TITLE: Norm of $\mathbb{Q}_2(i)^\times$
QUESTION [5 upvotes]: 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 $\mathbb{Z}_2[i]^\times$.

REPLY [3 votes]: 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.