2.4 Separability and the Galois Criterion

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Let f(x) 2 F[x] be an irreducible polynomial. We say that f(x) is separable if

f(x) has no repeated roots in a splitting field. In general, a polynomial (not

necessarily irreducible) is called separable if each is its irreducible factors is

separable.

Next, if F _ K is a field extension, and if _ 2 K, we say that _ is separable

over F if _ is algebraic over F, and if m_,F(x) is a separable polynomial.

Finally we say that the extension F _ K is a separable extension K is

algebraic over F and if every element of K is separable over F.

The following two results relate algebraic Galois extensions and separable

extensions:

Theorem 2.4.1 Let F _ K be an algebraic extension of fields. Then K is

Galois over F if and only if K is the splitting field over F for some set of

separable polynomials in F[x].

Corollary 2.4.1.1 Let K _ F be generated over F by a set of separable

elements. Then K is a separable extension of F.

Proposition 2.4.2 Let F _ K _ E be an algebraic extension with K separable

over F. If _ 2 E is separable over K, then _ is separable over F.

If F _ E is an algebraic extension of fields such that no element of E−F

is separable over F, then we say that E is a purely inseparable extension of

F. If _ 2 E is such that F(_) is a purely inseparable extension of F, we say

that _ is a purely inseparable element over F.

Theorem 2.4.3 Let F _ E be an algebraic extension. Then there exists a

unique maximal subfield Esep _ E such that Esep is separable over F and E

is purely inseparable over Esep.

Given f(x) =

P

aixi 2 F[x], we may define its (formal) derivative by

setting f0(x) =

P

iaixi−1. One has the usual product rule: (f(x)g(x))0 =

f(x)g0(x) + f0(x)g(x).

The following is quite simple.

Lemma 2.4.4 Let f(x) 2 F[x].

56 CHAPTER 2. FIELD AND GALOIS THEORY

(i) If g.c.d.(f(x), f0(x)) = 1, then f(x) has no repeated roots in any splitting

field. (Note: this is stronger than being separable.)

(ii) If f(x) is irreducible, then f(x) is separable if and only if f0(x) 6= 0.

(iii) If F has characteristic p > 0, and if f(x) 2 F[x] is irreducible but not

separable, then f(x) = g(xp) for some irreducible g(x) 2 F[x].

Obviously, it follows that if char F = 0 then every polynomial f(x) 2 F[x]

is separable.

Lemma 2.4.5 Let F _ K be an algebraic extension of fields where F has

characteristic p > 0. If _ 2 K, then _ is separable over F if and only if

F(_) = F(_p).

Proposition 2.4.6 Let F _ K be an algebraic extension, where F is a field

of characteristic p > 0. Let _ 2 K be an inseparable element over F. The

following are equivalent:

(i) _ is purely inseparable over F.

(ii) The minimal polynomial has the form m_(x) = xpe

−a 2 F[x], for some

positive integer e and for some a 2 F.

(iii) The minimal polynomial m_(x) 2 F[x] has a unique root in any splitting

field, viz., _.

Let F be a field of characteristic p > 0. We may define the p-th power

map (·)p : F ! F, _ 7! _p. Clearly (·)p is a monomorphism of F into itself.

We say that the field F is perfect if one of the following holds:

(i) F has characteristic 0, or

(ii) F has characteristic p > 0 and (·)p : F ! F is an automorphism of F.

Corollary 2.4.6.1 Let F be a perfect field. Then any algebraic extension

of F is a separable extension.

We can apply the above discussion to extensions of finite fields. Note

first that if F is a finite field, it obviously has positive characteristic, say p.

Thus F is a finite dimensional vector space over the field Fp ( alternatively

denoted Z/(p), the integers, modulo p). From this it follows immediately

2.4. SEPARABILITY AND THE GALOIS CRITERION 57

that if n is the dimension of F over Fp, then |F| = pn. Note furthermore that

by Corollary 2.1.5.1 of Section 2.1, F× is a cyclic group, and so the elements

of F are precisely the roots of xq − x, where q = pn. In other words,

F is a splitting field over Fp for the polynomial xq − x. From this we

infer immediately the following.

Proposition 2.4.7 Two finite fields F1 and F2 are isomorphic if and only

if they have the same order.

The only issue left unsettled by the above is whether for any prime p

and any integer n, there really exists a finite field of order pn. The answer

is yes, and is very easily demonstrated. Indeed, let q = pn, and let f(x) =

xq − x 2 Fp[x]. By Lemma 2.4.4, part (i) f(x) is separable. Thus if F _ Fp

is a splitting field, then it’s easy to see that F consists wholly of the q roots

of f(x). Thus:

Proposition 2.4.8 For any prime p, and any integer n, there exists a field

of order pn.

Thus, for any prime power q = pn there exists a unique (up to isomorphism)

field of order q. We denote such a field simply by Fq.

Finally, we’ll say a few words about Galois groups in this setting. Let

F = Fq be the finite field of order q, and let K = Fqn be an extension of

degree n. Since K is the splitting field over F for the separable polynomial

xqn

− x, we conclude that K is a Galois extension of F.

Define the map

F : K −! K,

_ 7−! _q.

Then F is easily seen to be an F-automorphism of K, often called the Frobenius

automorphism of K. The following is easy to prove.

Theorem 2.4.9 In the notation above, Gal(K/F) is cyclic of order n and

is generated by the Frobenius automorphism F.

Exercises 2.4

58 CHAPTER 2. FIELD AND GALOIS THEORY

1. Let F _ E be an algebraic Galois extension and let f(x) 2 F[x] be a

separable polynomial. Let K _ E be the splitting field for f(x) over

E. Prove that K is Galois over F.

2. Let f(x) = x3 + x2 − 2x− 1 2 Q[x], and let K _ Q be a splitting field

for f(x) over Q (cf. Exercise 3 of Section 2.2). Compute Gal(K/F).

3. Let K = Q(

p

2 +

p

2)

(a) Show that K is a Galois extension of Q.

(b) Show that Gal(K/Q) _= Z4.

4. Let K = Q(

p

2,

p

3, u), where u2 = (9 − 5

p

2)(2 −

p

2).

(a) Show that K is a Galois extension of Q.

(b) Compute Gal(K/Q).

5. Let b be an even positive integer of the form 2m, m odd, and set

a = 1

2 b2. Set K = Q(

p

b −

p

a). Compute Gal(K/Q).

6. Let q be a prime power and let [E : Fq] = n. Let F be the Frobenius

automorphism of E, given by F(_) = _q. Define the norm map

N = NE/Fq : E −! Fq

by setting

N(_) = _ · F(_) · F2(_) · · · Fn−1(_).

Note that N restricts to a mapping

N : E× −! F×q

.

(a) Show that N : E× ! F×qis a group homomorphism.

(b) Show that |kerN| = qn−1

q−1 .

7. Let p be a prime and let r be a positive integer. Prove that there

exists an irreducible polynomial of degree r over Fp.

8. Let p be prime, n a positive integer and set q = pn. If f(x) 2 Fp[x]

is irreducible of degree m, show that f(x)|xq − x. More generally,

show that if f(x) is irreducible of degree n, where n|m, then again,

f(x)|xq − x.

2.4. SEPARABILITY AND THE GALOIS CRITERION 59

9. Let f(x) 2 F[x] and assume that f(xn) is divisible by (x − a)k, where

0 6= a 2 F. Prove that f(xn) is also divisible by (xn − an)k. (Hint:

If F(x) = f(xn) is divisible by (x − a)k, then F0(x) = f0(xn)nxn−1 is

divisible by (x−a)k−1, i.e., f0(xn) is divisible by (x−a)k−1. Continue

in this fashion to argue that that f(k−1)(xn) is divisible by x−a, from

which one concludes that f(x) is divisible by (x − an)k.)

10. This, and the next exercise are devoted to finding a formula for the

number of irreducible polynomials over Fq, where q is a prime power.

The key rests on the so-called Inclusion-Exclusion Principle of combinatorial

theory. To this end, define the function μ : N ! Z by

setting

μ(n) =

n (−1)k if n factors into k distinct primes,

0 if not.

(i) Show that if k, l are integers with k|l, then

X

k|n|l

μ(n) =

n 1 if k = l,

0 if not.

(ii) Now let f, g : N ! R be real-valued functions, and assume that

for each n 2 N, we have

f(n) =

X

k|n

g(k).

Prove that for each n 2 N,

g(n) =

X

k|n

μ(n

k

)f(k).

(Hint: For any m|n we have

f(m) =

X

k|m

g(k).

Next, multiply the above by μ( n

m) and sum over m|n:

X

m|n

μ( n

m

)f(m) =

X

k|m|n

μ( n

m

)g(k) = g(n).)

60 CHAPTER 2. FIELD AND GALOIS THEORY

11. For any integer n, let Dn be the number of irreducible polynomials of

degree n in Fq[x]. Prove that

Dn =

1

n

X

k|n

μ(n

k

)qk.

(Hint: Simply note that, by Exercise 8, qn =

P

k|n k · Dk.)

2.5. BRIEF INTERLUDE: THE KRULL TOPOLOGY 61

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