Cutting the circle into roots

"What are roots of unity?" That is, what numbers can we take to some integer power to get good old $1\in\Z$. Surely, $1$ is such a root, but if we are in $\mathbb{C}$, we can get it in a couple more ways. It turns out that these ways follow a very regular construction and this is the topic of this post! Due to all these roots necessarily lying on the unit circle, we call this procedure circle cutting. An even fancier word is cyclotomy.

We will have a look at the result due to Vandermonde (later refined by Gauss). It proves that every "primitive root of unity" has a non-trivial radical expression. All this might be a bit cryptic without some knowledge of Galois theory. I might write a more introductory treatment at some point. Good reference is the Stewart's book1.

Definition. An $\alpha\in\mathbb{C}$ has a nontrivial radical expression if $\alpha$ belongs to a radical extension of $\mathbb{Q}$ formed by successive adjunction of $k$th roots, where $k$ is less than or equal to the degree of the minimal polynomial of $\alpha$ over $\mathbb{Q}$. We like this definition, as it

A rabbit out of the hat

Following Vandermonde, we want to obtain $4$ independent linear combinations $\alpha$ of powers of $\zeta$ over $\mathbb{Q}(i)$ so that $\alpha^4$ also lies in $\mathbb{Q}(i)$, and then solve for $\zeta$. Such equations are as follows

$$\zeta+\zeta^2+\zeta^4+\zeta^3 = -1 \\ \zeta+i\zeta^2-\zeta^4-i\zeta^3 = \sqrt[4]{-15+20i} \\ \zeta-\zeta^2+\zeta^4-\zeta^3 = \sqrt{5} \\ \zeta-i\zeta^2-\zeta^4+i\zeta^3 = \sqrt[4]{-15-20i}$$

We solve them by summing and cancelling powers of $\zeta$, and hence obtain a radical expression


A more refined rabbit

The question we aim to answer is when do these combinations exist and how to find them, in general. Let's take a closer look at 5th root case first

  1. Exponents ${1,2,3,4}$ form a group $\mathbb{Z}^*_5$ under multiplication modulo $5$
  2. $2$ is a generator of $\mathbb{Z}^*_5$ and has order $4$ in it
  3. $i$ is a primitive 4th root of unity and has order $4$ in the multiplicative group of 4th roots of unity i.e. ${1,i,-1,-i}$
  4. Consider the Galois group of $\mathbb{Q}(i\zeta):\mathbb{Q}$ where $i\zeta$ is the primitive 20th root of unity as 4 and 5 are coprime
  5. Use structure of $\mathbb{Z}^*_{20}$ to get that Galois group of $\mathbb{Q}(i\zeta):\mathbb{Q}(i)$ is generated by a $\mathbb{Q}(i)$-automorphism $\rho_2$ that sends $\zeta$ to $\zeta^2$ and fixes $\mathbb{Q}(i)$
  6. We can show that $\alpha^4$ is fixed by $\rho_2$ and so it lies in its fixed field and so in the fixed field of this Galois group
  7. As the extension is normal and seperable, this fixed field is $\mathbb{Q}(i)$

This algorithm uses the radical expression for the preceding root, and also guarantees that the roots used will be of right orders, in above case, the expression only uses square roots. Neat!


Stewart, Ian (2015). Galois Theory (4th ed.). Chapman and Hall/CRC.

04 December, 2020