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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 $L$.

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  • $\begingroup$ Extend it to monomials and then to $f(x)=a_nx^n+\cdots +a_0$. $\endgroup$ – Dietrich Burde Mar 4 at 16:32
  • $\begingroup$ You may try first with $f(x) = 2 x^2+1$ $\endgroup$ – reuns Mar 4 at 16:34
  • $\begingroup$ 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 $Q$. $\endgroup$ – Cardioid_Ass_22 Mar 4 at 17:29
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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}

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  • $\begingroup$ Beautiful proof! Thanks so much! $\endgroup$ – RandomThinker Mar 6 at 1:02

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