Let $L/Q$ be a field extension. Let $\sigma\in\textrm{Aut}_Q(L)$. Let $f(x) \in Q[x]$ be a polynomial. Show that $f(σ(α)) = σ(f(α))$ for all $α ∈ L.$

Question

Let $L/Q$ be a field extension. Let $\sigma\in\textrm{Aut}_Q(L)$. Let $f(x) \in Q[x]$ be a polynomial. Show that $f(σ(α)) = σ(f(α))$ for all $α ∈ L.$

The statement is obviously true for $α ∈ Q$ because $\text{Aut}_Q(L)$ fixes $Q$. However, I don't know how to extend the conclusion to other elements in $L$.

Answer 1

Write $f(x)=\sum_{i=0}^na_ix^i$ with $n\in\Bbb N$ and $a_i\in Q$. Then for every $\alpha\in L$ we have \begin{align} \sigma(f(\alpha)) &=\sigma\left(\sum_{i=0}^na_i\alpha^i\right)\\ &=\sum_{i=0}^na_i\sigma(\alpha)^i\\ &=f(\sigma(\alpha)) \end{align}