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

Question

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

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

and

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

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

and

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

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

Answer 1

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 $n=0,1$.

Answer 2

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.

Answer 3

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 $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.