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Calculate the degree of the extension $[\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}]$ where $p$ is a prime number.

My thoughts are:

  • I am lost
  • My intuition says it has to be $ \frac{p-1}{2}$ and that the base vectors are probably $\{\cos(\frac{2n\pi}{p}) \mid 0 \le n \le p-1 \}$, notice $\cos(\frac{2n\pi}{p})=\cos(\frac{2(p-n)\pi}{p})$.

I don't know how to prove/calculate it.

egreg
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Daniel
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    Let $\zeta = \exp(2\pi i/p)$ so that $\zeta + \zeta^{-1} = 2\cos \frac{2\pi}{p}$. Now we know that $\zeta$ has degree $p-1$ over the rationals. This gives us a way to start. – Nicolas Bourbaki May 08 '15 at 19:38
  • This doesn't help with your proof, but note that generally $\left[\mathbb{Q}(\cos(2\pi/n)):\mathbb{Q}\right]=\frac{\varphi(n)}{2}$. – tc1729 May 08 '15 at 22:46
  • @SiddharthPrasad So is that the fact that in general, $cos(\frac{2\pi}{n})$ is rational iff $n=1,2,3$ or otherwise $n=2p$ for some prime $p$? – Y.X. May 29 '17 at 12:29
  • I think the conclusion should be $cos(2\pi/n)$ is rational iff $n=1,2,3,4,6$,and no more. – 王李远 Jun 12 '17 at 04:54

1 Answers1

13

If we accept as given the fact that

$[\Bbb Q(\zeta): \Bbb Q] = p - 1, \tag{1}$

where $\zeta = \cos (2\pi / p) + i\sin (2\pi/p)$, we may proceed as follows:

since, as the comments point out,

$\cos \dfrac{2\pi}{p} = \dfrac{\zeta + \zeta^{-1}}{2}, \tag{2}$

we infer that $\cos (2\pi / p) \in \Bbb Q(\zeta)$; thus $Q(\cos (2\pi / p))$ is a real subfield of $\Bbb Q(\zeta)$; we have the tower of fields

$\Bbb Q \subset \Bbb Q(\cos \dfrac{2\pi}{p}) \subset \Bbb Q(\zeta); \tag{3}$

from (3) we conclude that

$[\Bbb Q(\zeta): \Bbb Q(\cos\dfrac{2\pi}{p})][\Bbb Q(\cos \dfrac {2\pi}{p}):\Bbb Q] = [\Bbb Q(\zeta): \Bbb Q] = p - 1; \tag{4}$

next, observe:

$\zeta = \cos \dfrac{2\pi}{p} + i \sin \dfrac{2\pi}{p}. \tag{5}$

whence

$(\zeta - \cos \dfrac{2\pi}{p})^2 = -\sin^2 \dfrac{2\pi}{p} = \cos^2 \dfrac{2\pi}{p} - 1, \tag{6}$

or

$\zeta^2 - 2\cos \dfrac{2\pi}{p} \zeta + 1 = 0; \tag{7}$

$\zeta$ thus satisfies the quadratic polynomial

$q_\zeta(x) = x^2 - 2\cos \dfrac{2\pi}{p} x + 1 \in \Bbb Q(\cos \dfrac{2\pi}{p})[x]; \tag{8}$

the zeroes of $q_\zeta(x)$ are in fact $\zeta$ and $\bar \zeta$, as may easily seen by taking the complex conjugate of (7); neither lies in $\Bbb Q (\cos (2\pi/p))$ since it is a real subfield of $\Bbb Q(\zeta)$; thus $q_\zeta(x)$ is irreducible over $\Bbb Q(\cos (2\pi /p))$ and hence

$[\Bbb Q(\zeta): \Bbb Q(\cos \dfrac{2\pi}{p})] = 2; \tag{9}$

using this fact in (4) yields

$[\Bbb Q(\cos \dfrac{2\pi}{p}): \Bbb Q] = \dfrac{p - 1}{2}, \tag{10}$

the answer sought.

Robert Lewis
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