Algebra, Field extension - group stuff question, much appriciated.?

Question

Consider the extension E=F(x_1) of F=Q(a_0,...,a_4). Show that if E is contained in a radical extension of F then so is F(x_i) for each i and ehence so is F(x_1,...,x_5) = Q(x_1,...,x_5).
Any help is greatly appriciated.

Q is the rational field. a_i and x_i come from a polynomial x^5+a_4*x^4+a_3*x^3+...+a_0
a_i are algebraic over Q. x_i come from the identity
x^5+a_4*x^4+a_3*x^3+...+a_0 =
(x-x_1)...(x-x_n).
I am actually a bit lost here, so as far as my understanding goes, x_i are roots of the polynomial, and the field Q(x_1,...,x_5) is the 'root field' of the polynomial.
Thanks a lot for help.

Answer 1

Most important what are x_i ?
Be more clear, what are a_i? Algebraic elements, transcendental?
Is Q the rational field?
edit: ok, I'll think about it. It's not a trivial problem.
edit: ok, the proof is long
Definition: Let K a field. L is radical ext. of K if L=K(r_1,...r_n) such that for any i, there exists k _i such that r_i^{k_i} belongs to K(r_1,...,r_{i-1}). The sequence {r_i}_i is called radical sequence for L over K.
Proposition:
(i) If K Proof: Note that if {r_i}_i is a radical sequence of L over K and {s_i}_i is a radical sequence of M over L, then { r_i, s_i} is a radical sequence of M over K.
(ii) If K Proof: If {r_i}_i is a radical sequence of L_1 over K, then it remains a radical sequence of L_1L_2 over L_2.
(iii) If L_1, L_2 are both radical extensions of K, then their compositum field L_1L_2 is radical ext. of K.
Proof: Results from (ii) and (i).

Theorem: If K Proof: If K It results further that Gal(M|K) is finite.
If we prove that M=union U s(L), where the union is over Gal(M|K), the proof follows. Indeed, s(L) is isomorphic with L, therefore is radical ext. of K. By (iii), we will get that the union is radical ext. of K.

So let us prove: M=union U s(L).
M=K(Z), where Z are roots of the minimal polynomials over K of elements z in L. But from Galois theory we know that each such root is the image of a polynomial in Gal(M|K), s(a) =z.
Therefore U s(L) <= M< = U s(L) and we get equality.

Now we return to our problem.
For that we should adapt the proposition (iii) and the theorem:
(iv) If L_1, L_2 are contained in radical extensions of K, then their compositum field L_1L_2 is contained in a radical ext. of K.
Proof: If L_1 and L_2 are contained in the radical ext. M_1, M_2 respectively, then by (iii) M_1 M_2 is radical ext. of K and contains L_1 L_2.
Corollary to theorem: K Proof: The proof is the same with the theorem except a slight modification, using (iv) instead of (iii).

Now we have the theoretical tool.
F=Q(a_0,...,a_4) Suppose E is contained in a radical extension of F.
By corollary, we obtain that the normal closure of E is contained in a radical extension of K.
The roots of the minimal polynomial of x_1 over F (i.e., its conjugates) are x_1,x_2,...,x_5.
The normal closure of E=F(x_1) is F(x_1,...,x_5),
because a normal closure contains along with an element, all its conjugates.
Hence F(x_1,...,x_5) is contained in a radical ext. of K.
In particular, all F(x_i) are contained in a radical ext. of K.
We also show Q(x_1,...,x_5)=F(x_1,...,x_5):
Clearly Q(x_1,...,x_5)<= F(x_1,...,x_5)
Since Viete's formula show that a_i are symmetrical combinations of x_i, we obtain the reverse inclusion.