# Roots of Unity

Gauss generalized his method to find an expression using radicals for any root of unity. (Compare with Vandermonde’s method.)

Suppose we want to find an expression for a primitive $$p$$th root of unity $$\zeta$$ for a prime $$p$$, and assume we have done so for smaller primes. Let $$d, D$$ be factors of $$p-1$$ such that $$D = q d$$ for some $$q$$. Let $$g$$ be a generator of $$\mathbb{Z}_p^*$$. Let $$\beta$$ be a primitive $$q$$th root of unity.

For any expression $$\gamma$$ containing $$\zeta$$, define $$S \gamma$$ to be the same expression with each $$\zeta$$ replaced by $$\zeta^g$$.

Suppose $$\gamma$$ satisfies $$S^D \gamma = \gamma$$. Then define

$t = \gamma + \beta S^d \gamma + \beta^2 S^{2d} \gamma + ... + \beta^{q-1} S^{(q-1)d} \gamma$

Then replacing $$\zeta$$ by $$\zeta^{g^d}$$ in this expression yields $$\beta^{-1} t$$. Since $$t^q = (\beta^{-1} t)^q$$ we can equate the coefficients of the powers of $$\zeta$$ as before to argue $$t^q$$ can be expressed in terms of $$\beta$$.

Example: Take $$d=1, D=2, p=17$$. Then $$q = 2, \beta = -1$$. If we take $$\gamma = x_1$$ as defined earlier discussing the 17-gon, we see $$S x_1 = x_2, S x_2 = x_1$$, thus $$S^2 \gamma = \gamma$$. Then the expression $$t$$ is simply

$t = \gamma + \beta S \gamma = x_1 - x_2$

and we saw before $$t^2$$ must be an integer.

To continue, we took $$d=2, D=4$$. Again $$q=2, \beta=-1$$ and we can take $$\gamma = y_1$$ as defined earlier. Then note $$S^4 \gamma = \gamma$$ and we see $$(y_1 - y_2)^2$$ is an integer.

Now define $$t_i$$ to be $$t$$ where each $$\beta$$ has been replaced by $$\beta^i$$. Then we have

$\gamma = \frac{t_1 + ... + t_q}{q}$

(much cancellation occurs since the sum of the $$k$$th roots of unity is zero for any $$k> 1$$). By a similar argument, each $$t_i^q$$ is known, and thus if we choose $$q$$th roots correctly, then

$\gamma = \frac{1}{q} \sum_{i=1}^q \sqrt[q]{t_i^q}$

(the $$\sqrt{}$$ symbol does not have its usual meaning here because the particular $$q$$th roots we need may not be real).

Instead of trying every possible root until the resulting $$\gamma$$ is correct, we consider the expression $$t_i t_1^{q-i}$$. If we change each $$\zeta$$ to $$\zeta^{g^d}$$ (that is apply $$S^d$$) then $$t_i$$ changes to $$\beta^{-i} t_i$$, while from before we know $$t_1^{q-i}$$ becomes $$\beta^{-(q-i)}t_1^{q-i}$$, thus their product is unchanged.

Arguing as before, $$t_i t_1^{q-i}$$ is known for all $$i$$, so once we have made a choice for the value of $$t_1$$ we can easily find the values for each $$t_i$$ without guesswork.

Example: Let $$\zeta$$ be a primitive fifth root of unity. We shall derive an expression for $$\zeta$$ in terms of a primitive fourth root of unity.

Set $$d= 1,D=4, p=5$$. Take $$g = 2$$, since $$2$$ generates $$\mathbb{Z}_5^*$$. Then $$q=4,\beta = i$$. Set $$\gamma$$ to simply $$\zeta$$, so the $$t_i$$s are:

\begin{aligned} t_1 &=& \zeta + i \zeta^2 - \zeta^4 - i\zeta^3 \\ t_2 &=& \zeta - \zeta^2 + \zeta^4 - \zeta^3 \\ t_3 &=& \zeta - i \zeta^2 - \zeta^4 + i\zeta^3 \\ t_4 &=& \zeta + \zeta^2 + \zeta^4 + \zeta^3 = -1 \end{aligned}

We compute $$t_1^4$$ and choose a fourth root of the result, from which we work out $$t_2, t_3, t_4$$. To make the computation easier we notice

$t_2^2 = (\zeta + \zeta^2 + \zeta^3 + \zeta^4) + 2(-\zeta^3 + 1 - \zeta^4 - \zeta^1 + 1 - \zeta^2)$
$= (-1) + 2(2-(-1)) = 5$

(I’ve omitted shortcuts I took for clarity, e.g. since $$2 \in \mathbb{Z}_5^*$$ the squares of different powers of $$\zeta$$ will be different powers of $$\zeta$$, and they will add up to $$-1$$.)

Now

$t_1^2 = (\zeta^2 - \zeta^4 + \zeta^3 - \zeta) +2(i\zeta^3 - 1 - i\zeta^4 - i\zeta + 1 + i\zeta^2)$
$= (-t_2) + 2i(-t_2)$

Thus $$t_1^4 = 5(1+2i)^2$$ whence $$t_1 = \alpha{(\sqrt[4]{5}\sqrt{1+2i})}$$ where $$\alpha$$ is a fourth root of unity.

Now that we have found the solutions of $$t_1$$, we compute

\begin{aligned} t_1^2 t_2 &=& -(t_2^2) (1+2i) = -5(1+2i) \\ t_1 t_3 &=& (\zeta-\zeta^4)^2 - (i(\zeta^2 - \zeta^3))^2 \\ &=& (\zeta^2 + \zeta^3 + \zeta^4 + \zeta) + 2(-1 -1) = -5 \\ t_4 &=& -1 \end{aligned}

(Actually first equation is unnecessary since we already have $$t_2$$ in terms of $$t_1$$ from before.)

Thus after some algebraic manipulation we find

\begin{aligned} t_1 &=& \alpha{(\sqrt[4]{5}\sqrt{1+2i})} \\ t_2 &=& -\alpha^2\sqrt{5} \\ t_3 &=& -\alpha^3{(\sqrt[4]{5}\sqrt{1-2i})} \\ t_4 &=& -1 \end{aligned}

Finally, we have all four of the primitive fifth roots of unity:

$\zeta = \frac{ -\alpha^2\sqrt{5}-1 + \alpha(\sqrt[4]{5})(\sqrt{1+2i} - \alpha^2\sqrt{1-2i}) }{4}$

where $$\alpha = \pm 1, \pm i$$.

If instead we had chosen $$d=1, D=2$$, and then $$d=2, D=4$$ (i.e. mirror the process used for the 17th roots of unity] we have $$\zeta$$ expressed in terms of a primitive square root of unity (i.e. over the rationals, since $$-1$$ is rational):

$\zeta = \frac{\sqrt{5}-1 \pm \sqrt{-2\sqrt{5}-10}}{4}, \frac{-\sqrt{5}-1 \pm \sqrt{2\sqrt{5}-10}}{4}$

which can be verified to be the same solutions.

Ben Lynn blynn@cs.stanford.edu 💡