# 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 $α ∈ L.$

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 $$L$$.

• Extend it to monomials and then to $f(x)=a_nx^n+\cdots +a_0$. – Dietrich Burde Mar 4 at 16:32
• You may try first with $f(x) = 2 x^2+1$ – reuns Mar 4 at 16:34
• Start from the R.H.S.- use the fact that $\sigma(xy)=\sigma(x)\sigma(y)$ and $\sigma(x+y)=\sigma(x)+\sigma(y)$- then, finally, use the fact that $\sigma$ fixes $Q$. – Cardioid_Ass_22 Mar 4 at 17:29

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}