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 1,...,n. The order of GalF(f ) is m. Let be algebraic over F, n = deg(IrrF(,X)). Then
(1) deg(F()/F) = n.
(2) {1,,2,...,n-1} is a basis for F() over F.
Let f = a0+a1X+...+anXn Z[X] be primitive and let p be prime. Assume that p|ai (0 < i < n-1), pan, p2a0. Then f is irreducible in Z[X]. 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 = .
The elements 1,...,n are called the conjugates of over F.

The set of real numbers. The complex numbers. The set residue classes mod n. The set of rational numbers. A function f : AB is said to be surjective (or onto) if for every y B there exists x A such that f(x) = y. 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 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) A function f : AB 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 E be such that E = F(). Further, let 1,...,n be the conjugates of over F and let i G(E/F) be such that i() = i. Then i Gal(E/F) if and only if i E. In particular, if E is a normal extension of F, then all i 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 F[X]. 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 f 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 F[X]

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

f = (X - i)

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

Ef = F(1,...,n).

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

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

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

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

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

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

P =

is a permutation of the n distinct zeros {1,...,n}. It is trivial to verify that the mapping

: GalF(f ) Sn,
(1)
() = P

is a homomorphism. Moreover, since is completely determined by (1),...,(n), we see that is an injection. Therefore, we have proved:

Theorem 4: GalF(f ) is isomorphic to a subgroup of the group of permutations of the zeros 1,...,n.

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 of (1) will not generally be surjective, so that GalF (f ) will not generally contain all the permutations on 1,...,n. 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( and GalF(f ) consists of the mappings

,   .

Thus, GalF(f ) Z2.

Example 2: f = (X2 - 2)(X2 - 3). Here Ef = Q(,) and 1 = , 2 = - , 3 = , 4 = - , Then GalF(f ) consists of the mappings

,    ,
,    .

Moreover, G1 = {0,1} and G2 = {0,2} are subgroups of GalF(f ) isomorphic to Z2 and

GalF( f ) G1 G2 Z2 Z2.

Example 3: f = X3 - 2. Then 1 = , 2 = , 3 = 2, where denotes a real cube root of 2 and is a primitive cube root of unity. Let us show that GalQ( f ) S3, so that all permutations of the roots 1, 2, 3 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 ) 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, 3, 4, and 5. Let 1 = a + b C be a nonreal zero of f.

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

If = x + y (x,y R) is a complex number, let us define the complex conjugate of of by

= x - y .

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

0 = 0 =
=
= 15 - 6 1 +3
= f(1).

Therefore, 2 = 1 is a zero of f and Ef = Q(1,2,3,4,5).

The mapping : EfEf defined by () = is a Q-automorphism of Ef and therefore GalQ( f ). Moreover,

(1) = 2,
(2) = 1,
(i) = i, (i = 3,4,5)

Therefore, when is viewed as a permutation, is just the permutation (12), which interchanges 1 and 2 and leaves all other i 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(1)/Q) = 5 by Theorem 3 of the section on algebraic numbers. Therefore, since

deg(Ef /Q) = deg(Ef /Q(1))·deg(Q(1)/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 1,...5 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 of order 5. Let

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

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

= (1i2i3i4i5).

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

= (12i3i4i5).

However, by permuting and renumbering the 3,4,5, we may assume that

= (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 be a primitive mth root of 1 and let F be a field containing . Let a F, f = Xm - a 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,b,...,bm-1,

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

(b) = ba(),   0 < a() < m - 1.

Let : GalF( f ) Zm be defined by

() = a() (mod m).

Then is an injection. Moreover,

()(b) = (ba())
= (b) · a() (since F and is an F-automorphism of E)
= ba()+b().

On the other hand, ()(b) = ba()+b(), so that a()+b()-a() = 1 and a() +b() - a() 0 (mod m). Thus,

() = () + (),

and 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 ) Zm.