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Fix a prime $p$. A p-adic field is a finite extension of $\mathbb{Q}_p$.

Question 1: Let $K$ be a $p$-adic field and fix $n$. Is there $m$ such that if $\alpha \in \mathbb{Q}_p$ is an $m$th power in $K$ then $\alpha$ is an $n$th power in $\mathbb{Q}_p$? $\quad$ (Probably $m$ is a multiple of $n$.)

I think that you'd approach this by decomposing $K/\mathbb{Q}_p$ into a tower of extensions of some sort and then go up the tower. So we probably want to actually prove the following.

Question 2: Suppose that $K/L$ is an extension of $p$-adic fields and fix $n$. Is there $m$ such that if $\alpha \in K$ is an $m$th power in $L$ then $\alpha$ is an $n$th power in $K$?

I can prove Question 1 in the case when $K/\mathbb{Q}_p$ is unramified, but I don't know about the general case. I think that one might be able to use class field theory, or maybe it's already well-known, or maybe I am just missing something elementary.

I have good motivation for this question, but it would take a bit of work to explain, the motivation comes from logic.

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    $\begingroup$ Small comment: this is "true up to roots of unity". Set $d=[K:L]$ to be the degree of the field extension and let $m=nd$. The norm map $N\colon K\to L$ is multiplicative and has the property that $N(x)=x^d$ if $x\in L$. Thus, if $\alpha\in L$ and $\alpha=y^m$ for some $y\in K$, then taking norms we obtain $\alpha^d=N(x)=N(y)^m=N(y)^{nd}$. So $\alpha = \zeta\cdot N(y)^n$ for some $d$-th root of unity $\zeta\in L$. $\endgroup$
    – Piotr Achinger
    Commented Jan 24, 2022 at 20:38
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    $\begingroup$ fixed it, thanks @LSpice $\endgroup$
    – Erik Walsberg
    Commented Jan 24, 2022 at 23:55

1 Answer 1

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Yes.

I'll solve the more general question 2.

Let $v_p(n)$ be the highest power of $p$ dividing $n$. Then if $\alpha \in L$ is congruent to $1$ modulo $p^{v_p(n)+1}$, then $\alpha$ is an $n$th power in $L$.

Every $\alpha \in L$ can be written as $\pi^j u$ where $j\in \mathbb Z$ and $u$ is a unit in $L$. It suffices to find $m$ such that if $\pi^j u$ is an $m$th power in $K$ then $j$ is divisible by $n$ and $u$ is congruent to $1$ modulo $p^{v_p(n)+1}$.

Let $e$ be the ramification degree of $K$ over $L$, i.e. the greatest $e$ such that there is an element of $K$ whose $e$'th power is $\pi$ times a unit and let $k$ be any natural number. Then if $\pi^j u$ is a $ke$'th power in $K$ then $j$ is divisible by $k$, and hence $u = \pi^j u / (\pi^{j/k})^k$ is also a $k$th power in $K$.

So it suffices to find $k$ such that if $u$ is a unit that is a $k$th power in $K$ then $u$ is congruent to $1$ mod $p^{v_p(n)+1}$.

Let $q$ be the order of the residue field in $k$. If $u$ is a $k$th power, then it is a $k$th power of some unit $v$, and $v^{q-1}$ is congruent to $1$ modulo the uniformizer $\pi'$ of $K$. So $$ v^{ (q-1) p^r} = (1 + (v^{q-1}-1))^{p^r} = \sum_{i=0}^{P^r} \binom{p^r}{i} (v^{q-1}-1)^i$$ where the $i$th term is divisible by $\binom{p^r}{i} \pi'^{i}$.

Taking $r$ sufficiently large, we can ensure that $\binom{p^r}{i} \pi'^{i}$ is divisible by $p^{ v_p(n)+1}$ for all $i>0$, so $v^{ (q-1) p^r}$ is congruent to $1$ mod $p^{ v_p(n)+1}$, so we may take $k= (q-1) p^r$.

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  • $\begingroup$ how do you get the divisibility in the penultimate line? $\endgroup$
    – Erik Walsberg
    Commented Jan 25, 2022 at 23:01
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    $\begingroup$ @ErikWalsberg Here's a crude argument: For $i$ sufficiently large depending on $K,n$, $\pi'^{i}$ is divisible by $p^{v_p(n)+1}$ so there are only finitely many $i$s to consider. For each such $i$, for $r$ sufficiently large, $\binom{p^r}{i}$ is divisible by $p^{v_p(n)+1}$ by Kummer's theorem, since adding $i$ to $p^r-i$ has at least $r - \log_p(i)$ carries. $\endgroup$
    – Will Sawin
    Commented Jan 26, 2022 at 0:28
  • $\begingroup$ ok, cool. I'll accept this as an answer once I convince myself it all works. this will take a bit because i know nothing about these things. $\endgroup$
    – Erik Walsberg
    Commented Jan 26, 2022 at 19:36
  • $\begingroup$ As part of this you essentially show that if $U \subseteq \mathbb{Z}^\times_p$ is a nbhd of $1$ then there is $m$ such that if $\alpha \in \mathbb{Z}^\times_p$ is an $m$th power in $K$ then $\alpha \in U$. (This is what the binomial coefficient stuff does.) Now, we know that the topology on $\mathbb{Q}_p$ agrees with that induced by the topology on $K$. So this boils down to saying that the collection of sets of $m$th powers in $V^\times$ forms a nbhd basis at $1$. (Here $V$ is the valuation ring of $K$.) $\endgroup$
    – Erik Walsberg
    Commented Feb 2, 2022 at 19:33
  • $\begingroup$ As $V^\times$ is abelian any finite index subgroup contains the $m$th powers for some $m$. So we just need to know that the collection of finite index subgroups in $V^\times$ forms a nbhd basis at $1$. This follows as $V^\times$ is a profinite group. $\endgroup$
    – Erik Walsberg
    Commented Feb 2, 2022 at 19:35

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