Here is my problem: Let $k\subseteq F$ be a finite separable extension. Consider all homomorphisms $\phi:F\rightarrow \bar{k}$ extending the inclusion $k\subseteq \bar{k}$. Prove that $k\subseteq F$ is Galois if and only if the image of $\phi$ is independent of the homomorphism $\phi$.
I'm having a lot of trouble with this problem. I believe I have an outline for an argument, but feel like I'm overlooking some subtleties. I'd appreciate any feedback.
First note that the extension $k\subseteq F$ must be simple, as it is finite and separable. So, we may write $F=k(\alpha)$ for some element $\alpha$. Now, the embeddings $\phi$ described above are determined by where they send $\alpha$, and $\phi(\alpha)$ must be a root of the minimal polynomial $p(x)$ of $\alpha$ over $k$ (a separable polynomial).
For the 'if' direction, suppose that $\phi(F)=\phi'(F)$ for all such embeddings $\phi, \phi'$. Mapping $\alpha$ to itself shows that (perhaps up to some 'identification') $\phi(F)=F$ for all $\phi$. Let $\beta$ be any other root of $p(x)$. Take $\phi$ to be the homomorphism determined by $\alpha \mapsto \beta$. Then $\beta = \phi(\alpha)\in F$, so $F$ contains all roots of $p(x)$. It follows that $F=k(\alpha)$ is a splitting field for the separable polynomial $p(x)$, so $F$ is Galois over $k$, as needed.
For the converse, assume that $k\subseteq F$ is Galois. Then, in particular, this extension is normal. We'll show that $\phi(F)=F$ for all $\phi$ as above. Let $\beta$ be a root of the irreducible polynomial $f(x)\in k[x]$. For any $\phi$, we have
$$f(\beta)=0 \implies \phi(f(\beta))=f(\phi(\beta))=0,$$
which shows that $\phi(\beta)$ is a root of $f$. By normality of the extension, $\phi(\beta)\in F$, so we get $\phi(F)\subseteq F$. Dimension considerations give equality. So, the image of $\phi$ does not depend on $\phi$.
Sorry for the lengthy post; I'd appreciate any help.
