Degree of the extension $\mathbb{Q}(\zeta_3,\zeta_7)$ over $\mathbb{Q}$

Question

Question is to compute the degree of the extension $\mathbb{Q}(\zeta_3,\zeta_7)$ over $\mathbb{Q}$.

We have $[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}]=[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}(\zeta_7)][\mathbb{Q}(\zeta_7):\mathbb{Q}]$

As $[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}(\zeta_7)]$ can be at most $2$ and $[\mathbb{Q}(\zeta_7):\mathbb{Q}]=6$ we have $[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}]=6$ or $12$.

I am not able to compute $[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}(\zeta_7)]$..

I believe this is dead simple but i could not make it...

Hint would be sufficient..

Note : $\zeta_3$ and $\zeta_7$ are primitive third and seventh roots of unity respectively..

Answer 1

As is stated correctly, $[\mathbb{Q}(\zeta_3,\zeta_7):\mathbb{Q}(\zeta_7)]$ can be at most $2$. It suffices to show that $\zeta_3 \notin \mathbb{Q}(\zeta_7)$. Assume it were in this field. Then $\mathbb{Q}(\zeta_7)$ would contains $\zeta_3 \zeta_7$, which is a primitive $21$-st root of unity. The degree of such a root over $\mathbb{Q}$ is $\varphi(21)= 12$, a contradiction to the degree of $\mathbb{Q}(\zeta_7)$ over $\mathbb{Q}$ being $6$.

Indeed, you could argue like this right away: $\mathbb{Q}(\zeta_3,\zeta_7) =\mathbb{Q} (\zeta_{21}) $.

Answer 2

An alternative proof:

Assuming $\zeta_3\in\mathbb{Q}(\zeta_7)$, we have $\sqrt{-3}\in\mathbb{Q}(\zeta_7)$, so, by Cauchy's theorem, $x^2+3$ splits over any finite field $\mathbb{F}$ such that $7\mid \#\,\mathbb{F}^*$. However, $x^2+3$ is irreducible over $\mathbb{F}_{29}$, since $29\equiv -1\pmod{3}$ and $-3$ is a quadratic residue $\!\!\pmod{p}$ only if $p\equiv 1\pmod{3}$.