D+F prop. 14.1.2

[(Prop. 14.1.2)] Given the conditions that $K/F$ is a field extension, $\alpha \in{K}$ is algebraic over $F$ and $\sigma \in{Aut(K/F)}$ then prove that any polynomial with coefficients in $F$ having $\alpha$ as a root also has $\sigma(\alpha)=\sigma\alpha$ as a root.

Some notes: $\alpha\in{K}$ albegraic over $F$ means that $\alpha$ is the root of some nonzero polynomial in $F$.
$\sigma \in{Aut(K/F)}$ means that $\sigma$ is an operation preserving permutation on the elements of $K$ and $\sigma$ fixes $F$ i.e. $\sigma(a)=a \quad \forall a\in{F}$.


$\textit{Proof}$ :
Let $\alpha$ be a root $f(x)\in{F}[x]$. Dividing a polynomial through by a nonzero constant to make it monic does not change the roots. So, we may write $$\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=0 \quad a_i\in{F}$$ Now applying $\sigma$ to both sides, we get $$\sigma(\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=\sigma(0)=0 \quad (\star)$$ $$=\sigma(\alpha^n) + \sigma(a_{n-1}\alpha^{n-1} + \cdots + \sigma(a_0)=0$$ since $\sigma$ is a homomorphism with respect to addition. $$=(\sigma(\alpha))^n+ \sigma(a_{n-1})\sigma(\alpha)^{n-1} + \cdots + \sigma(a_0)$$ since $\sigma$ is also a homomorphism with respect to multiplication. Now, since $\sigma$ fixes $F$, $$=(\sigma\alpha)^n + a_{n-1}(\sigma\alpha)^{n-1}+ \cdots + a_0$$ Which states exactly that $\sigma\alpha$ is also a root of $f(x)$.

$(\star): \sigma(0)=0$ since $\sigma$ is an automorphism: $\sigma(0)=\sigma(0+0)=\sigma(0)+\sigma(0) \Longrightarrow \sigma(0)=0$. $\blacksquare$