Question:
Let $k$ be a infinite field. If $F = k(\alpha_1,...,\alpha_r)$, with each $\alpha_i$ separable over $k$, prove that there exist $c_1,...,c_r \in k$ such that $F = k(c_1\alpha_1+...+c_r\alpha_r)$ .
I want to prove it by embedding $F$ into $\mathcal{M}_n(k)$. I think the proof would be neater than the ones using Galois Theory or explicitly constructing irreducible polynomials.
Let $[F:k] = n$. $c_1\alpha_1+...+c_r\alpha_r$ can be embedded into $\mathcal{M}_n(k)$ and all I have to prove is that the corresponding matrix has rank $n$, but do not know how to proceed my proof. Please help.
The standard proof for this uses Galois theory, I hope you are familiar with this. The theorem you're trying to prove is more or less the Primitive Element Theorem, and it seems a difficult problem coming up with the proof on your own. Here is a sketch. I simplify to $F=k(\alpha,\beta)$, the general case follows by induction.
Lemma. $F/k$ has only finitely many intermediate fields.
Proof. As $F/k$ is finite and separable, since $\alpha$ and $\beta$ are separable, we may take the Galois closure to get a finite Galois extension $L/k$ with $F\subset L$. This has a finite Galois group, and from the fundamental theorem of Galois theory, we know that intermediate fields correspond to subgroups of the Galois group, of which there must be finitely many. As $L/k$ has finitely many intermediate fields, then certainly $F/k$ must as well. //
Now, form the subfields $k(\alpha+c\beta)$ for $c\in k$. These are clearly intermediate fields. As $k$ is infinite, but there are only finitely many intermediate fields by the lemma, we see that $k(\alpha+c\beta) = k(\alpha+c'\beta)$ for some $c\neq c'$. We then see $(c-c')\beta$ must lie in this field, and as $c-c'$ is non-zero, thus invertible, so must $\beta$, and then so must $\alpha$. Thus, we get $k(\alpha+c\beta)=k(\alpha,\beta)=F$, and we are done.