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deg(E/F) = 1 if and only if E = F. Let E = F(alpha,beta) be an algebraic extension of F. Then there exists gamma element of E such that E = F(gamma). Thus, E is a simple extension of F. The subfield F(alpha1,...,alphan) 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 beta element of E be such that E = F(beta). Further, let sigma be an F-isomorphism of E into F. Then sigma has the form (1), where sigma(beta) is a conjugate of beta over F. In particular, there are at most n F-isomorphisms of E into F. Let E1 and E2 be fields, f :E1maps to E2 a ring homomorphism. Then either f(x) = 0 for all x element of E1 or f is an isomorphism. Let F be a field, F(alpha) an algebraic extension of F of degree n, K an algebraically closed field, f=IrrF(alpha,X). Suppose that an isomorphism sigma:FmapsK is given. If f=a0+a1X+...+an-1Xn-1+Xnelement ofF[X], let us define f sigma=sigma(a0)+sigma(a1)X+...+sigma(an-1)Xn-1+Xn element of K[X], Let beta be any zero of f sigma in K. Then there exists an isomorphism eta:F(alpha)mapsK such that (1) the restriction of eta to F coincides with sigma and (2) eta(alpha)=beta. In other words, eta is an extension of sigma to F(alpha) which maps alpha to beta. Let f element of F[X] be irreducible and of degree > 1. Then all zeros of f in F are simple. The algebraic closure of the field Let alpha be algebraic over F, n = deg(IrrF(alpha,X)). Then - (1) deg(F(alpha)/F) = n and (2) {1,alpha,alpha2,...,alphan-1} is a basis for F(alpha) over F. Let F(alpha1,alpha2,...,alphan) be an algebraic extension of F. Then there exists gamma element of F(alpha1,...,alphan) such that F(alpha1,...,alphan) = F(gamma). The set of rational numbers. We say that alpha is algebraic over F if there exists a nonzero polynomial f element of F[X] such that f(alpha) = 0. If alpha is not algebraic over F, then we say that alpha is transcendental over F. Let alpha element of E be algebraic over F. alpha the zero of a monic irreducible polynomial p element F[X]. The polynomial p is called the irreducible polynomial of alpha over F denoted IrrF(alpha,X)

Conjugates and Normal Extensions

Let F be a field, F its algebraic closure, alpha element of F. Then alpha is algebraic over F. Let f = IrrF(alpha,X). Then

f = (X - alpha1)(X - alpha2)...(X - alphan),    alphai element of F, alpha1 = alpha.

Definition 1: The elements alpha1,...,alphan are called the conjugates of alpha over F.

Example 1: If alpha element of F, then IrrF(alpha,X) = X - alpha and the only conjugate of alpha over F is alpha itself.

Example 2: Let F = Q, alpha = square root of 5. Then IrrQ(alpha,X) = X2 - 5, and the conjugates of alpha over Q are square root of 5 and -square root of 5.

Example 3: Let F = Q, alpha = (-1 + square root of 3i)/2. Then IrrQ(alpha,X) = X2 + X + 1, so that the conjugates of alpha over Q are (-1 + square root of 3i)/2, (-1 - square root of 3i)/2.

Example 4: Let F = Q, alpha = zeta, a primitive nth root of 1. Then IrrQ(zeta,X) = phin(X), the nth cyclotomic polynomial. Therefore, the conjugates of zeta over Q are zetaa [0 < a < n - 1, (a,n) = 1].

Example 5: Let F = Q, alpha = cube root of 2. Then IrrQ(alpha,X) = X3 - 2. Therefore, the conjugates of cube root of 2 over Q are cube root of 2, cube root of 2omega, cube root of 2omega2, where omega = (-1 + square root of 3i)/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 alpha1,...,alphan element of E such that E = F(alpha1,...,alphan). By Corollary 4 of our restrictive assumption, there exists beta element of E such that E = F(beta). Moreover, by Theorem 3 of the section on algebraic numbers, every element x of E can be written uniquely in the form

x = a0 + a1beta + ... + an-1betan-1,  ai element of F, n = deg(E/F).

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

(1)
sigma(x) = sigma(a0) + sigma(a1)sigma(beta) + ... + sigma(an-1)sigma(beta)n-1
= a0 + a1sigma(beta) + ... + an-1sigma(beta)n-1.

Thus, sigma is completely determined once sigma(beta) is specified. We assert that sigma(beta) is a conjugate of beta over F. Let f = b0 + b1X + ... + bn-1Xn-1 + Xn = IrrF(beta,X) Then

0 = b0 + b1beta + ... + bn-1betan-1 + betan
implies0 = sigma(0) = sigma(b0 + b1beta + ... + bn-1betan-1 + betan)
= sigma(b0) + sigma(b1)sigma(beta) + ... + sigma(bn-1)sigma(beta)n-1 + sigma(beta)n
= b0 + b1sigma(beta) + ... + bn-1sigma(beta)n-1 + sigma(beta)n
= f(sigma(beta)).

Therefore, sigma(beta) is a zero of f, and hence is a conjugate of beta 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 beta element of E be such that E = F(beta). Further, let sigma be an F-isomorphism of E into F. Then sigma has the form (1), where sigma(beta) is a conjugate of beta 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(alpha) an algebraic extension of F of degree n, K an algebraically closed field, f = IrrF(alpha,X). Suppose that an isomorphism sigma: FmapsK is given. If

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

let us define

f sigma = sigma(a0) + sigma(a1)X + ... + sigma(an-1)Xn-1 + Xn element of K[X],

Let beta be any zero of f sigma in K. Then there exists an isomorphism eta: F(alpha) mapsK such that (1) the restriction of eta to F coincides with sigma and (2) eta(alpha) = beta. In other words, eta is an extension of sigma to F(alpha) which maps alpha to beta.

Proof: Every element x of F(alpha) can be uniquely written in the form

(2)
x = a0 + a1alpha + ... + an-1alphan-1,   aielement of F.

Therefore, let us define eta:F(alpha)mapsK by

(3)
eta(x) = sigma(a0) + sigma(a1)beta + ... + sigma(an-1)betan-1.

It is clear that eta restricted to F equals sigma and that eta(alpha) = beta. Thus, it remains to prove that eta is an isomorphism. Let x be defined by (2) and let

y = b1 + b1alpha + ... + bn-1alphan-1,   bi element of F.

It is straightforward to verify that eta(x + y) = eta(x) + eta(y). Let us show that eta(x · y) = eta(x) · eta(y). Set g = sum to n-1aiXi, h = sum to n-1bjXj. By the division algorithm in F[X]. there exist polynomials q,r element of F[X] such that

(4)
g · h = q · f + r,   r = do + d1X + .. + dn-1Xn-1

Since f(alpha) = 0, (4) implies that

x · y = g(alpha)h(alpha) = r(alpha).

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

ksigma = sigma(e0) + sigma(e1)X + ... + sigma(es)Xs element of K[X],

Then by the definitions of eta and the fact that f sigma(beta) = 0, we have

eta(x · y) = rsigma(beta)
= f sigma(beta) · qsigma(beta) + rsigma(beta)
=(f · q + r)sigma(beta)
= (g · h)sigma(beta)
= gsigma(beta) · hsigma(beta)
= eta(x) · eta(y).

Thus, eta is a homomorphism. But since eta(1) = 1 not equal0, Proposition 2 in the introduction to the theory of fields implies that eta 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 beta element of E be such that E = F(beta). Further, let beta1,...,betan be the conjugates of beta over F. Then

(1) beta1,...,betan are distinct.

(2) For each i (1 < i < n), the exists an F-isomorphism sigma of E into F such that sigmai(beta) = betai. Moreover, sigmai is unique and is given by

(5)
sigmai(b0 + b1beta + ... + bn-1betan-1)
= b0 + b1betai + ... + bn-1betain-1,   bj element of F.

(3) There are exactly n F-isomorphisms of E into F and these are sigma1,...,sigman.

Proof: (1) beta1,...,betan are the zeros of Irrf(beta,X), Therefore, beta1,...,betan are distinct by Theorem 1 of the restrictive assumption.

(2) Let sigma:FmapsF be the isomorphism defined by sigma(x) = x (x element of F). Then f sigma = f and Theorem 3 implies that there exists an F-isomorphism sigmai such that sigmai(beta) = betai. Moreover, by (1), sigma1 is given by (5).

(3) By Proposition 2, the number of F-isomorphisms of E into F is at most n. Since the betai are all distinct by part (1), the sigmai are all distinct. Therefore, we have exhibited n different F-isomorphisms of E into F. This proves part (3).

The isomorphisms sigmai (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(square root of 2), then

G(E/F) = {sigma1,sigma2},

where sigma1(square root of 2) = square root of 2 and sigma2(square root of 2) = -square root of 2. If F = Q, E = Q(cube root of 2), then

G(E/F) = {lambda1,lambda2,lambda3},

where lambda1(cube root of 2) = cube root of 2, lambda2(cube root of 2) = cube root of 2omega, lambda3(cube root of 2) = cube root of 2omega2.

Note that the extension F(alpha) may not contain all the conjugates of alpha. For example, Q(cube root of 2) subset of R, but cube root of 2omega not an element of 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 element of F[X].

For example, if f element of 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 element of F[X].

(2) E is a normal extension of F.

(3) If sigma element of G(E/F), then sigma(E) = E.

(4) If x element of E, then all conjugates of x over F lie in E.

Proof: We will prove the theorem by proving the series of implications

(1)implies(2)implies(3)implies(4)implies(1).

(1)implies(2). We observed this fact above.

(2)implies(3). Suppose that E is a normal extension of F and that sigma element of G(E/F). Then E is obtained from F by adjoining all zeros {alpha1,...,alphas} of a finite collection of polynomials {f1,...,fm}, fj element of F[X]. In particular, if alpha element of {alpha1,...,alphas}, then all conjugates of alpha over F belong to {alpha1,...,alphas}, and hence to E = F(alpha1,...,alphas). But sigma(alphai) is one of the conjugates of alphai over F. Therefore, sigma(alphai) element of E. Hence, since sigma is an F-isomorphism, we see that sigma(E) subset of E. Let deg(E/F) = n. Then, since sigma is an F-isomorphism, deg(sigma(E)/F) = n. Therefore, since F subset of sigma(E) subset of E, and since deg(E/F) = deg(E/sigma(E)) · deg(sigma(E)/F), we see that deg(E/sigma(E)) = 1. Therefore, E = sigma(E) by Proposition 2 of the section on algebraic extensions.

(3)implies(4). Suppose the sigma element of G(E/F) implies that sigma(E) = E, and assume that x element of E. Let x' be a conjugate of x over F. We wish to show that x' element of E. By Theorem 3, there exists and F-isomorphism sigma:F(x)mapsF such that sigma(x) = x'. By Theorem 3 of the section on the restrictive assumption, there exists beta element of E such that E = F(x)(beta). Then, by Theorem 3, sigma can be extended to an F-isomorphism eta:EmapsF. [Map beta onto any one of the zeros of f sigma, f = IrrF(x)(beta,X).] But then eta(x) = x', so that by our assumption, x' element of eta(E) = E.

(4)implies(1). By Theorem 3 of the section of the restrictive assumption, there exist beta element of E such that E = F(beta). Let beta1,...,betan be the conjugates of beta over F. By (4), beta1,...,betan all belong to E, so that E = F(beta1,...,betan). Thus, E can be gotten by adjoining to F all zeros of IrrF(beta,X), so that E is a splitting field of IrrF(beta,X) over F.

As an immediate consequence of the above result we deduce that Q(cube root of 2) is not a normal extension of Q, since the conjugates cube root of 2omega and cube root of 2omega2 of cube root of 2 over Q are not in Q(cube root of 2), which violates condition (4). Moreover, if zetam is a primitive mth root of 1, then Q(zetam) is a normal extension of Q, since it is the splitting field over Q of phim(X). [The zeros of phim(X) are zetama, 0 < a< m - 1, (a,m) = 1.]