
Conjugates and Normal Extensions
Let F be a field, F its algebraic closure, F. Then is algebraic over F. Let f = Irr_{F}(,X). Then
f = (X  _{1})(X  _{2})...(X  _{n}), _{i} F, _{1} = .
Definition 1: The elements _{1},...,_{n} are called the conjugates of over F.
Example 1: If F, then Irr_{F}(,X) = X  and the only conjugate of over F is itself.
Example 2: Let F = Q, = . Then Irr_{Q}(,X) = X^{2}  5, and the conjugates of over Q are and .
Example 3: Let F = Q, = (1 + i)/2. Then Irr_{Q}(,X) = X^{2} + X + 1, so that the conjugates of over Q are (1 + i)/2, (1  i)/2.
Example 4: Let F = Q, = , a primitive nth root of 1. Then Irr_{Q}(,X) = _{n}(X), the nth cyclotomic polynomial. Therefore, the conjugates of over Q are ^{a} [0 < a < n  1, (a,n) = 1].
Example 5: Let F = Q, = . Then Irr_{Q}(,X) = X^{3}  2. Therefore, the conjugates of over Q are , , ^{2}, where = (1 + i)/2 is a primitive cube root of 1.
Let E be a finite extension of F. Let us use the notion of conjugate elements to determine all the Fisomorphisms of E into F. Since E is a finite extension of F, E is an algebraic extension of F and there exist _{1},...,_{n} E such that E = F(_{1},...,_{n}). By Corollary 4 of our restrictive assumption, there exists E such that E = F(). Moreover, by Theorem 3 of the section on algebraic numbers, every element x of E can be written uniquely in the form
x = a _{0} + a _{1} + ... + a _{n1}^{n1}, a _{i} F, n = deg(E/F).
Therefore, if is an Fisomorphism of E into F, then
(1)
(x) = (a _{0}) + (a _{1}) ( ) + ... + (a _{n1}) ( ) ^{n1}
= a _{0} + a _{1}( ) + ... + a _{n1}( ) ^{n1}.
Thus, is completely determined once () is specified. We assert that () is a conjugate of over F. Let f = b_{0} + b_{1}X + ... + b_{n1}X^{n1} + X^{n} = Irr_{F}(,X) Then
0 = b _{0} + b _{1} + ... + b _{n1}^{n1} + ^{n}
0 = (0) = (b _{0} + b _{1} + ... + b _{n1}^{n1} + ^{n})
= (b _{0}) + (b _{1}) ( ) + ... + (b _{n1}) ( ) ^{n1} + ( ) ^{n}
= b _{0} + b _{1}( ) + ... + b _{n1}( ) ^{n1} + ( ) ^{n}
= f( ( )).
Therefore, () is a zero of f, and hence is a conjugate of over F. We may summarize our findings in the following proposition.
Proposition 2: Let E be an extension of F of degree n and let E be such that E = F(). Further, let be an Fisomorphism of E into F. Then has the form (1), where () is a conjugate of over F. In particular, there are at most n Fisomorphisms of E into F.
We will refine Proposition 2 by determining precisely all the Fisomorphisms of E into F. As a first step in this direction, let us establish the following general result.
Theorem 3: Let F be a field, F() an algebraic extension of F of degree n, K an algebraically closed field, f = Irr_{F}(,X). Suppose that an isomorphism : FK is given. If
f = a _{0} + a _{1}X + ... + a _{n1}X ^{n1} + X ^{n} F[X],
let us define
f^{ } = (a _{0}) + (a _{1})X + ... + (a _{n1})X ^{n1} + X ^{n} K[X],
Let be any zero of f^{ } in K. Then there exists an isomorphism : F() K such that (1) the restriction of to F coincides with and (2) () = . In other words, is an extension of to F() which maps to .
Proof: Every element x of F() can be uniquely written in the form
(2)
x = a _{0} + a _{1} + ... + a _{n1}^{n1}, a _{i} F.
Therefore, let us define :F()K by
(3)
(x) = (a _{0}) + (a _{1}) + ... + (a _{n1}) ^{n1}.
It is clear that restricted to F equals and that () = . Thus, it remains to prove that is an isomorphism. Let x be defined by (2) and let
y = b _{1} + b _{1} + ... + b _{n1}^{n1}, b _{i} F.
It is straightforward to verify that (x + y) = (x) + (y). Let us show that (x · y) = (x) · (y). Set g = a_{i}X^{i}, h = b_{j}X^{j}. By the division algorithm in F[X]. there exist polynomials q,r F[X] such that
(4)
g · h = q · f + r, r = d_{o} + d_{1}X + .. + d_{n1}X^{n1}
Since f() = 0, (4) implies that
Fo any polynomial k = e_{0} + e_{1}X + .. + e_{s}X^{s} F[X]. set
k ^{} = (e _{0}) + (e _{1})X + ... + (e _{s})X ^{s} K[X],
Then by the definitions of and the fact that f^{ }() = 0, we have
(x · y) = r ^{}( )
= f^{ }( ) · q ^{}( ) + r ^{}( )
=( f · q + r) ^{}( )
= (g · h) ^{}( )
= g ^{}( ) · h ^{}( )
= (x) · (y).
Thus, is a homomorphism. But since (1) = 1 0, Proposition 2 in the introduction to the theory of fields implies that is an isomorphism.
It is now easy to determine all Fisomorphisms of E into F.
Theorem 4: Let E be an extension of F of degree n, and let E be such that E = F(). Further, let _{1},...,_{n} be the conjugates of over F. Then
(1) _{1},...,_{n} are distinct.
(2) For each i (1 < i < n), the exists an Fisomorphism of E into F such that _{i}() = _{i}. Moreover, _{i} is unique and is given by
(5)
_{i}(b _{0} + b _{1} + ... + b _{n1}^{n1})
= b _{0} + b _{1}_{i} + ... + b _{n1}_{i}^{n1}, b _{j} F.
(3) There are exactly n Fisomorphisms of E into F and these are _{1},...,_{n}.
Proof: (1) _{1},...,_{n} are the zeros of Irr_{f}(,X), Therefore, _{1},...,_{n} are distinct by Theorem 1 of the restrictive assumption.
(2) Let :FF be the isomorphism defined by (x) = x (x F). Then f^{ } = f and Theorem 3 implies that there exists an Fisomorphism _{i} such that _{i}() = _{i}. Moreover, by (1), _{1} is given by (5).
(3) By Proposition 2, the number of Fisomorphisms of E into F is at most n. Since the _{i} are all distinct by part (1), the _{i} are all distinct. Therefore, we have exhibited n different Fisomorphisms of E into F. This proves part (3).
The isomorphisms _{i} (1 < i < n) are called conjugation mappings for the extension E of F. Let us denote the set of all conjugation mappings by G(E/F). For example, if F = Q, E = Q(), then
G(E/F) = { _{1}, _{2}},
where _{1}() = and _{2}() = . If F = Q, E = Q(), then
G(E/F) = { _{1}, _{2}, _{3}},
where _{1}() = , _{2}() = , _{3}() = ^{2}.
Note that the extension F() may not contain all the conjugates of . For example, Q() R, but R. The extensions which are obtained by adjoining to F all of the conjugates of a set of elements of F are a very important class of extensions, called normal extensions. More precisely, let us make the following definition.
Definition 5: Let F be a field. A normal extension of F is an extension of F obtained by adjoining all zeros of a finite set of polynomials {f_{1},...,f_{m}}, f_{j} F[X].
For example, if f F[X] is a nonconstant polynomial, then the splitting field E_{f} of f is a normal extension of F. Since a normal extension of F is gotten by adjoining a finite number of algebraic elements to F, a normal extension is finite and algebraic. Let us derive a number of equivalent conditions for an extension of F to be normal.
Theorem 6: Let F be a field, E a finite algebraic extension of F. The the following conditions are equivalent:
(1) E is the splitting field of a polynomial f F[X].
(2) E is a normal extension of F.
(3) If G(E/F), then (E) = E.
(4) If x E, then all conjugates of x over F lie in E.
Proof: We will prove the theorem by proving the series of implications
(1)(2)(3)(4)(1).
(1)(2). We observed this fact above.
(2)(3). Suppose that E is a normal extension of F and that G(E/F). Then E is obtained from F by adjoining all zeros {_{1},...,_{s}} of a finite collection of polynomials {f_{1},...,f_{m}}, f_{j} F[X]. In particular, if {_{1},...,_{s}}, then all conjugates of over F belong to {_{1},...,_{s}}, and hence to E = F(_{1},...,_{s}). But (_{i}) is one of the conjugates of _{i} over F. Therefore, (_{i}) E. Hence, since is an Fisomorphism, we see that (E) E. Let deg(E/F) = n. Then, since is an Fisomorphism, deg((E)/F) = n. Therefore, since F (E) E, and since deg(E/F) = deg(E/(E)) · deg((E)/F), we see that deg(E/(E)) = 1. Therefore, E = (E) by Proposition 2 of the section on algebraic extensions.
(3)(4). Suppose the G(E/F) implies that (E) = E, and assume that x E. Let x' be a conjugate of x over F. We wish to show that x' E. By Theorem 3, there exists and Fisomorphism :F(x)F such that (x) = x'. By
Theorem 3 of the section on the restrictive assumption, there exists E such that E = F(x)(). Then, by Theorem 3, can be extended to an Fisomorphism :EF. [Map onto any one of the zeros of f^{ }, f = Irr_{F(x)}(,X).] But then (x) = x', so that by our assumption, x' (E) = E.
(4)(1). By Theorem 3 of the section of the restrictive assumption, there exist E such that E = F(). Let _{1},...,_{n} be the conjugates of over F. By (4), _{1},...,_{n} all belong to E, so that E = F(_{1},...,_{n}). Thus, E can be gotten by adjoining to F all zeros of Irr_{F}(,X), so that E is a splitting field of Irr_{F}(,X) over F.
As an immediate consequence of the above result we deduce that Q() is not a normal extension of Q, since the conjugates and ^{2} of over Q are not in Q(), which violates condition (4). Moreover, if _{m} is a primitive mth root of 1, then Q(_{m}) is a normal extension of Q, since it is the splitting field over Q of _{m}(X). [The zeros of _{m}(X) are _{m}^{a}, 0 < a< m  1, (a,m) = 1.]
