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braically closed it is an algebraic closure of K.
(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

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