Prove that complex conjugation of $\mathbb{C}$ preserves the splitting field

Question

Q) Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial over $\mathbb{Q}$ and let $K\subset \mathbb{C}$ be the splitting field of $f(x)$ in $\mathbb{C}$. Prove that the complex conjugation of $\mathbb{C}$ preserves $K$.

My attempt: Consider the polynomial $f(x) = x^n + 1$ which is irreducible over $\mathbb{Q}$. It's roots are $\{\sqrt[n]{-1},w\sqrt[n]{-1},...,w^{n-1}\sqrt[n]{-1}\}$ where $w$ is a primitive $n^{th}$ root of unity. These roots are in $K$. Let $\phi\in Aut(\mathbb{C})\implies \phi(w^k \sqrt[n]{-1}) = \phi(w)^{k} \phi(\sqrt[n]{-1})$ since $\phi$ is a homomorphism. I can see that $\phi(w)^k$ is also an $n^{th}$ root of unity but am not sure what $\phi(\sqrt[n]{-1})$ is so as to check that $\phi(w^k \sqrt[n]{-1})$ is an $n^{th}$ root of $-1$? Appreciate a hint.

Answer 1

Since $f(x)$ is a polynomial with real coefficients, for each root $r$ of $f(x)$, $\overline r$ is also a root of $f(x)$. On the other hand, $K$ is the smallest subfield of $\mathbb C$ which contains the roots of $f(x)$. But, since the set of the roots of $f(x)$ is preserved by the conjugation, so is the field $K$.

Answer 2

Well, complex conjugation ${\Bbb C}\rightarrow{\Bbb C}:z=a+ib \mapsto\overline z = a-ib$ is a field automorphism which leaves ${\Bbb R}$ fixed. Thus for each polynomial $f(x)\in{\Bbb R}[x]$, if $z$ is a zero of $f(x)$, then $$f(\overline z) = \overline{f(z)} = \overline 0 = 0$$ and so $\overline z$ is a zero of $f(x)$. It follows that the splitting field $K$ of $f(x)$ is kept fixed.