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Greetings Mathematics Community.

I am having much difficulty in solving the following problem:

If $m\equiv 2$ (mod 4), show that $\mathbb{Q(\zeta_m)}=\mathbb{Q(\zeta_{\frac{m}{2}})}$ where $\zeta$ is a primitive root of unity. I know that $\zeta_m=e^{\frac{2\pi i}{m}}$ and by Euler's equation, we have $\cos(\frac{2 \pi }{m})+i\sin(\frac{2\pi }{m})$.

I have tried to compare $[\mathbb{Q(\zeta_{\frac{m}{2}})}:\mathbb{Q}]$ with $[\mathbb{Q(\zeta_m)}:\mathbb{Q}]$ by substituting $\frac{m}{2}$ in Euler's equation to get $\cos(\frac{4 \pi }{m})+i\sin(\frac{4\pi }{m})$. My intuition is telling me that I should somehow use the given fact that $m\equiv 2$ (mod 4), but I am not sure how.

Furthermore, I do know that this extension degree can be found using Euler's $\phi-$function on $m$. But I am unsure about how to apply it in my situation.

As always, any help is greatly appreciated. Thanks in advance.

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    $\begingroup$ Here's one way to proceed. Show $\zeta_{m} \in \mathbb{Q}(\zeta_{m/2})$ and $\zeta_{m/2} \in \mathbb{Q}(\zeta_{m})$. $\endgroup$
    – Alex Wertheim
    Nov 8, 2014 at 21:10
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    $\begingroup$ Can you show that $\phi(t)=\phi(2t)$, is $t$ is odd? Apply that to $t=m/2$. Mind you, Rene's way is probably simpler. $\endgroup$
    – Jyrki Lahtonen
    Nov 8, 2014 at 21:16
  • $\begingroup$ And to start seeing the light: Try this out in the special cases $m/2=3$ and $m/2=5$. $\endgroup$
    – Jyrki Lahtonen
    Nov 8, 2014 at 21:17

2 Answers 2

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Let $m=4k+2$ $\zeta_m$ is a root of $x^{4k+2}=1$ and $\zeta_{\frac{m}{2}}$ a root of $y^{2k+1}=1$. It is now clear to see that we can take $x=-y$.

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  • $\begingroup$ Ok, I think I understand the substitutions, but how come there is a $\zeta_m$ in the $m$ expression, but not in the polynomial equation? $\endgroup$
    – Jamil_V
    Nov 8, 2014 at 21:20
  • $\begingroup$ Also, why exactly is $x=-y$? I have been manipulating the expressions and end up with $x^{2(2k+1)}=y^{2k+1}$. If I take $x=-y,$ then wouldn't I have $y^{2(2k+1)}=y^{2k+1}$? $\endgroup$
    – Jamil_V
    Nov 11, 2014 at 14:49
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Since M and M/2 make up a 'twice-odd' pair, it would be helpful to consider the same question for 2N and N where N is odd. The heart of the matter is:

if $\zeta $ is a primitive root of ${{z}^{{{2n}^{{}}}}}=1$, then ${{\zeta }^{2{{n}^{{}}}}}=1$ implies ${{\zeta }^{n}}=-1$ and when n is odd ${{(-\zeta )}^{n}}=1$

This equality of cyclotomic fields for a twice-odd pair is actually 'iff' but it is not trivial to prove the converse.

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