Galois group is determined by action on roots of polynomial This is a very simple question but I can't give a good answer to it. If we have a field $K$ and a Galois extension $L/K$ where $L=K(\alpha_1,\ldots,\alpha_n)$ and $\alpha_1,\ldots,\alpha_n$ are the roots of some separable polynomial in $K[x]$, then any automorphism of $L/K$ is uniquely determined by its action on the $\alpha_i$. I suppose it is 'clear' because any element of $L$ is formed by combining elements of $K$ and the $\alpha_i$ with field operations but this is not really rigorous enough for me. I'm sure there's a better explanation but I can't see it.
(Explanation in suggested problem didn't explain the specific point I did not understand, but I understand now.)
 A: If you have a base field $F$ and $K/F$ is an extension, say $K = F(\alpha)$ then $K \cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + \dots + a_1X + a_0$ is an irreducible polynomial having $\alpha$ as a root. Now in $K$ you have $\alpha^n = -(a_{n-1}\alpha^{n-1} + \dots + a_1\alpha + a_0)$ so $\lbrace 1, \alpha, \dots, \alpha^{n-1}\rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}\alpha^{n-1} + \dots + a_1\alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + \dots + a_1X + a_0 \equiv 0 \bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + \dots + a_1 X + a_0$. But $\deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $\alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
