Number fields without universal quadratic forms of small rank exist in most degrees

Abstract

We prove that in each degree divisible by 2 or 3, there are infinitely many totally real number fields that require universal quadratic forms to have arbitrarily large rank.

1. Introduction

In 1770 Lagrange proved that every positive integer is the sum of four squares, opening up the study of universal quadratic forms. These were then first investigated over the integers $\mathbb{Z}$ , leading to the celebrated 15- and 290-Theorems [ Bh, BH ], and also over number fields, starting with Maaß [ Ma ] and Siegel [ Si ] in the 1940s. To be precise, let $\mathcal{O}_F$ be the ring of integers in a totally real number field F. A totally positive quadratic form Q with $\mathcal{O}_F$ -coefficients is universal over F if it represents all the totally positive elements of $\mathcal{O}_F$ .

Universal forms exist over every F thanks to the asymptotic local-global principle [ HKK ]. Of particular interest is thus the smallest possible rank m ^′(F) of a universal form over F. For example, we have $m'(\mathbb{Q})=4$ by the Four-Square Theorem. Among real quadratic fields, $m'(\mathbb{Q}(\sqrt D))=3$ for $D=2,3,5$ , and these are the only real quadratic fields that admit a ternary universal form that is moreover classical (i.e., has all its cross-terms divisible by 2) [ CKR ]. This provides interesting evidence towards Kitaoka’s Conjecture that there are only finitely many number fields F with $m'(F)=3$ .

Further, the ranks $m'(\mathbb{Q}(\sqrt D))$ can be arbitrarily large [ BK1 ], [ Ka ], ditto for multiquadratic fields of a given degree [ KS ]. Despite a number of other exciting results obtained in the last 25 years [ BK2, CL+, EK, KT, KY, Ki, KKP, KTZ, Ya ], ranks of universal forms over number fields, especially of higher degree, remain mysterious.

The aim of this short paper is to extend the previous special results on unbounded ranks m ^′(F) to number fields of most degrees:

Theorem 1. Let d and m be positive integers such that d is divisible by 2 or 3. Then there are infinitely many totally real number fields F of degree $[F\;:\;\mathbb{Q}]=d$ over which every universal quadratic form has rank at least m.

If $d=2$ , this was proved by the author [ Ka , Theorem 1·1]. The key idea was to use continued fractions to construct quadratic fields that have many indecomposable elements, which are hard to represent by a quadratic form. Constructing such elements in higher degrees is more difficult, nevertheless, the author and Svoboda [ KS , Theorem 1] extended the result to all degrees $d=2^h$ . In the cubic case $d=3$ , this theorem was proved by Yatsyna [ Ya , Theorem 5] using interlacing polynomials and elements of trace one.

Our argument will use Schur’s trace bound [ Sch ] to show a general Theorem 4: in certain cases, suitable elements from a cyclic number field L force a quadratic form that represents them to have many variables, even in an overfield. We will then prove Theorem 1 by choosing L to be a real quadratic, or simplest cubic [ Sh ] number field.

Theorem 1 also holds for quadratic lattices that are not necessarily free; in fact, we will formulate the rest of the paper in lattice-theoretic language. Also note that we do not assume the quadratic forms to be classical, although this is a very common assumption and there are only very few results available without it (e.g., [ De ]).

2. Preliminaries

Let F be a totally real number field of degree $[F\;:\;\mathbb{Q}]=N$ over $\mathbb{Q}$ , i.e., there are N real embeddings $\sigma_1,\dots,\sigma_N\;:\;F\rightarrow \mathbb{R}$ . We denote $\mathcal{O}_F$ the ring of algebraic integers. An element $\alpha\in F$ is totally positive (denoted $\alpha\succ0$ ) if $\sigma_i(\alpha)>0$ for all $1\leq i\leq N$ . Further, $\alpha\succeq\beta$ if $\alpha-\beta\succ 0$ or $\alpha=\beta$ . The set of all totally positive algebraic integers is $\mathcal{O}_F^+$ .

For $\alpha\in F$ we have its trace $\mathrm{Tr}_{F/\mathbb{Q}}(\alpha)=\sum_{1\leq i\leq N}\sigma_i(\alpha)$ and discriminant $\Delta_{F/\mathbb{Q}}(\alpha)$ , which is the square of the determinant of the matrix $(\sigma_i(\alpha^{j-1}))_{1\leq i,j\leq N}$ . The discriminant of F, i.e., the discriminant of an integral basis for $\mathcal{O}_F$ , will be denoted $\mathrm{disc}_F$ . We have $\mathrm{disc}_F\mid\Delta_{F/\mathbb{Q}}(\alpha)$ for each $\alpha\in\mathcal{O}_F$ .

A totally positive quadratic $\mathcal{O}_F$ -lattice of rank r (an $\mathcal{O}_F$ -lattice for short) is a pair $(\Lambda,Q)$ , where $\Lambda$ is a finitely generated $\mathcal{O}_F$ -submodule of $F^r$ such that $\Lambda F=F^r$ , $Q\;:\;F^r\rightarrow F$ is a quadratic form, and $Q(v)\in\mathcal{O}_F^+$ for all $v\in \Lambda, v\neq 0$ . We also have the attached symmetric bilinear form $B(v,w)=(Q(v+w)-Q(v)-Q(w))/2$ . An $\mathcal{O}_F$ -lattice $(\Lambda,Q)$ is universal (over F) if for each $\alpha\in\mathcal{O}_F^+$ there is $v\in\Lambda$ with $Q(v)=\alpha$ .

Let m(F) denote the minimal rank of a universal $\mathcal{O}_F$ -lattice. Note that to each quadratic form Q (as considered in the Introduction) corresponds the $\mathcal{O}_F$ -lattice $(\mathcal{O}_F^r,Q)$ , and so $m(F)\leq m'(F)$ (it is interesting to note that no example of strict inequality is known here).

Take non-zero vectors $v_1,\dots,v_n\in\Lambda$ . The corresponding Gram matrix is the $n\times n$ matrix $A=(B(v_i,v_j))_{1\leq i,j\leq n}$ . Note that we have $B(v_i,v_i)=Q(v_i)=a_i$ and $B(v_i,v_j)=b_{ij}/2$ for all $i\neq j$ and suitable $a_i, b_{ij}\in\mathcal{O}_{F}$ . As the lattice $\Lambda$ is totally positive, $a_i\succ 0$ and we have a version of the Cauchy–Schwarz inequality $4a_ia_j \succeq b_{ij}^2$ for all $i\neq j$ , equivalently, $Q(v_i)Q(v_j)\succeq B(v_i,v_j)^2$ (this quickly follows from the positive-definiteness of the quadratic form $\sigma_h(Q)$ on $\mathbb{R}^r$ for $h=1,\dots,N$ ).

Also note that the rank of A (as a matrix over the field F) is at most the rank r of the lattice $\Lambda$ (for the rank of A equals the rank of the $\mathcal{O}_F$ -sublattice of $\Lambda$ spanned by $v_1,\dots,v_n$ ). For more background on quadratic lattices, see [ OM ].

Further, we will crucially use the following lower bound due to Schur.

Proposition 2. ([ Sch , Section 2·II.]). Let F be a totally real number field of degree $[F\;:\;\mathbb{Q}]=N$ .

If $\beta\in\mathcal{O}_F$ , then

\begin{equation*}\mathrm{Tr}_{F/\mathbb{Q}} \beta^2\geq c_{N}\Delta_{F/\mathbb{Q}}(\beta)^{2/(N^2-N)}\text{ with }c_{N}=\frac{N^2-N}{\left(2^2\cdot3^3\cdot 4^4\cdots(N-1)^{N-1}\cdot N^N\right)^{2/(N^2-N)}}.\end{equation*}

Proof. Schur’s bound [ Sch , Section 2·II.] states that if $x_1,\dots,x_N$ are real numbers such that $x_1^2+\dots+x_N^2\leq 1$ , then the discriminant $\prod_{1\leq i<j\leq N}(x_i-x_j)^2\leq c_N^{-(N^2-N)/2}$ . Setting $x_i=\sigma_i(\beta)/(\mathrm{Tr}_{F/\mathbb{Q}} \beta^2)^{1/2}$ gives the inequality we need.

For an integer $k\geq 2$ , we will denote $S_k$ and $A_k$ the symmetric and alternating groups on the set $\{1,2,\dots,k\}$ . The cyclic group of order k is denoted $C_k$ (considered multiplicatively).

We will work with extensions of a given number field by an $S_k$ -number field, whose existence is given by the following proposition.

Proposition 3. Let $D, X, k\geq 2$ . There are infinitely many totally real number fields K of degree $[K\;:\;\mathbb{Q}]=k$ whose discriminant $\mathrm{disc}_K>X$ is coprime with D and whose Galois closure $\widetilde K$ has Galois group $\mathrm{Gal}(\widetilde K/\mathbb{Q})\simeq S_k$ .

Proof. This is well known. The most straightforward proof is probably using Hilbert’s irreducibility theorem (see, e.g., [ Kal , Theorem 4·2·3]).

Much more strongly, Kedlaya [ Ke , Theorem 1·1] proved that one can even impose the additional condition that the $\mathrm{disc}_K$ is squarefree. Further, Bhargava, Shankar and Wang [ BSW ] proved that the polynomials $f(x)=x^k+a_1x^{k-1}+\dots+a_{k-1}x+a_k$ , whose rupture fields K have the required properties (including squarefree $\mathrm{disc}_K$ ), have positive density when ordered by $\max\{|a_i|^{1/i}\mid 1\leq i\leq k\}$ .

3. The proof

To prove Theorem 1, we will use the following general theorem that we will then apply to suitable fields L (of degrees $\ell=2,3$ ).

Theorem 4. Let $k,\ell,m,n$ be positive integers such that $k=3$ or $k\geq 5$ .

Assume that there is a totally real Galois number field L of degree $[L\;:\;\mathbb{Q}]=\ell$ whose Galois group is $\mathrm{Gal}(L/\mathbb{Q})\simeq C_\ell$ and that contains elements $a_1,\dots,a_n\in\mathcal{O}_L^+$ such that if an $\mathcal{O}_L$ -lattice represents $a_1,\dots,a_n$ , then it has rank $\geq m$ .

There is $B>0$ (depending on $k,\ell,L,a_i$ ) with the following property: for every totally real number field K of degree $[K\;:\;\mathbb{Q}]=k$ whose discriminant $\mathrm{disc}_K>B$ is coprime with $\mathrm{disc}_L$ and whose Galois closure $\widetilde K$ has Galois group $\mathrm{Gal}(\widetilde K/\mathbb{Q})\simeq S_k$ , we have $[KL\;:\;\mathbb{Q}]=k\ell$ and

\begin{equation*}m(KL)\geq m.\end{equation*}

Proof. Let $L,K,\widetilde K$ be as in the statement (with B to be specified later).

As $\mathrm{disc}_K$ and $\mathrm{disc}_L$ are coprime, we have $K\cap L=\mathbb{Q}$ (for $\mathrm{disc}_{K\cap L}$ is a common divisor of $\mathrm{disc}_K$ and $\mathrm{disc}_L$ by the formula for the discriminant of a tower of number fields [ Neu , Corollary III·2·10]). Thus $[KL\;:\;\mathbb{Q}]=k\ell$ . Let us use Galois theory to describe all subfields $\mathbb{Q}\subset M\subset KL$ (without giving references for all the theorems that we use – see any good textbook on Abstract Algebra).

First, consider $H=\widetilde K\cap L$ . Since H is a subfield of L, we have $\mathrm{Gal}(H/\mathbb{Q})\simeq C_t$ for some $t\mid\ell$ , and $\mathrm{Gal}(\widetilde K/H)$ is a normal subgroup of $\mathrm{Gal}(\widetilde K/\mathbb{Q})\simeq S_k$ . The only such subgroups are $S_k$ and $A_k$ (as $k\neq 4$ ), and so correspondingly, $H=\mathbb{Q}$ or $H=\mathbb{Q}(\sqrt{\mathrm{disc}_K})$ . But the latter case is impossible, as $\mathrm{disc}_K$ and $\mathrm{disc}_L$ are coprime, and so $\sqrt{\mathrm{disc}_K}\not\in L$ .

Thus $\widetilde K\cap L=\mathbb{Q}$ , and so $\widetilde KL$ is Galois with

\begin{equation*}\mathrm{Gal}(\widetilde KL/\mathbb{Q})\simeq S_k\times C_\ell.\end{equation*}

Further,

\begin{equation*}\mathrm{Gal}(\widetilde KL/KL)\simeq S_{k-1}\times \{1\}\end{equation*}

(on the right-hand side, we will view $S_{k-1}$ as the subgroup of $S_k$ that consists of permutations fixing the element k).

Thus by Galois correspondence, the fields $\mathbb{Q}\subset M\subset KL$ correspond to subgroups

\begin{equation*}S_k\times C_\ell\supset G\supset S_{k-1}\times \{1\}.\end{equation*}

We claim that for each such subgroup, we have $G\subset S_{k-1}\times C_\ell$ or $G\supset S_k\times \{1\}$ .

For if $G\not\subset S_{k-1}\times C_\ell$ , then there is an element $(\sigma,u)\in G$ with $\sigma\not\in S_{k-1}$ . Considering the decomposition of $\sigma\in S_k$ into disjoint cycles, we can write $\sigma=\tau (i_1\cdots i_jk)$ where $\tau\in S_{k-1}, j\geq 1,$ and $(i_1\cdots i_jk)$ denotes the cycle of length $j+1$ that permutes $i_1,\cdots,i_j,k$ in the given order. Multiplying $(\sigma,u)\in G$ from the left by $((i_1\cdots i_j)^{-1}\tau^{-1},1)\in S_{k-1}\times \{1\}\subset G$ , we obtain $((i_jk),u)\in G$ . If we finally conjugate this element by $((1i_j),1)\in S_{k-1}\times \{1\}\subset G$ , we get $((1k),u)\in G$ .

If the order o of u in $C_\ell$ is odd, then $((1k),1)=((1k),u)^o\in G$ . If o is even, then also

\begin{align*} G\ni ((12),1)[((1k),u)((12),1)((1k),u)((1k),u)^{o-2}]((12),1)&\\ =((12),1)((2k),1)((12),1)&=((1k),1).\end{align*}

Thus $G\supset S_k\times \{1\}$ (as $((1k),1)\in G$ and $G\supset S_{k-1}\times \{1\}$ ), as we wanted to show.

Correspondingly, each intermediate field $\mathbb{Q}\subset M\subset KL$ satisfies

\begin{equation*}M\supset K\text{ or } M\subset L.\end{equation*}

Let us finally specify that

\begin{equation*}\mathrm{disc}_K>B \, \text{for} B=\text{max}_M\left(\left(\frac{keT}{\ell c_{ke}}\right)^{(k^2e-k)/2} \right),\end{equation*}

where the maximum is taken over all fields M such that $K\subset M\subset KL$ , $e=[M\;:\;K]$ , $T=4\max\{\mathrm{Tr}_{L/\mathbb{Q}}(a_ia_j)\mid 1\leq i<j\leq n\}$ and $c_{ke}$ are the constants from Proposition 2.

Let $(\Lambda,Q)$ be a universal $\mathcal{O}_{KL}$ -lattice. As $L\subset KL$ , the lattice $\Lambda$ represents all the elements $a_1,\dots,a_n\in\mathcal{O}_L^+$ ; fix vectors $v_i\in \Lambda$ such that $Q(v_i)=a_i$ . We will show that $\Lambda$ has rank $\geq m$ by showing that the Gram matrix $(B(v_i,v_j))_{1\leq i,j\leq n}$ corresponding to the vectors $v_i$ has rank $\geq m$ .

We have $B(v_i,v_i)=Q(v_i)=a_i$ and let $B(v_i,v_j)=b_{ij}/2$ for all $i\neq j$ and suitable $b_{ij}\in\mathcal{O}_{KL}$ . We will now show that $b_{ij}\in L$ for all $i\neq j$ .

Assume that this is not the case for some $i\neq j$ and let $M=\mathbb{Q}(b_{ij})$ . By the description of possible fields $\mathbb{Q}\subset M\subset KL$ obtained above, we have that $M\supset K$ . Let $[M\;:\;K]=e$ ; then $\mathrm{disc}_M\geq\mathrm{disc}_K^e$ (again by the formula for the discriminant of a tower).

As the $\mathcal{O}_{KL}$ -lattice $\Lambda$ is totally positive, we have the Cauchy–Schwartz inequality $4a_ia_j \succeq b_{ij}^2$ (see Section 2).

Taking traces and applying Proposition 2 for the field M of degree $[M\;:\;\mathbb{Q}]=ke$ , we get

\begin{equation*}kT\geq \mathrm{Tr}_{KL/\mathbb{Q}}(4a_ia_j)\geq \mathrm{Tr}_{KL/\mathbb{Q}}(b_{ij}^2) = \frac \ell e \mathrm{Tr}_{M/\mathbb{Q}}(b_{ij}^2) \geq \frac \ell e c_{ke}\Delta_{M/\mathbb{Q}}(b_{ij})^{2/((ke)^2-ke)}.\end{equation*}

As $b_{ij}$ does not lie in a proper subfield of M, we have $\Delta_{M/\mathbb{Q}}(b_{ij})\neq 0$ , and so $\Delta_{M/\mathbb{Q}}(b_{ij})\geq \mathrm{disc}_M\geq \mathrm{disc}_K^e>B^e$ . Thus

\begin{equation*}kT>\frac \ell e c_{ke}B^{2/(k^2e-k)},\end{equation*}

contradicting the choice of B.

We proved that $b_{ij}\in L$ for all $i\neq j$ .

Therefore all the entries of the Gram matrix $(B(v_i,v_j))_{1\leq i,j\leq n}$ lie in L, and so this matrix corresponds to an $\mathcal{O}_L$ -lattice $\Lambda'$ that represents all the elements $a_1,\dots,a_n$ over L. By the assumption of the theorem, every such lattice has rank $\geq m$ .

Accordingly, the Gram matrix $(B(v_i,v_j))_{1\leq i,j\leq n}$ has rank $\geq m$ , which finally implies that the rank of $\Lambda$ is also $\geq m$ .

We can now finally use the preceding result to prove our main theorem.

Theorem (Theorem 1, lattice-theoretic formulation). Let d and m be positive integers such that d is divisible by 2 or 3. Then there are infinitely many totally real number fields F of degree $[F\;:\;\mathbb{Q}]=d$ over which every totally positive universal quadratic $\mathcal{O}_F$ -lattice has rank at least m, i.e., $m(F)\geq m$ .

Proof. For $d=2,4,8$ , this was proved by Kala–Svoboda [ KS , Theorem 1], and for $d=3$ by Yatsyna [ Ya , Theorem 5].

If $d=6$ or $d\geq 10$ is even, choose $\ell=2$ and $k=d/2$ ; we have $k=3$ or $k\geq 5$ .

By [ Ka , Section 4], there are (infinitely many) real quadratic fields $L=\mathbb{Q}(\sqrt D)$ that contain $n=m$ elements ( $\alpha_1,\alpha_3,\dots,\alpha_{2M+1}$ in the notation of [ Ka ] for $M=m-1$ ) such that their corresponding Gram matrix is diagonal by [ Ka , Proposition 4·1], and so every $\mathcal{O}_K$ -lattice that represents these elements has rank $\geq m$ .

By Proposition 3, there are infinitely many fields K of degree $k=d/2$ with the properties required by Theorem 4, and so for each of them we have $[KL\;:\;\mathbb{Q}]=d$ and $m(KL)\geq m$ , as needed.

If $d=9$ or $d\geq 15$ is divisible by 3, choose $\ell=3$ and $k=d/3$ ; again $k=3$ or $k\geq 5$ .

Let $n=\max(9m^2,240)$ and let us consider Shanks’ simplest cubic fields [ Sh ] $L=\mathbb{Q}(\rho)$ (where $\rho$ is a root of the polynomial $x^3-ax^2-(a+3)x-1$ for some $a\in\mathbb{Z}_{\geq -1}$ ). Each simplest cubic field L is Galois with $\mathrm{Gal}(L/\mathbb{Q})\simeq C_3$ .

Kala–Tinková [ KT , Subsection 7·2] proved that there are (infinitely many) such fields L that contain n elements $a_1, a_2, \dots, a_n\in\mathcal O_L^+$ and an element $\delta\succ 0$ in the codifferent $\{\alpha\in L\mid \mathrm{Tr}_{L/\mathbb{Q}}(\alpha\beta)\in\mathbb{Z}\text{ for all }\beta\in\mathcal{O}_L\}$ with $\mathrm{Tr}_{L/\mathbb{Q}}(\delta a_i)=1$ for all i. By [ KT , Subsection 7·2 and Proof of Proposition 7·4], if an $\mathcal{O}_L$ -lattice represents all the elements $a_1,\dots,a_n$ , then it has rank $\geq \sqrt {n}/3=m$ .

It again just remains to use Proposition 3 and Theorem 4.

This covers all the positive integers d that are divisible by 2 or 3, finishing the proof.

In Theorem 4, one can directly claim that $m(KL)\geq m(L)$ instead of assuming the existence of suitable elements $a_1,\dots,a_n$ , for such elements with $m=m(L)$ always exist:

Theorem 5 ([ CO , Corollary 5·8]). Let F be a totally real number field. There is a finite set $S\subset \mathcal{O}_F^+$ such that if an $\mathcal{O}_F$ -lattice represents all the elements of S, then it is universal.

Over the rationals $\mathcal{O}_F=\mathbb{Z}$ this is just a weak version of the famous 290-theorem [ BH ]. More generally, Kim–Kim–Oh [ KKO ] proved a similar result for representations of quadratic forms by quadratic forms (over $\mathbb{Z}$ ), and remarked that their theorem should also hold over number fields. This was indeed recently established by Chan–Oh [ CO ] (a similar result was also announced by L. Sun [ Su ]).

Note that Theorem 4 also holds with different Galois groups than $C_\ell$ and $S_k$ [ Do ] (for example, already [ KS ] dealt with multiquadratic fields F, i.e., $\mathrm{Gal}(F/\mathbb{Q})\simeq C_2^h$ ). However, the present formulation is sufficient for the proof of our main Theorem 1, and a more general statement probably would not bring more clarity.

Finally, let us comment that the most direct way of extending Theorem 1 to all degrees $d>1$ does not work: We would need to know the existence of (infinitely many) totally real cyclic fields L of any prime degree $\ell\geq 5$ that contain suitable elements $a_1,\dots,a_n$ . To ensure the existence of these elements, it is very helpful for L to have a power integral basis and units of all signatures (e.g., for then Yatsyna’s “Condition (A)” [ Ya ] is satisfied). However, Gras [ Gr ] showed that the only totally real cyclic fields L of prime degree $\ell\geq 5$ with a power integral basis are the maximal real subfields of a cyclotomic field, i.e., there is only one such field in a given degree. Thus one would need to work with fields without power integral basis, or with different Galois structure.