Abstract Algebra:The Galois Group of a Polynomial

G contains a p-Sylow subgroup A group is considered cyclic if it is generated by a single element. Gal_F(f ) is isomorphic to a subgroup of the group of permutations of the zeros alpha

₁,...,

_n. The order of Gal_F(f ) is m. Let alpha

be algebraic over F, n = deg(Irr_F( alpha

,X)). Then
(1) deg(F( alpha

)/F) = n.
(2) {1, alpha

²,...,

^n-1} is a basis for F( alpha

) over F. Let f = a₀+a₁X+...+a_nXⁿ element of

Z[X] be primitive and let p be prime. Assume that p|a_i (0 < i < n-1), p does not divide

a_n, p²

a₀. Then f is irreducible in Z[X]. Let F be a field, F its algebraic closure, alpha

F. Then

is algebraic over F. Let f = Irr_F( alpha

,X). Then

f = (X -

₁)(X -

₂)...(X -

_n),

₁ =

The elements alpha

₁,...,

_n 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 : A maps

B 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

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

(

) is a conjugate of beta

over F. In particular, there are at most n F-isomorphisms of E into F. Let alpha

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 Irr_F( alpha

,X) A function f : A maps

B 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

E be such that E = F( beta

). Further, let beta

₁,...,

_n be the conjugates of beta

over F and let sigma

G(E/F) be such that sigma

_i(

) =

_i. Then

Gal(E/F) if and only if beta

E. In particular, if E is a normal extension of F, then all beta

_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,...,f_m}, f_j element of

F[X]. The subfield F( alpha

₁,...,

_n) 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 + a₁X^n-1 + ... + a_n 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 =

(X -

_i)

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

E_f = F(

₁,...,

_n).

It is clear that E_f 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 Gal_F(f ), is the finite group Gal(E_f/F).

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

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

A typical element sigma of Gal_F(f ) is an F-automorphism of E_f. But since E_f is generated over F by alpha ₁,..., alpha _n, sigma is completely determined once sigma ( alpha _i) (i = 1,...,n) is specified. Let us try to pin down what sigma ( alpha _i) can be.

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

(2) If alpha _j(i) = alpha _j(i'), then i = i'.

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

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

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

P =

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

: Gal_F(f ) maps

S_n,

(1)

(

) = P

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

Theorem 4: Gal_F(f ) is isomorphic to a subgroup of the group of permutations of the zeros alpha ₁,..., alpha _n.

Corollary 5: Gal_F(f ) is isomorphic to a subgroup of S_n.

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

Note that the isomorphism psi of (1) will not generally be surjective, so that Gal_F (f ) will not generally contain all the permutations on alpha ₁,..., alpha _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 E_f = Q( square root of 2 and Gal_F(f ) consists of the mappings

Thus, Gal_F(f ) isomorphic to Z₂.

Example 2: f = (X² - 2)(X² - 3). Here E_f = Q( square root of 2 , square root of 3 ) and alpha ₁ = , alpha ₂ = - , alpha ₃ = , alpha ₄ = - , Then Gal_F(f ) consists of the mappings

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

Gal_F( f ) isomorphic with

G₁

G₂

Z₂

Z₂.

Example 3: f = X³ - 2. Then alpha ₁ = cube root of 2 , alpha ₂ = omega , alpha ₃ = omega ², where denotes a real cube root of 2 and omega is a primitive cube root of unity. Let us show that Gal_Q( f ) isomorphic S₃, so that all permutations of the roots alpha ₁, alpha ₂, alpha ₃ correspond to a Galois group. By Theorem 4, Gal_Q( f ) is isomorphic to a subgroup of S₃. Moreover, E_f/Q is a Galois extension, so that by Theorem 4 of the section the Galois group of an extension, the order of Gal_Q( f ) equals deg(E_f/Q). Therefore, it suffices to show that deg(E_f/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 Gal_Q( 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, alpha ₃, alpha ₄, and alpha ₅. Let alpha ₁ = a + b square root of -1 element of C be a nonreal zero of f.

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

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

= x - y

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

0 = 0 =

₁⁵ - 6

₁ +3

= f(

₁).

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

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

(

₁) =

₂,

(

₂) =

₁,

(

_i) =

_i, (i = 3,4,5)

Therefore, when sigma is viewed as a permutation, sigma is just the permutation (12), which interchanges alpha ₁ and alpha ₂ and leaves all other alpha _i fixed. Therefore, we have located one specific element of Gal_Q( f ).

By the Eisenstein irreducibility criterion, f is irreducible in Q[X], so that deg(Q( alpha ₁)/Q) = 5 by Theorem 3 of the section on algebraic numbers. Therefore, since

deg(E_f/Q) = deg(E_f/Q( alpha

₁))·deg(Q( alpha

₁)/Q),

we see that deg(E_f /Q) is divisible by 5. Therefore, by Proposition 2 and Theorem 4, Gal_Q( f ) is a subgroup of the group of permutations of alpha ₁,... alpha ₅ and that the order of Gal_Q( f ) is divisible by 5. Moreover, by the first Sylow Theorem, Gal_Q( f ) contains a subgroup H of order 5. In particular, Gal_Q( f ) contains an element eta of order 5. Let

= (i₁...i_r)(j₁...j_s)(k₁...k_t)...

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

= (1i₂i₃i₄i₅).

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

= (12i₃i₄i₅).

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

= (12345).

Therefore, we have shown that Gal_Q( 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, Gal_Q( f ) contains the transpositions (12), (23), (34), (45), and (51). And these transpositions generate all transpositions. Therefore, since S₅ is generated by transpositions, Gal_Q( 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 = X^m - 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

,...,b

^m-1,

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

(b) = b

^a(), 0 < a( sigma

) < m - 1.

Let psi : Gal_F( f ) maps Z_m be defined by

(

) = a(

) (mod m).

Then psi is an injection. Moreover,

(

)(b) =

^a())

(b) ·

^a() (since zeta

F and

is an F-automorphism of E)

= b

^a()+b().

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

(

) =

(

) +

(

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