Galois Theory question

[Metric_Space]

Homework Statement

Let L/K be a Galois extension with Galois group isomorphic to A4. Let g(x) ϵ K [x] be an irreducible polynomial that is degree 3 that splits in L. Show that the Galois group of
g(x) over K is cyclic.

Homework Equations

The Attempt at a Solution

I know, Galois means normal and separable.

Irrdeducible poly of degree 3 that splits in L means it splits into linear factors.

ie. g(x)=(x-a1)*(x-a2)*(x-a3)

I am thinking Galois correspondence might come in handy, but not sure how...any ideas?

I'm not sure what information Cyclic gives.

 

[mighty2000]

I know that cyclic groups are abelian and have a generator that generates the entire group, but I'm not sure how that relates to the Galois group of a polynomial.

First, let's review some basic definitions and properties of Galois extensions and groups. A Galois extension L/K is a field extension that is both normal and separable. This means that every irreducible polynomial in K[x] that has a root in L splits into linear factors in L[x]. The Galois group of L/K is the group of automorphisms of L that fix K pointwise, denoted as Gal(L/K). In this case, we are given that the Galois group of L/K is isomorphic to A4, the alternating group of degree 4. This means that the Galois group has order 12 and can be generated by a single element of order 3 and a single element of order 2.

Now, let's consider the polynomial g(x) in K[x] that is irreducible of degree 3 and splits in L. This means that g(x) has 3 distinct roots in L, say a1, a2, and a3. Since L/K is a Galois extension, we know that the Galois group acts transitively on the roots of g(x). This means that there exists an element of the Galois group that sends a1 to a2 and a2 to a3, and another element that sends a1 to a3 and a3 to a2. These elements must have order 3 since they are sending each root to a different root.

Now, we can use the fact that the Galois group is isomorphic to A4 to show that the Galois group of g(x) over K is cyclic. Since A4 has a subgroup of order 3, we know that the Galois group of g(x) has a subgroup of order 3 as well. This subgroup is generated by the element of the Galois group that sends a1 to a2 and a2 to a3. Since this element has order 3, it generates a cyclic subgroup of order 3.

Therefore, the Galois group of g(x) over K is a cyclic group of order 3, which means that it is generated by a single element of order 3. This element corresponds to an automorphism of L that sends a1 to a2 and a2 to a3, and fixes a3.