Let $L$ be a splitting field of $f$ over $K$, with $\text{deg}(f) = n$. Prove that every $K$-automorphism of $L$ permutes the roots of $f$ in L.
I am a little confused with this question. I know that a $K$-automorphism is a map $\psi: L \rightarrow L$ which is a bijective isomoprhism. Furthermore I know that $f = \prod_{i=1}^{n} (X - \alpha_{i}) \in L[X]$. I have read in another post that under the automorphism it must hold that $\psi(\prod_{j=1}^{n} (\alpha_{j})) = \prod_{j=1}^{n} \alpha_{j}$, and from this one can deduce that the automorphism permutes the roots of $f$ in $L$. However, I don't really understand this step and why this equality must hold. If anybody could help me into the right direction I would appreciate that.
I can't see why $\psi(\prod_{j=1}^{n} (\alpha_{j})) = \prod_{j=1}^{n} \alpha_{j}$ implies that the automorphism permutes the roots in $L$ (though it is correct). However, the standard way to argue as follows: if $\alpha$ is a root of $f \in K[x]$, then $0 = f(\alpha) = \sum_i a_i \alpha^i$ for some $a_i \in K$. Then $$ 0 = \psi(f(\alpha)) = \sum_i a_i \psi(\alpha)^i = f(\psi(\alpha)) $$ so $\psi(\alpha)$ is too a root of $f$.