Diagonal forms of higher degree over function fields of $p$-adic curves

TL;DR

This paper studies diagonal forms of higher degree over function fields of p-adic curves, showing isotropy over certain fields when isotropic over completions, and relates this to bounds on the higher u-invariant.

Contribution

It introduces a method to analyze isotropy of diagonal forms over specific fields using patching, extending understanding of higher u-invariants for these forms.

Findings

Isotropic forms over completions imply isotropy over related fields.

The work complements existing bounds on the higher u-invariant.

Focus on diagonal forms of degree d in characteristic not dividing d!.

Abstract

We investigate diagonal forms of degree $d$ over the function field $F$ of a smooth projective $p$-adic curve: if a form is isotropic over the completion of $F$ with respect to each discrete valuation of $F$, then it is isotropic over certain fields $F_U$, $F_P$ and $F_p$. These fields appear naturally when applying the methodology of patching; $F$ is the inverse limit of the finite inverse system of fields $\{F_U,F_P,F_p\}$. Our observations complement some known bounds on the higher $u$-invariant of diagonal forms of degree $d$. We only consider diagonal forms of degree $d$ over fields of characteristic not dividing $d!$.

Equations

$$\theta (v_1,\dots ,v_d)=\frac{1}{d!}\sum _{1\le i_1<\cdots <i_l\le d}(-1{)}^{d-l}\phi (v_{i_1}+\cdots +v_{i_l})$$

$$({\theta }_1\perp {\theta }_2)(u_1+v_1,\dots ,u_d+v_d)={\theta }_1(u_1,\dots ,u_d)+{\theta }_2(v_1,\dots ,v_d)$$

$$u_{diag}(d,k)\le \min \{n|\mathrm{all}\mathrm{forms}\mathrm{of}\mathrm{degree}d\mathrm{over}k\mathrm{of}\mathrm{dimension}\ge n\mathrm{are}\mathrm{universal}\}$$

$${\rho }_0\perp s{\rho }_1\perp \cdots \perp s^{d-1}{\rho }_{d-1}$$

$${\phi }_1\perp x{\phi }_2\perp y{\phi }_3\perp xy{\phi }_4\perp x^2{\phi }_5\perp y^2{\phi }_6\perp x^2{y}^2{\phi }_7\perp x^{2}y{\phi }_8\perp x{y}^2{\phi }_9\perp \cdots \perp x^{d-1}{y}^{d-1}{\phi }_{d^2},$$

$$({\phi }_1\perp x{\phi }_2\perp x^2{\phi }_5\perp \dots )\perp y({\phi }_3\perp x{\phi }_4\perp \dots )\perp y^2({\phi }_6\perp \dots )\perp \cdots \perp y^{d-1}(\cdots \perp x^{d-1}{\phi }_{d^2})$$

$$({\phi }_1\perp x{\phi }_2\perp x^2{\phi }_5\perp \dots ),({\phi }_3\perp x{\phi }_4\perp \dots ),\cdots ,$$

Full text

We only consider diagonal forms of degree $d$ over fields of characteristic not dividing $d!$.

Introduction

The fact that Springer’s Theorem holds for diagonal forms of higher degree over fields of characteristic not dividing $d!$ [9] guarantees that on occasion diagonal forms of higher degree defined over function fields behave similarly to quadratic forms. For a survey on the behaviour of (diagonal) forms of higher degree in general the reader is referred to [12].

In this note we consider diagonal forms of degree $d$ over function fields $F=K(X)$ where $X$ is a smooth, projective, geometrically integral curve over $K$ and $K$ is the fraction field of a complete discrete valuation ring with a residue field $k$ of characteristic not dividing $d!$. Let $v$ be a rank one discrete valuation of $F$, and $F_{v}$ the completion of $F$ with respect to $v$. It was shown by Colliot-Thélène, Parimala and Suresh [2, Theorem 3.1] that a quadratic form which is isotropic over $F_{v}$ for each $v$ is already isotropic over $F$, using the methodology of patching developed by Habater and Hartmann [4], i.e. viewing $F$ as the inverse limit of a finite inverse system of certain fields $\{F_{U},F_{P},F_{p}\}$.

Given a nondegenerate diagonal form $\varphi$ over $F$ of degree $d>2$ and dimension $>2$, it is not clear, however, whether the isotropy of $\varphi$ over $F_{v}$ for each $v$ implies that $\varphi$ is isotropic.

Our main result proves that the isotropy of a nondegenerate diagonal form $\varphi$ over $F_{v}$ for each $v$ implies that at least over the field extensions $F_{U}$, $F_{P}$ and $F_{p}$ of $F$, $\varphi$ is isotropic as well (Theorem 6). These fields depend on the choice of the form $\varphi=\langle a_{1},\dots,a_{n}\rangle$, more precisely on the choice of the regular proper model ${\mathcal{X}}$ (over the complete discrete valuation ring $A$) of the curve $X$ over $K$, which depends on $\varphi$: ${\mathcal{X}}$ is selected such that there exists a reduced divisor $D$ with strict normal crossings, which contains both the support of the divisor of all the entries $a_{i}$, $1\leq i\leq n$, and the components of the special fibre of $X/A$. Since nondegenerate diagonal forms of degree $d\geq 3$ have finite automorphism groups [6, p. 137], we are not able to apply [5, Theorem 3.7] to conclude that the isotropy of $\varphi$ over the $F_{U}$’s and $F_{P}$’s implies that $\varphi$ is also isotropic over $F$, however. This is only possible for $d=2$.

After collecting the terminology and some basic results in Section 1, in particular defining diagonal $u$-invariants of degree $d$ over $k$, we consider diagonal forms of higher degree over valued fields in Section 2 and then study diagonal forms of higher degree over function fields of $p$-adic curves using some of the ideas of [2] in Section 3. Recall that a $p$-adic field is a finite field extension of $\mathbb{Q}_{p}.$

As a consequence of Springer’s Theorem for diagonal forms, any diagonal form of degree $d$ and dimension $>d^{3}+1$ over a function field in one variable $F=K(t)$, where $K$ is a $p$-adic field with residue field $k$, ${\rm char}(k)\nmid d!$, is isotropic over $F_{v}$ for every discrete valuation $v$ with residue field either a function field in one variable over $k$ or a finite extension of $K$. Moreover, it is isotropic over $F_{U}$ for each reduced, irreducible component $U\subset Y$ of the complement of $S$ in the special fibre $Y={\mathcal{X}}\times_{A}k$ of $X/A$, and isotropic over $F_{P}$ for each $P\in S$ (Corollary 9), and thus isotropic over $F_{p}$ for each $p=(U,P)$. Here, $S$ is the inverse image under a finite $A$-morphism $f:{\mathcal{X}}\to\mathbb{P}_{A}^{1}$ of the point at infinity of the special fibre $\mathbb{P}_{k}^{1}$.

1. Preliminaries

Let $k$ be a field such that ${\rm char}(k)$ does not divide $d!$.

1.1. Forms of higher degree

Let $V$ be a finite-dimensional vector space over $k$ of dimension $n$. A $d$-linear form over $k$ is a $k$-multilinear map $\theta:V\times\dots\times V\to k$ ($d$-copies) which is symmetric, i.e. $\theta(v_{1},\dots,v_{d})$ is invariant under all permutations of its variables. A form of degree $d$ over $k$ (and of dimension $n$) is a map $\varphi:V\to k$ such that $\varphi(av)=a^{d}\varphi(v)$ for all $a\in k$, $v\in V$ and such that the map $\theta:V\times\dots\times V\to k$ defined by

[TABLE]

($1\leq l\leq d$) is a $d$-linear form over $k$. By fixing a basis $\{e_{1},\dots,e_{n}\}$ of $V$, any form $\varphi$ of degree $d$ can be viewed as a homogeneous polynomial of degree $d$ in $n={\rm dim}\,V$ variables $x_{1},\dots,x_{n}$ via $\varphi(x_{1},\dots,x_{n})=\varphi(x_{1}e_{1}+\dots+x_{n}e_{n})$ and, vice versa, any homogeneous polynomial of degree $d$ in $n$ variables over $k$ is a form of degree $d$ and dimension $n$ over $k$. Any $d$-linear form $\theta:V\times\dots\times V\to k$ induces a form $\varphi:V\to k$ of degree $d$ via $\varphi(v)=\theta(v,\dots,v)$. We can hence identify $d$-linear forms and forms of degree $d$.

Two $d$-linear spaces $(V_{i},\theta_{i})$, $i=1,2$, are isomorphic (written $(V_{1},\theta_{1})\cong(V_{2},\theta_{2})$ or just $\theta_{1}\cong\theta_{2}$) if there exists a bijective linear map $f:V_{1}\to V_{2}$ such that $\theta_{2}(f(v_{1}),\dots,f(v_{d}))=\theta_{1}(v_{1},\dots,v_{d})$ for all $v_{1},\dots,v_{d}\in V_{1}.$ A $d$-linear space $(V,\theta)$ (or the $d$-linear form $\theta$) is nondegenerate if $v=0$ is the only vector such that $\theta(v,v_{2},\dots,v_{d})=0$ for all $v_{i}\in V$. A form of degree $d$ is called nondegenerate if its associated $d$-linear form is nondegenerate. A form $\varphi$ over $k$ is called anisotropic, if it does not have any non-trivial zeroes, otherwise it is called isotropic.

The orthogonal sum $(V_{1},\theta_{1})\perp(V_{2},\theta_{2})$ of two $d$-linear spaces $(V_{i},\theta_{i})$, $i=1,2$, is the $k$-vector space $V_{1}\oplus V_{2}$ together with the $d$-linear form

[TABLE]

($u_{i}\in V_{1}$, $v_{i}\in V_{2}$) [7].

A $d$-linear space $(V,\theta)$ is called decomposable if $(V,\theta)\cong(V,\theta_{1})\perp(V,\theta_{2})$ for two non-zero $d$-linear spaces $(V,\theta_{i})$, $i=1,2$. If $\varphi$ is represented by the homogeneous polynomial $a_{1}x_{1}^{d}+\ldots+a_{m}x_{m}^{d}$ ($a_{i}\in k^{\times}$) we use the notation $\varphi=\langle a_{1},\ldots,a_{n}\rangle$ and call $\varphi$ diagonal. A diagonal form $\varphi=\langle a_{1},\ldots,a_{n}\rangle$ over $k$ is nondegenerate if and only if $a_{i}\in k^{\times}$ for all $1\leq i\leq n$.

If $d\geq 3$, $a_{i},b_{j}\in k^{\times}$, then $\langle a_{1},\ldots,a_{n}\rangle\cong\langle b_{1},\ldots,b_{n}\rangle$ if and only if there is a permutation $\pi\in S_{n}$ such that $\langle b_{i}\rangle\cong\langle a_{\pi(i)}\rangle$ for every $i$. This is a special case of [6, Theorem 2.3].

Note that for quadratic forms ($d=2$), the automorphism group of $\varphi$ is infinite, whereas for $d\geq 3$, the automorphism group of $\varphi$ usually is finite, for instance if $\varphi$ is is nonsingular in the sense of algebraic geometry [13]. In particular, nondegenerate diagonal forms of degree $d\geq 3$ have finite automorphism groups [6, p. 137], which creates a problem when trying to imitate patching arguments as it is not possible to apply [5, Theorem 3.7].

1.2. Higher degree $u$-invariants

The $u$-invariant (of degree $d$) of $k$ is defined as $u(d,k)=\sup\{\dim_{k}\varphi\}$, where $\varphi$ ranges over all the anisotropic forms of degree $d$ over $k$. The diagonal $u$-invariant (of degree $d$) of $k$ is defined as $u_{diag}(d,k)=\sup\{\dim\varphi\}$, where $\varphi$ ranges over all the anisotropic diagonal forms over $k$.

Thus the diagonal $u$-invariant $u_{diag}(d,k)$ is the smallest integer $n$ such that all diagonal forms of degree $d$ over $k$ of dimension greater than $n$ are isotropic, and the $u$-invariant $u(d,k)$ is the smallest integer $n$ such that all forms of degree $d$ over $k$ of dimension greater than $n$ are isotropic. Obviously, $u_{diag}(d,k)\leq u(d,k).$ If $u=u(d,k)$ then each anisotropic form of degree $d$ over $k$ of dimension $u$ is universal. If $u=u_{diag}(d,k)$ then each diagonal anisotropic form of degree $d$ over $k$ of dimension $u$ is universal. We have

[TABLE]

with the understanding that the “minimum” of an empty set of integers is $\infty$, cf. [12]. For $d=2$, $u_{diag}(d,k)=u(d,k)$ is the $u$-invariant of quadratic forms.

For an algebraically closed field $k$, $|k^{\times}/k^{\times d}|=1$ and hence $u_{diag}(d,k)=u(d,k)=1.$ For a formally real field $k$, the diagonal $u$-invariant is infinite for even $d$: since $-1\not\in\sum k^{2}$, also $-1\not\in\sum k^{d}$ for any even $d$. Thus the form $m\times\langle 1\rangle$ of degree $d$ is anisotropic for each integer $m$, implying $u_{diag}(d,k)=u(d,k)=\infty.$

The strong diagonal $u$-invariant of degree $d$ of $k$, denoted $u_{diag,\,s}(d,k)$, is the smallest real number $n$ such that

1. (1)

every finite field extension $E/k$ satisfies $u_{diag}(d,E)\leq n$, and 2. (2)

every finitely generated field extension $E/k$ of transcendence degree one satisfies $u_{diag}(d,E)\leq dn$.

If these $u$-invariants are arbitrarily large, put $u_{diag,\,s}(d,k)=\infty.$

Analogously as observed in [5] for $d=2$, $u_{diag,\,s}(d,k)\leq n$ if and only if every finitely generated field extension $E/k$ of transcendence degree $l\geq 1$ satisfies $u_{diag}(d,k)\leq d^{l}n$. Thus if $u_{diag,\,s}(d,k)$ is finite, it is at least 1 and lies in $\frac{1}{d}\mathbb{N}$.

1.3. $C_{r}^{0}$ fields

Let $r\geq 1$ be an integer. A field $F$ is a $C_{r}$-field if for all $d\geq 1$ and $n>d^{r}$, every homogeneous form of degree $d$ in $n$ variables over $F$ has a non-trivial solution in $F$. In particular, then $F$ satisfies $u(d,F)\leq d^{r}$. Moreover, every finite extension of $F$ is a $C_{r}$-field, and every one-variable function field over $F$ a $C_{r+1}$-field [14, II.4.5]. Hence $u_{diag,\,s}(d,F)\leq d^{r}$ for a $C_{r}$-field $F$.

A field $F$ is a $C_{r}^{0}$-field if the following holds: For any finite field extension $F^{\prime}$ of $F$ and any integers $d\geq 1$ and $n>d^{r}$, for any homogeneous form over $F^{\prime}$ of degree $d$ in $n$ variables, the greatest common divisor of the degrees of finite field extensions $F^{\prime\prime}/F^{\prime}$ over which the form acquires a nontrivial zero is one. This amounts to requiring that the $F^{\prime}$-hypersurface defined by the form has a zero-cycle of degree 1 over $F^{\prime}$.

Assume ${\rm char}(F)=0$. For each prime $l$, let $F_{l}$ be the fixed field of a pro-$l$-Sylow subgroup of the absolute Galois group of $F$. Any finite subextension of $F_{l}/F$ is of degree coprime to $l$. The field $F$ is $C_{r}^{0}$ if and only if each of the fields $F_{l}$ is $C_{r}$. A finite field extension of a $C_{r}^{0}$-field is $C_{r}^{0}$. If $F$ is $C_{r}^{0}$ then a function field $E=F(x_{1},\dots,x_{s})$ in $s$ indeterminates $x_{1},\dots,x_{s}$ over $F$ is $C_{r+s}^{0}$ [2, 2.1].

It is not known if $p$-adic fields have the $C^{0}_{2}$-property.

Remark 1**.**

Assume that $p$-adic fields have the $C^{0}_{2}$-property. Let $K(X)$ be any function field of transcendence degree $r$ over a $p$-adic field $K$ (here we do not need to assume $p\not=2,3$). Suppose that there is $\ell\not=2$ such that there exists a finite subextension of $K_{\ell}(X)/K(X)$ of degree 2. Then any cubic form over $K(X)$ in strictly more than $3^{2+r}$ variables has a nontrivial zero: If the $p$-adic field $K$ is $C_{3}^{0}$, then the function field $E=K(X)$ in $r$ variables over $K$ is $C_{3+r}^{0}$ [2, Lemma 2.1]. Thus a cubic form over $E=K(X)$ in strictly more than $3^{2+r}$ has a nontrivial zero in each of the fields $K_{\ell}(X)$, $l$ a prime, hence in a finite extension of $K(X)$ of degree coprime to $\ell$, for each $\ell$ prime. Pick $\ell\not=2$, then $[K_{\ell}(X):K(X)]$ is even. Moreover, pick $l\not=2$ such that there exists a finite subextension of $E_{\ell}/K(X)$ of degree 2 then the cubic form has a zero over it. By Springer’s Theorem for cubic forms and their behaviour under quadratic field extensions [8, VII], thus the cubic form has a nontrivial zero in $K(X)$. This is the analogue of [2, Proposition 2.2].

2. Diagonal forms over Henselian valued fields

2.1.

Let $K$ be a valued field with valuation $v$, valuation ring $R$ and maximal ideal $m$. Let $\Gamma$ be the value group. Assume that $d!$ is not divisible by the characteristic of the residue field $k=R/m$. For $u\in R$, denote by $\bar{u}$ the image of $u$ in $k$. For a polynomial $f\in R[X]$, $f=a_{n}x^{n}+\dots+a_{1}x+a_{0}$, define the polynomial $\overline{f}=\bar{a}_{n}x^{n}+\dots+\bar{a}_{1}x+\bar{a}_{0}$ over $k$. If $\varphi=\langle a_{1},\dots,a_{n}\rangle$ is a nondegenerate diagonal form with entries $a_{i}\in R$, define the diagonal form $\overline{\varphi}=\langle\bar{a}_{1},\dots,\bar{a}_{n}\rangle$ over $k$. $\varphi$ is called a unit form, if $\overline{\varphi}$ is nondegenerate. Choose a set $\{\pi_{\gamma}\in R\;|\;\gamma\in I\}$ such that the values of the $\pi_{\gamma}$’s represent the distinct cosets in $\Gamma/d\Gamma$. We may decompose a diagonal form $\varphi$ as $\varphi=\perp\varphi^{\prime}_{\gamma}$ by taking $\varphi^{\prime}_{\gamma}$ to be the diagonal form whose entries comprise all $a_{i}$ with $v(a_{i})=v(\pi_{\gamma})\,{\rm mod}\,d\,\Gamma$. By altering the slots by $d$-powers if necessary, we may then write $\varphi^{\prime}_{\gamma}=\pi_{\gamma}\varphi_{\gamma}$ with each $\varphi_{\gamma}$ a diagonal unit form. There are only finitely many non-trivial $\varphi_{\gamma}$ [9]. If $\Gamma=\mathbb{Z}$, the set $\{\pi_{\gamma}\;|\;\gamma\in I\}$ can be chosen to be $\{\pi^{i}\;|\;i=0,\dots,d-1\}$ and $|\Gamma/d\Gamma|=d$ is finite.

If $R$ satisfies Hensel’s Lemma then $(K,v)$ is called a Henselian valued field and $R$ a Henselian valuation ring. Every complete discretely valued field is Henselian.

Let $\varphi$ be a diagonal form over a Henselian valued field $(K,v)$. Write $\varphi=\pi_{1}\varphi_{1}\perp\dots\perp\pi_{r}\varphi_{r}$ with each $\varphi_{i}$ a diagonal unit form and the $\pi_{i}$ having distinct values in $\Gamma/d\Gamma$. Then $\varphi$ is isotropic if and only if some $\overline{\varphi_{i}}$ is isotropic [9, Proposition 3.1]. This is because for a diagonal unit form $\varphi$ over a Henselian valued field $(K,v)$, $\varphi$ is isotropic if and only if $\overline{\varphi}$ is isotropic [9, Lemma 2.3].

Theorem 2**.**

*([9] or [12, Theorem 4, Corollary 2]) Suppose that ${\rm char}(k)\nmid d!$.

(i) Let $(K,v)$ be a Henselian valued field. Then $u_{diag}(d,K)=|\Gamma/d\Gamma|\,u_{diag}(d,k).$

(ii) Let $(K,v)$ be a Henselian valued field. If every diagonal form of degree $d$ of dimension $n+1$ over $k$ is isotropic, then every diagonal form of degree $d$ and dimension $dn+1$ over $K$ is isotropic. If $k$ has an anisotropic form of degree $d$ and dimension $n$, then $K$ has an anisotropic form of degree $d$ and dimension $dn$.

(iii) Let $K$ be a discretely valued field. Then $u_{diag}(d,K)\geq d\,u_{diag}(d,k).$

(iv) Let $F$ be a field extension of finite type over $k$ of transcendence degree $n$. Then $u_{diag}(d,F)\geq d^{n}u_{diag}(d,k^{\prime})$ for a suitable finite field extension $k^{\prime}/k$.

The (in)equalities in (i), (iii), (iv) also hold when the values are infinite.*

For $d=2$, (ii) is Springer’s Theorem for quadratic forms over Henselian valued fields [15]. Springer’s Theorem does not hold for non-diagonal forms of higher degree than 2 [9, 2.7]. Theorem 2 is a major ingredient in our proofs, for instance we can show:

Proposition 3**.**

*Let $A$ be a discrete valuation ring with fraction field $K$ and residue field $k$ such that ${\rm char}(k)\nmid d!$.

(i) $u_{diag}(d,K)\geq d\,u_{diag}(d,k)$ and $u_{diag,\,s}(d,K)\geq d\,u_{diag,\,s}(d,k)$.

(ii) If $A$ is Henselian then $u_{diag}(d,K)=d\,u_{diag}(d,k)$.

(iii) If $A$ is a complete discrete valuation ring then every finite extension of $K$ has diagonal $u$-invariant at most $d\,u_{diag}(d,k)$.*

The first assertions of (i) as well as (ii) and (iii) follow from Theorem 2. The proof of the second claim in (i) is analogous to the one of [5, 4.9], employing Theorem 2 instead of Springer’s Theorem.

2.2.

A field $K$ is called an $m$-local field with residue field $k$ if there is a sequence of fields $k_{0},\dots,k_{m}$ with $k_{0}=k$ and $k_{m}=K$, and such that $k_{i}$ is the fraction field of an excellent Henselian discrete valuation ring with residue field $k_{i-1}$ for $i=1,\dots,m$. Recall that a discrete valuation ring $R$ is called excellent, if the field extension ${\widehat{K}}/K$ is separable, where ${\widehat{K}}$ denotes the quotient field of $R$ and $K$ is its completion. (This condition is trivially satisfied if $K$ has characteristic 0 or $R$ is complete.)

Proposition 3 implies (compare the next two results with [5, Corollary 4.13, 4.14] for quadratic forms):

Corollary 4**.**

*Suppose that $K$ is an $m$-local field whose residue field $k$ is a $C_{r}$-field with ${\rm char}(k)\nmid d!$. Let $F$ be a function field over $K$ in one variable.

(i) $u_{diag}(d,k)=u_{diag,s}(d,k)=d^{r}$ and $u_{diag}(d,K)=d^{r+m}$.

Moreover, if some normal $K$-curve with function field $F$ has a $K$-point, then

$u_{diag}(d,F)\geq d^{r+m+1}$.

(ii) If $u_{diag}(d,k^{\prime})=d^{r}$ for every finite extension $k^{\prime}/k$, then $u_{diag}(d,F)\geq d^{r+m+1}$.*

Proof.

(i) Since $k$ is a $C_{r}$-field, $u_{diag}(d,k)=d^{r}$, thus $u_{diag}(d,k)\leq u_{diag,s}(d,k)\leq d^{r}$ and the first two equations follow. Applying Proposition 3 and induction yields that $u_{diag}(d,K)\geq d^{m}u_{diag}(d,k)=d^{r+m}$. Let $X$ be a normal $K$-curve with function field $F$ and let $\xi$ be a $K$-point on $X$. The local ring at $\xi$ has fraction field $F$ and residue field $K$. So Proposition 3 implies that $u_{diag}(d,K)=d^{m}u_{diag}(d,k)$ and $u_{diag}(d,F)\geq du_{diag}(d,K)=d^{r+m+1}$.

(ii) Choose a normal or equivalently a regular $K$-curve $X$ with function field $F$, and a closed point $\xi$ on $X$. Let $R$ be the local ring of $X$ at $\xi$ with residue field $\kappa(\xi)$. Then the fraction field of $R$ is $F$, and $\kappa(\xi)$ is a finite extension of $K$. Hence $\kappa(\xi)$ is an $m$-local field whose residue field $k^{\prime}$ is a finite extension of $k$. By assumption, $u_{diag}(d,k^{\prime})=d^{r}$ and $k^{\prime}$ is a $C_{r}$-field since $k$ is. So applying part (ii) to $k^{\prime}$ and $\kappa(\xi)$, it follows that $u_{diag}(d,\kappa(\xi))\geq d^{r+m}$. Proposition 3 yields $u_{diag}(d,F)\geq d^{r+m+1}$. ∎

Corollary 5**.**

*(i) Let $F$ be a one-variable function field over an $m$-local field $K$ with residue field $k$ such that ${\rm char}(k)\nmid d!$ and $k$ is algebraically closed. Then $u_{diag}(d,F)\geq d^{m+1}$.

(ii) If $k$ is a finite field and $u_{diag}(d,k)=d^{r}$ with $r\in\{0,1\}$, then $u_{diag}(d,k)=u_{diag,s}(d,k)=d^{r}$ and $u_{diag}(d,K)\geq d^{r+m}$. Moreover, if some normal $K$-curve with function field $F$ has a $K$-point, then $u_{diag}(d,F)\geq d^{r+m+1}$.*

Proof.

(i) This is a special case of Corollary 4 using that an algebraically closed field $k$ is $C_{0}$, satisfies $u_{diag}(d,k)=1$, and has no non-trivial finite extensions.

(ii) A finite field is $C_{1}$ [14, II.3.3(a)], hence $u_{diag}(d,k)\leq d$. Here $u_{diag}(d,k)=u_{diag,\,s}(d,k)=d^{r}$ and Corollary 4 yields the assumption. ∎

In general, for any finite field $k={\mathbb{F}}_{q}$ we obviously do not have $u_{diag}(d,k)=d^{r}$, $r\in\{0,1\}$: for instance, if $-1\in{\mathbb{F}}_{q}^{\times d}$ and $d\geq 4$ then $u_{diag}(d,{\mathbb{F}}_{q})\leq d-1$ by [10]. Or, if $d^{*}={\rm gcd}(d,q-1)$, then $u_{diag}(d,\mathbb{F}_{q})\leq d^{*}.$ This implies that $u_{diag}(d,\mathbb{F}_{q})=1$, if $d$ is relatively prime to $q-1$ and that for $q>(d^{*}-1)^{4}$, $u_{diag}(d,\mathbb{F}_{q})=2$ [12, 5.1].

3. The behaviour of diagonal forms of higher degree over function fields of $p$-adic curves

Whenever we write ‘discrete valuation ring’ and ‘discrete valuation’ we mean a discrete valuation ring of rank one and a valuation with value group $\mathbb{Z}$.

3.1.

Let $A$ be a complete discrete valuation ring with fraction field $K$ and residue field $k$ with ${\rm char}(k)\nmid d!$. Let $X$ be a smooth, projective, geometrically integral curve over $K$ and $F=K(X)$ be the function field of $X$. Let $t$ denote a uniformizing parameter for $A$. For each (rank one) discrete valuation $v$ of $F$, let $F_{v}$ denote the completion of $F$ with respect to $v$.

We will adapt some ideas from [2] to diagonal forms of higher degree: take a nondegenerate form $\varphi=\langle a_{1},\dots,a_{n}\rangle$ of degree $d$ over $F$. Then choose a regular proper model ${\mathcal{X}}/A$ of $X/K$, such that there exists a reduced divisor $D$ with strict normal crossings which contains both the support of the divisor of all the entries $a_{i}$, $1\leq i\leq n$, and the components of the special fibre of $X/A$. (Note that this implies that the regular proper model ${\mathcal{X}}/A$ depends on the form $\varphi$, and thus so do $Y,Y_{i}$, $S_{0}$, $S$, $F_{P},F_{U},\dots$ as defined in the following.)

Let $Y={\mathcal{X}}\times_{A}k$ be the special fibre of $X/A$. Let $x_{i}$ be the generic point of an irreducible component $Y_{i}$ of $Y$. Then there is an affine Zariski neighbourhood $W_{i}\subset{\mathcal{X}}$ of $x_{i}$, such that the restriction of $Y_{i}$ to $W_{i}$ is a principal divisor. Let $S_{0}$ be a finite set of closed points of $Y$ containing all singular points of $D$, and all the points that lie on some $Y_{i}$, but not in the corresponding $W_{i}$.

Choose a finite $A$-morphism $f:{\mathcal{X}}\to\mathbb{P}_{A}^{1}$ as in [4, Proposition 6.6]. Let $S$ be the inverse image under $f$ of the point at infinity of the special fibre $\mathbb{P}_{k}^{1}$. Then the set $S_{0}$ is contained in $S$. All the intersection points of two components $Y_{i}$ are in $S$. Each component $Y_{i}$ contains at least one point of $S$. Let $U\subset Y$ run through the reduced irreducible components of the complement of $S$ in $Y$. Then each $U$ is a regular affine irreducible curve over $k$ and we define $k[U]$ to be its ring of regular functions and $k(U)$ to be its fraction field. $k[U]$ is a Dedekind domain and $U={\rm Spec}\,k[U]$. Each $U$ is contained in an open affine subscheme ${\rm Spec}\,R^{U}$ of ${\mathcal{X}}$ and is a principal effective divisor in ${\rm Spec}\,R^{U}$. Moreover, $R_{U}$ is the ring of elements in $F$ which are regular on $U$ and also a regular ring, since it is the direct limit of regular rings. The ring $R_{U}$ is a localisation of $R^{U}$ and so $U$ is a principal effective divisor on ${\rm Spec}\,R_{U}$ given by the vanishing of an element $s\in R_{U}$. The $t$-adic completion ${\widehat{R}}_{U}$ of $R_{U}$ is a domain and coincides with the $s$-adic completion of $R_{U}$, since $t=us^{r}$ for some integer $r\geq 1$ and a unit $u\in R_{U}^{\times}$. By definition, $F_{U}$ is the field of fractions of ${\widehat{R}}_{U}$. We have $k[U]=R_{U}/s={\widehat{R}}_{U}/s$. For $P\in S$, the completion ${\widehat{R}}_{P}$ of the local ring $R_{P}$ of $\mathcal{X}$ at $P$ is a domain and $F_{P}$ is the field of fractions of ${\widehat{R}}_{P}$. Let $p=(U,P)$ be a pair with $P\in S$ in the closure of an irreducible component $U$ of the complement of $S$ in $Y$. Then let $R_{p}$ be the complete discrete valuation ring which is the completion of the localisation of ${\widehat{R}}_{P}$ at the height one prime ideal corresponding to $U$. Then $F_{p}$ is the field of fractions of ${\widehat{R}}_{p}$ and $F$ is the inverse limit of the finite inverse system of fields $\{F_{U},F_{P},F_{p}\}$ by [4, Proposition 6.3].

The following can be seen as a weak generalization of [2, Theorem 3.1] to diagonal forms of higher degree. Here we are not able to conclude that under the given assumptions, $\varphi$ is isotropic over $F$, only over the $F_{U}$’s and $F_{P}$’s:

Theorem 6**.**

*Let $\varphi$ be a nondegenerate diagonal form of degree $d$ over $F$. If $\varphi$ is isotropic over the completion $F_{v}$ of $F$ with respect to each discrete valuation $v$ of $F$ with residue field either a function field in one variable over $k$ or a finite extension of $K$, then:

(i) $\varphi$ is isotropic over $F_{U}$ for each reduced irreducible component $U\subset Y$ of the complement of $S$ in $Y$,

(ii) $\varphi$ is isotropic over $F_{P}$ for each $P\in S$.*

Proof.

Suppose $\varphi=\left<a_{1},\dots,a_{n}\right>$.

(i) Each entry $a_{i}$ of $\varphi$ is supported only along $U$ in ${\rm Spec}\,R_{U}$, thus has the form $us^{j}$ where $u\in R_{U}^{\times}$. We sort the entries $a_{i}=u_{i}s^{j}$ by the power $j$ of $s$ and use them to define new diagonal forms $\rho_{j}$ which have all the $u_{i}$’s belonging to those $a_{i}$ where $s$ occurred in the $j$th power as their diagonal entries. Hence $\varphi$ is isomorphic to the diagonal form

[TABLE]

over $F$, where the $\rho_{i}$ are nondegenerate diagonal forms of degree $d$ over $R_{U}$. Note that if for some $j\in\{0,1,\dots,d-1\}$ there is no $a_{i}$ with $a_{i}=u_{i}s^{j}$, then there is no corresponding form $\rho_{j}$ and a $\rho_{j}$ does not appear as a component in the sum.

By hypothesis, $\varphi$ is isotropic over the field of fractions of the completed local ring of $\mathcal{X}$ at the generic point of $U$. By Theorem 2, this implies that the image of at least one of the forms $\rho_{0}$, $\rho_{1}$ or $\rho_{d-1}$ under the composite homomorphism $R_{U}\to k[U]\to k(U)$ is isotropic over $k(U)$. Since the residue characteristic $p$ does not divide $d!$, the forms $\rho_{0},\rho_{1},\dots,\rho_{d-1}$ define a smooth projective variety over $R_{U}$. In particular, all of them define a smooth variety over $k[U]$. Since $k[U]$ is a Dedekind domain, if such a projective variety has a point over $k(U)$, it has a point over $k[U]$. Since the variety is smooth over $R_{U}$, a $k[U]$-point lifts to an ${\widehat{R}}_{U}$-point (cf. the discussion after [5, Lemma 4.5]). Thus $\varphi$ has a nontrivial zero over $F_{U}$.

(ii) Let $P\in S$. The local ring $R_{P}$ of $\mathcal{X}$ at $P$ is regular. Its maximal ideal is generated by two elements $(x,y)$ with the property that any $a_{i}$ is the product of a unit, a power of $x$ and a power of $y$. Thus over $F$, the fraction field of $R_{P}$, $\varphi$ is isomorphic to

[TABLE]

where each $\varphi_{i}$ is a nondegenerate diagonal form over $R_{P}$. Let $R_{y}$ be the localization of $R_{P}$ at the prime ideal $(y)$. $R_{y}$ is a discrete valuation ring with fraction field $F$. The residue field $E$ of $R_{y}$ is the field of fractions of the discrete valuation ring $R_{P}/(y)$. By hypothesis, the form

[TABLE]

is isotropic over the field of fractions of the completion of $R_{y}$. By Theorem 2, the reduction of one of the forms

[TABLE]

is isotropic over $E$. Since $x$ is a uniformizing parameter for $R_{P}/(y)$, by Theorem 2 this implies that over the residue field of $R_{P}/(y)$, the reduction of one of the forms $\varphi_{1},\varphi_{2},\varphi_{3}\dots,\varphi_{{d}^{2}}$ is isotropic. But then one of these forms is isotropic over ${\widehat{R}}_{P}$, hence over the field $F_{P}$ which is the fraction field of ${\widehat{R}}_{P}$. ∎

Remark 7**.**

(i) In the proof of Theorem 6, one of the forms is isotropic over ${\widehat{R}}_{P}$, and since $R_{p}$ is the complete discrete valuation ring which is the completion of the localisation of ${\widehat{R}}_{P}$ at the height one prime ideal corresponding to $U$ when $p=(U,P)$, this form is also isotropic over $R_{p}$ and therefore over the field of fractions $F_{p}$ of ${\widehat{R}}_{p}$. This implies that if $\varphi$ is isotropic over the completion of $F$ with respect to each discrete valuation of $F$, then $\varphi$ is isotropic over $F_{U}$ for each reduced irreducible component $U\subset Y$ of the complement of $S$ in $Y$, over $F_{P}$ for each $P\in S$ and over $R_{p}$ for each $p=(U,P)$. Since $F$ is the inverse limit of the finite inverse system of fields $\{F_{U},F_{P},F_{p}\}$, $\varphi$ is isotropic over all overfields used in the inverse limit.

(ii) The discrete valuation rings used in the above proof are the local rings at a point of codimension 1 on a suitable regular proper model $\mathcal{X}$ of $X$ determined by the choice of $\varphi$ (analogously as noted in [2, Remark 3.2]).

Given a nondegenerate diagonal form $\varphi$ of degree $d$ and dimension greater than two over $F$, it is not clear whether the isotropy of $\varphi$ over $F_{v}$ for each $v$ (respectively, of $\varphi$ over all $F_{U}$, $F_{P}$ and $F_{p}$) implies that $\varphi$ is isotropic (the fact that ${\rm dim}\,\varphi>2$ is necessary: it is easy to adjust the example in [2, Appendix] to two-dimensional diagonal forms of even degree).

Corollary 8**.**

*Let $r\geq 1$ be an integer and $d\geq 3$. Assume that any diagonal form in strictly more than $dr$ variables over any function field in one variable over $k$ is isotropic. Then:

(i) Any diagonal form of degree $d$ and dimension $>d^{2}r$ over the function field $F=K(X)$ of a curve $X/K$ is isotropic over $F_{v}$, for every discrete valuation $v$ with residue field either a function field in one variable over $k$ or a finite extension of $K$.

(ii) Any diagonal form of degree $d$ and dimension $>d^{2}r$ over the function field $F=K(X)$ of a curve $X/K$ is isotropic over $F_{U}$ for each reduced, irreducible component $U\subset Y$ of the complement of $S$ in $Y$ and is isotropic over $F_{P}$ for each $P\in S$.*

Note that $Y$ and $S$ depend on $\varphi$.

Proof.

(i) Let $L$ be a finite field extension of $K$. This is a complete discretely valued field with residue field a finite extension $\ell$ of $k$. The assumption made on diagonal forms of degree $d$ over functions fields in one variable over $k$, in particular diagonal forms of degree $d$ over the field $\ell(t)$, and Theorem 2 applied to $\ell((t))$ show that any diagonal form of dimension $>r$ over $\ell$ has a zero. A second application of Theorem 2 yields that any diagonal form of degree $d$ of dimension $>dr$ over $L$ is isotropic. Let $\varphi$ be a diagonal form of dimension $n$ over $F$ with $n>d^{2}r$. By the assumption and Theorem 2, $\varphi$ is isotropic over $F_{v}$ for every discrete valuation $v$ with residue field either a function field in one variable over $k$ or a finite extension of $K$.

(ii) follows from Theorem 6. ∎

This shows that trying to extend [2, Corollary 3.4] from quadratic to diagonal forms of higher degree results in a much weaker version.

3.2.

Let $K$ be a $p$-adic field with residue field $k$ such that ${\rm char}(k)\nmid d!$.

Corollary 9**.**

*Any diagonal form of degree $d$ and dimension $>d^{3}+1$ over a function field in one variable $F=K(t)$ is

(i) isotropic over $F_{v}$, for every discrete valuation $v$ with residue field either a function field in one variable over $k$ or a finite extension of $K$;

(ii) isotropic over $F_{U}$ for each reduced, irreducible component $U\subset Y$ of the complement of $S$ in $Y$ and isotropic over $F_{P}$ for each $P\in S$.*

Proof.

Every finite field $k$ is $C_{1}$ and so every diagonal form of degree $d$ and dimension $>d^{2}$ over any function field in one variable over $k$ (which is $C_{2}$) is isotropic. Assertion (i) is a direct consequence of Theorem 2 and (ii) follows from Corollary 8 (ii). ∎

So if $\varphi$ is a diagonal form of degree $d$ in at least $d^{3}+1$ variables over $\mathbb{Q}(t)$ then $\varphi$ is isotropic over $(\mathbb{Q}_{p}(t))_{U}$ for any $p\nmid d!$, for each reduced, irreducible component $U\subset Y$ of the complement of $S$ in $Y$, and isotropic over $(\mathbb{Q}_{p}(t))_{P}$ for each $P\in S$.

Remark 10**.**

Let us compare Corollary 9 with the Ax-Kochen-Ersov Transfer Theorem [1]: given a degree $d$, for almost all primes $p$, a form of degree $d$ over $\mathbb{Q}_{p}$ of dimension greater than or equal to $d^{2}+1$ is isotropic [G, (7.4)]. Moreover, for any form $\varphi$ of degree $d\geq 2$ and dimension greater than $d^{3}$ over $\mathbb{Q}(t)$, for almost all primes $p$ the form $\varphi$ is isotropic over $\mathbb{Q}_{p}(t)$ ([16] for $d=2$, [12] for $d\geq 3$). The model-theoretic proofs of both results do not allow for a more concrete observation on which primes exactly are included here, nor can they be extended to other base fields.

Stronger upper bounds on $u_{diag}(d,\mathbb{F}_{q}(t))$ will yield stronger results on its dimension, since we only used the upper bound in the well known inequality $d\cdot u_{diag}(d,\mathbb{F}_{q})\leq u_{diag}(d,\mathbb{F}_{q}(t))\leq d^{2}$ to prove Corollary 9, for instance we obtain:

Corollary 11**.**

*Assume that $u_{diag}(d,k(t))=dr<d^{2}$ for some $r\in\{1,\dots,d-1\}$. Let $\varphi$ be a diagonal form of degree $d$ and dimension $>d^{2}r+1$ over a function field in one variable $F=K(t)$. Then:

(i) $\varphi$ is isotropic over $F_{v}$, for every discrete valuation $v$ with residue field either a function field in one variable over $k$ or a finite extension of $K$;

(ii) $\varphi$ is isotropic over $F_{U}$ for each reduced, irreducible component $U\subset Y$ of the complement of $S$ in $Y$ and over $F_{P}$ for each $P\in S$.*

It is well known that $u_{diag}(d,K)\leq d^{2}$ for a $p$-adic field $K$ with residue field $k=\mathbb{F}_{q}$ [3]. Indeed, $u_{diag}(d,K)=d\,u_{diag}(d,\mathbb{F}_{q})$ by Theorem 2, assuming that ${\rm char}\,\mathbb{F}_{q}=p\nmid d!$ as before, which shows that clearly $u_{diag}(d,K)$ can be smaller than $d^{2}$. On the other hand, Artin’s conjecture that $\mathbb{Q}_{p}$ is a $C_{2}$-field is false for instance for forms of degree 4.