# produce primitive p-th root of unity

Dwork is one of the mathematicians that I admire so much.

Recently I learned how he produced primitive p-th roots of unity in $\mathbb{C}_p$ “by hand”, which is interesting enough for me to write it down here.

Let $\pi \in \mathbb{C}_p$ satisfies $\pi^{p-1} = -p$. Consider the following function $\theta(z) = Exp(\pi z - \pi z^p)$. Then after plugging $\{1, \zeta, \zeta^2, \ldots, \zeta^{p-2}\}$ in $\theta(z)$, we will miraculously get all the primitive p-th roots of unity in $\mathbb{C}_p$. Here $\zeta$ is a primitive $(p-1)$-st root of unity. The function $\theta(z)$ bears the name “Dwork’s exponential”, I believe.

To prove the above statement, it suffices to prove it for $z=1$. So we need to prove $\theta(1)^p = 1$ yet $\theta(1) \not= 1$.

One thing to notice at the first place is that the convergence range of exponential function is the open disc of radius $|\pi|$, and the polynomial $\pi z - \pi z^p$ has norm exactly $|\pi|$ on the closed disc of radius $1$. So the value $\theta(1)$ cannot be computed by plugging $1$ into the polynomial, which is $0$, and then plug the output in the exponential function.

However, $\theta(z)^p = Exp(\pi z - \pi z^p)^p = Exp(p\pi z - p \pi z^p)$. Now the polynomial $p (\pi z - \pi z^p)$ has smaller norm on the closed disc, therefore we may compute the value at $1$ by means in the previous paragraph. Hence we see that $\theta(1)^p = 1$.

Now we need to see that $\theta(1) \not= 1$. In the following we will show that $\theta(1) - 1 = \pi \mod(\pi^2)$. To achieve this, let us rewrite $\theta(z) = E_p(z) F(z)$. Here $E_p(z) = Exp(x + x^p/p + x^{p^2}/p^2 + \ldots)$, and $F(z) = Exp(-x^{p^2}/p^2 - x^{p^3}/p^3 - \ldots)$, where $x = \pi z$. The function $E_p(z)$ is called the Artin-Hasse exponential. There’s a lemma of Dwork telling us that $E_p(z) \in \mathbb{Z}_p[[\pi z]]$, hence $E_p(1) = 1 + \pi \mod (\pi^2)$. On the other hand it’s easy to show directly that $F(1) = 1 \mod(\pi^2)$. Put everything together, we see that $\theta(1) - 1 = (E_p(1) F(1)) - 1 = (1+\pi) - 1 = \pi \mod(\pi^2)$ which is what we want to show.

In fact, before doing everything we probably have to show the convergence radius of $\theta(z)$ is strictly bigger than $1$. This actually easily follows from the estimate of $E_p(z) \text{ and } F(z)$.

From the above construction, we see that there is a bijection between sets from $(p-1)$-st roots of $-p$ and primitive p-th roots of unity. This is of course well known, since if we let $w = \zeta_p - 1$, then we’ll have $w^{p-1} = -(p + \binom{p}{2} w + \ldots + p w^{p-2})$.