(ii) Every non-constant polynomial in (L)alg [X] has a root in (L) and by Proposition 3.16, all

of its roots are in fact algebraic over K since (L)alg is. Hence these roots lie in (L)alg , which

shows that it is algebraically closed.

For example, we have Q = Calg and R = C.

There is a stronger result than Theorem 3.44(ii), the Monomorphism Extension Theorem,

which we will ¬nd useful. Again the proof uses Zorn™s Lemma which we will state. First we

need some de¬nitions.

Definition 3.46. A partially ordered set (X, ) consists of a set X and a binary relation

such that whenever x, y, z ∈ X,

• x x;

• if x y and y z then x z;

• if x y and y x then x = y.

(X, ) is totally ordered if for every pair x, y ∈ X, at least one of x y or y x is true.

Definition 3.47. Let (X, ) be a partially ordered set and Y ⊆ X. An element y ∈ X

is an upper bound for Y if for every y ∈ Y , y y. An upper bound for X itself is a maximal

element of X.

Axiom 3.48 (Zorn™s Lemma). Let (X, ) be a partially ordered set in which every totally

ordered subset has an upper bound. Then X has a maximal element.

Theorem 3.49 (Monomorphism Extension Theorem). Let M/K be an algebraic extension

and L/K M/K. Suppose that •0 : L ’’ K is a monomorphism ¬xing the elements of K.

3.4. ALGEBRAIC CLOSURES 39

Then there is an extension of •0 to a monomorphism • : M ’’ K.

8@ K

‘‘

•

‘‘

‘‘

‘‘

‘‘ •0

M

‘‘

‘‘

‘

‘‘

L

= /K

K

™

Proof. We consider the set X consisting of all pairs (F, θ), where F/L M/L and

™™¦

™

θ : F ’’ K extends •0 . We order X using the relation for which (F1 , θ1 ) (F2 , θ2 ) whenever

F1 F2 and θ2 extends θ1 . Then (X, ) is a partially ordered set.

Suppose that Y ⊆ X is a totally ordered subset. Let

F= F.

(F,θ)∈Y

Then F /L M/L. Also there is a function θ : F ’’ K de¬ned by

θ(u) = θ(u)

whenever u ∈ F for (F, θ) ∈ Y . It is routine to check that if u ∈ F for (F , θ ) ∈ Y then

θ (u) = θ(u),

so θ is well-de¬ned. Then for every (F, θ) ∈ Y we have (F, θ) (F , θ), so (F , θ) is an upper

bound for Y . By Zorn™s Lemma there must be a maximal element of X, (M0 , θ0 ).

Suppose that M0 = M , so there is an element u ∈ M for which u ∈ M0 . Since M is algebraic

/

over K it is also algebraic over M0 , hence u is algebraic over M0 . If

minpolyM0 ,u (X) = a0 + · · · + an’1 X n’1 + X n ,

then the polynomial

f (X) = θ0 (a0 ) + · · · + θ0 (an’1 )X n’1 + X n ∈ (θ0 M0 )[X]

is also irreducible and so it has a root v in K (which is also an algebraic closure of θ0 M0

K). The Homomorphism Extension Property 1.21 of the polynomial ring M0 [X] applied to

the monomorphism θ0 : M0 ’’ K yields a homomorphism θ0 : M0 [X] ’’ K extending θ0

and for which θ0 (u) = v. This factors through the quotient ring M0 [X]/(minpolyM0 ,u (X)) to

give a monomorphism θ0 : M0 (u) ’’ K extending θ0 . But then (M0 , θ0 ) (M0 (u), θ0 ) and

(M0 , θ0 ) = (M0 (u), θ0 ), contradicting the maximality of (M0 , θ0 ). Hence M0 = M and so we

can take • = θ0 .

Example 3.50. Let u ∈ K and suppose that p(X) = minpolyK,u (X) ∈ K[X]. Then for any

other root of p(X), v ∈ K say, there is a monomorphism •v : K(u) ’’ K with •v (u) = v. This

extends to a monomorphism • : K ’’ K.

Definition 3.51. Let u, v ∈ K. Then v is conjugate to u over K or is a conjugate of u over

K if there is a monomorphism • : K ’’ K for which v = •(u).

Lemma 3.52. If u, v ∈ K, then v is conjugate to u over K if and only if minpolyK,u (v) = 0.

Proof. Suppose that v = •(u) for some • ∈ MonoK (K, K). If

minpolyK,u (X) = a0 + a1 X + · · · + ad’1 X d’1 + X d ,

40 3. ALGEBRAIC EXTENSIONS OF FIELDS

then

a0 + a1 u + · · · + ad’1 ud’1 + ud = 0

and so

a0 + a1 v + · · · + ad’1 v d’1 + v d = •(a0 + a1 u + · · · + ad’1 ud’1 + ud ) = 0.

The converse follows from Example 3.50.

3.5. Multiplicity of roots and separability

Let K be a ¬eld. Suppose that f (X) ∈ K[X] and u ∈ K is a root of f (X), i.e., f (u) = 0.

Then we can factor f (X) as f (X) = (X ’ u)f1 (X) for some f1 (X) ∈ K[X].

Definition 3.53. If f1 (u) = 0 then u is a multiple or repeated root of f (X). If f1 (u) = 0

then u is a simple root of f (X).

We need to understand more clearly when an irreducible polynomial has a multiple root

since this turns out to be important in what follows. Consider the formal derivative on K[X],

i.e., the function ‚ : K[X] ’’ K[X] given by

‚(f (X)) = f (X) = a1 + 2a2 X + · · · + dad X d’1 ,

where f (X) = a0 + a1 X + a2 X 2 + · · · + ad X d with aj ∈ K.

Proposition 3.54. The formal derivative ‚ : K[X] ’’ K[X] has the following properties.

(i) ‚ is K-linear.

(ii) ‚ is a derivation, i.e., for f (X), g(X) ∈ K[X],

‚(f (X)g(X)) = ‚(f (X))g(X) + f (X)‚(g(X)).

(iii) If char K = 0, then ker ‚ = K and ‚ is surjective.

(iv) If char K = p > 0, then

ker ‚ = {h(X p ) : h(X) ∈ K[X]}

and im ‚ is spanned by the monomials X k with p (k + 1).

Proof. (i) This is routine.

(ii) By K-linearity, it su¬ces to verify this for the case where f (X) = X r and g(X) = X s with

r, s 0. But then

‚(X r+s ) = (r + s)X r+s’1 = rX r’1 X s + sX r X s’1 = ‚(X r )X s + X r ‚(X s ).

(iii) If f (X) = a0 + a1 X + a2 X 2 + · · · + ad X d then

‚(f (X)) = 0 ⇐’ a1 = 2a2 = · · · = dad = 0.

So ‚(f (X)) = 0 if and only if f (X) = a0 ∈ K. It is also clear that every polynomial g(X) ∈ K[X]

has the form g(X) = ‚(f (X) where f (X) is an anti-derivative of g(X).

(iv) For a monomial X m , ‚(X m ) = mX m’1 and this is zero if and only if p | m. Using this we

see that

‚(a0 + a1 X + a2 X 2 + · · · + ad X d ) = 0 ⇐’ am = 0 whenever p m.

Also, im ‚ is spanned by the monomials X k for which ‚(X k+1 ) = 0, which are the ones with

p (k + 1).

We now apply the formal derivative to detect multiple roots.

Proposition 3.55. Let f (X) ∈ K[X] have a root u ∈ L where L/K is an extension. Then

u is a multiple root of f (X) if and only if f (X) and f (X) have a common factor of positive

degree in K[X] which vanishes at u.

Proof. Working in L[X], let f (X) = (X ’ u)f1 (X). Then

f (X) = f1 (X) + (X ’ u)f1 (X),

so f (u) = f1 (u). Hence u is a multiple root if and only if f (X) and f (X) have a common

factor in L[X] and hence in K[X] which vanishes at u.

3.5. MULTIPLICITY OF ROOTS AND SEPARABILITY 41

Corollary 3.56. If f (X) is irreducible in K[X] then a root u is a multiple root if and only

if f (X) = 0. In particular, this can only happen if char K > 0.

Corollary 3.57. If char K = 0 and f (X) is irreducible in K[X], then every root of f (X)

is simple.

1, show that each of the roots of f (X) = X n ’ 1 in C is simple.

Example 3.58. For n

Solution. We have f (X) = ‚(X n ’ 1) = nX n’1 , so for any root ζ of f (X),

f (ζ) = nζ n’1 = 0.

Example 3.59. Show that 2i is a multiple root of f (X) = X 4 + 8X 2 + 16.

Solution. We have f (X) = 4X 3 +16X. Using Long Division and the Euclidean Algorithm

we ¬nd that gcd(f (X), f (X)) = X 2 + 4, where 2i is also a root of X 2 + 4. Hence 2i is a multiple

root of f (X). In fact, X 4 + 8X 2 + 16 = (X 2 + 4)2 , so this is obvious.

Example 3.60. Let p > 0 be a prime and suppose that L/Fp is an extension. Show each of

that the roots of f (X) = X p ’ 1 in L is multiple.

Solution. We have f (X) = ‚(X p ’ 1) = pX p’1 = 0, so if ζ is any root of f (X) then

f (ζ) = 0. Later we will see that 1 is the only root of X p ’ 1.

Definition 3.61. An irreducible polynomial p(X) ∈ K[X] is separable over K if every root

of p(X) in an extension L/K is simple. By Corollary 3.56, this is equivalent to requiring that

p (X) = 0. If u ∈ L is a multiple root of p(X), then the multiplicity of u in p(X) is the maximum

m such that p(X) = (X ’ u)m q(X) for some q(X) ∈ L[X].

Proposition 3.62. Let K be a ¬eld and let K be an algebraic closure. If the irreducible

polynomial p(X) ∈ K[X] has distinct roots u1 , . . . , uk ∈ K, then the multiplicities of the uj are

equal. Hence in K[X],

p(X) = c(X ’ u1 )m · · · (X ’ uk )m ,

where c ∈ K and m 1.

Proof. Let u ∈ K be a root of p(X) and suppose that it has multiplicity m, so we can

write p(X) = (X ’ u)m p1 (X) where p1 (X) ∈ K(u)[X] and p1 (u) = 0.

Now let v ∈ K be any other root of p(X). By Proposition 3.34, there is a monomorphism

•v : K(u) ’’ K for which •v (u) = v. When p(X) is viewed as an element of K(u)[X], the

coe¬cients of p(X) are ¬xed by •v . Then

•v ((X ’ u)m p1 (X)) = (X ’ u)m p1 (X),

and so

(X ’ v)m p1 (X) = (X ’ u)m p1 (X),

where p1 (X) ∈ K[X] is obtained applying •v to the coe¬cients of p1 (X). Now by Corollary 1.33,

(X ’ v)m must divide p1 (X) in K[X], and therefore the multiplicity of v must be at least m.

Interchanging the rˆles of u and v we ¬nd that the multiplicities of u and v are in fact equal.

o

Corollary 3.63. Let K be a ¬eld and let K be an algebraic closure. If the irreducible

polynomial p(X) ∈ K[X] has distinct roots u1 , . . . , uk ∈ K which are all simple then in K[X],

p(X) = c(X ’ u1 ) · · · (X ’ uk ),

where c ∈ K and k = deg p(X).

Corollary 3.64. Let K be a ¬eld and let u ∈ K. Then the number of distinct conjugates

of u is

deg minpolyK,u (X)

,

m

where m is the multiplicity of u in minpolyK,u (X).

42 3. ALGEBRAIC EXTENSIONS OF FIELDS

Definition 3.65. An algebraic element u ∈ L in an extension L/K is separable if its minimal

polynomial minpolyK,u (X) ∈ K[X] is separable.

Definition 3.66. An algebraic extension L/K is called separable if every element of L is

separable over K.

Example 3.67. An algebraic extension L/K of a ¬eld of characteristic 0 is separable by

Corollary 3.57.

Definition 3.68. Let L/K be a ¬nite extension. The separable degree of L over K is

(L : K) = | MonoK (L, K)|.

Lemma 3.69. For a ¬nite simple extension K(u)/K,

(K(u) : K) = | Roots(minpolyK,u , K)|.

If K(u)/K is separable, then [K(u) : K] = (K(u) : K).

Proof. This follows from Proposition 3.34 applied to the case L = K.

Any ¬nite extension L/K can be built up from a succession of simple extensions

(3.1) K(u1 )/K, K(u1 , u2 )/K(u1 ), · · · , L = K(u1 , . . . , uk )/K(u1 , . . . , uk’1 ).

So we can use the following to compute (L : K) = (K(u1 , . . . , uk ) : K).

Proposition 3.70. Let L/K and M/L be ¬nite extensions. Then

(M : K) = (M : L)(L : K).

Proof. For ± ∈ MonoK (M, K) let ±L ∈ MonoK (L, K) be its restriction to L. By the

Monomorphism Extension Theorem 3.49, each element of MonoK (L, K) extends to a monomor-

phism M ’’ K, so every element β ∈ MonoK (L, K) has the form β = ±L for some ± ∈

MonoK (M, K); since (L : K) = | MonoK (L, K)|, we need to show that the number of such ± is

always (M : L) = | MonoK (M, K)|.

So given β ∈ MonoK (L, K), choose any extension to a monomorphism β : K ’’ K; by

Proposition 3.39, β is an automorphism. Of course, restricting to M K we obtain a monomor-

phism M ’’ K. Now for any extension β : M ’’ K of β we can form the composition

β ’1 —¦ β : M ’’ K; notice that if u ∈ L, then

β ’1 —¦ β (u) = β ’1 (β(u)) = u,

hence β ’1 —¦ β ∈ MonoL (M, K). Conversely, each γ ∈ MonoL (M, K) gives rise to a monomor-

phism β —¦ γ : M ’’ K which extends β. In e¬ect, this shows that there is a bijection

extensions of β to monomorphism a M ’’ K ←’ MonoL (M, K),

so (M : L) = | MonoL (M, K)| agrees with the number of extensions of β to a monomorphism

M ’’ K. Therefore we have the desired formula (M : K) = (M : L)(L : K).

Corollary 3.71. Let L/K be a ¬nite extension. Then (L : K) | [L : K].

Proof. If L/K is a simple extension then by Propositions 3.62 and 3.34 we know that this

is true. The general result follows by building up L/K as a sequence of simple extensions as

in (3.1) and then using Theorem 2.6(ii) which gives

[L : K] = [K(u1 ) : K] [K(u1 , u2 ) : K(u1 )] · · · [K(u1 , . . . , uk ) : K(u1 , . . . , uk’1 )].

For each k, (K(u1 , . . . , uk ) : K(u1 , . . . , uk’1 )) divides [K(u1 , . . . , uk ) : K(u1 , . . . , uk’1 )], so the

desired result follows.

Proposition 3.72. Let L/K be a ¬nite extension. Then L/K is separable if and only if

(L : K) = [L : K].

3.6. THE PRIMITIVE ELEMENT THEOREM 43

Proof. Suppose that L/K is separable. If K E L, then for any u ∈ L, u is alge-

braic over E, and in the polynomial ring E[X] we have minpolyE,u (X) | minpolyK,u (X). As

minpolyK,u (X) is separable, so is minpolyE,u (X), and therefore L/E is separable. Clearly E/K

is also separable. We have (L : K) = (L : E) (E : K) and [L : K] = [L : E] [E : K], so to

verify that (L : K) = [L : K] it su¬ces to show that (L : E) = [L : E] and (E : K) = [E : K].

Expressing L/K in terms of a sequence of simple extensions as in (3.1), we have

(L : K) = (K(u1 ) : K) · · · (L : K(u1 , . . . , uk’1 )),

[L : K] = [K(u1 ) : K] · · · [L : K(u1 , . . . , uk’1 )].

Now we can apply Lemma 3.69 to each of these intermediate separable simple extensions to

obtain (L : K) = [L : K].

For the converse, suppose that (L : K) = [L : K]. We must show that for each u ∈ L, u is

separable. For the extensions K(u)/K and L/K(u) we have (L : K) = (L : K(u)) (K(u) : K)

and [L : K] = [L : K(u)] [K(u) : K]. By Corollary 3.71, there are some positive integers r, s for

which [L : K(u)] = r(L : K(u)) and [K(u) : K] = s(K(u) : K). Hence

(L : K(u))(K(u) : K) = rs(L : K(u))(K(u) : K),

which can only happen if r = s = 1. Thus (K(u) : K) = [K(u) : K] and so u is separable.

Proposition 3.73. Let L/K and M/L be ¬nite extensions. Then M/K is separable if and

only if L/K and M/L are separable.

Proof. If M/K is separable then [M : K] = (M : K) and so by Proposition 3.70,

[M : L][L : K] = (M : L)(L : K).

This can only happen if [M : L] = (M : L) and [L : K] = (L : K), since (M : L) [M : L] and

(L : K) [L : K]. By Proposition 3.72 this implies that L/K and M/L are separable.

Conversely, if L/K and M/L are separable then [M : L] = (M : L) and [L : K] = (L : K),

hence

[M : K] = [M : L][L : K] = (M : L)(L : K) = (M : K).

Therefore M/K is separable.

3.6. The Primitive Element Theorem

Definition 3.74. For a ¬nite simple extension L/K, an element u ∈ L is called a primitive

element for the extension if L = K(u).

Theorem 3.75 (Primitive Element Theorem). Let L/K be a ¬nite separable extension.

Then L has a primitive element.

Proof. The case where K is a ¬nite ¬eld will be dealt with in Proposition 5.16. So we will

assume that K is in¬nite.

Since L is built up from a sequence of simple extensions it su¬ces to consider the case

L = K(u, v). Let p(X), q(X) ∈ K[X] be the minimal polynomials of u and v over K. Suppose

that the distinct roots of p(X) in K are u = u1 , . . . , ur , while the distinct roots of q(X) are

v = v1 , . . . , vs . By the separability assumption, r = deg p(X) and s = deg q(X).

Since K is in¬nite, we can choose an element t ∈ K for which

u ’ ui

t=

vj ’ v

whenever j = 1. Then taking w = u + tv ∈ L, we ¬nd that w = ui + tvj whenever j = 1. De¬ne

the polynomial (of degree r)

h(X) = p(w ’ tX) ∈ K(w)[X] ⊆ L[X].

Then h(v) = p(u) = 0, but h(vj ) = p(ui ) = 0 for any j = 1 by construction of t, so none of the

other vj is a zero of h(X).

Now since the polynomials h(X), q(X) ∈ K(w)[X] have exactly one common root in K,

namely v, by separability their greatest common divisor in K(w)[X] is a linear polynomial which

44 3. ALGEBRAIC EXTENSIONS OF FIELDS