Sage

Again, our competence at examining fields with Sage will allow us to study the main concepts of Galois Theory easily. We will thoroughly examine Example 7 carefully using our computational tools.

We will repeat and analyze carefully the splitting field of the polynomial \(p(x)=x^4-2\). We begin with an initial field extension containing at least one root.

The .galois_closure() method will create an extension containing all of the roots of the defining polynomial of a number field.

From the factorization, it is clear that L is the splitting field of the polynomial, even if the factorization is not pretty. It is easy to then obtain the Galois group of this field extension.

We can examine this group, and identify it. Notice that since the field is a degree \(8\) extension, the group is described as a permutation group on \(8\) symbols. (It is just a coincidence that the group has \(8\) elements.) With a paucity of nonabelian groups of order \(8\), it is not hard to guess the nature of the group.

That's it. But maybe not very satisfying. Let us dig deeper for more understanding. We will start over and create the splitting field of \(p(x)=x^4-2\) again, but the primary difference is that we will make the roots extremely obvious so we can work more carefully with the Galois group and the fixed fields. Along the way, we will see another example of linear algebra enabling certain computations. The following construction should be familiar by now.

The important thing to notice here is that we have arranged the splitting field so that the four roots, a, -a, b, -b, are very simple functions of the generators. In more traditional notation, a is \(2^{\frac{1}{4}}=\sqrt[4]{2}\), and b is \(2^{\frac{1}{4}}i=\sqrt[4]{2}i\) (or their negatives).

We will find it easier to compute in the flattened tower, a now familiar construction.

We can return to our original polynomial (over the rationals), and ask for its roots in the flattened tower, custom-designed to contain these roots.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

In a previous Sage exercise, we computed the fixed fields of single field automorphisms for finite fields. This was easy in the sense that we could just test every element of the field to see if it was fixed, since the field was finite. Now we have an infinite field extension. How are we going to determine which elements are fixed by individual automorphisms, or subgroups of automorphisms?

The answer is to use the vector space structure of the flattened tower. As a degree \(8\) extension of the rationals, the first \(8\) powers of the primitive element c form a basis when the field is viewed as a vector space with the rationals as the scalars. It is sufficient to know how each field automorphism behaves on this basis to fully specify the definition of the automorphism. To wit, \[\begin{aligned}\tau(x)&=\tau\left(\sum_{i=0}^7\,q_ic^i\right)&&q_i\in{\mathbb Q} \\ &=\sum_{i=0}^7\,\tau(q_i)\tau(c^i)&&\tau\text{ is a field automorphism} \\ &=\sum_{i=0}^7\,q_i\tau(c^i)&&\text{rationals are fixed}\end{aligned}\] So we can compute the value of a field automorphism at any linear combination of powers of the primitive element as a linear combination of the values of the field automorphism at just the powers of the primitive element. This is known as the power basis, which we can obtain simply with the .power_basis() method. We will begin with an example of how we can use this basis. We will illustrate with the fourth automorphism of the Galois group. Notice that the .vector() method is a convenience that strips a linear combination of the powers of c into a vector of just the coefficients. (Notice too that \(\tau\) is totally defined by the value of \(\tau(c)\), since as a field automorphism \(\tau(c^k)=(\tau(c))^k\). However, we still need to work with the entire power basis to exploit the vector space structure.)

The last line expresses the fact that tau_matrix is a matrix representation of the field automorphism, viewed as a linear transformation of the vector space structure. As a representation of an invertible field homomorphism, the matrix is invertible. As an order \(2\) permutation of the roots, the inverse of the matrix is itself. But these facts are just verifications that we have the right thing, we are interested in other properties.

Each row of the basis matrix is a vector representing an element of the field, specifically 1, c + (1/38)*c^5, c^2 - (1/22)*c^6, c^3 + (1/278)*c^7. Let's take a closer look at these fixed elements, in terms we recognize.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

The entire subfield structure of our splitting field is determined by the subgroup structure of the Galois group (), which is isomorphic to a group we know well. What are the subgroups of our Galois group, expressed as permutation groups? (For brevity, we just list the generators of each subgroup.)

tau above is the fourth element of the automorphism group, and the fourth permutation in elements is the permutation (2,3), the generator (of order 2) for the second subgroup. So as the only nontrivial element of this subgroup, we know that the corresponding fixed field is \({\mathbb Q}(\sqrt[4]{2}i)\).

Let us analyze another subgroup of order 2, without all the explanation, and starting with the subgroup. The sixth subgroup is generated by the fifth automorphism, so let us determine the elements that are fixed.

The first element indicates that the rationals are fixed (we knew that). Scaling the second element gives b - a as a fixed element. Scaling the third and fourth fixed elements, we recognize that they can be obtained from powers of b - a.

So the fixed field of this subgroup can be formed by adjoining b - a to the rationals, which in mathematical notation is \(\sqrt[4]{2}i - \sqrt[4]{2} = (1-i)\sqrt[4]{2}\), so the fixed field is \({\mathbb Q}(\sqrt[4]{2}i - \sqrt[4]{2}) = {\mathbb Q}((1-i)\sqrt[4]{2})\).

We can create this fixed field, though as created here it is not strictly a subfield of L. We will use an expression for b - a that is a linear combination of powers of c.

The .subfield() method returns a pair. The first item is a new number field, isomorphic to a subfield of L. The second item is an injective mapping from the new number field into L. In this case, the image of the primitive element c0 is the element we have specified as the generator of the subfield. The primitive element of the new field will satisfy the defining polynomial \(x^4+8\) you can check that \((1-i)\sqrt[4]{2}\) is indeed a root of the polynomial \(x^4 + 8\).

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Consider the third subgroup in the list above, generated by the permutation (1,4). As a subgroup of order \(2\), it only has one nontrivial element, which here corresponds to the seventh automorphism. We determine the fixed elements as before.

As usual, ignoring rational multiples, we see powers of a and recognize that a alone will be a primitive element for the fixed field, which is thus \({\mathbb Q}(\sqrt[4]{2})\). Recognize that a was our first root of \(x^4-2\), and was used to create the first part of original tower, N. So N is both \({\mathbb Q}(\sqrt[4]{2})\) and the fixed field of \(H=\langle(1,4)\rangle\).

\({\mathbb Q}(\sqrt[4]{2})\) contains at least one root of the irreducible \(x^4-2\), but not all of the roots (witness the factorization above) and therefore does not qualify as a normal extension. By part (4) of the automorphism group of the extension is not normal in the full Galois group.

As expected.