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Let $$ L=\mathbb{Q}(\sqrt{-1})\otimes_\mathbb{Q} \mathbb{Q}_3 $$ where $\mathbb{Q}_3$ denotes de $3$-adic rational numbers. Then $L$ is a quadratic extension of the local field $\mathbb{Q}_3$. Furthermore, the valuation ring of $L$ is $B:=\mathbb{Z}[\sqrt{-1}] \otimes \mathbb{Z}_3$.

It is known that the norm map $N$ maps the set $U_L$ of units in $B$ onto the set $U_{\mathbb{Q}_3}$ of units in $\mathbb{Z}_3$ (see Serre's book "Local Fields", Chapter V, Prop. 3). This implies that there is an element $x\in U_L$ such that $N(x)=-1$ since $-1$ is a unit in $\mathbb{Q}_3$. Could someone specify this element $x$?

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  • $\begingroup$ The fact that $-1$ is in the image of the norm map from $\mathbf{Q}_3(\sqrt{-1})^\times$ down to $\mathbf{Q}_3^\times$ is a particular case of the following : if $p$ is an odd prime and $u,v\in\mathbf{Z}_p^\times$, then $u$ is in the image of the norm map from $\mathbf{Q}_p(\sqrt{v})^\times$ down to $\mathbf{Q}_p^\times$. See for examples Serre's Course in arithmetic, Chapter III. $\endgroup$ – Chandan Singh Dalawat Dec 30 '11 at 6:44
  • $\begingroup$ The correct label for this kind of questions is nt.number-theory. $\endgroup$ – Chandan Singh Dalawat Dec 30 '11 at 6:53
  • $\begingroup$ For the image of the norm map $K^\times\to\mathbf{Q}_2^\times$ for a quadratic extension $K$ of $\mathbf{Q}_2$, see mathoverflow.net/questions/55390/…. $\endgroup$ – Chandan Singh Dalawat Dec 30 '11 at 10:35
  • $\begingroup$ Chandan, thank you for your comments. I changed the label as you suggested. $\endgroup$ – emiliocba Dec 30 '11 at 14:51
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One approach is by Hensel's Lemma: consider $N(x+i)=x^2+1=-1$, for example. Since $1^2+1^2=-1 \mod 3$ while $2\cdot 1=2\not= 0 \mod 3$, the equation $x^2+1=-1$ has a solution in $\mathbb Z_3$. Hensel's lemma gives a sequence of integers approaching the solution. Note that there are many things in $\mathbb Q_3(i)$ with norm $1$, so there's no unique solution to $N(x+iy)=-1$.

Edit: As @Lubin notes, Hensel (and also exponential/logarithm) considerations demonstrate a $\sqrt{-2}$ in $\mathbb Q_3$, call it $\beta$. Then $N(\beta+i)=\beta^2+1=-2+1=-1$, yet again.

A point neglected in my earlier answer was that the $3$-adic exponential and log can give other ways to express things such as $\sqrt{-2}$, as alternative to Hensel. The outcomes have different utilities.

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    $\begingroup$ Or, to make your response crystal clear, you could point out that Hensel or almost anything else shows that ${\mathbb{Q}}_3$ has in it a square root $\alpha$ of $-2$, so that $\alpha +i $ has norm $-1$. $\endgroup$ – Lubin Dec 29 '11 at 23:54

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