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

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.

• I used the same approach here. I'm sure you can also cook up suitable trigonometric identities that allow you to settle the question (with a specific choice of $\alpha$). The key is that the exponents $1,2,4$ are exactly the quadratic residues modulo $7$. That allows an approach similar to yours work with any odd prime $p$ instead of $7$. There will be $(p-1)/2$ terms in the sum in the general case. Wikipedia on Gauss sums. Aug 10, 2021 at 9:40
• Aug 10, 2021 at 9:41
• So I think this is a duplicate of those two. Can't decide which :-) Aug 10, 2021 at 9:42
• en.wikipedia.org/wiki/Gaussian_period#Example gives a generalized version
– user581023
Aug 10, 2021 at 13:45
• zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf and many, many examples in Reuschle 1875: google.com/books/edition/Tafeln_complexer_Primzahlen/… Aug 10, 2021 at 18:06

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.

• how do you go from that quartic to the result? also aren't we looking for a quadratic?
– user581023
Aug 10, 2021 at 10:24
• 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. 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 $$\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$$.

• 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.)
– dxiv
Aug 24, 2021 at 22:01