Norm of $\mathbb{Q}_2(i)^\times$ I am trying to compute the norm group of $\mathbb{Q}_2(i)^\times$. If i'm not mistaken we have $\mathbb{Q}_2(i)^\times = (1+i)^\mathbb{Z}\mathbb{Z}_2[i]^\times$ so $N(\mathbb{Q}_2(i)^\times) = N(1+i)^\mathbb{Z}N(\mathbb{Z}_2[i]^\times) = 2^\mathbb{Z} \{ a^2 + b^2 \;|\; a,b\in \mathbb{Z}_2 \text{ and } a^2 + b^2 \in \mathbb{Z}_2^\times \}$.
So, if what was written above is right, the question becomes to what exactly corresponds $\{ a^2 + b^2 \;|\; a,b\in \mathbb{Z}_2 \text{ and } a^2 + b^2 \in \mathbb{Z}_2^\times \}$ in $\mathbb{Z}_2[i]^\times$.
 A: Presumably you are doing this to work out a local classfield theory example explicitly? That is, one anticipates that the index is 2, for this (totally ramified) quadratic extension.
On a finite-index subgroup of the units, the $2$-adic logarithm converges (as does the $2$-adic exponential on a finite-index subgroup of $\mathbb Z_2$), so you can reduce to the analogous question on some $\mathbb Z/2^N$, with $N=3$ or so, a finite computation.
EDIT: more explicitly, the exponential $e^x=\sum_{n\ge 0} x^n/n!$ converges for $2$-adic $x$ with ${\rm ord}_2 x>2$, at worst. One can (re-) do the estimate on powers of $2$ dividing exponentials to recover the fact. For $p\not=2$ convergence is better: for ${\rm ord}_p x>0$. A similar discussion of logarithm gives an inverse from $1+8\mathbb Z_2$ (or better) back to $8\mathbb Z_2$. There are similar exp and log on the quadratic extension, with somewhat worse convergence because the extension is ramified... and exp and log behave reasonable with respect to norm/trace. Thus, a specific ("sufficiently small" ... but precisely) finite-index subgroup of $\mathbb Z_2^\times$ is demonstrably hit by norm. Thus, with "8" replaced by whatever it really should be, it suffices to determine the image of norm of $1+8\mathbb Z_2[i]$ in $1+8\mathbb Z_2$ (both of these conventional expressions for finite-index subgroups of the local units).
A: If you already know that the norm group is of index two in $\mathbb Q_2^*$, then it’s not so hard just to guess something that’s not a norm. You do know, I guess, that $\mathbb Q_2^*$ is $\langle 2,-1,1+4\mathbb Z_2\rangle$. From that point, there shouldn’t be much more to do.
If you don’t know the index of the norms in the downstairs multiplicative group, I guess @paulgarrett’s strategy is best.
