I am trying to show that if $K$ is a field and $f \in K[x] $ has exactly $n$ distinct roots, say $\alpha_1 ,..., \alpha_n \in L $ where $L$ is a splitting field (so $L=K(\alpha_1 , ..., \alpha_n ) $) and $\sigma \in \text{Aut}(L/K) $ then $\sigma $ is completely determined by $\sigma (\alpha_1) ,... ,\sigma(\alpha_n ).$
Proposition
Let $K_i =K_{i-1}(\alpha_i )$ for each $i\geq 1 $ with $K_0=K$. Then $K_i =K(\alpha_1,...,\alpha_i).$
Let $\sigma \in \text{Aut}(L/K)$. Then $\sigma $ is completely determined on $K_j $ by $\sigma(\alpha_1),...,\sigma(\alpha_j).$
My attempt of a proof
For $j=1$, $K_1=K(\alpha_1)$ which is a simple finite extension ($\alpha_1$ is a root of $f$ after all ) so $K_1$ has $K-$basis $1,\alpha_1 ,..., \alpha_1 ^{m-1} $ where $m$ is the degree of the minimal polynomial of $\alpha_1 $ over $K$.
So if $\beta \in K_1 , \beta =a_0+a_1 \alpha_1 +...+a_{m-1} \alpha_1 ^{m-1}$ with $a_i \in K$ and so $\sigma (\beta) = a_0 +a_1 \sigma(\alpha_1)+...+a_{m-1} (\sigma (\alpha_1 ))^{m-1}$ and so is determined by $\sigma(\alpha_1 ).$
Suppose now it holds for some $j$ with $1\leq j \leq n-1.$ That is $\sigma \in \text{Aut}(L/K)$ is completely determined on$K_j$ by $\sigma(\alpha_1),...,\sigma(\alpha_j).$
Then $K_{j+1}=K_j (\alpha_{j+1})$ and similar to base case, $K_{j+1} $ has $K_j $-basis $1, \alpha_{j+1},...,\alpha_{j+1}^{m-1}$ where $m$ is the degree of the minimal polynomial of $\alpha_{j+1}$ over $K_j$.
If $\beta \in K_{j+1}$ then $\beta = a_0 +...+a_{m-1}\alpha_{j+1}^{m-1}$ with $a_i \in K_j$ and so $\sigma(\beta)= \sigma(a_0) +...+\sigma(a_{m-1})(\sigma(\alpha_{j+1}))^{m-1}$.
Now each $a_i \in K_j$ and so its value is determined by $\sigma(\alpha_1),...,\sigma(\alpha_n)$ and so $\sigma $ is completely determined on $K_{j+1}$ by additionally knowing the value of $\sigma(\alpha_{j+1})$. i.e. $\sigma $ on $K_{j+1}$ is determined completely by $\sigma(\alpha_1), ..., \sigma(\alpha_{j+1}).$
So my claim is proved by induction (I think).
Then we can just set $j=n$ and $\sigma \in \text{Aut}(L/K)$ is completely determined on $K_n=L$ by $\sigma(\alpha_1),...,\sigma(\alpha_n)$.
Is this valid? I think it is ok.
Thank you in advance.
That looks fine. As I understand, formally you prove by induction that if two automorphisms $\sigma, \tau$ agree of $\alpha_1,...,\alpha_j$ then they agree on $K_j$. This is correct, though it is easier to prove this statement without induction. First of all, a general element of $K(\alpha_1,...,\alpha_n)$ has the form $\frac{f(\alpha_1,...,\alpha_n)}{g(\alpha_1,...,\alpha_n)}$ where $f,g\in K[x_1,...,x_n]$ and the denominator doesn't vanish. So you could compute everything in one step instead of using induction.
And there is an even easier proof. If $\sigma, \tau\in\text{Gal}(L/K)$ agree on $\alpha_1,...,\alpha_n$ then we can define:
$M=\{x\in L: \sigma(x)=\tau(x)\}$
Clearly this is a subfield of $L$ which contains $K$ and the elements $\alpha_1,...,\alpha_n$. Since $L=K(\alpha_1,...,\alpha_n)$, by definition we must have $M=L$.