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Let $F$ be an unramified extension of $Q_2$. How can I compute the Hilbert symbol (a,b) for $a,b \in F^*$. Here, (a,b) is 1 if $ax^2+by^2=z^2$ has a nontrivial solution, and -1 otherwise.

In the case of $F=Q_2$, I can do this by finding representatives of $Q_2^*/(Q_2^*)^2$, and calculating the Hilbert symbol for each possible pair. However, this becomes a bit unmanageable for larger fields $F$. Is there a better way of approaching this?

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If the degree of $F/\mathbb{Q}_2$ is odd, then I believe that the symbol in question is equal to the symbol for their norms (in $mathbb{Q}_2$). Besides, there are some explicit formulas for the Hilber symbol. – Mikhail Bondarko Nov 17 '11 at 10:25
There is an old paper of Hasse which gives a general formula, even when the extension $F$ of ${\bf Q}_2$ is not unramified. If I'm able to figure out what he does, I will come back to post an answer here. The Hilbert symbol is also related to the Weil index, but I cannot find the reference right now. – Chandan Singh Dalawat Nov 17 '11 at 11:43
Hi Soroosh -- see for results of Vostokov and Letsko, for the Hilbert symbol in extensions of 2-adic fields. I generally remember Vostokov's name for explicit formulas for Hilbert symbols - he's written a lot on this. – Marty Nov 17 '11 at 16:34
Explicit reciprocity laws were also proved by Guy Henniart in Sur les lois de réciprocité explicites. I. J. Reine Angew. Math. 329 (1981), 177–203. – Chandan Singh Dalawat Nov 17 '11 at 16:52

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