Section8.13

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## 8.13 Computation of Galois Groups

As a sample use of Axiom's algebraic number facilities, group:Galois we compute Galois:group the Galois group of the polynomial p(x)=x 5-5 x+12.

p := x**5 - 5*x + 12

x 5-5 x+12

Type: Polynomial Integer

We would like to construct a polynomial f(x) such that the splitting field:splitting field splitting field of p(x) is generated by one root of f(x). First we construct a polynomial r=r(x) such that one root of r(x) generates the field generated by two roots of the polynomial p(x). (As it will turn out, the field generated by two roots of p(x) is, in fact, the splitting field of p(x).)

From the proof of the primitive element theorem we know that if a and b are algebraic numbers, then the field Q(a, b) is equal to Q(a+kb) for an appropriately chosen integer k. In our case, we construct the minimal polynomial of a i-a j, where a i and a j are two roots of p(x). We construct this polynomial using resultant. The main result we need is the following: If f(x) is a polynomial with roots a i…a m and g(x) is a polynomial with roots b i…b n, then the polynomial h(x)=resultant(f(y),g(x-y),y) is a polynomial of degree m*n with roots a i+b j, i=1…m, j=1…n.

For f(x) we use the polynomial p(x). For g(x) we use the polynomial -p(-x). Thus, the polynomial we first construct is resultant(p(y),-p(y-x),y).

q := resultant(eval(p,x,y),-eval(p,x,y-x),y)

x 25-50 x 21-2375 x 17+90000 x 15-5000 x 13+2700000 x 11+250000 x 9+18000000 x 7+64000000 x 5

Type: Polynomial Integer

The roots of q(x) are a i-a j, i≤1, j≤5. Of course, there are five pairs (i, j) with i=j, so 0 is a 5-fold root of q(x).

Let's get rid of this factor.

q1 := exquo(q, x**5)

x 20-50 x 16-2375 x 12+90000 x 10-5000 x 8+2700000 x 6+250000 x 4+18000000 x 2+64000000

Type: Union(Polynomial Integer,...)

Factor the polynomial q1.

factoredQ := factor q1

(x 10-10 x 8-75 x 6+1500 x 4-5500 x 2+16000)*(x 10+10 x 8+125 x 6+500 x 4+2500 x 2+4000)

Type: Factored Polynomial Integer

We see that q1 has two irreducible factors, each of degree 10. (The fact that the polynomial q1 has two factors of degree 10 is enough to show that the Galois group of p(x) is the dihedral group of order 10.See McKay, Soicher, Computing Galois Groups over the Rationals, Journal of Number Theory 20, 273-281 (1983). We do not assume the results of this paper, however, and we continue with the computation. Note that the type of factoredQ is FR POLY INT, that is, Factored Polynomial Integer. Factored This is a special data type for recording factorizations of polynomials with integer coefficients.

We can access the individual factors using the operation nthFactornthFactorFactored.

r := nthFactor(factoredQ,1)

x 10-10 x 8-75 x 6+1500 x 4-5500 x 2+16000

Type: Polynomial Integer

Consider the polynomial r=r(x). This is the minimal polynomial of the difference of two roots of p(x). Thus, the splitting field of p(x) contains a subfield of degree 10. We show that this subfield is, in fact, the splitting field of p(x) by showing that p(x) factors completely over this field.

First we create a symbolic root of the polynomial r(x). (We replaced x by b in the polynomial r so that our symbolic root would be printed as b.)

beta:AN := rootOf(eval(r,x,b))

b

Type: AlgebraicNumber

We next tell Axiom to view p(x) as a univariate polynomial in x with algebraic number coefficients. This is accomplished with this type declaration.

p := p::UP(x,INT)::UP(x,AN)

x 5-5 x+12

Type: UnivariatePolynomial(x,AlgebraicNumber)

Factor p(x) over the field . (This computation will take some time!)

algFactors := factor(p,[beta])

(x+(-85 b 9-116 b 8+780 b 7+2640 b 6+14895 b 5-8820 b 4-127050 b 3-327000 b 2-405200 b+2062400)1339200)(x+-17 b 8+156 b 6+2979 b 4-25410 b 2-14080 66960)(x+143 b 8-2100 b 6-10485 b 4+290550 b 2-334800 b-960800 669600)(x+143 b 8-2100 b 6-10485 b 4+290550 b 2+334800 b-960800 669600)(x+(85 b 9-116 b 8-780 b 7+2640 b 6-14895 b 5-8820 b 4+127050 b 3-327000 b 2+405200 b+2062400)1339200)

Type: Factored UnivariatePolynomial(x,AlgebraicNumber)

When factoring over number fields, it is important to specify the field over which the polynomial is to be factored, as polynomials have different factorizations over different fields. When you use the operation factor, the field over which the polynomial is factored is the field generated by

1. the algebraic numbers that appear in the coefficients of the polynomial, and
2. the algebraic numbers that appear in a list passed as an optional second argument of the operation.

In our case, the coefficients of p are all rational integers and only beta appears in the list, so the field is simply .

It was necessary to give the list [beta] as a second argument of the operation because otherwise the polynomial would have been factored over the field generated by its coefficients, namely the rational numbers.

factor(p)

x 5-5 x+12

Type: Factored UnivariatePolynomial(x,AlgebraicNumber)

We have shown that the splitting field of p(x) has degree 10. Since the symmetric group of degree 5 has only one transitive subgroup of order 10, we know that the Galois group of p(x) must be this group, the dihedral group group:dihedral of order 10. Rather than stop here, we explicitly compute the action of the Galois group on the roots of p(x).

First we assign the roots of p(x) as the values of five root variables.

We can obtain an individual root by negating the constant coefficient of one of the factors of p(x).

factor1 := nthFactor(algFactors,1)

x+(-85 b 9-116 b 8+780 b 7+2640 b 6+14895 b 5-8820 b 4-127050 b 3-327000 b 2-405200 b+2062400)1339200

Type: UnivariatePolynomial(x,AlgebraicNumber)
root1 := -coefficient(factor1,0)

(85 b 9+116 b 8-780 b 7-2640 b 6-14895 b 5+8820 b 4+127050 b 3+327000 b 2+405200 b-2062400)1339200

Type: AlgebraicNumber

We can obtain a list of all the roots in this way.

roots := [-coefficient(nthFactor(algFactors,i),0) for i in 1..5]

[(85 b 9+116 b 8-780 b 7-2640 b 6-14895 b 5+8820 b 4+127050 b 3+327000 b 2+405200 b-2062400)1339200, 17 b 8-156 b 6-2979 b 4+25410 b 2+14080 66960,-143 b 8+2100 b 6+10485 b 4-290550 b 2+334800 b+960800 669600,-143 b 8+2100 b 6+10485 b 4-290550 b 2-334800 b+960800 669600,(-85 b 9+116 b 8+780 b 7-2640 b 6+14895 b 5+8820 b 4-127050 b 3+327000 b 2-405200 b-2062400)1339200]

Type: List AlgebraicNumber

The expression

- coefficient(nthFactor(algFactors, i), 0)}

is the i-th root of p(x) and the elements of roots are the i-th roots of p(x) as i ranges from 1 to 5.

Assign the roots as the values of the variables a1,...,a5.

(a1,a2,a3,a4,a5) := (roots.1,roots.2,roots.3,roots.4,roots.5)

(-85 b 9+116 b 8+780 b 7-2640 b 6+14895 b 5+8820 b 4-127050 b 3+327000 b 2-405200 b-2062400)1339200

Type: AlgebraicNumber

Next we express the roots of r(x) as polynomials in beta. We could obtain these roots by calling the operation factor: factor(r,[beta]) factors r(x) over . However, this is a lengthy computation and we can obtain the roots of r(x) as differences of the roots a1,...,a5 of p(x). Only ten of these differences are roots of r(x) and the other ten are roots of the other irreducible factor of q1. We can determine if a given value is a root of r(x) by evaluating r(x) at that particular value. (Of course, the order in which factors are returned by the operation factor is unimportant and may change with different implementations of the operation. Therefore, we cannot predict in advance which differences are roots of r(x) and which are not.)

Let's look at four examples (two are roots of r(x) and two are not).

eval(r,x,a1 - a2)

0

Type: Polynomial AlgebraicNumber
eval(r,x,a1 - a3)

(47905 b 9+66920 b 8-536100 b 7-980400 b 6-3345075 b 5-5787000 b 4+75572250 b 3+161688000 b 2-184600000 b-710912000)4464

Type: Polynomial AlgebraicNumber
eval(r,x,a1 - a4)

0

Type: Polynomial AlgebraicNumber
eval(r,x,a1 - a5)

405 b 8+3450 b 6-19875 b 4-198000 b 2-588000 31

Type: Polynomial AlgebraicNumber

Take one of the differences that was a root of r(x) and assign it to the variable bb.

For example, if eval(r, x, a1-a4) returned 0, you would enter this.

bb := a1 - a4

(85 b 9+402 b 8-780 b 7-6840 b 6-14895 b 5-12150 b 4+127050 b 3+908100 b 2+1074800 b-3984000)1339200

Type: AlgebraicNumber

Of course, if the difference is, in fact, equal to the root beta, you should choose another root of r(x).

Automorphisms of the splitting field are given by mapping a generator of the field, namely beta, to other roots of its minimal polynomial. Let's see what happens when beta is mapped to bb.

We compute the images of the roots a1,...,a5 under this automorphism:

aa1 := subst(a1,beta = bb)

-143 b 8+2100 b 6+10485 b 4-290550 b 2+334800 b+960800 669600

Type: AlgebraicNumber
aa2 := subst(a2,beta = bb)

(-85 b 9+116 b 8+780 b 7-2640 b 6+14895 b 5+8820 b 4-127050 b 3+327000 b 2-405200 b-2062400)1339200

Type: AlgebraicNumber
aa3 := subst(a3,beta = bb)

(85 b 9+116 b 8-780 b 7-2640 b 6-14895 b 5+8820 b 4+127050 b 3+327000 b 2+405200 b-2062400)1339200

Type: AlgebraicNumber
aa4 := subst(a4,beta = bb)

-143 b 8+2100 b 6+10485 b 4-290550 b 2-334800 b+960800 669600

Type: AlgebraicNumber
aa5 := subst(a5,beta = bb)

17 b 8-156 b 6-2979 b 4+25410 b 2+14080 66960

Type: AlgebraicNumber

Of course, the values aa1,...,aa5 are simply a permutation of the values a1,...,a5.

Let's find the value of aa1 (execute as many of the following five commands as necessary).

(aa1 = a1) :: Boolean

false

Type: Boolean
(aa1 = a2) :: Boolean

false

Type: Boolean
(aa1 = a3) :: Boolean

true

Type: Boolean
(aa1 = a4) :: Boolean

false

Type: Boolean
(aa1 = a5) :: Boolean

false

Type: Boolean

Proceeding in this fashion, you can find the values of aa2,...aa5. You have represented the automorphism beta->bb as a permutation of the roots a1,...,a5. If you wish, you can repeat this computation for all the roots of r(x) and represent the Galois group of p(x) as a subgroup of the symmetric group on five letters.

Here are two other problems that you may attack in a similar fashion:

1. Show that the Galois group of p(x)=x 4+2 x 3-2 x 2-3 x+1 is the dihedral group of order eight. group:dihedral (The splitting field of this polynomial is the Hilbert class field Hilbert class field of field:Hilbert class the quadratic field Q(145).)
2. Show that the Galois group of p(x)=x 6+108 has order 6 and is isomorphic to S 3, the symmetric group on three letters. group:symmetric (The splitting field of this polynomial is the splitting field of x 3-2.)

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