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 [ Reference O’MearaOM ].

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

Proposition 2. ([ Reference SchurSch , 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 [ Reference SchurSch , 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.

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

Much more strongly, Kedlaya [ Reference KedlayaKe , Theorem 1·1] proved that one can even impose the additional condition that the $\mathrm{disc}_K$ is squarefree. Further, Bhargava, Shankar and Wang [ Reference Bhargava, Shankar and WangBSW ] 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 [ Reference NeukirchNeu , 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.

Proof. For $d=2,4,8$ , this was proved by Kala–Svoboda [ Reference Kala and SvobodaKS , Theorem 1], and for $d=3$ by Yatsyna [ Reference YatsynaYa , 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 [ Reference KalaKa , 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 [ Reference KalaKa ] for $M=m-1$ ) such that their corresponding Gram matrix is diagonal by [ Reference KalaKa , 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 [ Reference ShanksSh ] $L=\mathbb{Q}(\rho)$ (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 [ Reference Kala and TinkováKT , 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 [ Reference Kala and TinkováKT , 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:

Over the rationals $\mathcal{O}_F=\mathbb{Z}$ this is just a weak version of the famous 290-theorem [ Reference Bhargava and HankeBH ]. More generally, Kim–Kim–Oh [ Reference Kim, Kim and OhKKO ] 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 [ Reference Chan and OhCO ] (a similar result was also announced by L. Sun [ Reference SunSu ]).

Note that Theorem 4 also holds with different Galois groups than $C_\ell$ and $S_k$ [ Reference DoleƝálekDo ] (for example, already [ Reference Kala and SvobodaKS ] 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)” [ Reference YatsynaYa ] is satisfied). However, Gras [ Reference GrasGr ] 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.