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Let $\zeta_7= e^{i2\pi/7}$ be the $7$th root of unity.Find the quadratic sub field of $\mathbb Q(\zeta_7)$ which can be expressed in the form $\mathbb Q(\sqrt D)$,where $D$ is an integer.

My try:let $\omega=\zeta_7+\zeta_7^{-1}\in\mathbb{Q}(\zeta_7)$ is real which is $\omega=2\cos(2\pi/7)$. Thus $K=\Bbb{Q}(\omega)$ is a proper subfield of $\Bbb{Q}(\zeta_7)$. And the polynomial $$ p(x)=(x-\zeta_7)(x-\zeta_7^{-1})=x^2-\omega x+1 $$ has its coefficients in the subfield $K$. Thus it has to be the minimal polynomial of $\zeta_7$, and $[K(\zeta_7):K]=2$.

But the problem is that what is $D?$ and how can i find that?Thank you

MatheMagic
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    See https://math.stackexchange.com/questions/376827/quadratic-subfield-of-cyclotomic-field. – lhf Apr 27 '17 at 13:10
  • Every quadratic field is of the form $\mathbb{Q}(\sqrt{D})$ for some integer $D$. – reuns Apr 27 '17 at 13:11
  • $\mathbb{Q}(\omega)$ is (contained in) the fixed field of the complex conjugaison, a subgroup with two elements of $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q})$, and $[\mathbb{Q}(\zeta_7):\mathbb{Q}] = 6$, which means $[\mathbb{Q}(\omega):\mathbb{Q}] = 3$ so $\mathbb{Q}(\omega)$ is a cubic field. – reuns Apr 27 '17 at 13:14
  • As others have pointed out, you have the extension of index $2$, whereas you need to find the extension of degree $2$, i.e., you're supposed to find $K$ with $[K: \mathbb{Q}] = 2$ instead of $[\mathbb{Q}(\zeta_7) : K] = 2$. You can find a very classical answer by reading about Gauss sums. – Viktor Vaughn Apr 27 '17 at 13:24
  • Garrett's work is good : http://www-users.math.umn.edu/~garrett/m/algebra/notes/19.pdf

    He has some elementary computations there that might help you, including finding a quadratic intermediate extension.

     – Chickenmancer Apr 27 '17 at 13:28
  • Thanks @lhf for that wonderful link which generalizes my answer below. – Paramanand Singh Apr 29 '17 at 04:42

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You can note that the degree of the cyclotomic extension $\mathbb{Q}(\zeta_{7})\supset \mathbb{Q}$ is $\phi(7) = 6$ and also note that $g = 3$ is a primitive root of $7$ and thus we can write the roots of $\Phi_{7}(z) = (z^{7} - 1)/(z - 1)$ as $$z_{k} = \zeta_{7}^{3^{k}}$$ for $k = 0, 1, 2, 3, 4, 5$. Next we need to use the magic created by Gauss (details available here and here).

Since $3\mid 6 = \phi(7)$ and $6/3 = 2$ we can form a Gaussian period of 3 terms which will be root of a quadratic polynomial with coefficients in $\mathbb{Q}$. Thus $$y_{0} = z_{0} + z_{2} + z_{4} = \zeta_{7} + \zeta_{7}^{2} + \zeta_{7}^{4}$$ will be root of a quadratic polynomial. It is easy to find this quadratic whose other root is another period of 3 terms namely $$y_{1} = z_{1} + z_{3} + z_{5} = \zeta_{7}^{3} + \zeta_{7}^{6} + \zeta_{7}^{5}$$ Together we can see that $y_{0} + y_{1} = -1$ and clearly $$y_{0}y_{1} = \zeta_{7}^{4} + 1 + \zeta_{7}^{6} + \zeta_{7}^{5} + \zeta_{7} + 1 + 1 + \zeta_{7}^{3} + \zeta_{7}^{2} = 2$$ and hence the desired polynomial is $x^{2} + x + 2$ and the roots are $(-1\pm i\sqrt{7})/2$. It follows that $\mathbb{Q}(\sqrt{-7})$ is the desired extension and we have $\mathbb{Q}(\zeta_{7})\supset\mathbb{Q}(\sqrt{-7})\supset\mathbb{Q}$. Using the same technique one can show that $\zeta_{7}$ (which is a period of $1$ term) is the root of a cubic polynomial with coefficients in $\mathbb{Q}(\sqrt{-7})$.

Paramanand Singh
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