Powers of a uniformizer in a totally ramified extension of $\mathbb{Q}_p$

Question

Consider the cyclotomic extension $\mathbb{Q}_p(\zeta_{p^n})/\mathbb{Q}$. This is a totally ramified extension of degree $\phi(p^n) = p^{n-1}(p-1)$ ($\phi$ is the Euler totient) and a particular choice of uniformizer is $\pi = \zeta_{p^n}-1$. As this is a uniformizer, we have an equation $\pi^{p^{n-1}(p-1)} = pu$ for some $u\in\mathbb{Z}_p[\zeta_{p^n}]^\times$.

Question: When is $u^k\in\mathbb{Z}_p$ for some $k\geq 1$?

In other words, when is some power of the uniformizer contained in $\mathbb{Z}_p^\times$? Does this depend on the choice of uniformizer?

An example where this happens is $\mathbb{Q}_2(\zeta_4)$ where we have $\pi^2 = -2\zeta_4$ so that $\pi^4 = -4\in\mathbb{Z}_2$. However, for $\mathbb{Q}_2(\zeta_8)$, I don't think there is any such power of $\zeta_8-1$ that lies in $\mathbb{Z}_2^\times$.

Answer 1

This only happens when $p^n \in \{2,3,4\}$.

Write $\pi = \zeta - 1$. If $\pi^n \in \mathbf{Q}_p$, then $(\sigma \pi)^n = \sigma \pi^n = \pi^n$, and so $(\sigma \pi/\pi)^n = 1$, and so $\sigma \pi/\pi$ is a root of unity. At this point one can forget about $\mathbf{Q}_p$ entirely.

If $p$ is odd, take $\sigma \pi = \zeta^2 - 1$ and then $\zeta + 1$ is a root of unity, which implies that $\zeta$ has order exactly three. (Note that $z=1+\zeta$ lies on the intersection of $|z|=1$ and $|z-1| = 1$.) One sees that $(\zeta-1)^6 = -27$ in this case.

If $p = 2$, take $\sigma \pi = \zeta^3 - 1$ and then $1+\zeta+\zeta^2$ is a root of unity. But then $\zeta^{-1} + 1 + \zeta$ will be a real root of unity and real, and hence equal $+1$ or $-1 $. In the former case $\zeta = \pm i$, and the latter case $\zeta = -1$. Certainly $(-1-1)^1 = -2$ and $(\pm i - 1)^4 = -4$.

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Added: A remark on the extent to which this depends on the choice of uniformizer. As KCd mentioned, $\pi = (-p)^{1/(p-1)}$ is a well-known uniformizer in $\mathbf{Q}_p(\zeta_p)$ with $\pi^{p-1} \in \mathbf{Q}_p$. But this is more or less the only example. Namely, write $K = \mathbf{Q}_p(\zeta_{p^n})$. Then I claim that no choice of uniformizer $\pi$ has a power of $\pi$ being in $\mathbf{Q}_p$ unless either $n = 1$, in which case one can take $\pi = (-p)^{1/(p-1)}$, or $n = 2$ and $p = 2$.

Write $K = \mathbf{Q}_p(\zeta_{p^n})$, and $G = \mathrm{Gal}(K/\mathbf{Q}_p) = (\mathbf{Z}/p^n \mathbf{Z})^{\times}$, where the last isomorphism sends $[a]$ to $\zeta \mapsto \zeta^a$. We know that, under the assumption that a power of $\pi$ lies in $\mathbf{Q}_p$, that for any $\sigma \in G$ we have

$$\sigma \pi = \eta \pi,$$

where $\eta \in K$ is a root of unity. Now the roots of unity in $K$ (easy exercise) all have order dividing $p^n(p-1)$. It follows that there exists a map:

$$G \rightarrow \mu_{p^n(p-1)} = \mu_{p^n} \oplus \mu_{p-1}, \quad \sigma \mapsto \frac{\sigma \pi}{\pi}$$

and this is (almost by definition) a cocycle in $H^1(G,\mu_{p^n} \oplus \mu_{p-1})$ where the coefficients have the natural action of $G$. The key point is now that this cohomology group is very small and one can compute it explicitly. By Sah's lemma (since $G$ is abelian), if $[c]$ is any non-trivial cocycle then $[c]$ is annihilated by $g-1$ for any $g \in G$.

Assume first that $p$ is odd. Then there exists an $[a] \in G$ with $a-1 \not\equiv 0 \bmod p$. But $(a-1)$ acts by an automorphism on $\mu_{p^n}$. That means that, up to a coboundary, the class $[c]$ lies in $H^1(G,\mu_{p-1})$. In more direct language, that means that after replacing $\pi$ by $\varpi = \pi \xi$ for a root of unity $\xi$ (corresponding to the coboundary $\sigma \mapsto \sigma \xi/\xi$), it must be the case that $\sigma \varpi/\varpi \in \mu_{p-1}$ for all $\sigma$. But since $K$ is totally ramified we have $K = \mathbf{Q}_p(\varpi)$, and thus there are $p^{n-1}(p-1)$ conjugates of $\sigma$ and so this is a contradiction as soon as $n > 1$.

Now assume that $p=2$. We now take $[a]=[-1]$, and now by Sah's lemma we deduce that $2[c]$ is trivial (there is no $\mu_{p-1}$ component anymore). The cocycle associated to $2[c]$ is $\sigma \mapsto \sigma \pi^2/\pi^2$, and hence it follows that for some root of unity $\xi$ we have $\pi^2 \xi$ is $G$-invariant and so lies in $\mathbf{Q}_p$. But just by looking at valuations, this is only possible if $e_2 \le 2$, and so we obtain a contradiction as soon as $n > 2$.

Answer 2

A "famous" alternate description of $\mathbf Q_p(\zeta_p)$ is $\mathbf Q_p(\sqrt[p-1]{-p})$, so inside this field there is a prime $\pi$ such that $\pi^{p-1} = -p$. This $\pi$ is not $1-\zeta_p$, and with this prime we get $u = -1$.

All such things would be uniformizer dependent, and generally it is not realistic to think they should happen at all.

Although you're asking about particular totally ramified extensions, let me address a related point about general finite extensions $K/\mathbf Q_p$, where the prime $\pi$ in $K$ has $p = \pi^e{u}$ where $u \in \mathcal O_K^\times$ and $e$ is the ramification index of $K$ over $\mathbf Q_p$. We usually do not expect to have $u \in \mathbf Z_p^\times$: if there were a prime $\pi$ where $u \in \mathbf Z_p^\times$, then $\pi$ would be the root of $X^e - p{u}^{-1}$, which is Eisenstein at $p$ in $\mathbf Z_p[X]$, so $\mathbf Q_p(\pi)$ would be a degree $e$ totally ramified extension of $\mathbf Q_p$ inside $K$ and we could realize $K$ as the composite of an unramified and totally ramified extension of $\mathbf Q_p$. That's simply not how things work in general: always $K$ is a totally ramified extension of an unramified extension of $\mathbf Q_p$, but we aren't normally able to break that tower apart to come from totally ramified and unramified extensions of $\mathbf Q_p$.

Example. Let $F$ be the quadratic unramified extension of $\mathbf Q_p$, so $p$ is prime in $F$. Being quadratic and unramified, $F = \mathbf Q_p(\sqrt{u})$ for some $u \in \mathbf Z_p^\times$. Set $K = F(\sqrt{p\sqrt{u}})$. Since $p$ is prime in $F$, $p\sqrt{u}$ is not a square in $F$, so $K/F$ is a ramified quadratic extension. Then $[K:\mathbf Q_p] = 4$ with $e \geq 2$ and $f \geq 2$, so $e = f = 2$. It turns out that the only quadratic intermediate field between $K$ and $\mathbf Q_p$ is $F$, so $K$ is not a composite of a totally ramified and unramified extension of $\mathbf Q_p$. When $p = 5$ and we use $u = 2$, $K = \mathbf Q_5(\sqrt{5\sqrt{2}}) = \mathbf Q_5(\sqrt[4]{50})$ is a Kummer extension of $\mathbf Q_5$, so this is an example where $K/\mathbf Q_5$ is a Galois (even abelian) extension.