Four-dimensional quadratic forms over $\mathbb C(\!(t)\!)(X)$

Abstract

For quadratic forms in 4 variables defined over the rational function field in one variable over $\mathbb C(\!(t)\!)$, the validity of the local-global principle for isotropy with respect to different sets of discrete valuations is examined.

1 Introduction

Let E be a field of characteristic different from 2 and let E(X) denote the rational function field in one variable over E.

For $E = \mathbb C(\!(t)\!)$, the field of Laurent series in one variable over the complex numbers, the quadratic form

$$\begin{aligned} Y_1^2+tY_2^2+tY_3^2+X(Y_1^2+Y_2^2+ t Y_4^2) \end{aligned}$$

in the variables $Y_1, Y_2,Y_3,Y_4$ over E(X) has no non-trivial zero, but it has a non-trivial zero over the completion of E(X) with respect to any non-trivial valuation on E(X) that is trivial on E. This is in contrast to the situation when E is a finite field, by the Hasse-Minkowski theorem (see [6, Chapter VI, Theorem 66.1]). Note that, in both cases, the field E has a unique extension of each degree in a fixed algebraic closure.

By a $\mathbb Z$-valuation we mean a valuation with value group $\mathbb Z$. A quadratic form is isotropic if it has a non-trivial zero, otherwise it is anisotropic. Without any restrictions on the base field E other than $\mathsf {char}(E) \ne 2$, any anisotropic quadratic form over E(X) of dimension at most 3 remains anisotropic over the completion of E(X) with respect to some $\mathbb Z$-valuation on E(X) that is trivial on E; this follows for example from Milnor’s exact sequence [4, Theorem IX.3.1]. The case of 4-dimensional quadratic forms is the first case over E(X) where the validity of such a local-global principle for isotropy depends on the base field E.

When E is a non-dyadic local field, using a result of Lichtenbaum [5], one obtains that a 4-dimensional anisotropic quadratic form over E(X) remains anisotropic over the completion of E(X) with respect to some $\mathbb Z$-valuation on E(X) that is trivial on E (see [1, Remark 3.8]). This resembles the case where E is a finite field.

In contrast to the situations where E is a finite field or a local field, for $E=\mathbb C(\!(t)\!)$, the example of the quadratic form above shows that the local-global principle for isotropy of 4-dimensional quadratic forms over E(X) fails with respect to $\mathbb Z$-valuations that are trivial on E. However, anisotropy of this quadratic form can be detected over the larger field $\mathbb C(X)(\!(t)\!)$ by using Springer’s theorem (see [4, Proposition VI.1.9]).

Consider the more general situation where the field E is complete with respect to a non-dyadic $\mathbb Z$-valuation v. In this case, a local-global principle for isotropy was obtained in [1] using a geometric setup. Let $\mathcal {O}_v$ denote the valuation ring of v. By a model for E(X) over $\mathcal {O}_v$ we mean a two-dimensional integral normal projective flat $\mathcal {O}_v$-scheme $\mathscr {X}$ whose function field is isomorphic to E(X). Codimension-one points on a model of E(X) over $\mathcal {O}_v$ correspond to certain $\mathbb Z$-valuations on E(X). For a model $\mathscr {X}$ of E(X) over $\mathcal {O}_v$, let $\Omega _\mathscr {X}$ denote the set of $\mathbb Z$-valuations given by codimension-one points of $\mathscr {X}$. Consider the set $\Omega = \bigcup _{\mathscr {X} } \Omega _\mathscr {X}$ where the union is taken over all models $\mathscr {X}$ of E(X) over $\mathcal {O}_v$. It follows from [1, Theorem 3.1 and Remark 3.2] that an anisotropic quadratic form over E(X) remains anisotropic over the completion of E(X) with respect to some $\mathbb Z$-valuation in $\Omega$. One may ask whether this remains true if one replaces $\Omega$ by $\Omega _{\mathscr {X}}$ for some well-chosen model $\mathscr {X}$ of E(X) over $\mathcal {O}_v$.

The aim of this note is to show that this is not the case: if the residue field of v is separably closed, then, for any model $\mathscr {X}$ of E(X) over $\mathcal {O}_v$, there exists an anisotropic 4-dimensional quadratic form over E(X) which is isotropic over the completion of E(X) with respect to any $w\in \Omega _{\mathscr {X}}$ (Corollary 2). Let $\pi \in \mathcal {O}_v$ be a uniformiser of v. For any model $\mathscr {X}$ of E(X) over $\mathcal {O}_v$, the set $\{w(\pi ) \mid w \in \Omega _{\mathscr {X}} \}$ is finite and hence it has an upper bound. However, for any positive integer r, the quadratic form

$$\begin{aligned} \varphi _r = (X^r-\pi )Y_1^2+ (X^{r+1}+\pi )Y_2^2 + \pi X Y_3^2+ X(X^r+\pi )Y_4^2 \end{aligned}$$

is anisotropic over E(X), but it is isotropic over the completion of E(X) with respect to any $\mathbb Z$-valuation w on E(X) with $w(\pi ) <r$ (Theorem). The construction of $\varphi _r$ is inspired by the example in [1, Remark 3.6] of an anisotropic 6-dimensional quadratic form over $\mathbb Q_p(X)$ where p is an odd prime.

2 Results

We assume some familiarity with basic quadratic form theory over fields, for which we refer to [4]. We first fix some notation and recall some results.

By a quadratic form or simply a form we mean a regular quadratic form. Let E always be a field of characteristic different from 2 and let ${E}^{\times }$ denote its multiplicative group. For $a_1, \ldots , a_n \in {E}^{\times }$, the diagonal form $a_1X_1^2+\dots +a_nX_n^2$ is denoted by $\langle a_1, \ldots ,a_n\rangle$.

Let v be a $\mathbb Z$-valuation on E. We denote the corresponding valuation ring, its maximal ideal, and its residue field respectively by $\mathcal {O}_v, \mathfrak {m}_v$, and $\kappa _v$. For an element $a\in \mathcal {O}_v$, let $\overline{a}$ denote the residue class $a+\mathfrak {m}_v$ in $\kappa _v$. The completion of E with respect to v is denoted by $E_v$. We say that v is henselian if it extends uniquely to every finite field extension of E. Complete discretely valued fields are henselian (see [2, Theorem 1.3.1 and Theorem 4.1.3]). We recall a consequence of Hensel’s lemma:

Lemma

Let v be a henselian $\mathbb Z$-valuation on E such that $v(2) =0$. Then:

• (a) For $u_1,u_2\in {\mathcal {O}}^{\times }_v$, the quadratic form $\langle u_1, u_2 \rangle$ over E is isotropic if and only if $\overline{u_1u_2} \in -{\kappa }^{\times 2}_v$.

• (b) If $\kappa _v$ is separably closed, then every 3-dimensional form over E is isotropic.

Proof

(a) For $u_1,u_2\in {\mathcal {O}}^{\times }_v$, since v is henselian and $v(2)=0$, it follows by [2, Theorem 4.1.3(4)] that $u_1u_2 \in -{E}^{\times 2}$ if and only if $\overline{u_1u_2} \in -{\kappa }^{\times 2}_v$.

(b) Every 3-dimensional form over E contains a 2-dimensional form isometric to $\lambda \langle 1 ,u\rangle$ for some $u \in {\mathcal {O}}^{\times }_v$ and $\lambda \in {E}^{\times }$. If $\kappa _v$ is separably closed, then $\overline{u} \in -{\kappa }^{\times 2}_v$ and hence $\langle 1, u\rangle$ is isotropic by (a). $\square$

The set of all $\mathbb Z$-valuations on E(X) is denoted by $\Omega _{E(X)}$. For $r\in \mathbb {N}$, we define

$$\begin{aligned} \Omega _r = \{ w\in \Omega _{E(X)} \mid w({E}^{\times }) =i\mathbb Z \text{ for } \text{ some } 0\le i \le r \}. \end{aligned}$$

With this notation, $\Omega _0$ is the set of all E-trivial $\mathbb Z$-valuations on E(X). We recall that any monic irreducible polynomial $p \in E[X]$ determines a unique $\mathbb Z$-valuation $v_p$ on E(X) which is trivial on E and such that $v_p(p)=1$. There is further a unique $\mathbb Z$-valuation $v_\infty$ on E(X) such that $v_\infty (f)=-\deg (f)$ for any $f\in E[X]\setminus \{0\}$. Moreover, every $\mathbb Z$-valuation w on E(X) trivial on E is either equal to $v_\infty$ or to $v_p$ for some monic irreducible polynomial $p \in E[X]$ (see [2, Theorem 2.1.4]), and in either of the two cases the residue field is a finite field extension of E.

Theorem

Let v be a henselian $\mathbb Z$-valuation on E such that $v(2) =0$. Assume that $\kappa _v$ is separably closed. Let $\pi \in {E}^{\times }$ be such that $v(\pi ) =1$ and let $r \in \mathbb {N}$. Then the quadratic form

$$\begin{aligned} \varphi _r = \langle X^r-\pi , X^{r+1}+\pi , \pi X, X(X^r+\pi ) \rangle \end{aligned}$$

is isotropic over $E(X)_w$ for every $\mathbb Z$-valuation $w \in \Omega _{r-1}$, but anisotropic over $E(X)_w$ for some $w\in \Omega _r$.

Proof

Set $F=E(X)$. We first show that $\varphi _r$ is isotropic over $F_w$ for all $w\in \Omega _{r-1}$. Consider $w\in \Omega _{r-1}$.

Case 1: $w(\pi ) =0 = w(X)$. Then $\kappa _w$ is a finite extension of E. Since v is henselian, there is a unique extension $v'$ of v to $\kappa _w$, and $v'$ is again henselian. Furthermore, it follows by [2, Theorem 3.3.4] that $v'({\kappa }^{\times }_w)$ is isomorphic to $\mathbb Z$ and $\kappa _{v'}$ is separably closed. It follows by part (b) of the Lemma that every 3-dimensional quadratic form over $\kappa _w$ is isotropic. We have that $w =v_p$ for some monic irreducible polynomial $p \in E[X]$ such that $p \ne X$. Note that, in this case, at least three diagonal coefficients of $\varphi _r$ are units in $\mathcal {O}_w$. It follows by Springer’s theorem [4, Proposition VI.1.9] that $\varphi _r$ is isotropic over $F_w$.

Case 2: $0 \le w (\pi )< r$ and $1\le w(X)$. Let $u = (X^r\pi ^{-1}-1)(X^{(r+1)}\pi ^{-1}+1)$. Then $w(u) =0$ and $\overline{u} =-1 \in - {\kappa }^{\times 2}_w$. It follows by part (a) of the Lemma that the form $\pi ^{-1} \langle X^r-\pi , X^{r+1}+\pi \rangle$ is isotropic over $F_w$. Thus $\varphi _r$ is isotropic over ${F}_w$.

Case 3: $w(X)< 0 \le w (\pi )< r$. Note that, if $w(\pi ) =0$, then $w =v_\infty$ and $\kappa _w =E$, and otherwise $w|_E$ is equivalent to v and $\kappa _v \subseteq \kappa _w$; since $-1\in {\kappa }^{\times 2}_v$, we get in either case that $-1 \in {\kappa }^{\times 2}_w$. Consider $u = (1+ \pi X^{-{(r+1)}})(1+\pi X^{-r})$. We have that $w(u) =0$ and $\overline{ u } = 1 \in {\kappa }^{\times 2}_w = -{\kappa }^{\times 2}_w$. It follows by part (a) of the Lemma that the form $X^{-(r+1)} \langle X^{r+1}+\pi , X(X^{r}+\pi )\rangle$ is isotropic over $F_w$. Thus $\varphi _r$ is isotropic over $F_w$.

We have thus shown that $\varphi _r$ is isotropic over $F_w$ for every $w \in \Omega _{r-1}$. Now we show that $\varphi _r$ is anisotropic over $F_w$ for some $w\in \Omega _F$.

Let $E' = E(s)$, where $s =\root r \of {\pi }$. Then v extends uniquely to a valuation on $E'$ which we again denote by v. Note that $s^r = \pi$ in $E'$ and hence $v (\pi ) = rv(s)$. Then $v' = rv$ is a $\mathbb Z$-valuation on $E'$.

Let $L= E'(X)$ and let $Y =\frac{X}{s}$. Note that $L =E'(Y)$. By [2, Corollary 2.2.2], there exists a unique extension of $v'$ to L such that $v(Y) =0$ and $\overline{Y}$ is transcendental of $\kappa _{v'}$; we further have that $\kappa _w = \kappa _{v'}(\overline{Y})$ and $w({L}^{\times }) = v'({E'}^{\times }) = \mathbb Z$. Since $w(Y) =0$, we have that $w(X) =w(s) =1$. We get that

$$\begin{aligned} \varphi _r = \langle s^r(Y^r-1), s^r(sY+1), s^{r+1}Y, s^{r+1}Y(Y^r+1) \rangle . \end{aligned}$$

Consider the forms $\varphi _1 = \langle Y^r-1,~ sY+1\rangle$ and $\varphi _2 = \langle Y,~ Y(Y^r+1) \rangle$.

Since $\overline{Y}^r-1, \overline{Y}^{r} +1 \notin - {\kappa }^{\times 2}_w$, it follows by Springer’s theorem [4, Proposition VI.1.9] that the quadratic form $s^{-r}\varphi _r$ is anisotropic over $L_w$. Hence $\varphi _r$ is anisotropic over $L_w$. We obtain that $\varphi _r$ is anisotropic over $F_{w|_F}$. Note that $w(\pi ) = w(s^r) =rw(s) =r$, thus $w\in \Omega _r$. $\square$

We now provide a different perspective to the above theorem. For a subset $\Omega \subseteq \Omega _{E(X)}$, we say that $\Omega$ has the finite support property if for every $f \in {E(X)}^{\times }$, the set $\{w\in \Omega \mid w(f) \ne 0\}$ is finite. It is well-known that $\Omega _{0}$ has the finite support property. When E carries a discrete valuation, the set $\Omega _{E(X)}$ does not have the finite support property. However, for any model $\mathscr {X}$ of E(X) over $\mathcal {O}_v$, the set $\Omega _\mathscr {X}$ contains $\Omega _0$ and has the finite support property. We show the following:

Corollary 1

Let v be a henselian $\mathbb Z$-valuation on E with $v(2) =0$. Assume that $\kappa _v$ is separably closed. Let $\Omega \subseteq \Omega _{E(X)}$ be a subset with the finite support property. Then there exists an anisotropic 4-dimensional quadratic form over E(X) which is isotropic over $E(X)_w$ for every $w\in \Omega$.

Proof

Let $\pi \in {E}^{\times }$ be such that $v(\pi ) =1$. Since $\Omega$ has the finite support property, the set $\{w\in \Omega \mid w(\pi ) \ne 0 \}$ is finite. Set $r= 1+\max \{w(\pi ) \mid w\in \Omega \}$. Clearly $\Omega \subseteq \Omega _{r-1}$. Then the form $\varphi _r$ in the Theorem is isotropic over $E(X)_w$ for every $w\in \Omega$, but anisotropic over E(X). $\square$

Corollary 2

Let v be a henselian $\mathbb Z$-valuation on E with $v(2) =0$. Assume that $\kappa _v$ is separably closed. Let $\mathscr {X}$ be a regular model of E(X) over $\mathcal {O}_v$. Then there exists an anisotropic 4-dimensional quadratic form over E(X) which is isotropic over $E(X)_w$ for every $w\in \Omega _{\mathscr {X}}$.

Proof

By [3, Chapter  II, Lemma 6.1], for every element $f\in {E(X)}^{\times }$, the set $\{w \in \Omega _{\mathscr {X}} \mid w(f) \ne 0 \}$ is finite, hence the statement follows by Corollary 1.

$\square$