Showing $\zeta_5 \notin \mathbb{Q}(\zeta_7)$

Question

I was assigned this problem as homework, and got it wrong. I have not gotten a chance to ask the teacher about the solution.

Can someone tell me why I am wrong, and how to do this correctly?

Let $\zeta_n = e^{2\pi i/n}.$ Prove that $\zeta_5 \notin \mathbb{Q}(\zeta_7).$

Since $\zeta_5 = e^{2\pi i/5}$ and $\mathbb{Q}(\zeta_7)$ is a ring, we have $\zeta_5 \cdot \zeta_7 = e^{2\pi i/5} \cdot e^{2\pi i/7} = e^{24\pi i/35} \in \mathbb{Q}(\zeta_7).$ This means $\left(e^{24\pi i/35}\right)^7 = 1.$ But $\left(e^{24\pi i/35}\right)^7 = e^{24\pi i/5} \neq 1.$ So $e^{24\pi i/35} \notin \mathbb{Q}(\zeta_7),$ a contradiction. So it must be that $e^{2\pi i/5} = \zeta_5 \notin \mathbb{Q}(\zeta_7).$

Answer 1

You know (hopefully!) that the degree of $\mathbb Q(\zeta_5)$ over $\mathbb Q$ is $4$, and that of $\mathbb Q(\zeta_7)$ over $\mathbb Q$ is $6$. As $4$ does not divide $6$, it follows that $\mathbb Q(\zeta_5)$ is not contained in $\mathbb Q(\zeta_7)$.

By the way, what you wrote is not correct because that $\zeta_5\zeta_7$ be an element of $\mathbb Q(\zeta_7)$ does not mean/imply at all that $(\zeta_5\zeta_7)^7=1$. For example, $8$ is also an element of $\mathbb Q(\zeta_7)$ and you surely see that $8^7\neq1$.

Answer 2

To show that $\zeta_5\notin \mathbb Q(\zeta_7)$ we need to show only that $[\mathbb Q(\zeta_5,\zeta_7):\mathbb Q(\zeta_7)]>1$.Now consider the following diagram:

It is evident that $[\mathbb Q(\zeta_5):\mathbb Q]=4$ and $\mathbb Q(\zeta_7):\mathbb Q]=6$. So by tower theorem, $12$ divides $[\mathbb Q(\zeta_5,\zeta_7):\mathbb Q]\implies [\mathbb Q(\zeta_5,\zeta_7):\mathbb Q]=12k\implies 6[\mathbb Q(\zeta_5,\zeta_7):\mathbb Q(\zeta_7)]=12k$ $\implies [\mathbb Q(\zeta_5,\zeta_7):\mathbb Q(\zeta_7)]=2k>1$ ,thus we are done.