Writing elements of a finitely generated field extension in terms of the generators I have a question that arose in the process of solving a different problem in Galois theory. That problem asked to show that if $\alpha_1, \dots, \alpha_n$ are the generators of an extension $K/F$, then any automorphism $\sigma$ of $K$ which fixes $F$ is uniquely determined by $\sigma(\alpha_1), \dots, \sigma(\alpha_n)$.
I originally thought this question would be easy because it seemed intuitively obvious from the minimality of $K$ as an extension that any element in $K$ could be written in terms of the generators. However, it isn't given that each of the $\alpha_i$ are algebraic, so writing an explicit expression for $k \in K$ in terms of the $\alpha_i$'s proved to be difficult.
Is must be possible to write any $k \in K$ in terms of the $\alpha_i$ because the $\sigma \in \mathrm{Aut}(K/F)$ are indeed determined by what they do on the generators (I proved this using induction and casework). Is there an easy way to see it, though?

The sophisticated way of doing this is the following:
Take two automorphisms $\sigma, \tau$ of $K$, which fix $F$ and agree on $\{\alpha_1, \dotsc, \alpha_n\}$. The set $M := \{x \in K | \sigma(x)=\tau(x)\} \subset K$ is a field, which contains $F$ and all $\alpha_i$. By the definition (minimality of $K$ as you mentioned) of $K$, we obtain $K \subset M$, hence $\sigma = \tau$.
