Special elements of fields extensions I was wondering if there is a method to find all elements $w\in F(\alpha_1,\ldots,\alpha_n)$ such that $F(w)=F(\alpha_1,\ldots,\alpha_n)$, where $\alpha_1,\ldots,\alpha_n$ are algebraic over the field $F$ of degrees $m_1,\ldots,m_n$, and $\alpha_1,\ldots,\alpha_n$ are algebraically independent over $F$? 
This means that $B=\{\prod_{i=1}^{n}\alpha_{i}^{k_i}, 0\leq k_i\leq m_i-1 \text{ for all } 1\leq i \leq n\}$ is a basis of $F(\alpha_1,\ldots,\alpha_n)$. 
Would this be easier if $\gcd(m_1,\ldots,m_n)=1?$ Or if $\operatorname{char}(F)=0$?
Could you show this on a specific example, e.g. $\mathbb{Q}(\sqrt2,\sqrt[3]3)?$ 
Thank you.
 A: Here's a slightly abstract answer:
Suppose we have found one primitive element $w$ with minimal polynomial $m(x) = x^n + \ldots \in F[x]$. Any element $y\in F(w)$ may be expressed as a rational polynomial of $w$ of degree at most $n-1$. Thus, if $F(y) = F(w)$ then $y = f(w)$ and $w = g(y)$. Hence, $w = (g\circ f)(w)$. Recall that we have an isomorphism,
$$
F(w) \cong F[x]/(m(x)).
$$
Then there is a correspondence between primitive elements and polynomials $h(x) \equiv x \bmod m(x)$ that decompose as $h = g\circ f$. The correspondence is certainly not one-to-one. However, I think the general intuition about when a polynomial decomposes as a composition is that its rare for both $g$ and $f$ to be non-linear. Hence, most other primitive roots come from composition with a linear factor.
Finding some primitive root is in practice not too difficult. I think most linear combinations of the $\alpha_i$ (as in your question) will result in primitive elements following an inductive argument like that suggested by Jared. 
A: (I'm assuming characteristic zero, although this applies to a wide variety of positive characteristic situations too)
You're thinking about it backwards: nearly every element $w$ of $E = F(\alpha_1, \cdots, \alpha_n)$ has the property that $F(w) = E$.
It's the elements $w$ such that $F(w) \neq E$ that are special. But there is an easy characterization:


*

*$F(w) \neq E$ if and only if $w$ is contained in an intermediate field $L \subsetneq E$.


So, if you can find all of the maximal subfields of $E$, then you have your description of the set of $w$'s such that $F(w) = E$: they are precisely the $w$'s that do not lie in any of those maximal subfields.
