Prove the minimal polynomial $a$ over $\mathbb{Q}$ is equal to the minimal polynomial $\overline{a}$ over $\mathbb{Q}$ Let $a \in \mathbb{C}$ be such that $a \notin \mathbb{R}$ and $a$ is algebraic over $\mathbb{Q}$. Let $\overline{a}$ be the complex conjugate of $a$.
Let $f(x)$ be the minimal polynomial $a$ over $\mathbb{Q}$ and $g(x)$ be the minimal polynomial $\overline{a}$ over $\mathbb{Q}$.
(1) Prove that $f(x) = g(x)$.
(2) Prove or disprove that $[\mathbb{Q}(a) : \mathbb{Q}]$ is an even integer.
$\textbf{My Attempt:}$
Question (1):
Since, $f(x)$ is the minimal polynomial $a$ over $\mathbb{Q}$, if $f(x)$ is monic polynomial and irreducible over $\mathbb{Q}$ and $f(a) = 0$.
Similarly $g(x)$ is the minimal polynomial $\overline{a}$ over $\mathbb{Q}$, if $g(x)$ is monic polynomial and irreducible over $\mathbb{Q}$ and $g(\overline{a}) = 0$.
Then, $f(a) = 0$ and $g(\overline{a}) = 0$.
So, $f(a) = g(\overline{a})$ which means $f(x)$ and $g(x)$ has exactly same roots.
Therefore, $f(x) = g(x)$.
Question (2):
Want to prove that $[\mathbb{Q}(a) : \mathbb{Q}]$ is an even integer.
Since, notice that $a \in \mathbb{C}$ be such that $a \notin \mathbb{R}$ and $a$ is algebraic over $\mathbb{Q}$.
Which means $a$ must equal to some terms with adding, substracting, multiplying or dividing $i = \sqrt{-1}$ somewhere.
So, in order to get $f(x)$, we must square both side to get rid of $i = \sqrt{-1}$.
So, by square both side, $f(x)$ will have degree of a even number.
Since, notice that $[\mathbb{Q}(a) : \mathbb{Q}] = $ the degree of $f(x)$.
So, $[\mathbb{Q}(a) : \mathbb{Q}]$ is even.
$\textbf{My Question:}$
For question (1): I don't think the prove is correct, but I have no other idea on proving $f(x) = g(x)$.
For question (2): Is this correct ? (if so, are there ways to make this prove more theoretical ?).
 A: Your logic going from $f(a)=g(\bar{a})$ and concluding that they must have the same roots is flawed. How do you know that? The way to prove the first part is to take the minimal polynomial of $a$, $f(x)$, and showing that
$$
0=f(a)=\overline{f(a)}=f(\overline{a})\ ,
$$
and concluding that $f$ is also the minimal polynomial of $\overline{a}$. You can go figure out the details, in particular, why we can say $\overline{f(a)}=f(\overline{a})$ and why $f(\overline{a})$ is not only a polynomial with $\overline{a}$ as a root, but must also be its minimal polynomial.
EDIT: My original answer to the second part was completely wrong. Here is a counterexample which shows that the answer to part $2$ of the question is 'No'.
Consider the polynomial $f(X)=X^3-2$ over $\mathbb{Q}$. This is the go-to example for a non-normal extension of $\mathbb{Q}$. Let $\alpha=\sqrt[3]{2}$ and $\omega=e^{2\pi i/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then the roots of $f$ are $\alpha, \omega\alpha$ and $\omega^2\alpha$. It can be shown that $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega\alpha)$ and $\mathbb{Q}(\omega^2\alpha)$ are all isomorphic and that all their degrees over $\mathbb{Q}$ are $3$. Taking $a=\omega\alpha$ gives the desired result.
