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G contains a p-Sylow subgroup A group is considered cyclic if it is generated by a single element. GalF(f ) is isomorphic to a subgroup of the group of permutations of the zeros alpha1,...,alphan. The order of GalF(f ) is m. Let alpha be algebraic over F, n = deg(IrrF(alpha,X)). Then
(1) deg(F(alpha)/F) = n.
(2) {1,alpha,alpha2,...,alphan-1} is a basis for F(alpha) over F.
Let f = a0+a1X+...+anXn element of Z[X] be primitive and let p be prime. Assume that p|ai (0 < i < n-1), pdoes not dividean, p2does not dividea0. Then f is irreducible in Z[X]. 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.
The elements alpha1,...,alphan are called the conjugates of alpha over F.

The set of real numbers. The complex numbers. The set residue classes mod n. The set of rational numbers. A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. 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 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) A function f : AmapsB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. 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 and let sigmai element of G(E/F) be such that sigmai(beta) = betai. Then sigmai element of Gal(E/F) if and only if betai element of E. In particular, if E is a normal extension of F, then all betai belong to E and Gal(E/F) contains n elements. An F-automorphism of E is an F-isomorphism of E onto itself. The set of all F-automorphisms of E will be denoted Gal(E/F). Gal(E/F) is called the Galois group of the extension E over F 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]. 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 f element of F[X] be irreducible and of degree > 1. Then all zeros of f in F are simple.

An algebraic closure of a field F is an extension F of F such that (1) F is algebraically closed, and (2) F is an algebraic extension of F.

The Galois Group of a Polynomial

In the introduction we promised that we would associate to a polynomial a finite group which reflects the algebraic properties of the polynomial. We have now reached the point where that promise is very simple to fulfill.

Let F be a field and let

f = Xn + a1Xn-1 + ... + an element of F[X]

be a nonconstant polynomial. Let us fix an algebraic closure F of F. Then, in F[X], we can write

f = sum over i(X - alphai)

where alpha1,...,alphan are the zeros of f in F. By Theorem 1 of the section on the restrictive assumption, alpha1,...,alphan are all distinct. The splitting field Ef of f over F is given by

Ef = F(alpha1,...,alphan).

It is clear that Ef is a normal extension of F and is therefore a Galois extension of degree m, say.

Definition 1: The Galois group of f over F denoted GalF(f ), is the finite group Gal(Ef /F).

By Theorem 4 of the section on Galois group of an extension, we have

Proposition 2: The order of GalF(f ) is m.

A typical element sigma of GalF(f ) is an F-automorphism of Ef. But since Ef is generated over F by alpha1,...,alphan, sigma is completely determined once sigma(alphai) (i = 1,...,n) is specified. Let us try to pin down what sigma(alphai) can be.

Lemma 3: (1) The image of alphai is an alphaj for some j depending on i. In other words sigma(alphai) = alphaj(i) for some j(i) (1 < j(i) < n),

(2) If alphaj(i) = alphaj(i'), then i = i'.

Proof: (1) Clearly sigma(alphai) equals some F-conjugate of alphai, so that sigma(alphai) is a zero of IrrF(alphai,X) (by Proposition 2 of the section on conjugates). But since f element of F[X] has alphai as a zero, IrrF(alphai,X)|f. Therefore, sigma(alphai) is a zero of f and sigma(alphai) = alphaj(i) for some j(i) (1 < j(i) < n).

(2) If alphaj(i) = alphaj(i'), then sigma(alphai) = sigma(alphai'), But since sigma is an injection, this implies that alphai = alphai'. Therefore, i = i', since all alphai are distinct.

Lemma 3 asserts that the set {sigma(alpha1),...,sigma(alphan)} coincides with the set {alpha1,...,alphan}. For, indeed {sigma(alpha1),...,sigma(alphan)} subset of {alpha1,...,alphan} by (1); and {sigma(alpha1),...,sigma(alphan)} contains n distinct elements by (2). In other words,

Psugma> = permutation of sigma

is a permutation of the n distinct zeros {alpha1,...,alphan}. It is trivial to verify that the mapping

psi: GalF(f ) maps Sn,
(1)
psi(sigma) = Psigma

is a homomorphism. Moreover, since sigma is completely determined by sigma(alpha1),...,sigma(alphan), we see that psi is an injection. Therefore, we have proved:

Theorem 4: GalF(f ) is isomorphic to a subgroup of the group of permutations of the zeros alpha1,...,alphan.

Corollary 5: GalF(f ) is isomorphic to a subgroup of Sn.

Corollary 6: GalF(f ) has order of at most n!

Note that the isomorphism psi of (1) will not generally be surjective, so that GalF (f ) will not generally contain all the permutations on alpha1,...,alphan. The reader is referred to the examples below for instances of this phenomenon. In Examples 1-4, let F = Q

Example 1: f = X2 - 2. Then Ef = Q(square root of 2 and GalF(f ) consists of the mappings

,   .

Thus, GalF(f ) isomorphic to Z2.

Example 2: f = (X2 - 2)(X2 - 3). Here Ef = Q(square root of 2,square root of 3) and alpha1 = square root of 2, alpha2 = - square root of 2, alpha3 = square root of 3, alpha4 = - square root of 3, Then GalF(f ) consists of the mappings

,    ,
,    .

Moreover, G1 = {psi0,psi1} and G2 = {psi0,psi2} are subgroups of GalF(f ) isomorphic to Z2 and

GalF( f ) isomorphic with G1 cross G2 isomorphic with Z2 cross Z2.

Example 3: f = X3 - 2. Then alpha1 = cube root of 2, alpha2 = cube root of 2omega, alpha3 = cube root of 2omega2, where cube root of 2 denotes a real cube root of 2 and omega is a primitive cube root of unity. Let us show that GalQ( f ) isomorphic S3, so that all permutations of the roots alpha1, alpha2, alpha3 correspond to a Galois group. By Theorem 4, GalQ( f ) is isomorphic to a subgroup of S3. Moreover, Ef/Q is a Galois extension, so that by Theorem 4 of the section the Galois group of an extension, the order of GalQ( f ) equals deg(Ef /Q). Therefore, it suffices to show that deg(Ef /Q) > 6. and this was proved in Example 3 of the section of examples of splitting fields.

Example 4: Let f = X5 - 6X + 3. We will show that GalQ( f ) isomorphic to S5. Let us begin by considering the function f(x) = x5 - 6x + 3 for real x. From elementary calculus, we see that this function has a maximum for x = -(6/5)1/4 and a minimum at (6/5)1/4. Moreover, these are the only extrema of f(x). Thus, we may sketch the graph of f(x) as in figure 1, and wee see immediately that f has exactly three real zeros, alpha3, alpha4, and alpha5. Let alpha1 = a + bsquare root of -1 element of C be a nonreal zero of f.

figure 1
Figure 1: Graph of f(x) = x5 - 6x + 3.

If alpha = x + ysquare root of -1 (x,y element of R) is a complex number, let us define the complex conjugate of alpha bar of alpha by

alpha conjugate = x - ysquare root of -1 .

The mapping alphamapsalpha conjugate is an automorphism of C. From this fact it follows that

0 = 0 =
=
= alpha conjugate15 - 6 alpha conjugate1 +3
= f(alpha conjugate1).

Therefore, alpha2 = alpha conjugate1 is a zero of f and Ef = Q(alpha1,alpha2,alpha3,alpha4,alpha5).

The mapping sigma: EfmapsEf defined by sigma(alpha) = alpha conjugate is a Q-automorphism of Ef and therefore sigma element of GalQ( f ). Moreover,

sigma(alpha1) = alpha2,
sigma(alpha2) = alpha1,
sigma(alphai) = alphai, (i = 3,4,5)

Therefore, when sigma is viewed as a permutation, sigma is just the permutation (12), which interchanges alpha1 and alpha2 and leaves all other alphai fixed. Therefore, we have located one specific element of GalQ( f ).

By the Eisenstein irreducibility criterion, f is irreducible in Q[X], so that deg(Q(alpha1)/Q) = 5 by Theorem 3 of the section on algebraic numbers. Therefore, since

deg(Ef /Q) = deg(Ef /Q(alpha1))·deg(Q(alpha1)/Q),

we see that deg(Ef /Q) is divisible by 5. Therefore, by Proposition 2 and Theorem 4, GalQ( f ) is a subgroup of the group of permutations of alpha1,...alpha5 and that the order of GalQ( f ) is divisible by 5. Moreover, by the first Sylow Theorem, GalQ( f ) contains a subgroup H of order 5. In particular, GalQ( f ) contains an element eta of order 5. Let

eta = (i1...ir)(j1...js)(k1...kt)...

be the decomposition of eta into a product of disjoint cycles. The order of eta is the least common multiple of r,s,t, ..., Therefore, since eta has order 5, eta must be a 5-cycle, say

eta = (1i2i3i4i5).

By replacing eta by some power of eta, we may assume eta has the form

eta = (12i3i4i5).

However, by permuting and renumbering the alpha3,alpha4,alpha5, we may assume that

eta = (12345).

Therefore, we have shown that GalQ( f ) contains the permutations (12), (12345). However, it is easy to check that for r = 0,1,2,3,

(12345)-r(12)(12345)r = (r + 1, r + 2),
(12345)-4(12)(12345)4 = (51).

Therefore, GalQ( f ) contains the transpositions (12), (23), (34), (45), and (51). And these transpositions generate all transpositions. Therefore, since S5 is generated by transpositions, GalQ( f ) = S5.

Example 5: Let zeta be a primitive mth root of 1 and let F be a field containing zeta. Let a element of F, f = Xm - a element of F[X], and let E = the splitting field of f over F. If b is a zero of f in F, then the zeros of f ar given by

b,bzeta,...,bzetam-1,

so that f = (X - b)...(X - bzetam-1) in F[X], and E = F(b) since zeta element of F. If sigma element of GalF( f ), then there exists a unique integer a(sigma) such that

sigma(b) = bzetaa(sigma),   0 < a(sigma) < m - 1.

Let psi: GalF( f ) maps Zm be defined by

psi(sigma) = a(sigma) (mod m).

Then psi is an injection. Moreover,

(etasigma)(b) = eta(bzetaa(sigma))
= eta(b) · zetaa(sigma) (since zeta element of F and eta is an F-automorphism of E)
= bzetaa(sigma)+b(sigma).

On the other hand, (etasigma)(b) = bzetaa(sigma)+b(sigma), so that zetaa(sigma)+b(sigma)-a(etasigma) = 1 and a(sigma) +b(sigma) - a(etasigma) congruent to 0 (mod m). Thus,

psi(etasigma) = psi(eta) + psi(sigma),

and psi is an isomorphism. We have proved that GalF( f ) is isomorphic to a subgroup of Zm. In particular, since Zm is a cyclic group, GalF( f ) is cyclic. It is not usually true that GalF( f ) isomorphic to Zm.