Isomorphism between additive and multiplicative ideals In algebraic number theory, consider a non-archimedean completion of a number field $F$, say $F_v$, with ring of integers $O_v$ and maximal ideal $p_v$. Why is it true that
$$O_v^\times / (1+p_v^e) \simeq (O_v / p_v^e)^\times ? $$
And is there a natural or intuitive way to think about it? It looks like a way to transpose multiplicative thinks into additive things. Topologically, $1+p_v^e$ is a neighborhood of $1$ (in $O_v^\times$), while $p_v^e$ is a neighborhood of zero (in $O_v$), is there an analogue of the isomorphism above in this light? an analogous situation in the archimedean setting?
 A: To prove the isomorphism you can define a map $\mathcal O_v^\times \to (\mathcal O_v/\mathfrak p_v^e)^\times$ by just sending $u$ to $u + \mathfrak p_v^e$. It is not hard to see that this is surjective and the kernel is given by $1+\mathfrak p_v^e$. Hence the isomorphism follows.
On the other side, note that both groups are multiplicative. Indeed, $\mathcal O_v^\times$ is the group of units and $1+\mathfrak p_v^e$ is an ideal in it. Clearly, $(\mathcal O_v/\mathfrak p_v^e)^\times$ is also  multiplicative.
An interesting observation is to note that $\mathfrak p_v^e$ and $1+\mathfrak p_v^e$ are isomorphic for $e$ large enough, where $\mathfrak p_v^e$ is considered as an additive group. You can see that by defining $\mathfrak p$-adic log and exp functions, by simply using their well-known Taylor expansions. However, you have to worry about the convergence, which is why we want $e$ to be large enough. For example, we have $1+p\mathbb Z_p \simeq p\mathbb Z_p$ if $p$ is an odd prime, but $1+4\mathbb Z_2 \simeq 4\mathbb Z_2$, as exp doesn't converge on $2\mathbb Z_2$.
