Roots of Unity

Gauss generalized his method to to find an expression (using radicals) for any root of unity.

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 $D=qd$ 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Now define $t_i$ to be $t$ where each $\beta$ has been replaced by ${\beta }^i$.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 $$\gamma =\frac{t_1+...+t_q}{q}$$ (a lot of cancellation occurs, from the fact that 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). We can use brute force: try every possible root until the resulting $\gamma$ is correct.

A much better way is to consider the expression $t_i{t}_1^{q-i}$. Then 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, this means $t_i{t}_1^{q-i}$ 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: $$\begin{array}{ccc}{t}_1& =& \zeta +i{\zeta }^2-{\zeta }^4-i{\zeta }^3\\ t_2& =& \zeta -{\zeta }^2+{\zeta }^4-{\zeta }^3\\ t_3& =& \zeta -i{\zeta }^2-{\zeta }^4+i{\zeta }^3\\ t_4& =& \zeta +{\zeta }^2+{\zeta }^4+{\zeta }^3=-1\end{array}$$ 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 do not blindly proceed using the most obvious method.

Instead 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 $$\begin{array}{ccc}{t}_1^2{t}_2& =& -(t_2^2)(1+2i)=-5(1+2i)\\ t_1{t}_3& =& (\zeta -{\zeta }^4{)}^2-(i({\zeta }^2-{\zeta }^3){)}^2\\ & =& ({\zeta }^2+{\zeta }^3+{\zeta }^4+\zeta )+2(-1-1)=-5\\ t_4& =& -1\end{array}$$ (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 $$\begin{array}{ccc}{t}_1& =& \alpha (\sqrt[4]{5}\sqrt{1+2i})\\ t_2& =& -{\alpha }^2\sqrt{5}\\ t_3& =& -{\alpha }^3(\sqrt[4]{5}\sqrt{1-2i})\\ t_4& =& -1\end{array}$$ 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.