If $K/F$ is Galois and $E$ is a subextension then $E$ is generated by roots of a polynomial over $F$? Let $K/F$ be finite Galois field extension, then $K$ is the splitting
field of a separable polynomial $p$ over $F$, i.e. $K=F(a_{1},..a_{n})$
where $p=(x-a_{1})...(x-a_{n})$.
My question is: is it true that if $E$ is a subextension of $K/F$
then $E$ is also of the form $E=F(b_{i_1},..b_{i_t})$ where $g=(x-b_{1})...(x-b_{k})$ 
is in $F[x]$ ?
 A: The first statement is not even necessarily correct. Take $K = \Bbb{Q}(\sqrt[3]{2},\zeta_3)$ that  is the splitting field of $p(x) = x^3 - 2 = (x - \sqrt[3]{2})(x- \sqrt[3]{2}\zeta_3)(x- \sqrt[3]{2}\zeta_3^2) \in \Bbb{F}[x]$ where $F = \Bbb{Q}$. So just because $K$ is the splitting field of a separable polynomial $p$ over $F$, it is true that $K$ is $F$ adjoined a finite number of elements but they need not be roots of the polynomial $p$.
A: A finite Galois extension is a finite, separable, normal extension. Any subextension of a separable extension is separable, similarly for finite extensions (mostly obviously). However, it is certainly not true for normality.
Consider arbitrary $F$, $K$ the splitting field of a minimal (separable) polynomial in $F[X]$ with at least two distinct roots. Then $F[\alpha]$, where $\alpha$ is just one of the roots of the polynomial is not normal.
But if you want $E$ to be generated by some roots of some polynomial, then of course this is true. You don't even need $K$ to be Galois, just finite (and hence algebraic). Any finite field extension is algebraic, so for arbitrary $F\subseteq E\subseteq K$ we have $E=F[\alpha_1,\ldots,\alpha_n]$ (this is true for example because $E$ is a subspace of $K$ as a vector field over $F$), an then $\alpha_j$ are algebraic, so if you take for $p$ the product of their respective minimal polynomials over $F$, they will all be roots of $p$.
Furthermore, if $K$ was separable, then if you take only distinct minimal polynomials of $\alpha_j$ (so you won't take $x^2+1$ twice for $\alpha_1=i,\alpha_2=-i$, for instance), $p$ will also be separable (because any $\alpha_j\in K$ is separable).
A: Now that we have the question nailed down: Let $E/F$ be a finite extension. The first question is whether $E$ can be written as $F(b_1, \ldots, b_n)$. This is true. You could, for example, take $\{b_i\}$ to be a basis for $E$ as a vector space over $F$. Each $b_i$ has some minimal polynomial $p_i \in F[x]$ over $F$, and $p_1 \cdots p_r$ seems to do what you want.
Note that the above choice of $\{b_i\}$ was probably not very efficient with regard to $n$. The primitive element theorem says that if $E/F$ is separable then $E = F(\alpha)$.
A: It is true is that for any normal extension $\,A/B\,$ , $\,A\,$ is the splitting field of a familiy of polynomials over $\,B\,$ , so if the extension is finite we can take one polynomial s.t. $\,A\,$ is its splitting field (as there must be a finite number of roots involved and thus we can take the product of a finite number of polynomials. This is stressed in, for example, Lang's "Algebra").
Thus, it is true in our case that $\,K=F(a_1,...,a_n)\,$ , as you wrote.
As for the second question the answer is no: for example, the Galois extension of the polynomial $\,x^4-2\in\Bbb Q[x]\,$ contains the subextension $\,\Bbb Q(\sqrt[4] 2)/\Bbb Q\,$ is not normal...
