Properties of Galois Extensions

[jk1100]

Homework Statement

Suppose F -- K is a Galois extension and f(x) in F[x] is an irreducible
polynomial that has a root in K. Show that f(x) factors into a product
of linear factors in K[x].

Homework Equations

I read on Wolfram this fact is equivalent to being a Galois extension, but I am drawing a complete blank on how to show that.

The Attempt at a Solution

Idea 1:
I know that if K is an extension of F and f(x) is a polynomial with coefficients in F that any F-automorphism of K will map roots on f(x) to another roots of f(x). Now I know at least one root of f(x) is in K; so the image of that root under any F-automorphism will also be in K and will also be a root of f(x)

Idea 2:
I also know I can build a tower of fields by adjoining the element a st f(a)=0 to obtain
F -- F(a) -- K
Because the root a must be in K so that field F(a) must sit between F and K. Since F -- K is a Galois extension and F(a) is an intermediate field I know F(a) -- K must be a Galois extension as well

Idea 3:
I also know since K is an extension of F and it is Galois that then it is the splitting field for some polynomial in F[x].

I just need some way to bring these ideas together (or take one all the way). I feel like I am half way there in bunch a different ways, but I am just blanking of finishing the proof.

 

[mighty2000]

Thank you for your post. Your ideas are on the right track, but let me offer a more complete solution for you.

First, we know that since F -- K is a Galois extension, then K is the splitting field for some polynomial p(x) in F[x]. This means that all the roots of p(x) are in K.

Now, since f(x) is irreducible and has a root in K, then we know that f(x) must divide p(x) in F[x]. This is because if p(x) factors into two polynomials, one of which is f(x), then f(x) must have a root in K.

Next, we will use the fact that any F-automorphism of K will map roots of f(x) to other roots of f(x). Since K is the splitting field for p(x), any F-automorphism of K will map roots of p(x) to other roots of p(x). But since f(x) divides p(x), then any F-automorphism of K will map roots of f(x) to other roots of f(x). This means that all the roots of f(x) are in K, and since f(x) is irreducible, it must have a root in K.

Finally, since all the roots of f(x) are in K, then f(x) must factor into a product of linear factors in K[x]. This is because if f(x) has a root a in K, then we can write f(x) = (x-a)g(x), where g(x) is a polynomial in K[x]. We can continue this process until we have factored f(x) completely into linear factors in K[x].

I hope this helps clarify the proof for you. Let me know if you have any further questions.