For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$

My solution :

let $$p = \alpha + \alpha^2 + \alpha^4$$ and $$q = \alpha^3 + \alpha^5 + \alpha^6.$$

We know $$1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 = 0,$$

$$p + q = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$$ $$ = -1$$


$$pq = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6)$$ $$= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^7 + \alpha^9 + \alpha^{10}$$ $$= 2.$$

Therefore, $p$ and $q$ are the two roots of the following equation : $$x^2 + x + 2 = 0$$


$$p = \alpha + \alpha^2 + \alpha^4$$ $$= \frac{-1 ± \sqrt{7} i}{2}.$$

Would there be other ways of evaluating? I'm thinking of polar forms but not sure how to do this with it.


3 Answers 3


Let $p=\alpha+\alpha^2+\alpha^4$. Squaring both sides we obtain $$p^2= \alpha^2+\alpha^4+\alpha^8+2\alpha^3+2\alpha^6+2\alpha^5.$$ Re arrange the terms we get $$p^2= \alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha^3+\alpha^5+\alpha^6.$$ Which is equal to $p^2=-1+ \alpha^3+\alpha^5+\alpha^6$ same as $p^2+1=\alpha^3+\alpha^5+\alpha^6$. Again squaring both sides and re arrange the terms we obtain $(p^2+1)^2=(\alpha^3+\alpha^5+\alpha^6)^2.$ $(p^2+1)^2=\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha+\alpha^2+\alpha^4$. Which is same as $(p^2+1)^2=-1+p \Rightarrow p^4+2p^2-p+2=0.$ The root of this equation gives the desired result.

  • 2
    $\begingroup$ how do you go from that quartic to the result? also aren't we looking for a quadratic? $\endgroup$
    – user581023
    Aug 10, 2021 at 10:24
  • $\begingroup$ More work than necessary. Once you find $p^2=-1+ \alpha^3+\alpha^5+\alpha^6$, all you have to do is add $p$ and $1$. Though this isn't really different from the OP's approach. You find the minimal polynomial and use the quadratic formula to find the roots. Also it's not clear how you would find the roots of that quartic. $\endgroup$ Aug 11, 2021 at 3:45

The extension $\mathbb Q(\alpha)/\mathbb Q$ has Galois group $(\mathbb Z/7\mathbb Z)^\times\cong\mathbb Z/6\mathbb Z$, generated by $3\in(\mathbb Z/7\mathbb Z)^\times$. Denote the image of $a$ under the isomorphism $(\mathbb Z/7\mathbb Z)^\times\to\mathrm{Gal}(\mathbb Q(\alpha)/\mathbb Q):a\mapsto (\alpha\mapsto\alpha^a)$ as $\sigma_a$.

Now, since $u=\alpha+\alpha^2+\alpha^4$ is invariant under $\sigma_2$, it must be in the fixed field of $\langle\sigma_2\rangle$, which has index $2$ in $\mathrm{Gal}(\mathbb Q(\alpha)/\mathbb Q)$. Thus, under the Galois correspondence, it must be contained in the degree-$2$ extension $\mathbb Q(\sqrt{-7})$. Since $u$ is integral, it is contained in $\mathbb Z\big[\frac{1+\sqrt{-7}}2\big]$.

Since $\sigma_3$ is the generator of the Galois group of $\mathbb Q(\sqrt{-7})/\mathbb Q$, it is complex conjugation. We have $u+\sigma_3(u)=\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6=-1$, so $u$ has real part $-\frac12$. We conclude $u=-\frac12+\frac{2n-1}2{\sqrt{-7}}$ for some integer $n$.

We are almost done at this point, and we have several ways to finish off:

  1. Look at the absolute values. $|-\frac12+\frac{2n-1}2{\sqrt{-7}}|^2=\frac14+\frac{(2n-1)^2}4\cdot7$, while $|u|^2\le(|\alpha|+|\alpha|^2+|\alpha|^4)^2=9$. We thus need $n=0,1$, giving the desired result.

  2. Calculate the norm of $u\in\mathbb Q(\sqrt{-7})$ which is $u\cdot\sigma_3(u)$, which turns out to be $2$. Comparing with the formula $u=-\frac12+\frac{2n-1}2\sqrt{-7}$, we again obtain $n=0,1$.


Yet another way (albeit not too different from OP's), using that $\,|\alpha| = 1\,$, $\,\overline \alpha=\dfrac{1}{\alpha}\,$:

  • $\require{cancel}\displaystyle \;p+\overline p = \alpha + \alpha^2+\alpha^4 + \overline \alpha + \overline \alpha^2 + \overline \alpha^4 =\alpha + \alpha^2+\alpha^4 + \frac{1}{\alpha}+\frac{1}{\alpha^2}+\frac{1}{\alpha^4} \\\displaystyle =\frac{\cancel{\alpha^5+\alpha^6+\alpha+\alpha^3+\alpha^2+1\color{red}{+\alpha^4}}-\color{red}{\alpha^4}}{\alpha^4}=\frac{-\alpha^4}{\alpha^4}=-1$

  • $\displaystyle \;|p|^2=\cancel{|\alpha|^2}\,|1+\alpha+\alpha^3|^2=\left(1+\alpha+\alpha^3\right)\left(1+\overline\alpha+\overline\alpha^3\right)= \left(1+\alpha+\alpha^3\right)\left(1+\frac{1}{\alpha}+\frac{1}{\alpha^3}\right) \\\displaystyle = \frac{\left(1+\alpha+\alpha^3\right)\left(1+\alpha^2+\alpha^3\right)}{\alpha^3}=\frac{\cancel{1+\alpha^2+\alpha^3+\alpha}+\alpha^3+\cancel{\alpha^4}+\alpha^3+\cancel{\alpha^5+\alpha^6}}{\alpha^3}=2$

The two relations give the real part $\dfrac{-1}{2}$ and magnitude $\sqrt{2}$ of $\,p\,$, so $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.

  • $\begingroup$ Wish the downvoter had left a comment why. (Or, in case they are who I think they are given the timing of this, I wish they had just minded their other business.) $\endgroup$
    – dxiv
    Aug 24, 2021 at 22:01

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