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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} \in \mathbb{Q}$$\xi +\xi^{6} \notin \mathbb{Q}$. This is the point I´m stuck. My first approach was computing the third power and trying to find a relation, but i don´t get anywhere. So I´ve thought about this one: a basis of $\mathbb{Q}(\xi)$ over $\mathbb{Q}$ is $\{1, \xi, \xi^{2}, \xi^{3}, \xi^{4}, \xi^{5}\}$, so the relation $$ 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?

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 $\xi +\xi^{6} \in \mathbb{Q}$. 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?

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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asked Jul 11 at 10:42

MrGran

Compute all subextensions of $\mathbb{Q}(\xi)$

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 $\xi +\xi^{6} \in \mathbb{Q}$. 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?