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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 $e^{2\pi i/5} = \zeta_5 \notin \mathbb{Q}(\zeta_7).$

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  • $\begingroup$ Why $(\zeta_5\zeta_7)^7 = 1$? There is no reason that the (multiplicative) order of elements in $\Bbb{Q}(\zeta_7)$ are not 5. (It holds, but you need a proof.) $\endgroup$ – Hanul Jeon May 3 '15 at 2:35
  • $\begingroup$ Showing that a primitive $35$th root of unity lies in the 7th cyclotomic extension, contradicting the degree calculation is the easiest way. You are on the right track. $\endgroup$ – P Vanchinathan May 3 '15 at 2:37
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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 $8^7\neq1$.

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  • $\begingroup$ Yes, I follow the first line. Could you please clarify the second? $\endgroup$ – user85362 May 3 '15 at 2:45
  • $\begingroup$ You probably know that if $E/F$ is an extension of fields and $K$ is a subfield of $E$ containing $F$, then $[E:F]=[E:K][K:F]$. $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 2:47
  • $\begingroup$ Thank you. I guess it was a matter of putting together the ingredients. $\endgroup$ – user85362 May 3 '15 at 2:49
  • $\begingroup$ Well, it almost always is! $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 2:55

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