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.

  • 1
    $\begingroup$ What is a local dyadic number field? $\endgroup$ Sep 2, 2016 at 4:59
  • $\begingroup$ Looks like he means 2-adic local field. $\endgroup$
    – Pig
    Sep 2, 2016 at 6:12
  • 3
    $\begingroup$ She means a finite extension of $\mathbf{Q}_2$. $\endgroup$ Sep 2, 2016 at 6:57

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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