This post is about the Galois group of over , where is the splitting field of , where is an odd prime.

First we show that the polynomial is irreducible over . This follows immediately by Eisenstein’s Criterion, since , and .

Next, we show that the splitting field of in is , where is a primitive p-th root of unity. The roots of are .

The splitting field contains and . Thus .

On the other hand, contains all the roots of , hence splits in . Thus , since is the smallest field that contains and all the roots of . All in all, we have that the splitting field .

The next part involves determining the Galois group of over . We have . Since (minimal polynomial ), and (minimal polynomial the cyclotomic polynomial ), thus . Here we have used the lemma that suppose and with , then .

What the Galois group does is it permutes the roots of . Let be an element of the Galois group. can possibly be , a total of choices. Similarly, , a total of choices. All these total up to elements, which is exactly the size of the Galois group.

The above Galois group is described by how its elements act on the generators. For a more concrete representation, we can actually prove that the Galois group above is isomorphic to the group of matrices , where , . We denote the group of matrices as .

To show the isomorphism, we define a map , mapping to .

Notation: is defined on the generators as follows, , .

We can clearly see that the map is bijective. To see it is a homomorphism, we compute .

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