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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 п¬Ѓnd useful. Again the proof uses ZornвЂ™s Lemma which we will state. First we
need some deп¬Ѓnitions.
Definition 3.46. A partially ordered set (X, ) consists of a set X and a binary relation
such that whenever x, y, z в€€ X,
вЂў x x;
вЂў if x y and y z then x z;
вЂў if x y and y x then x = y.
(X, ) is totally ordered if for every pair x, y в€€ X, at least one of x y or y x is true.
Definition 3.47. Let (X, ) be a partially ordered set and Y вЉ† X. An element y в€€ X
is an upper bound for Y if for every y в€€ Y , y y. An upper bound for X itself is a maximal
element of X.
Axiom 3.48 (ZornвЂ™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 П•0 : L в€’в†’ K is a monomorphism п¬Ѓxing the elements of K.
3.4. ALGEBRAIC CLOSURES 39

Then there is an extension of П•0 to a monomorphism П• : M в€’в†’ K.
8@ K
Г‘Г‘
П•
Г‘Г‘
Г‘Г‘
Г‘Г‘
Г‘Г‘ П•0
M
Г‘Г‘
Г‘Г‘
Г‘
Г‘Г‘
L

= /K
K
в™
Proof. We consider the set X consisting of all pairs (F, Оё), where F/L M/L and
в™Ґв™¦
в™Ј
Оё : F в€’в†’ K extends П•0 . We order X using the relation for which (F1 , Оё1 ) (F2 , Оё2 ) whenever
F1 F2 and Оё2 extends Оё1 . Then (X, ) is a partially ordered set.
Suppose that Y вЉ† X is a totally ordered subset. Let
F= F.
(F,Оё)в€€Y

Then F /L M/L. Also there is a function Оё : F в€’в†’ K deп¬Ѓned by
Оё(u) = Оё(u)
whenever u в€€ F for (F, Оё) в€€ Y . It is routine to check that if u в€€ F for (F , Оё ) в€€ Y then
Оё (u) = Оё(u),
so Оё is well-deп¬Ѓned. Then for every (F, Оё) в€€ Y we have (F, Оё) (F , Оё), so (F , Оё) is an upper
bound for Y . By ZornвЂ™s Lemma there must be a maximal element of X, (M0 , Оё0 ).
Suppose that M0 = M , so there is an element u в€€ M for which u в€€ M0 . Since M is algebraic
/
over K it is also algebraic over M0 , hence u is algebraic over M0 . If
minpolyM0 ,u (X) = a0 + В· В· В· + anв€’1 X nв€’1 + X n ,
then the polynomial
f (X) = Оё0 (a0 ) + В· В· В· + Оё0 (anв€’1 )X nв€’1 + X n в€€ (Оё0 M0 )[X]
is also irreducible and so it has a root v in K (which is also an algebraic closure of Оё0 M0
K). The Homomorphism Extension Property 1.21 of the polynomial ring M0 [X] applied to
the monomorphism Оё0 : M0 в€’в†’ K yields a homomorphism Оё0 : M0 [X] в€’в†’ K extending Оё0
and for which Оё0 (u) = v. This factors through the quotient ring M0 [X]/(minpolyM0 ,u (X)) to
give a monomorphism Оё0 : M0 (u) в€’в†’ K extending Оё0 . But then (M0 , Оё0 ) (M0 (u), Оё0 ) and
(M0 , Оё0 ) = (M0 (u), Оё0 ), contradicting the maximality of (M0 , Оё0 ). Hence M0 = M and so we
can take П• = Оё0 .
Example 3.50. Let u в€€ K and suppose that p(X) = minpolyK,u (X) в€€ K[X]. Then for any
other root of p(X), v в€€ K say, there is a monomorphism П•v : K(u) в€’в†’ K with П•v (u) = v. This
extends to a monomorphism П• : K в€’в†’ K.
Definition 3.51. Let u, v в€€ K. Then v is conjugate to u over K or is a conjugate of u over
K if there is a monomorphism П• : K в€’в†’ K for which v = П•(u).
Lemma 3.52. If u, v в€€ K, then v is conjugate to u over K if and only if minpolyK,u (v) = 0.

Proof. Suppose that v = П•(u) for some П• в€€ MonoK (K, K). If
minpolyK,u (X) = a0 + a1 X + В· В· В· + adв€’1 X dв€’1 + X d ,
40 3. ALGEBRAIC EXTENSIONS OF FIELDS

then
a0 + a1 u + В· В· В· + adв€’1 udв€’1 + ud = 0
and so
a0 + a1 v + В· В· В· + adв€’1 v dв€’1 + v d = П•(a0 + a1 u + В· В· В· + adв€’1 udв€’1 + ud ) = 0.
The converse follows from Example 3.50.

3.5. Multiplicity of roots and separability
Let K be a п¬Ѓeld. Suppose that f (X) в€€ K[X] and u в€€ K is a root of f (X), i.e., f (u) = 0.
Then we can factor f (X) as f (X) = (X в€’ u)f1 (X) for some f1 (X) в€€ 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 в€‚ : K[X] в€’в†’ K[X] given by
в€‚(f (X)) = f (X) = a1 + 2a2 X + В· В· В· + dad X dв€’1 ,
where f (X) = a0 + a1 X + a2 X 2 + В· В· В· + ad X d with aj в€€ K.
Proposition 3.54. The formal derivative в€‚ : K[X] в€’в†’ K[X] has the following properties.
(i) в€‚ is K-linear.
(ii) в€‚ is a derivation, i.e., for f (X), g(X) в€€ K[X],
в€‚(f (X)g(X)) = в€‚(f (X))g(X) + f (X)в€‚(g(X)).
(iii) If char K = 0, then ker в€‚ = K and в€‚ is surjective.
(iv) If char K = p > 0, then
ker в€‚ = {h(X p ) : h(X) в€€ K[X]}
and im в€‚ is spanned by the monomials X k with p (k + 1).
Proof. (i) This is routine.
(ii) By K-linearity, it suп¬ѓces to verify this for the case where f (X) = X r and g(X) = X s with
r, s 0. But then
в€‚(X r+s ) = (r + s)X r+sв€’1 = rX rв€’1 X s + sX r X sв€’1 = в€‚(X r )X s + X r в€‚(X s ).
(iii) If f (X) = a0 + a1 X + a2 X 2 + В· В· В· + ad X d then
в€‚(f (X)) = 0 в‡ђв‡’ a1 = 2a2 = В· В· В· = dad = 0.
So в€‚(f (X)) = 0 if and only if f (X) = a0 в€€ K. It is also clear that every polynomial g(X) в€€ K[X]
has the form g(X) = в€‚(f (X) where f (X) is an anti-derivative of g(X).
(iv) For a monomial X m , в€‚(X m ) = mX mв€’1 and this is zero if and only if p | m. Using this we
see that
в€‚(a0 + a1 X + a2 X 2 + В· В· В· + ad X d ) = 0 в‡ђв‡’ am = 0 whenever p m.
Also, im в€‚ is spanned by the monomials X k for which в€‚(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) в€€ K[X] have a root u в€€ 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 в€’ u)f1 (X). Then
f (X) = f1 (X) + (X в€’ 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 в€’ 1 in C is simple.
Example 3.58. For n
Solution. We have f (X) = в€‚(X n в€’ 1) = nX nв€’1 , so for any root О¶ of f (X),
f (О¶) = nО¶ nв€’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 п¬Ѓnd 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 в€’ 1 in L is multiple.
Solution. We have f (X) = в€‚(X p в€’ 1) = pX pв€’1 = 0, so if О¶ is any root of f (X) then
f (О¶) = 0. Later we will see that 1 is the only root of X p в€’ 1.
Definition 3.61. An irreducible polynomial p(X) в€€ 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 в€€ 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 в€’ u)m q(X) for some q(X) в€€ L[X].
Proposition 3.62. Let K be a п¬Ѓeld and let K be an algebraic closure. If the irreducible
polynomial p(X) в€€ K[X] has distinct roots u1 , . . . , uk в€€ K, then the multiplicities of the uj are
equal. Hence in K[X],
p(X) = c(X в€’ u1 )m В· В· В· (X в€’ uk )m ,
where c в€€ K and m 1.
Proof. Let u в€€ K be a root of p(X) and suppose that it has multiplicity m, so we can
write p(X) = (X в€’ u)m p1 (X) where p1 (X) в€€ K(u)[X] and p1 (u) = 0.
Now let v в€€ K be any other root of p(X). By Proposition 3.34, there is a monomorphism
П•v : K(u) в€’в†’ K for which П•v (u) = v. When p(X) is viewed as an element of K(u)[X], the
coeп¬ѓcients of p(X) are п¬Ѓxed by П•v . Then
П•v ((X в€’ u)m p1 (X)) = (X в€’ u)m p1 (X),
and so
(X в€’ v)m p1 (X) = (X в€’ u)m p1 (X),
where p1 (X) в€€ K[X] is obtained applying П•v to the coeп¬ѓcients of p1 (X). Now by Corollary 1.33,
(X в€’ v)m must divide p1 (X) in K[X], and therefore the multiplicity of v must be at least m.
Interchanging the rЛ†les of u and v we п¬Ѓnd that the multiplicities of u and v are in fact equal.
o
Corollary 3.63. Let K be a п¬Ѓeld and let K be an algebraic closure. If the irreducible
polynomial p(X) в€€ K[X] has distinct roots u1 , . . . , uk в€€ K which are all simple then in K[X],
p(X) = c(X в€’ u1 ) В· В· В· (X в€’ uk ),
where c в€€ K and k = deg p(X).
Corollary 3.64. Let K be a п¬Ѓeld and let u в€€ 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 в€€ L in an extension L/K is separable if its minimal
polynomial minpolyK,u (X) в€€ 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 п¬Ѓeld of characteristic 0 is separable by
Corollary 3.57.
Definition 3.68. Let L/K be a п¬Ѓnite extension. The separable degree of L over K is
(L : K) = | MonoK (L, K)|.
Lemma 3.69. For a п¬Ѓnite 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 п¬Ѓnite extension L/K can be built up from a succession of simple extensions
(3.1) K(u1 )/K, K(u1 , u2 )/K(u1 ), В· В· В· , L = K(u1 , . . . , uk )/K(u1 , . . . , ukв€’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 п¬Ѓnite extensions. Then
(M : K) = (M : L)(L : K).
Proof. For О± в€€ MonoK (M, K) let О±L в€€ 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 в€’в†’ K, so every element ОІ в€€ MonoK (L, K) has the form ОІ = О±L for some О± в€€
MonoK (M, K); since (L : K) = | MonoK (L, K)|, we need to show that the number of such О± is
always (M : L) = | MonoK (M, K)|.
So given ОІ в€€ MonoK (L, K), choose any extension to a monomorphism ОІ : K в€’в†’ K; by
Proposition 3.39, ОІ is an automorphism. Of course, restricting to M K we obtain a monomor-
phism M в€’в†’ K. Now for any extension ОІ : M в€’в†’ K of ОІ we can form the composition
ОІ в€’1 в—¦ ОІ : M в€’в†’ K; notice that if u в€€ L, then
ОІ в€’1 в—¦ ОІ (u) = ОІ в€’1 (ОІ(u)) = u,
hence ОІ в€’1 в—¦ ОІ в€€ MonoL (M, K). Conversely, each Оі в€€ MonoL (M, K) gives rise to a monomor-
phism ОІ в—¦ Оі : M в€’в†’ K which extends ОІ. In eп¬Ђect, this shows that there is a bijection
extensions of ОІ to monomorphism a M в€’в†’ K в†ђв†’ MonoL (M, K),
so (M : L) = | MonoL (M, K)| agrees with the number of extensions of ОІ to a monomorphism
M в€’в†’ K. Therefore we have the desired formula (M : K) = (M : L)(L : K).
Corollary 3.71. Let L/K be a п¬Ѓnite 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 )] В· В· В· [K(u1 , . . . , uk ) : K(u1 , . . . , ukв€’1 )].
For each k, (K(u1 , . . . , uk ) : K(u1 , . . . , ukв€’1 )) divides [K(u1 , . . . , uk ) : K(u1 , . . . , ukв€’1 )], so the
desired result follows.
Proposition 3.72. Let L/K be a п¬Ѓnite 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 в€€ 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 suп¬ѓces 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) В· В· В· (L : K(u1 , . . . , ukв€’1 )),
[L : K] = [K(u1 ) : K] В· В· В· [L : K(u1 , . . . , ukв€’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 в€€ 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 п¬Ѓnite 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 п¬Ѓnite simple extension L/K, an element u в€€ L is called a primitive
element for the extension if L = K(u).
Theorem 3.75 (Primitive Element Theorem). Let L/K be a п¬Ѓnite separable extension.
Then L has a primitive element.
Proof. The case where K is a п¬Ѓnite п¬Ѓeld will be dealt with in Proposition 5.16. So we will
assume that K is inп¬Ѓnite.
Since L is built up from a sequence of simple extensions it suп¬ѓces to consider the case
L = K(u, v). Let p(X), q(X) в€€ 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 inп¬Ѓnite, we can choose an element t в€€ K for which
u в€’ ui
t=
vj в€’ v
whenever j = 1. Then taking w = u + tv в€€ L, we п¬Ѓnd that w = ui + tvj whenever j = 1. Deп¬Ѓne
the polynomial (of degree r)
h(X) = p(w в€’ tX) в€€ K(w)[X] вЉ† 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) в€€ 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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