deg(E/F) = 1 if and only if E = F. Let E = F(,) be an algebraic extension of F. Then there exists E such that E = F(). Thus, E is a simple extension of F. The subfield F(1,...,n) of E gotten by adjoining all zeros of f(X) to F is called the splitting field of f. 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. Let E1 and E2 be fields, f :E1 E2 a ring homomorphism. Then either f(x) = 0 for all x E1 or f is an isomorphism. 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 :FK is given. If f=a0+a1X+...+an-1Xn-1+XnF[X], let us define f =(a0)+(a1)X+...+(an-1)Xn-1+Xn 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 . Let f F[X] be irreducible and of degree > 1. Then all zeros of f in F are simple. The algebraic closure of the field Let be algebraic over F, n = deg(IrrF(,X)). Then - (1) deg(F()/F) = n and (2) {1,,2,...,n-1} is a basis for F() over F. Let F(1,2,...,n) be an algebraic extension of F. Then there exists F(1,...,n) such that F(1,...,n) = F(). The set of rational numbers. We say that is algebraic over F if there exists a nonzero polynomial f F[X] such that f() = 0. If is not algebraic over F, then we say that is transcendental over F. Let E be algebraic over F. the zero of a monic irreducible polynomial p F[X]. The polynomial p is called the irreducible polynomial of over F denoted IrrF(,X)

### 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 = a0 + a1 + ... + an-1n-1,  ai F, n = deg(E/F).

Therefore, if is an F-isomorphism of E into F, then

(1)
(x) = (a0) + (a1)() + ... + (an-1)()n-1
= a0 + a1() + ... + an-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 = b0 + b1 + ... + bn-1n-1 + n
0 = (0) = (b0 + b1 + ... + bn-1n-1 + n)
= (b0) + (b1)() + ... + (bn-1)()n-1 + ()n
= b0 + b1() + ... + bn-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 : FK is given. If

f = a0 + a1X + ... + an-1Xn-1 + Xn F[X],

let us define

f = (a0) + (a1)X + ... + (an-1)Xn-1 + Xn 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 = a0 + a1 + ... + an-1n-1,   ai F.

Therefore, let us define :F()K by

(3)
(x) = (a0) + (a1) + ... + (an-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 = b1 + b1 + ... + bn-1n-1,   bi 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

x · y = g()h() = r().

Fo any polynomial k = e0 + e1X + .. + esXs F[X]. set

k = (e0) + (e1)X + ... + (es)Xs 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 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(b0 + b1 + ... + bn-1n-1)
= b0 + b1i + ... + bn-1in-1,   bj 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 :FF 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

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 {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 :EF. [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.]