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Let $K/\mathbb{Q}$ be a cyclic Galois extension of degree $n$ where $K=\mathbb{Q}(\theta)$ and $\theta$ is an algebraic integer. How can I describe the conjugates of $\theta$ in terms of the $\mathbb{Q}$-basis $\{1,\theta,\theta^2,\dots,\theta^{n-1}\}$?

Take $\theta=2\cos(\frac{2\pi}{7})$. Then $K/\mathbb{Q}$ is cyclic Galois of degree $3$ and the conjugates of $\theta$ are $\theta^2-2$ and $1-\theta-\theta^2$. This is given in page 239 in the textbook A First Course in Noncommutative Rings by T.Y. Lam.

If $\theta=\sqrt{2+\sqrt{2}}$, then $K/\mathbb{Q}$ is cyclic Galois of degree 4 and the conjugates of $\theta$ are $\theta^3-3\theta$, $-\theta$, and $3\theta-\theta^3$. I found this out because I knew the conjugates of $\theta$ were $\sqrt{2-\sqrt{2}}$, $-\sqrt{2+\sqrt{2}}$ and $-\sqrt{2-\sqrt{2}}$ so I asked Wolfram to solve

\begin{align} \sqrt{2-\sqrt{2}}=a+b\left(\sqrt{2+\sqrt{2}}\right)+c\left(\sqrt{2+\sqrt{2}}\right)^2+d\left(\sqrt{2+\sqrt{2}}\right)^3 \end{align} over the integers and it gave me $a=0$, $b=-3$, $c=0$, and $d=1$.

How can I find the representation of the conjugates of $\theta$ in terms of the $\mathbb{Q}$-basis $\{1,\theta,\theta^2,\dots,\theta^{n-1}\}$ in general? Is there literature on this? I feel like the minimal polynomial of $\theta$ would play a role (which is $X^3+X^2-2X-1$ and $X^4-4X^2+2$ respectively in the examples above) but I don't know how.

Ama
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  • $\theta=2\cos\left(\dfrac{2\pi}7\right)$ $=\exp\left(\dfrac{2\pi i}7\right)$ $+\exp\left(\dfrac{-2\pi i}7\right)$, so conjugates are $2\cos\left(\dfrac{4\pi}7\right)=\exp\left(\dfrac{4\pi i}7\right)+\exp\left(\dfrac{-4\pi i}7\right)=\theta^2-2$ and $2\cos\left(\dfrac{6\pi}7\right)=\exp\left(\dfrac{6\pi i}7\right)+\exp\left(\dfrac{-6\pi i}7\right)=\theta^3-3\theta=1-\theta-\theta^2$ – J. W. Tanner Nov 05 '24 at 04:25
  • If $\theta=2\cos\left(\dfrac{2 \pi}{16}\right)=\exp\left(\dfrac{2\pi i}{16}\right)+\exp\left(\dfrac{-2\pi i}{16}\right)$, then conjugates are $2\cos\left(\dfrac{2 \pi 3}{16}\right)=$ $\exp\left(\dfrac{2 \pi i3}{16}\right)+\exp\left(\dfrac{-2\pi i3}{16}\right)=\theta^3-3\theta$, $2\cos\left(\dfrac {2\pi 5}{16}\right)=-(\theta^3-3\theta)$ , and $2\cos\left(\dfrac{2 \pi 7}{16}\right)=-\theta$ – J. W. Tanner Nov 05 '24 at 04:49
  • @J.W.Tanner why are you multiplying the argument of cos by primes? In your first comment the argument is multiplied by 2 and then 3 and in your second comment the argument is multiplied by 3, 5, and then 7. Do you start at the prime one less than the degree of the minimal polynomial? What if one less than the degree is not prime? Additionally, how did you arrive at $\sqrt{2+\sqrt{2}}=2\cos\left(\frac{2\pi}{16}\right)$? – Ama Nov 05 '24 at 05:51

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As background, consider the cyclotomic field $\mathbb Q(\zeta_n)$,

where $\zeta_n=\exp(2\pi i/n)$ is a primitive $n^{th}$ root of unity.

The conjugates of $\zeta_n$ are $\zeta_n^m$, where $\gcd(m,n)=1$.

So for example $\zeta_7$ has $6$ conjugates ($\zeta_7^m, m=1,2,3,4,5,6)$,

whereas $\zeta_{16}$ has $8$ conjugates $(\zeta_{16}^m, m=1,3,5,7,9,11,13,15)$.

The complex conjugate of $\zeta$ is $\zeta^{-1}$,

so $\theta_n=\zeta_n+\zeta_n^{-1}=2\cos\left(\dfrac{2\pi}n\right)$ is in a real subfield of $\mathbb Q(\zeta_n)$.

So for example $\theta_7$ has $3=6/2$ conjugates, whereas $\theta_{16}$ has $4=8/2$ conjugates.

Because you said $\sqrt{2+\sqrt2}$ has $4$ conjugates, I recognized that $\sqrt{2+\sqrt2}$$=2\cos\left(\dfrac{2\pi}{16}\right)$;

that can be proved with the half-angle formula.

The conjugates of $\theta_7$ are $\zeta_7^2+\zeta_7^{-2}=\theta_7^2-2$ and $\zeta_7^3+\zeta_7^{-3}=\theta_7^{3}-3\theta_7=1-\theta_7-\theta_7^2;$

that last equality follows from $\theta_7$ being a root of its minimal polynomial.

The conjugates of $\theta_{16}$ are $\zeta_{16}^3+\zeta_{16}^{-3}=\theta_{16}^3-3\theta_{16}$, $\zeta_{16}^5+\zeta_{16}^{-5}=-\zeta_{16}^{-3}-\zeta_{16}^{3}=-(\theta_{16}^3-3\theta_{16})$,

and $\zeta_{16}^7+\zeta_{16}^{-7}=-\zeta_{16}^{-1}-\zeta_{16}^{1}=-\theta_{16}.$

J. W. Tanner
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  • Is there a way to know the coefficients of the basis representation of a conjugate for general $\theta$? That is, if $\theta_i$ is a conjugate, to know the $a_0,a_1,\dots,a_{n-1}$ such that $\theta_i=a_0+a_1\theta+\dots+a_{n-1}\theta^{n-1}$? Would Vieta's formulas help? I'm investigating the lengths of cyclic algebras and I think knowing these coefficients would really help. – Ama Nov 05 '24 at 15:34
  • As I indicated, the conjugates for $\theta_n=\zeta_n+\zeta_n^{-1}$ are $\zeta_n^m+\zeta_n^m$ with $\gcd(m,n)=1$, and these can be computed recursively using $\zeta_n^m+\zeta_n^{-m} =(\zeta_n^{m-1}+\zeta_n^{-(m-1)})(\zeta_n+\zeta_n^{-1})-(\zeta_n^{m-2}+\zeta_n^{-(m-2)})$ – J. W. Tanner Nov 05 '24 at 15:56
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    @Ama: Cf. also this question – J. W. Tanner Nov 05 '24 at 18:24