Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

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§ 8.6 Computing Galois Groups

 
Definition 8.6.1. Let G be a group acting on a set S. We say that G acts transitively on S if for each pair of elements x,y in S there exist an element g in G such that y = gx.
    If G is a subgroup of the symmetric group Sn, then G is called a transitive group if it acts transitively on the set { 1, 2, ... , n }.

 
Proposition 8.6.2. Let f(x) be a separable polynomial over the field K, with roots r1 , ... , rn in its splitting field F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively on the roots of f(x).

 
Lemma 8.6.3. Let p be a prime number, and let G be a transitive subgroup of Sp. Then any nontrivial normal subgroup of G is also transitive.

 
Lemma 8.6.4. Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G contains a cycle of length p.

 
Proposition 8.6.5. Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G is a subgroup of the normalizer in Sp of a cyclic subgroup of order p.

 
Let f(x) be a polynomial of degree n over the field K, and assume that f(x) has roots r1, r2, ... , rn in its splitting field F. The element of F defined by

= (ri - rj)2,

where the product is taken over all i,j with 1 i < j n, is called the discriminant of f(x).

It can be shown that the discriminant of any polynomial f(x) can be expressed as a polynomial in the coefficients of f(x), with integer coefficients. This requires use of elementary symmetric functions, and lies beyond the scope of what we have chosen to cover in the book.

    We have the following properties of the discriminant:

(i) 0 if and only if f(x) has distinct roots;

(ii) belongs to K;

(iii) If 0, then a permutation in Sn is even if and only if it leaves unchanged the sign of

(ri-rj) .

 
Proposition 8.6.6. Let f(x) be a separable polynomial over the field K, with discriminant , and let F be its splitting field over K. Then every permutation in Gal(F/K) is even if and only if is the square of some element in K.

 
We now restrict our attention to polynomials with rational coefficients. The next lemma shows that in computing Galois groups it is enough to consider polynomials with integer coefficients. Then a powerful technique is to reduce the integer coefficients modulo a prime and consider the Galois group of the reduced equation over the field GF(p).

 
Lemma 8.6.7. Let f(x) = xn + an-1 xn-1 + · · · + a1 x + a0 be a polynomial in Q[x], and assume that

ai = bi / d     for d, b0, b1, ... , bn-1 in Z.

Then dn f(x/d) is monic with integer coefficients, and has the same splitting field over Q as f(x).

 
If p is a prime number, we have the natural mapping : Z[x] -> Zp[x] which reduces each coefficient modulo p. We will use the notation ( f(x) ) = fp(x).

 
Theorem [Dedekind] Let f(x) be a monic polynomial of degree n, with integer coefficients and Galois group G over Q, and let p be a prime such that fp(x) has distinct roots. If fp(x) factors in Zp[x] as a product of irreducible factors of degrees n1, n2, ... , nk, then G contains a permutation with the cycle decomposition

(1,2, ... ,n1) (n1+1, n1+2, ... , n1+n2) · · · (n-nk+1, ... ,n),

relative to a suitable ordering of the roots.

 


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