Structure of unramified extension of $\mathbb{Q}_2$ Let $K$ be the unique unramfied extension of degree $2$ over $\mathbb{Q}_2$. In this case a well-known theorem, says that $\mathcal{O}_K^\times \cong \mathbb{Z}/(q-1)\mathbb{Z} \times \mathbb{Z}/2^a \mathbb{Z} \times \mathbb{Z}_2^2$, where $q$ is the size of the residue field and $a$ corresponds to all the $2$-power roots of unity. In this case, I think $q=4$ and $a=1$. My question is how do I explicitly write down this isomorphism? I know that there is an isomorphism $1+4\mathcal{O}_K$ with $\mathbb{Z}_2^2$ using the log and then picking a basis. But, I dont know how to use exp and log in this case, since after removing the torsion I am not in $(1+4\mathcal{O}_K)$.
In other words, what is the isomorphism $\mathcal{O}_K^\times/{torsion} \to \mathcal{O}_K\cong \mathbb{Z}_2^2$.
Thanks
 A: I'll abbreviate $U^{(n)}:= 1+2^n \mathcal{O}_K$, as multiplicative groups.
You already know that $U^{(2)} \simeq \mathbb Z_2 \oplus \mathbb Z_2$ (as topological groups) via exp and log, and convince yourself this also transports the $\mathbb Z_2$-module structure on the RHS to the left to any standard way one defines $\mathbb Z_2$-powers on that multiplicative group.
But those $\mathbb Z_2$-powers already work on $ U^{(1)}$, whose torsion part is $\pm1$. Now the quotient $U^{(1)}/U^{(2)}$ has order four, and modding out the torsion we are left with a quotient of order two. Choose one representative of the non-trivial coset, its square will be in $U^{(2)}$ and can serve as one of the two topological generators of that group.
(In this specific example, this works out nicely algebraically: $K$ is actually $\mathbb Q_2(\zeta_3)$ with a primitive third root of unity, and a nice representative of that coset is $1+2\zeta_3$: its square is $-3 \in U^{(2)}$.)
So if you insist that your isomorphism must restrict to an iso $U^{(2)} \simeq \mathbb Z_2 \oplus \mathbb Z_2$, then what that gives you is an iso $U^{(1)}/\pm1 \simeq (\frac12 \mathbb Z_2) \oplus \mathbb Z_2$. Otherwise, you shift that one summand one $p$-power up, and have an iso
$U^{(1)}/\pm1 \simeq \mathbb Z_2 \oplus \mathbb Z_2$
whose restriction to $U^{(2)}$, however, maps onto $(2\mathbb Z_2) \oplus \mathbb Z_2$.
What is psychologically annoying here is that the filtrations don't match. For $U^{(1)}/\pm1 \simeq \mathbb Z_2 \oplus \mathbb Z_2$, as said $U^{(2)} \simeq (2\mathbb Z_2) \oplus \mathbb Z_2$, then $U^{(3)} \simeq (4\mathbb Z_2) \oplus (2\mathbb Z_2)$, $U^{(4)} \simeq (8\mathbb Z_2) \oplus (4\mathbb Z_2)$ etc.
But to make it as explicit as possible: Every element $x \in U^{(1)}$ has a unique representation $x=\pm (1+2\zeta_3)^a (1+4\zeta_3)^b$ with $a,b \in \mathbb Z_2$, and one iso as demanded is $x \mapsto (a,b)$. Again, note that $(1+2\zeta_3)^2=-3 = 1+(-1)4$ is a topological generator of "the base field half" $1+4\mathbb Z_2$ of $U^{(2)}$.
