiplicity mi. A root of multiplicity

one is said to be simple.

We say that f has multiple roots if it has roots of multiplicity > 1 in some big field Ω.

It then has multiple roots in the subfield of Ω generated by its roots, and because any two

splitting fields are F-isomorphic, this shows that f will have roots of multiplicity > 1 in

every field containing F in which it splits.

If f has multiple factors in F[X], say f =

_

fi(X)mi with some mi > 1, then obviously

it will have multiple roots. If f =

_

fi with the fi distinct monic irreducible polynomials,

then the proposition shows that f can only have multiple roots if one of the fi has multiple

roots. Thus it remains to examine irreducible polynomials for multiple roots.

Example 3.2. Let F be of characteristic p, and assume that F has an element a that

is not a pth-power (e.g., F = Fp(T ); a = T ). Then Xp − a is irreducible in F[X], but

Xp − a = (X − α)p in its splitting field. Thus an irreducible polynomial can have multiple

roots.

We define the derivative f_(X) of a polynomial f(X) =

_

aiXi to be

_

iaiXi−1. When

F = R, this agrees with the usual definition. The usual rules for differentiating sums and

products still hold, but note that the derivative of Xp is zero in characteristic p.

Proposition 3.3. Let f be a (monic) irreducible polynomial in F[X]. The following

statements are equivalent:

FIELDS AND GALOIS THEORY 19

(a) f has at least one multiple root (in a splitting field);

(b) gcd(f, f_) _= 1;

(c) F has characteristic p _= 0 and f(X) = g(Xp), some g ∈ F[X];

(d) all the roots of f are multiple.

Proof. (a) =⇒ (b). Let α be a multiple root of f, and write f = (X−α)mg(X), m> 1,

in some splitting field. Then

f

_

(X) = m(X − α)m−1g(X) + (X − α)mg

_

(X).

Hence f_(α) = 0, and so gcd(f, f_) _= 1.

(b) =⇒ (c). Since f is irreducible and deg(f_) < deg(f),

gcd(f, f_) _= 1 =⇒ f_ = 0 =⇒ f = g(Xp).

(c) =⇒ (d). Suppose f(X) = g(Xp), and let g(X) =

_

(X −ai)mi in some splitting field.

Then

f(X) = g(Xp) =

_

(Xp − ai)mi =

_

(X − αi)pmi

where αp

i = ai (in some big field). Hence every root of f(X) has multiplicity at least p.

(d) =⇒ (a). Every root multiple =⇒ at least one root multiple (I hope).

Definition 3.4. A polynomial f ∈ F[X] is said to be separable if all its irreducible

factors have simple roots.

Note that the preceding discussion shows that f is not separable if and only if

(a) the characteristic of F is p _= 0, and

(b) at least one of the irreducible factors of f is a polynomial in Xp.

A field F is said to be perfect if all polynomials in F[X] are separable.

Proposition 3.5. A field F is perfect if and only if it either

• has characteristic 0, or

• it has characteristic p and F = Fp (i.e., every element of F is a pth power).

Proof. =⇒ : If char F = p and it contains an element a that is not a pth power, then

F[X] contains a nonseparable polynomial, namely, Xp − a.

⇐= : If char F = p and F = Fp, then every polyonomial in Xp is a pth _ power—

aiXp = (

_

biX)p if ai = bp

i—and so can’t be irreducible.

Example 3.6. (a) All finite fields are perfect (because a → ap is an injective homomorphism

F → F, which must be surjective if F is finite). In fact, any field algebraic over Fp is

perfect.

(b) If F0 has characteristic p, then F = F0(X) is not perfect (because X is not a pth

power).

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