As a sample use of Axiom's algebraic number facilities, group:Galois we compute Galois:group the Galois group of the polynomial .
We would like to construct a polynomial such that the splitting field:splitting field splitting field of is generated by one root of . First we construct a polynomial such that one root of generates the field generated by two roots of the polynomial . (As it will turn out, the field generated by two roots of is, in fact, the splitting field of .)
From the proof of the primitive element theorem we know that if and are algebraic numbers, then the field is equal to for an appropriately chosen integer . In our case, we construct the minimal polynomial of , where and are two roots of . We construct this polynomial using resultant. The main result we need is the following: If is a polynomial with roots and is a polynomial with roots , then the polynomial is a polynomial of degree with roots .
For we use the polynomial . For we use the polynomial . Thus, the polynomial we first construct is .
The roots of are . Of course, there are five pairs with , so is a 5-fold root of .
Let's get rid of this factor.
Factor the polynomial .
We see that has two irreducible factors, each of degree . (The fact that the polynomial has two factors of degree is enough to show that the Galois group of is the dihedral group of order .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 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.
Consider the polynomial . This is the minimal polynomial of the difference of two roots of . Thus, the splitting field of contains a subfield of degree . We show that this subfield is, in fact, the splitting field of by showing that factors completely over this field.
First we create a symbolic root of the polynomial . (We replaced by in the polynomial so that our symbolic root would be printed as .)
We next tell Axiom to view as a univariate polynomial in with algebraic number coefficients. This is accomplished with this type declaration.
Factor over the field . (This computation will take some time!)
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
In our case, the coefficients of are all rational integers and only appears in the list, so the field is simply .
It was necessary to give the list 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.
We have shown that the splitting field of has degree . Since the symmetric group of degree 5 has only one transitive subgroup of order , we know that the Galois group of must be this group, the dihedral group group:dihedral of order . Rather than stop here, we explicitly compute the action of the Galois group on the roots of .
First we assign the roots of as the values of five root variables.
We can obtain an individual root by negating the constant coefficient of one of the factors of .
We can obtain a list of all the roots in this way.
The expression
is the -th root of and the elements of are the -th roots of as ranges from to .
Assign the roots as the values of the variables .
Next we express the roots of as polynomials in . We could obtain these roots by calling the operation factor: factors over . However, this is a lengthy computation and we can obtain the roots of as differences of the roots of . Only ten of these differences are roots of and the other ten are roots of the other irreducible factor of . We can determine if a given value is a root of by evaluating 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 and which are not.)
Let's look at four examples (two are roots of and two are not).
Take one of the differences that was a root of and assign it to the variable .
For example, if returned , you would enter this.
Of course, if the difference is, in fact, equal to the root , you should choose another root of .
Automorphisms of the splitting field are given by mapping a generator of the field, namely , to other roots of its minimal polynomial. Let's see what happens when is mapped to .
We compute the images of the roots under this automorphism:
Of course, the values are simply a permutation of the values .
Let's find the value of (execute as many of the following five commands as necessary).
Proceeding in this fashion, you can find the values of . You have represented the automorphism as a permutation of the roots . If you wish, you can repeat this computation for all the roots of and represent the Galois group of as a subgroup of the symmetric group on five letters.
Here are two other problems that you may attack in a similar fashion: