Is Every Prime Degree Splitting Field a Simple Extension?

[PsychonautQQ]

1. The problem statement, all variafbles and given/known data
Problem: Let E be a splitting field of f over F. If [E:F] is prime, show that E=F(u) for some u in E (show that E is a simple extension of F)

Homework Equations

Things that might be useful:
If E>K>F are fields, where K and F are subfields of E and F is a subfield of K, then [E:F] = [E:K][K:F]

since E is a splitting field of f:
f = a(x-(u1))(x-(u2))...(x-(up))
E = F(u1,u2,...,up)
Did i write this correctly in the sense that if [E:F] is prime and E is a splitting field of f then f will have p roots in E?

The Attempt at a Solution

My most promising method of proving this is using the multiplication theorem stated above, noting that E = F(u1,u2...up)

so...
p = [E:F] = [F(u1,u2...up):F(u)] * [F(u1),F)
since p is prime, this would force [F(u1,u2...up):F(u1)] to equal one and so F(u1,u2...up) = F(u1).

I realize this isn't the most thorough argument, and possibly just straight up incorrect. Anybody that knows what they're talking about have any comments? Am I on the right track?

 

[mighty2000]

Thank you for bringing this problem to our attention. I would like to offer some insights and suggestions on how to approach and potentially solve this problem.

Firstly, your understanding of the problem seems to be correct. As you stated, if [E:F] is prime and E is a splitting field of f, then f will have p roots in E. This is because the degree of the splitting field is equal to the degree of the polynomial, and since p is prime, the only possible factorization is p = 1 * p.

Your approach of using the multiplication theorem is a good start, but it needs some refinement. Here are some suggestions on how to improve your solution:

1. Use the definition of a splitting field: A splitting field of a polynomial f over a field F is a field extension E of F such that f splits into linear factors over E. In other words, f can be factored as f = (x-u1)(x-u2)...(x-up) where u1,u2,...,up are elements of E.

2. Use the definition of a simple extension: A field extension E of a field F is said to be simple if E = F(u) for some element u in E. This means that every element in E can be written as a polynomial in u with coefficients in F.

3. Use the fact that E is a splitting field of f: Since E is a splitting field of f, we know that f = (x-u1)(x-u2)...(x-up) where u1,u2,...,up are elements of E. This also means that E = F(u1,u2,...,up).

4. Use the fact that [E:F] is prime: As you stated, this means that [E:F] = [F(u1,u2,...,up):F(u)] * [F(u):F]. Since [E:F] is prime, one of these factors must be equal to 1. Can you show which one?

5. Use the fact that [F(u1,u2,...,up):F(u)] is equal to the degree of the minimal polynomial of u over F: This is a well-known result in field theory. Can you use this fact to show that [F(u1,u2,...,up):F(u)] is equal to 1?

I hope these suggestions help you in solving this problem. Remember to