Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $x$?

share|improve this question
    
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. –  Chandan Singh Dalawat Dec 30 '11 at 6:44
    
The correct label for this kind of questions is nt.number-theory. –  Chandan Singh Dalawat Dec 30 '11 at 6:53
    
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/…. –  Chandan Singh Dalawat Dec 30 '11 at 10:35
    
Chandan, thank you for your comments. I changed the label as you suggested. –  emiliocba Dec 30 '11 at 14:51

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
2  
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 $-1$. –  Lubin Dec 29 '11 at 23:54
    
@Lubin: yes, thanks, will-do... –  paul garrett Dec 30 '11 at 0:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.