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Let $\xi \in \mathbb{C}$ be a primitive seventh root of unity. Determine all the subextensions of $\mathbb{Q}(\xi)$

I´ve found the subextensions but I need help finishing the argument. This is my attempt:

The Galois group of the extension is isomorphic to $\mathcal{U}(\mathbb{Z}_{7}) \cong \mathbb{Z}_{6}$, so the extension is cyclic and has degree $6$ over $\mathbb{Q}$ (it is Galois). The automorphisms $\tau_{k} \in Aut(\mathbb{Q}(\xi))$ are determined by $$ \tau_{k}(\xi) = \xi^{k},\ \ k = 1, 2, 3, 4, 5, 6. $$ Because the Galois group is cyclic of order 6, it contains one subgroup for every divisor of 6. By the Galois correspondence, there is one subextension for every divisor of $6$. To find the subextensions, we first analyze the action of the automorphism $\tau_{2}$: $$ \tau_{2}: \xi \mapsto \xi^{2} \mapsto \xi^{4} \mapsto \xi^{8} = \xi $$ So it has order $3$ and it must be in correspondence with a subextension of degree 2. Also $$ \tau_{2}(\xi +\xi^{2}+\xi^{4}) = \xi +\xi^{2} + \xi^{4}, $$ therefore $\mathbb{Q}(\xi+\xi^{2}+\xi^{4}) \leq \mathbb{Q}(\xi)^{\left<\tau_{2}\right>}$. Since the fixed field of $\left<\tau_{2}\right>$ has degree 2, either we have the equality or $\xi +\xi^{2}+\xi^{4} \in \mathbb{Q}$. Using the relations $$ \xi^{7} = 1,\ \ \xi^{6}+\xi^{5}+\xi^{4}+\xi^{3}+\xi^{2}+\xi+1 = 0 $$ we can see that $\xi + \xi^{2} + \xi^{4}$ is a root of $X^{2}+X+2 \in \mathbb{Q}[X]$. The roots of this polynomial are $\frac{1-\pm i\sqrt{7}}{2}$, and none of them is rational. Hence $\xi +\xi^{2}+\xi^{4} \notin \mathbb{Q}$ and $\mathbb{Q}(\xi+\xi^{2}+\xi^{4}) = \mathbb{Q}(\xi)^{\left<\tau_{2}\right>}$. The remaining extension is of degree $3$ and must correspond to a subgroup of order $2$. Consider the automorphism $\tau_{6}$ and note that $$ \tau_{6}(\xi) = \xi^{6} = \xi^{-1} = \bar{\xi} $$ that is, $\tau_{6}$ is the conjugation map and so it does have order $2$. Also, $$ \tau_{6}(\xi+\xi^{6}) = \xi + \xi^{6} $$ By the same reasoning as before, all that is left to prove is that $\xi +\xi^{6} \notin \mathbb{Q}$. This is the point I´m stuck. My first approach was computing the third power and trying to find a relation, but i don´t get anywhere. So I´ve thought about this one: a basis of $\mathbb{Q}(\xi)$ over $\mathbb{Q}$ is $\{1, \xi, \xi^{2}, \xi^{3}, \xi^{4}, \xi^{5}\}$, so the relation $$ 1+\xi+\xi^{2}+\xi^{3}+\xi^{4}+\xi^{5}+\xi^{6} = 0 $$ gives $$ \xi+\xi^{6} = -1-\xi^{2}-\xi^{3}-\xi^{4}-\xi^{5}. $$ Can I derive a contradiction from this?

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  • $\begingroup$ If my understanding is correct, in order to prove that $\mathbb{Q}(\xi+\xi^6)$ is the cubic subextension corresponding to the subgroup of order two $\langle\tau_6\rangle$. And we already know that the element fixed by $\tau_6$ (complex conjugation) is $\xi+\xi^6$, then we should prove $[\mathbb{Q}(\xi+\xi^6):\mathbb{Q}]=3$, which means $\xi+\xi^6 \notin \mathbb{Q}$, instead of proving it belongs. If anyone thinks I'm wrong, please help correct me, thank you. $\endgroup$ Commented Jul 11 at 11:29
  • $\begingroup$ With $z = \xi + \xi^{-1}$, I find $z^2 = \xi^2 - \xi^{-2} + 2$ and $z^3 = \xi^3 + \xi^{-3} + 3(\xi + \xi^{-1})$, thus $$z^3 + z^2 - 2z - 1 = 0\,.$$ $\endgroup$ Commented Jul 11 at 11:30
  • $\begingroup$ @QwQ yes, that was a typo. $\endgroup$ Commented Jul 11 at 12:15
  • $\begingroup$ That should have been $z^2 = \xi^2 + \xi^{-2} + 2$ in my previous comment. Muscle memory struck. $\endgroup$ Commented Jul 11 at 12:23
  • $\begingroup$ In general, even though there are only finite elements to analyze, the tower theorem might not always be the best way. It does give a good foundation to guess the degree of the minimal polynomial, then construct it the way @Dermot Craddock did $\endgroup$ Commented Jul 11 at 12:58

2 Answers 2

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@DermotCraddock gives the way to find the minimal polynomial in the comments, but that's actually more than we need. I find the following a bit easier: We can use this easy Linear Algebra lemma (I don't know if there is a common name for it, but it's usually given in the context of the Replacement Theorem/Steinitz Exchange Lemma):

Lemma Let $B=\{v_i\}_i$ be a basis of a vector speace $V$, and let $w=\sum_i\lambda_iv_i\in V$. Then $B':=\{v_i\}_i\cup\{w\}\setminus\{v_k\}$ is a basis of $V$ if and only if $\lambda_k\neq0$.

Proof Sketch: Since $v_k=\frac1{\lambda_k}(w-\sum_{i\neq k}\lambda_iv_i)\in\operatorname{span}(B')$, it's easy to see that $B'$ spans $V$. Independence is seen similarly. Alternatively, if $V$ is finite-dimensional, write down the base change matrix and note that its determinant is $-\lambda_k$.

Using the above lemma with the fact $\sum_{i=0}^6\xi^i=0$, we see that not only $\{1,\xi,\ldots,\xi^5\}$, but for example also $\{1,\xi,\xi^3,\xi^4,\xi^5,\xi^6\}$ is a basis of $\mathbb Q(\xi)/\mathbb Q$. And now your claim is easy: If $\xi+\xi^6=a\in\mathbb Q$, this would be a non-trivial linear combination of $1,\xi,\xi^6$, contradicting the fact that they are part of a basis. (the same argument works for $\xi+\xi^2+\xi^4$ as well.)

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I just saw the simplest, and what I think is a very fun proof that you might be interested in: $\xi$ is the 7th primitive root of unity, which can be written as: $\xi = \cos(2\pi/7) + i \sin(2\pi/7)$

$\xi^6$ is $\xi^{-1}$, which is the complex conjugate of $\xi$: $\xi^6 = \cos(2\pi/7) - i \sin(2\pi/7)$ So $z = \xi + \xi^6 = (\cos(2\pi/7) + i \sin(2\pi/7)) + (\cos(2\pi/7) - i \sin(2\pi/7))$, $z = 2 \cos(2\pi/7)$ There is a famous number theory theorem: The value of $\cos(2\pi/n)$ is rational only when $n = 1, 2, 3, 4, 6$. For all other integers $n$ greater than 2, $\cos(2\pi/n)$ is an irrational number.

Since $n=7$, $\cos(2\pi/7)$ is an irrational number. Thus $z = 2 \cos(2\pi/7)$ must also be an irrational number. Finally: $z$ cannot be in $\mathbb{Q}$.

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