Galois group. $K$-automorphisms take adjoined roots to other roots of minimal polynomial or take roots of $f$ to other roots of $f$ Say $f\in \Bbb Q[x]=K$ is irreducible, and $L/K$ is a field extension where $L=\Bbb Q(\alpha,\beta)$ where $\alpha,\beta$ are not in $\Bbb Q$ i.e. we need to adjoin both for $L$ to be the splitting field of $f$ since $\alpha,\beta$ are both roots of $f$. The minimal polynomials of $\alpha,\beta$ however aren't $f$.

Do 1) $K$-automorphisms of $L$ take roots of the minimal polynomial of $\alpha$ to other roots of the minimal polynomial of $\alpha$ and roots of the minimal polynomial of $\beta$ to other roots of the minimal polynomial of $\beta$

Or 2) Take roots of $f$ to other roots of $f$?

Or 3) These are somehow the same?



The first makes sense to me, but the second seems to be what we do in the examples. Thanks



I just mean that if we had $L=K(\alpha,\beta)$ is the splitting field of $f$, and we want to find $K$-automorphisms of $L$, in order to find elements of $\rm{Gal}(L/K)$, do we determine $\sigma\in\rm{Gal}(L/K)$ by seeing where it maps $\alpha,\beta$ in which we know these can only go to either, other roots of their respective minimal polynomials, or other roots of $f$.(Hopefully that makes sense)
Both are true. In general, let $g(x) = a_0 + a_1x + \cdots + a_nx^n$ be a polynomial with coefficients $a_i \in K$, $y \in L$ be a root of $g$, and $\phi$ be a $K$-automorphism of $L$. Then
$$0 = \phi(g(y)) = a_0 + a_1\phi(y) + \cdots + a_n\phi(y) = g(\phi(y)).$$
Thus $\phi$ permutes the roots of $g$.
Both are true. In general, let $g(x) = a_0 + a_1x + \cdots + a_nx^n$ be a polynomial with coefficients $a_i \in K$, $y \in L$ be a root of $g$, and $\phi$ be a $K$-automorphism of $L$. Then
$$0 = \phi(g(y)) = a_0 + a_1\phi(y) + \cdots + a_n\phi(y) = g(\phi(y)).$$
Thus $\phi$ permutes the roots of $g$.