Lemma: Suppose $K=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ is a field extension and $\sigma \in Aut(K)$ such that $\sigma$ fixes $F$ pointwise. Then $\sigma$ is uniquely determined by $\{\sigma(\alpha_i):1 \leq i \leq n\}$.
Proof:- Establishing uniqueness is straightforward. For any $\tau \in Aut(K)$ with the same properties, set $E=\{\alpha \in K: \sigma(\alpha)=\tau(\alpha)\}$. Then $E$ is a subfield of $K$ containing $F$ and all the $\alpha_i$, and hence $K$, thereby implying $K=E$, i.e., $\sigma = \tau$.
However how do we know the action of $\sigma$ on a general element of $K$?