Dyadic Hilbert symbols and higher unit groups 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.   
 A: 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.
