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For $p\neq 2$ it's easy to prove through the log/exp-correspondence that

$$(1+p\mathbb{Z}_p)^{p^k}=1+p^{k+1}\mathbb{Z}_p.$$

This gives an easy way to compute the groups $\mathbb{Q}_p^*/\mathbb{Q}_p^{*n}$ which are useful in a lot of computations involving local class field theory. I was wondering if there's some nice way to express the above equation in the case $p=2$?

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The log/exp correspondence for $p=2$ involves $4{\mathbf Z}_2$ and $1 + 4{\mathbf Z}_2$. Start with $(1+2{\mathbf Z}_2)^2 = ({\mathbf Z}_2^\times)^2 = 1 + 8{\mathbf Z}_2 = 1 + 2\cdot 4{\mathbf Z}_2$. –  KCd Nov 11 '11 at 3:51

1 Answer 1

Your equation is equivalent to the following statement: an element $x\in{\mathbb Z}_p$ is congruent to $1\pmod{p^{k+1}}$ if and only if one can write $x=y^{p^k}$, where $y\in{\mathbb Z}_p$ is congruent to $1\pmod p$. I believe you can prove this via Hensel's lemma, starting by looking $\pmod{p^{k+1}}$.

The analogous statement for $p=2$ can, I believe, be proved similarly; it is probably the same statement except with $k+1$ on the right-hand side replaced by $k+2$.

This is related to the fact that the multiplicative groups $({\mathbb Z}/p^k{\mathbb Z})^\times$ are cyclic when $p$ is odd, but $({\mathbb Z}/2^k{\mathbb Z})^\times$ isn't quite: it's the direct product of $\{\pm 1 \bmod 2^k\}$ with a cyclic group. This can be found in any number theory textbook that discusses primitive roots.

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