## Roots of Unity

Gauss generalized his method to 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 primitve $p$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Now define $t_i$ to be $t$ where each $\beta$ has been replaced by $\beta^i$.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 $k\gt 1$). 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:

$\array { t_1 &=& \zeta + i \zeta^2 - \zeta^4 - i\zeta^3 \\ t_2 &=& \zeta - \zeta^2 + \zeta^4 - \zeta^3 \\ t_3 &=& \zeta - i \zeta^2 - \zeta^4 + i\zeta^3 \\ t_4 &=& \zeta + \zeta^2 + \zeta^4 + \zeta^3 = -1 }$

We need to compute $t_1^4$, choose a fourth root of the result and then work out values for $t_2, t_3, t_4$ from there. 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

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

(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

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

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.