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Conjugates and Normal Extensions
Let F be a field, F its algebraic closure, F. Then is algebraic over F. Let f = IrrF( ,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 IrrF( ,X) = X - and the only conjugate of over F is itself.
Example 2: Let F = Q, = . Then IrrQ( ,X) = X2 - 5, and the conjugates of over Q are and - .
Example 3: Let F = Q, = (-1 + i)/2. Then IrrQ( ,X) = X2 + 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 IrrQ( ,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 IrrQ( ,X) = X3 - 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 F-isomorphisms 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 n-1 n-1, a i  F, n = deg(E/F).
Therefore, if is an F-isomorphism of E into F, then
(1)
 (x) =  (a 0) +  (a 1)  (  ) + ... +  (a n-1)  (  ) n-1
= a 0 + a 1 (  ) + ... + a n-1 (  ) n-1.
Thus, is completely determined once ( ) is specified. We assert that ( ) is a conjugate of over F. Let f = b0 + b1X + ... + bn-1Xn-1 + Xn = IrrF( ,X) Then
0 = b 0 + b 1 + ... + b n-1 n-1 + n
 0 =  (0) =  (b 0 + b 1 + ... + b n-1 n-1 + n)
=  (b 0) +  (b 1)  (  ) + ... +  (b n-1)  (  ) n-1 +  (  ) n
= b 0 + b 1 (  ) + ... + b n-1 (  ) n-1 +  (  ) 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 F-isomorphism of E into F. Then has the form (1), where ( ) is a conjugate of over F. In particular, there are at most n F-isomorphisms of E into F.
We will refine Proposition 2 by determining precisely all the F-isomorphisms 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 = IrrF( ,X). Suppose that an isomorphism : F K is given. If
f = a 0 + a 1X + ... + a n-1X n-1 + X n  F[X],
let us define
f  =  (a 0) +  (a 1)X + ... +  (a n-1)X n-1 + 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 n-1 n-1, a i F.
Therefore, let us define :F( ) K by
(3)
 (x) =  (a 0) +  (a 1)  + ... +  (a n-1) n-1.
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 n-1 n-1, b i  F.
It is straightforward to verify that (x + y) = (x) + (y). Let us show that (x · y) = (x) · (y). Set g = aiXi, h = bjXj. By the division algorithm in F[X]. there exist polynomials q,r F[X] such that
(4)
g · h = q · f + r, r = do + d1X + .. + dn-1Xn-1
Since f( ) = 0, (4) implies that
Fo any polynomial k = e0 + e1X + .. + esXs 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
=( f · q + r)  (  )
= (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 F-isomorphisms 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 F-isomorphism of E into F such that i( ) = i. Moreover, i is unique and is given by
(5)
i(b 0 + b 1 + ... + b n-1 n-1)
= b 0 + b 1 i + ... + b n-1 in-1, b j  F.
(3) There are exactly n F-isomorphisms of E into F and these are 1,..., n.
Proof: (1) 1,..., n are the zeros of Irrf( ,X), Therefore, 1,..., n are distinct by Theorem 1 of the restrictive assumption.
(2) Let :F F be the isomorphism defined by (x) = x (x F). Then f = f and Theorem 3 implies that there exists an F-isomorphism i such that i( ) = i. Moreover, by (1), 1 is given by (5).
(3) By Proposition 2, the number of F-isomorphisms 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 F-isomorphisms 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
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 {f1,...,fm}, fj F[X].
For example, if f F[X] is a nonconstant polynomial, then the splitting field Ef 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 {f1,...,fm}, fj 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 F-isomorphism, we see that (E) E. Let deg(E/F) = n. Then, since is an F-isomorphism, 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 F-isomorphism :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 F-isomorphism :E F. [Map onto any one of the zeros of f , f = IrrF(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 IrrF( ,X), so that E is a splitting field of IrrF( ,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 ma, 0 < a< m - 1, (a,m) = 1.]
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