2
$\begingroup$

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 ?).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ why $\overline{a}/|a|^2 \in \mathbb{Q}(a)$ implied $\overline{a} \in \mathbb{Q}(a)$? $|a|^2$ itself may not be not in $\mathbb{Q}(a)$ $\endgroup$ Feb 26 at 7:13
  • $\begingroup$ @PeopleMoutainPeopleSea You're right, my answer to the second part is completely wrong. $\endgroup$
    – Teddy38
    Feb 26 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.