Let be $\varepsilon$ one 9th primitive root of 1.
I have to calculate Irr($\varepsilon$,$\mathbb{Q}$); this is easy, because Irr($\varepsilon$,$\mathbb{Q}$)=$\Phi_{9}(x)=x^{6}+x^{3}+1$.
My problem is the following; I have to write the 9th primitive roots as linear combination of elements of one $\mathbb{Q}$-base of $\mathbb{Q}(\varepsilon)$. I saw in a post here, that the primitive roots have to have the form $e^{\frac{2ik\pi}{9}}$, with $k=\{0,\cdots,8\}$ and gcd($9,k$)=1, so this implies that the 9th primitives of 1 are, the complex numbers with the form that I have mentioned before, with $k=\{1,2,4,5,7,8\}$.
Well, what I tought is, as $[\mathbb{Q}(\varepsilon):\mathbb{Q}]=\Phi(9)=6$, I have that one $\mathbb{Q}$-base of $\mathbb{Q}(\varepsilon)$ is $\{1,\varepsilon,\cdots,\varepsilon^{5}\}$, so for $k=\{1,2,4,5\}$ I have no problem because they are elements of the base, so this is finished. So for the elements with $k=\{7,8\}$, as Irr$(\varepsilon,\mathbb{Q})=\Phi_{9}(x)$, from there I have that $\varepsilon^{6}+\varepsilon^{3}+1=0$, so $\varepsilon^{8}=-\varepsilon^{2}-\varepsilon^{5}$ and $\varepsilon^{7}=-\varepsilon-\varepsilon^{4}$. Is this correct?
The next thing that the exercise ask to me to do is calculate $[\mathbb{Q}(\varepsilon):\mathbb{Q}(\varepsilon^{3})]$. I have just calculated that $[\mathbb{Q}(\varepsilon):\mathbb{Q}]=6$, so, $[\mathbb{Q}(\varepsilon):\mathbb{Q}(\varepsilon^{3})]$ should divide 6, and as $[\mathbb{Q}(\varepsilon^{3}):\mathbb{Q})]=2$ (because $\mathbb{Q}(\varepsilon^{3})=\mathbb{Q}(\sqrt{3}i)$, and the irreducible here is $x^{2}+3$), then $[\mathbb{Q}(\varepsilon):\mathbb{Q}(\varepsilon^{3})]=3$. Is this correct?
The next part of the exercise is to show that $Irr(\varepsilon^{3},\mathbb{Q})=Irr(\varepsilon^{6},\mathbb{Q})$, this is not complicated because, they are both equal to $\mathbb{Q}(\sqrt{3}i)$. One, more time, is this correct?
The next following two parts are where I don't know how to proceed.
PART ONE
I have to justify, that exists one $\mathbb{Q}$-automorphism, $\sigma:\mathbb{Q}(\varepsilon)\rightarrow\mathbb{Q}(\varepsilon)$ that $\sigma(\varepsilon)=\varepsilon^{2}$.
PART TWO
I have to justify that for all $\mathbb{Q}$-automorphism $\tau:\mathbb{Q}(\varepsilon)\rightarrow\mathbb{Q}(\varepsilon)$ verifies that $\tau(\varepsilon)\neq\varepsilon^{3}$.
Can you give hints to continue please.
Thanks for all the answers.
Best Answer
As for part $1$, extend that map to the rest of $\mathbb{Q}(\epsilon)$ and just straight up check it. You have a representation of any element in $\mathbb{Q}(\epsilon)$. So check that by the axioms that if $\sigma(\sum a_i \epsilon^i) = \sum a_i \epsilon^{2i} $ , then $\sigma$ is a field automorphism. In particular verify that $\sigma \bigg(\sum a_i\epsilon^i \sum b_k \epsilon^k \bigg) = \sigma \bigg(\sum a_i\epsilon^i \bigg) \sigma \bigg( \sum b_k \epsilon^k \bigg)$ and check the other axioms.
As for part $2$, $\epsilon$ and $\epsilon^3$ have different minimal polynomials. Let $p(x)$ be the minimal polynomial of $\epsilon^3$. If $\tau$ was an automorphism then, then $\tau(p(\epsilon)) = p(\tau(\epsilon) )= 0 $ implying $\epsilon$ is a root of $p(x)$ by (injectivity of $\tau$). But the degree of $p(x)$ is small than the degree of the minimal polynomial of $\epsilon$, so this cannot be the case.