Galois extension of a field with Characteristic 0

[bham10246]

This is something I've been trying to work on on my own for the past few days but I'm not sure how to approach it.

My Question:
a. Let E be a Galois extension of a field F with characteristic 0. Prove that there is a unique smallest subfield K such that $F \subseteq K \subseteq E$, K is normal over F and E is subradical over K.
[One needs the following result: Let $A$ and $B$ be solvable subgroups of a group $G$ and suppose that $A$ is normal in $G$. Then $AB$ is solvable.]

b. Let f be an irreducible polynomial over $\mathbb{Q}$ which has degree 5 and at least two complex roots. Prove that $Gal(f)$ has order 10, 20, 60, or 120.

Thanks in advance for any kind of direction that you can provide me with...

 

[mathwonk]

what does smallest mean? the intersectiuoin of all such thigns? or smallest degree?

and are your extensions of finite degree?

so i guess you are claiming there is a unique largest normal solvable subgroup of a group. your description majkes it fairly obvious how to proceed.

 

[tacman]

Firstly, let us define the terms used in the question. A Galois extension of a field F is a field extension E such that every automorphism of E that fixes F is the identity map on E. The characteristic of a field is the smallest positive integer n such that n times the multiplicative identity of the field is equal to the additive identity. A subfield K of a field E is said to be normal over F if every irreducible polynomial in F[x] that has a root in K splits completely in K. A subfield E of a field K is said to be subradical over K if every element of E is algebraic over K and the splitting field of every polynomial in E[x] is contained in E.

a. To prove the existence of a unique smallest subfield K satisfying the given conditions, we will use the result stated in the question. Let A be the set of all automorphisms of E that fix F, and B be the set of all automorphisms of E that fix K. Since K is a subfield of E, every automorphism of E that fixes K will also fix F. Therefore, we have A \subseteq B.

Now, since E is a Galois extension of F, every element of E is algebraic over F. This implies that every element of E is also algebraic over K, since K is a subfield of E. This means that E is subradical over K.

Next, we prove that K is normal over F. Let f(x) \in F[x] be an irreducible polynomial with a root \alpha in K. Since \alpha is algebraic over K, there exists a monic polynomial g(x) \in K[x] such that g(\alpha) = 0. Since g(x) is monic and has a root \alpha in K, it must split completely in K. This implies that f(x) also splits completely in K, since K is a subfield of E. Therefore, K is normal over F.

Hence, K satisfies all the conditions required in the question, and it is the unique smallest subfield with these properties.

b. Let f(x) be an irreducible polynomial over \mathbb{Q} with degree 5 and at least two complex roots. By the fundamental theorem of algebra, f(x) has 5 roots in \mathbb{C}. Let