Let $p$ be an odd prime number.

As a corollary of Hensel's lemma, we know that for $m\in \mathbb{N}$ there is a primitive $m$-th root of unity iff $m|(p-1)$.

Hence, if $p\equiv 1$ mod 4, we have a $2^n$-th root of unity in $\mathbb{Q}_p$, for some $n\geq 2$.

My question is the following: once we have established the existence of such roots, how can we calculate the exact number of 2-power roots of unity in the $p$-adic field?

  • 2
    $\begingroup$ If $p \nmid 2n$ then $X^{2n}+1$ has $2n$ different roots in $\overline{\mathbb{Q}}$, $\overline{\mathbb{Q}_p}$ and $\overline{\mathbb{F}_p}$. Those in $\mathbb{Q}_p$ are the ones corresponding to the solutions in $\mathbb{F}_p$ : there are $gcd(p-1,2n)$ of them. $\endgroup$ – reuns Sep 4 '17 at 19:47

For $p$ an odd prime, the roots of unity in $\Bbb Q_p$ correspond to the nonzero elements of $\Bbb F_p$ and form a cyclic group of order $p-1$. So the number of $2$-power roots of unit in $\Bbb Q_p$ is the largest power of $2$ dividing $p-1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.