An Introduction to Galois Theory

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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)mp1(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
coefficients of p(X) are fixed by �v. Then
�v((X - u)mp1(X)) = (X - u)mp1(X),
and so
(X - v)mp1(X) = (X - u)mp1(X),
where p1(X) " K[X] is obtained applying �v to the coefficients of p1(X). Now by Corollary 1.34,
(X - v)m must divide p1(X) in K[X], and therefore the multiplicity of v must be at least m.
Interchanging the r�les of u and v we find that the multiplicities of u and v are in fact equal.
3.63. Corollary. Let K be a field 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).
3.64. Corollary. Let K be a field 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).
44 3. ALGEBRAIC EXTENSIONS OF FIELDS
3.65. Definition. An algebraic element u " L in an extension L/K is separable if its
minimal polynomial minpolyK,u(X) " K[X] is separable.
3.66. Definition. An algebraic extension L/K is called separable if every element of L is
separable over K.
3.67. Example. An algebraic extension L/K of a field of characteristic 0 is separable by
Corollary 3.57.
3.68. Definition. Let L/K be a finite extension. The separable degree of L over K is
(L : K) = | MonoK(L, K)|.
3.69. Lemma. For a finite 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 finite 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).
3.70. Proposition. Let L/K and M/L be finite 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 � " MonoK(L, K) has the form � = �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) = | MonoL(M, K)|.
So given � " MonoK(L, K), choose any extension to a monomorphism � : K -�! K; by
Proposition 3.39, � is an automorphism. Of course, restricting to M K we obtain a monomor-
phism M -�! K. Now for any extension � : M -�! K of � we can form the composition
�-1 �% � : M -�! K; notice that if u " L, then
�-1 �% � (u) = �-1(�(u)) = u,
hence �-1 �% � " MonoL(M, K). Conversely, each � " MonoL(M, K) gives rise to a monomor-
phism � �% � : M -�! K which extends �. In effect, this shows that there is a bijection
extensions of � to monomorphism a M -�! K �!�! MonoL(M, K),
so (M : L) = | MonoL(M, K)| agrees with the number of extensions of � to a monomorphism
M -�! K. Therefore we have the desired formula (M : K) = (M : L)(L : K).
3.71. Corollary. Let L/K be a finite 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.
3.72. Proposition. Let L/K be a finite extension. Then L/K is separable if and only if
(L : K) = [L : K].
3.6. THE PRIMITIVE ELEMENT THEOREM 45
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 suffices 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.
3.73. Proposition. Let L/K and M/L be finite extensions. Then M/K is separable if and
only if L/K and M/L are separable. [ Pobierz całość w formacie PDF ]

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