Determine primitive element of constructible points of polynomial

Question

I'm trying to determine the primitive element for the field $\Omega^{\text{constr}}\subset\Omega^f_{\mathbb{Q}}$ where $\Omega^f_{\mathbb{Q}}$ is the splitting field of the polynomial $$ f=x^9-2x^7+3x^2-6 $$ and where $\Omega^{\text{constr}}$ is subfield of elements $x\in\Omega^f_{\mathbb{Q}}$ that are constructible with compass and straight edge from the subset $\{0,1\}\in\mathbb{C}$.

First of all, we can factorise $f$, namely $$ f=(x^2-2)(x^7+3). $$ Hence, $\Omega^f_{\mathbb{Q}}=\mathbb{Q}(\sqrt{2},\xi_7\sqrt[7]{3})$ where $\xi_7$ is the seventh primitive root satisfying $x^7=1$. There is a theorem which says that the set of constructable points $C(X)$, where $X\subset\mathbb{C}$ contains the elements $0$ and $1$, is equal to the quadratic closure of $\mathbb{Q}(X,\overline{X})\subset\mathbb{C}$. In other words, if I show that $\xi_7\sqrt[7]{3}$ is not a square, will that provide me evidence that $\Omega^{\text{constr}}=\mathbb{Q}(\sqrt2)$?

Answer 1

The splitting field is $\Omega_\mathbf{Q}^f=\mathbf{Q}(\sqrt{2},\sqrt[7]{3},\zeta_7)$.

We determine the degree $[\Omega_\mathbf{Q}^f:\mathbf{Q}]$. We have $[\mathbf{Q}(\sqrt{2},\sqrt[7]{3}):\mathbf{Q}]=2\cdot 7=14$, $[\mathbf{Q}(\sqrt[7]{3},\zeta_7):\mathbf{Q}]=7\cdot 6=42$ since the degrees are coprime. For the degree $[\mathbf{Q}(\sqrt{2},\zeta_7):\mathbf{Q}]$, we ask if $X^2-2$ is irreducible over $\mathbf{Q}(\zeta_7)$. $\mathbf{Q}(\zeta_7)$ has a unique quadratic subfield $\mathbf{Q}(\sqrt{-7})$, so assume not, then these fields must be equal. Contradiction, so $[\mathbf{Q}(\sqrt{2},\zeta_7):\mathbf{Q}]=2\cdot 6=12$.

Now $[\Omega_\mathbf{Q}^f:\mathbf{Q}]\leqslant 2\cdot 7\cdot 6=84$ and $\underbrace{\operatorname{lcm}(14,12,42)}_{=\operatorname{lcm}(84,42)=84}\mid [\Omega_\mathbf{Q}^f:\mathbf{Q}]$, so $[\Omega_\mathbf{Q}^f:\mathbf{Q}]=84=2^2\cdot 3\cdot 3$.

A subfield of $\Omega_\mathbf{Q}^f$ with elements that are constructible has degree a power of $2$ over $\mathbf{Q}$, so the $[\Omega^{\text{constr}}:\mathbf{Q}]=2^2=4$. Since $\mathbf{Q}(\sqrt{2}),\mathbf{Q}(\sqrt{-7})$ are distinct quadratic subfields as noted before, we have $\mathbf{Q}(\sqrt{2},\sqrt{-7})=\color{red}{\mathbf{Q}(\sqrt{2}+\sqrt{-7})=\Omega^{\text{constr}}}$.

Answer 2

First of all the splitting field is $L = \mathbb{Q}(\sqrt{2},\zeta_7,\sqrt[7]{3})$ and not $\mathbb{Q}(\sqrt{2},\zeta_7\sqrt[7]{3})$.

Now note that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is the largest real subfield of $\mathbb Q(\zeta_7)$. Thus the largest real subfield of $L$ is $L' = \mathbb{Q}(\sqrt{2},\zeta_7+\zeta_7^{-1},\sqrt[7]{3})$. As the constructible numbers are real, by definition this subfield contains them. Furthermore the subfield of constructible numbers has extension degree $2^k$, while $[L':\mathbb Q] = 2 \cdot 3 \cdot 7 = 42$. So as $\Omega^{constr} \subset L'$ we have that $[\Omega^{constr} :\mathbb Q] = 1$ or $2$. Then it is not hard to conclude that $\Omega^{constr} = \mathbb Q(\sqrt{2})$