# 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 ?).

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.

• why $\overline{a}/|a|^2 \in \mathbb{Q}(a)$ implied $\overline{a} \in \mathbb{Q}(a)$? $|a|^2$ itself may not be not in $\mathbb{Q}(a)$ Feb 26 at 7:13
• @PeopleMoutainPeopleSea You're right, my answer to the second part is completely wrong. Feb 26 at 12:34