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Let $F$ be a local dyadic number field, $\mathfrak{p}$ its maximal ideal, $(*,*)_F$ its quadratic Hilbert symbol and $e$ its ramification index (i.e. $\mathfrak{p}^e$ is exact divisor of $2$). Fix an even $s\le 2e$. What is then the smallest $t\ge 0$ such that $(U_s,U_t)_F=1$ (i.e. such that for all $a\in U_s$ and $b\in U_t$, we have $(a,b)_F=1$). Here $U_s$ are the $s$-units, i.e. $U_s=1+\mathfrak{p}^s$ for $s\ge 1$, and $U_0$ are the units of $K$.

By the Local Square Theorem the elements of $U_{2e+1}$ are squares, hence such a $t$ exists. By the very last exercise in Serre's "Local Fields" we know that $t\le 2e-s$ (in fact, I do not have a proof for this and would be grateful for a complete reference).

Is always $t=2e-s$ (which is true for $\mathbb Q$), or are there fields and even $s<2e$ with $t<2e-s$?

I know that there are explicit formulas for dyadic Hilbert symbols (Vostokov/Letsko, Henniart, ...) which possibly would enable me to work out an answer to my question. However, to become comfortable with these formulas seems to be not so obvious and I would be happy for a reference or any hint to a more conceptual proof avoiding such formulas.

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    $\begingroup$ What is a local dyadic number field? $\endgroup$ – Qiaochu Yuan Sep 2 '16 at 4:59
  • $\begingroup$ Looks like he means 2-adic local field. $\endgroup$ – Pig Sep 2 '16 at 6:12
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    $\begingroup$ She means a finite extension of $\mathbf{Q}_2$. $\endgroup$ – Chandan Singh Dalawat Sep 2 '16 at 6:57
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More generally, allow $p$ to be any prime and let $K$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$ (which is automatic when $p=2$). The ramification index $e$ of $K$ over $\mathbf{Q}_p$ is divisible by $p-1$ (because $K$ contains $\mathbf{Q}_p(\zeta)$ by hypothesis); define $e_1$ by $e=(p-1)e_1$.

A little work as in Chapter 15 of Hasse's Number Theory (or as in Section V of Local discriminants) allows you to determine the structure of the filtered $\mathbf{F}_p$-space $K^\times\!/K^{\times p}$. The filtration on this quotient comes from the filtration $$ \cdots U_2\subset U_1\subset\mathfrak{o}_K^\times\subset K^\times $$ on $K^\times$, where $\mathfrak{o}_K$ is the ring of integers of $K$, with unique maximal ideal $\mathfrak{p}_K$, and, for every $i>0$, $U_i=1+\mathfrak{p}_K^i$ is the kernel of $\mathfrak{o}_K^\times\to(\mathfrak{o}/\mathfrak{p}_K^i)^\times$. Denote the image of $U_i$ in $\overline{K^\times}=K^\times\!/K^{\times p}$ by $\bar U_i$. Then the image of $\mathfrak{o}_K^\times$ is $\bar U_1$, we have $\bar U_{pe_1+1}=\{1\}$, and the filtration on $\overline{K^\times}$ looks like $$ \{1\} \subset_1\bar U_{pe_1} \subset_f\bar U_{pe_1-1} \cdots \subset_f\bar U_{pi+1} =\bar U_{pi} \subset_f\cdots \subset_f\bar U_1 \subset_1\overline{K^\times}. $$ Here, $i$ is any integer in the interval $[1,e_1[$ (which is empty when $e_1=1$), an inclusion $E\subset_rE'$ means that $E$ is a codimension-$r$ subspace of $E'$, and $f$ is the residual degree of $K$ over $\mathbf{Q}_p$.

We have the hilbertian pairing $\overline{K^\times}\times\overline{K^\times}\to{}_pK^\times$, where ${}_pK^\times$ is of course the group of $p$-th roots of $1$ in $K$.

The orthogonal complement of the subspace $\bar U_i$ for the hilbertian pairing is precisely $\bar U_{pe_1-i+1}$, for every $i\in[0,pe_1+1]$, provided we adopt the convention $\bar U_0=\overline{K^\times}$.

It is amusing to try to figure out the analogue of all this when $K$ is a finite extension of $\mathbf{F}_p((\pi))$, where $\pi$ is transcendental.

Addendum 1 Okay, here is a brief sketch of the proof. First, before the hilbertian pairing, there is the kummerian pairing : $$ \overline{K^\times}\times\mathrm{Gal}(M|K)\to{}_pK^\times, $$ where $M$ is the maximal abelian extension of $K$ of exponent $p$. The group $G=\mathrm{Gal}(M|K)$ comes with a natural filtration : the ramification filtration in the upper numbering. One may ask : how is the filtration on $\overline{K^\times}$ related to the filtration on $G$ ? Answer : The two filtrations are orthogonal to each other in an appropriate sense. See for example Section IX of Local discriminants.

Secondly, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$ (with a normalisation which doesn't affect anything here), and the hilbertian pairing is obtained from the kummerian pairing via this isomorphism. Moreover, the filtration on $G$ is the image of the filtration on $\overline{K^\times}$ by $\rho$. Putting these two things together gives you the result.

Addendum 2 What happens when the local field $K$ has characteristic $p$ ? Kummer theory has to be replaced by Artin-Schreier theory, so we have to first understand the filtration on $\overline{K^+}=K^+/\wp(K^+)$, where $K^+$ is the additive group of $K$ and $\wp(x)=x^p-x$. Denoting the image of $\mathfrak{p}_K^i$ by $\overline{\mathfrak{p}^i}$, it turns out that $\overline{\mathfrak{p}}=\{0\}$, and the analogous picture is $$ \{\bar0\}\subset_1 \overline{\mathfrak{p}^0}\subset_f \overline{\mathfrak{p}^{-1}} \cdots\subset_f \overline{\mathfrak{p}^{pj+1}} = \overline{\mathfrak{p}^{pj}} \subset_f \overline{\mathfrak{p}^{pj-1}} \cdots\subset K^+\!/\wp(K^+). $$ See for example Further remarks.

Let $M$ be the maximal abelian extension of $K$ [edit of exponent $p$] and $G=\mathrm{Gal}(M|K)$. We have the analogous pairing $$ \overline{K^+}\times G\to\mathbf{F}_p, $$ and we still have the ramfication filtration (in the upper numbering) on $G$. It turns out that the two filtrations are orthogonal to each other under this pairing.

As before, putting $\overline{K^\times}=K^\times\!/K^{\times p}$, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$, and it carries the filtration on $\overline{K^\times}$ (which is no longer finite) onto the filtration on $G$.

Putting these two facts together gives the analogous result in characteristic $p$. I leave for you the pleasure of working out the details.

Addendum 3 (2016/09/06) Still not convinced ? Some more details can be found in my Note arXiv:1609.01160.

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