# LINEAR RELATIONS AMONG GALOIS CONJUGATES OVER $\mathbb{F}_{q}(t)$

Will Hardt and John Yin

##  1\. Abstract

We classify the coefficients $(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ that
can appear in a linear relation $\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ among Galois
conjugates $\gamma_{i}\in\overline{\mathbb{F}_{q}(t)}$. We call such an
$n$-tuple a Smyth tuple. Our main theorem gives an affirmative answer to a
function field analogue of a 1986 conjecture of Smyth [4] over $\mathbb{Q}$.
Smyth showed that certain local conditions on the $a_{i}$ are necessary and
conjectured that they are sufficient. Our main result is that the analogous
conditions are necessary and sufficient over $\mathbb{F}_{q}(t)$, which we
show using a combinatorial characterization of Smyth tuples from [4]. We also
formulate a generalization of Smyth’s Conjecture in an arbitrary number field
that is not a straightforward generalization of the conjecture over
$\mathbb{Q}$ due to a subtlety occurring at the archimedean places.

##  2\. Introduction

The question of how prevalent linear relations among Galois conjugates are has
been studied from multiple angles. In [2] it is shown that for a Hilbertian
field $K$ and all but finitely many nonnegative integers $n$, there exists an
algebraic number $\alpha\in\overline{K}$ of degree $2^{n}n!$ whose conjugates
span a vector space of dimension $n$. In these cases, the dimension of
relations between conjugates is $2^{n}n!-n$, and so in this sense, linear
relations among conjugates are plentiful.

In an older paper, Smyth showed [4, Cor. 2] that for any
$a_{1},\dots,a_{n}\in{\mathbb{Z}}$ with $gcd(a_{1},\dots,a_{n})=1$, if there
exist Galois conjugates $\gamma_{1},\dots,\gamma_{n}\in\overline{\mathbb{Q}}$
such that $\sum_{i=1}^{n}a_{i}\gamma_{i}=0$, then the $a_{i}$ necessarily
satisfy the following two properties.

  1. (1)

$|a_{i}|\leq\sum_{j\neq i}|a_{j}|$ for all $i$

  2. (2)

Every prime $p$ divides at most $n-2$ of the $a_{i}$.

Smyth conjectured that the converse is true as well over $\mathbb{Q}$.

######  Conjecture 2.1 (Smyth’s Conjecture).

[4] If $(a_{1},\dots,a_{n})\in\mathbb{Q}^{n}$ satisfy (1) and (2), then there
exist Galois conjugates $\gamma_{1},\dots,\gamma_{n}$ so that
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$.

A natural way to generalize (1) and (2) to an arbitrary field $K$ is as
follows.

  * (1’)

For any archimedean absolute value $|\cdot|$ of $K$, we have
$|a_{i}|\leq\sum_{j\neq i}|a_{j}|$ for all $i$.

  * (2’)

For any nonarchimedean absolute value $|\cdot|$ of $K$, we have
$|a_{i}|\leq\max_{j\neq i}|a_{j}|$ for all $i$.

We call (1’) and (2’) the absolute value criteria over $K$.

Note that for any $a\in K^{*}$, a tuple $(a_{1},...,a_{n})\in K^{n}$ satisfies
the absolute value criteria if and only if $(aa_{1},...,aa_{n})$ does, and
similarly, $(a_{1},...,a_{n})$ is a Smyth tuple if and only if
$(aa_{1},...,aa_{n})$ is a Smyth tuple. Hence, when $K$ is the field of
fractions of a principal ideal domain, it is enough to look at coprime tuples
$(a_{1},...,a_{n})$.

Conjecture 2.1 remains open. Our main result is Theorem 4.2, which answers in
the affirmative the natural analogue of Smyth’s question over
${\mathbb{F}}_{q}(t)$. We will also formulate in Conjecture 10.1 a
generalization of Conjecture 2.1 for arbitrary number fields, which involves a
subtlety not present in the cases of $\mathbb{Q}$ or $\mathbb{F}_{q}(t)$. A
MathOverflow post of David Speyer [1] proves a special case of this
conjecture.

In Section 3 we record some terminology that we will use throughout the paper.
In Section 4 we set up Theorem 4.2, which we go on to prove in Section 5.
Section 6 examines how close the Galois relations constructed in our proof of
Theorem 4.2 are to “as small as possible.” In Section 7 we show that the
conjugates in our constructed linear relations for Smyth triples can be chosen
to have the full symmetric group as their Galois group and in Section 8 we
give heuristic reasoning for why we should expect the Galois group of linear
Galois relations to be large in general. In Section 9, we record some results
from Section 6 that also apply over $\mathbb{Q}$. Finally, in Section 10, we
formulate a number field generalization of Conjecture 2.1, and then reduce it
slightly to Conjecture 10.7.

##  3\. Preliminaries

We lay out some terminology and a background result.

We say that $\gamma_{1},...,\gamma_{n}$ are Galois conjugates over K if they
are all roots of the same irreducible polynomial over $K$. (When the base
field $K$ is clear from the context, we will often omit it from our
terminology.) Throughout the paper, we assume Galois conjugates are nonzero.

A linear Galois relation will mean a linear relation
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ among Galois conjugates
$\gamma_{1},...,\gamma_{n}\in\overline{K}$ with coefficients $a_{i}\in K$.

We will call a tuple $(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ coprime if
$a_{1},...,a_{n}$ generate the unit ideal (and similarly for
$(a_{1},...,a_{n})\in\mathbb{Z}^{n}$). The (logarithmic) height of a coprime
tuple $(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ is $\max_{i}\deg(a_{i})$.
The (logarithmic) height of a coprime tuple
$(a_{1},...,a_{n})\in\mathbb{Z}^{n}$ is $\max_{i}\log|a_{i}|$. A coprime tuple
is said to be a Smyth tuple if its coordinates appear as the coefficients of a
linear Galois relation.

A balanced multiset of tuples – which we will sometimes shorten to “balanced
multiset” – with respect to a tuple $(a_{1},...,a_{n})$ is a nonempty
collection of nonzero solutions $\\{(x_{i_{1}},...,x_{i_{n}})\in
K^{n}\\}_{i=1}^{N}$ to the equation $\sum_{i=1}^{n}a_{i}x_{i}=0$ such that the
multiset $\\{x_{i_{j}}\\}_{i=1}^{N}$ is independent of $j$; in other words,
for all $x\in K$, $x$ appears in each of the $N$ coordinate positions the same
number of times.

A balanced multiset $\\{(x_{i_{1}},...,x_{i_{n}})\in K^{n}\\}_{i=1}^{N}$ is a
1-factor if the multiset $\\{x_{i_{1}}\\}_{i=1}^{N}$ is in fact a set. (To see
the justification for this terminology, consider the hypergraph whose vertices
are elements of $K$ and whose edges are ordered tuples $(x_{1},...,x_{n})$ of
solutions to $\sum_{i=1}^{n}a_{i}x_{i}=0$. Then a 1-factor in this hypergraph
is precisely our definition of 1-factor.)

The degree of a linear Galois relation $\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ will
refer to the degree $[K(\gamma_{1}):K]$ of the conjugates in the linear
relation. Similarly the Galois group of a linear Galois relation
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ will mean $\operatorname{Gal}(E/K)$, where
$E$ is the Galois closure of $K(\gamma_{1})$.

Following standard terminology, we will call a Galois extension $L/K$ a
$G$-extension if $\operatorname{Gal}(L/K)\cong G$.

Our final preliminary is the fact that over any Hilbertian field, there exist
$S_{d}$-extensions for all $d\geq 1$. This is a special case of [2, Prop. 8].
We will apply this fact to the Hilbertian field $\mathbb{F}_{q}(t)$.

##  4\. Smyth’s Conjecture over ${\mathbb{F}}_{q}(t)$

Although our primary focus in this section (and this paper) is the function
field $\mathbb{F}_{q}(t)$, we will often state definitions and theorems over
an arbitrarily field $K$ when it is not substantially more complicated to do
so.

Our approach in this paper benefits from having several equivalent notions of
Smyth tuples. Most importantly, Smyth proved a combinatorial characterization
of Smyth tuples involving balanced multisets. We record this, along with one
other characterization of Smyth tuples, in Proposition 4.1.

######  Proposition 4.1.

Let $K$ be any field such that there exist $S_{d}$-extensions of $K$ for all
$d\geq 1$. The following are equivalent for $(a_{1},...,a_{n})\in K^{n}$.

  1. (1)

$(a_{1},...,a_{n})$ is a Smyth tuple

  2. (2)

There exists a balanced multiset of tuples with respect to $(a_{1},...,a_{n})$

  3. (3)

There exist permutation matrices $X_{1},...,X_{n}$ such that
$\det(\sum_{i=1}^{n}a_{i}X_{i})=0$

###### Proof.

Smyth proved (1) $\iff$ (2) for $K=\mathbb{Q}$ [4, Thm. 2], and the proof
works for any $K$ satisfying the hypotheses of this theorem.

(2) $\implies$ (3): Let $T=\\{(x_{i1},x_{i2},...,x_{in})\\}_{i=1}^{N}$ be a
balanced multiset of tuples of size $N$. For $j:1\leq j\leq n$, let
$v_{j}=(x_{ij})_{i=1}^{N}$ be the vector in $K^{N}$ obtained by taking the
$j^{th}$ entry from each tuple in $T$. By definition of balanced multiset,
there exist (not necessarily unique) $N\times N$ permutation matrices
$X_{1},...,X_{n}$ such that $X_{i}v_{n}=v_{i}$ for all $i$. Thus
$(\sum_{i=1}^{n}a_{i}X_{i})v_{n}=\sum_{i=1}^{n}a_{i}v_{i}=0$, so
$\sum_{i=1}^{n}a_{i}X_{i}$ has nontrivial kernel.

(3) $\implies$ (2): Reverse the previous argument as follows. Let $v_{n}$ be
any nonzero vector in the kernel of $\sum_{i=1}^{n}a_{i}X_{i}$ and let
$v_{i}:=X_{i}v_{n}$ for $1\leq i\leq n$. Then the coordinates of the vectors
$v_{1},...,v_{n}$ give a balanced multiset of tuples as above. ∎

Our main theorem is the following.

######  Theorem 4.2.

Let $n\geq 3$ be an integer. A coprime tuple
$(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ is a Smyth tuple if and only if
$a_{1},\dots,a_{n}$ satisfy the absolute value criteria over
${\mathbb{F}}_{q}(t)$.

The forward direction of Theorem 4.2 was established by Smyth over
$\mathbb{Q}$ in a proof that is valid over a general field $K$. We record this
result in Lemma 4.3.

######  Lemma 4.3.

[4, Cor. 2] Let $K$ be any field. If $(a_{1},...,a_{n})\in K^{n}$ is a Smyth
tuple, then $a_{1},...,a_{n}$ satisfy the absolute value criteria over $K$.

##  5\. Proof of Theorem 4.2

We now turn toward the proof of our main result, Theorem 4.2, which will
follow essentially as a corollary from Proposition 5.1.

Let $V_{N}\subset{\mathbb{F}}_{q}[t]$ be the set of polynomials of degree
$<N$. We will show that, surprisingly, for any coprime
$(a_{1},..,a_{n})\in\mathbb{F}_{q}[t]^{n}$ satisfying the absolute value
criteria and any $N\geq d=\max\deg_{i}a_{i}$, the set of all solutions
$(x_{1},...,x_{n})\in V_{N}^{n}$ to the equation $\sum_{i=1}^{n}a_{i}x_{i}=0$
is a balanced set.

######  Proposition 5.1.

Let $n\geq 3$ be an integer. Let
$(a_{1},\dots,a_{n})\in{\mathbb{F}}_{q}[t]^{n}$ of height $d$ be a coprime
tuple satisfying the absolute value criteria. Let $N\geq d$ be an integer and
let $j$ be an integer so that $1\leq j\leq n$. Fix $x_{j}\in V_{N}$. Then the
number of tuples $(x_{1},\dots,x_{j-1},x_{j+1},\dots x_{n})\in V_{N}^{n-1}$
satisfying $\sum_{i=1}^{n}a_{i}x_{i}=0$ is $q^{N(n-2)-d}$. In particular, this
count does not depend on $j$.

###### Proof.

Without loss of generality, we let $j=1$. By the absolute value criteria, the
maximum degree of $a_{1},\dots,a_{n}$ is achieved at least twice. Hence, some
$a_{i}$ with $i\neq 1$ has degree $d$; without loss of generality, assume that
$a_{n}$ does. Let $c=a_{1}x_{1}$. Define

$S=\\{(x_{2},\dots,x_{n})\in V_{N}^{n-1}:c+\sum_{i=2}^{n}a_{i}x_{i}=0\\}$

Our goal is to compute $\\#S$. To do so, we will project onto
${\mathbb{F}}_{q}[t]/a_{n}$, so we define

$\overline{S}=\\{(\overline{x_{2}},\dots,\overline{x_{n-1}})\in({\mathbb{F}}_{q}[t]/a_{n})^{n-2}:\overline{c}+\sum_{i=2}^{n-1}\overline{a_{i}x_{i}}=0\\}$

Reducing modulo $a_{n}$ in each coordinate and throwing out the last
coordinate gives a surjective $q^{(N-d)(n-2)}$-to-1 map $S\to\overline{S}$;
the pre-image of any
$(\overline{x}_{2},...,\overline{x}_{n-1})\in\overline{S}$ is

$\left\\{\left(x_{2}+h_{2}a_{n},...,x_{n-1}+h_{n-1}a_{n},-\left(\frac{c+\sum_{i=2}^{n-1}a_{i}x_{i}}{a_{n}}+\sum_{i=2}^{n-1}a_{i}h_{i}\right)\right):h_{i}\in
V_{N-d}\right\\},$

where $x_{i}$ is the unique polynomial of degree $<d$ equal to
$\overline{x}_{i}$ mod $a_{n}$. Thus, we have
$\\#S=q^{(N-d)(n-2)}\\#\overline{S}$.

So we want to count the number of solutions
$(\overline{x}_{2},...\overline{x}_{n-1})\in({\mathbb{F}}_{q}[t]/(a_{n}))^{n-2}$
to $\overline{c}+\sum_{i=2}^{n-1}\overline{a_{i}x_{i}}=0$. Let $a_{n}=\prod
p_{j}^{e_{j}}$ be the prime factorization of $a_{n}$. Let
$R_{j}:={\mathbb{F}}_{q}[t]/(p_{j}^{e_{j}})$; by the Chinese Remainder
Theorem, it will suffice to count the number of solutions in $R_{j}$ for each
$j$.

Specifically, let
$\overline{S}_{j}=\\{(\overline{x_{2}},...,\overline{x}_{n-1})\in
R_{j}^{n-2}:\bar{c}+\sum_{i=2}^{n-1}\overline{a_{i}}\overline{x_{i}}=0\\}$;
then the Chinese Remainder Theorem implies that

$\\#\overline{S}=\prod_{j}\\#\overline{S}_{j}.$

We now compute $\\#\overline{S}_{j}$. Recall that by the absolute value
criteria, for all $j$, there are at least two $a_{i}$ that are not divisible
by $p_{j}$. Of course $a_{n}$ is divisible by all $p_{j}$; hence, for all $j$,
there is at least one $a_{i}$, with $1<i<n$, such that $p_{j}\nmid a_{i}$, in
which case $\overline{a_{i}}$ is a unit in $R_{j}$. Thus, we can write
$\overline{x_{i}}=\frac{\overline{c}+\sum_{k\neq i,k\neq
1}\overline{a_{k}x_{k}}}{\overline{a_{i}}}$, and so any collection of choices
of $\overline{x_{k}}$ for $k\in\\{2,...,n-1\\}\setminus\\{i\\}$, will give a
unique choice of $\overline{x_{i}}$. There are
$\\#R_{j}=q^{\deg(p_{j}^{e_{j}})}$ choices for each $\overline{x}_{k}$, so
$\\#\overline{S}_{j}=q^{\deg(p_{j}^{e_{j}})(n-3)}$. Thus, since
$\sum_{j}\deg(p_{j}^{e_{j}})=\deg(a_{n})=d$, we have

$\\#S=q^{(N-d)(n-2)}\prod_{j}q^{\deg(p_{j}^{e_{j}})(n-3)}=q^{(N-d)(n-2)}q^{d(n-3)}=q^{N(n-2)-d},\text{
as desired.}$

∎

Now Theorem 4.2 follows easily.

###### Proof of Theorem 4.2.

($\Rightarrow$): This was Lemma 4.3.

($\Leftarrow$): Without loss of generality, we assume $(a_{1},\dots,a_{n})$ is
a coprime tuple in ${\mathbb{F}}_{q}[t]^{n}$. Let $T_{N}$ be the set of all
tuples $(x_{1},\dots,x_{n})\in V_{N}^{n}$ satisfying
$\sum_{i=1}^{n}a_{i}x_{i}=0$ and enumerate
$T_{N}=\\{(x_{i1},\dots,x_{in})\\}_{i=1}^{t}$ where $t=|T_{N}|$. In
Proposition 5.1 we showed that for every $x\in V_{N}$ and all $i:1\leq i\leq
n$, the number of tuples $(x_{1},\dots,x_{n})$ in $T_{N}$ with $x_{i}=x$ is
$q^{N(n-2)-d}$. This means that for each $j:1\leq j\leq n$, the multiset
$\\{(x_{ij})\\}_{i=1}^{t}$ is precisely $q^{N(n-2)-d}$ copies of $V_{N}$.
Thus, $T_{N}$ is a balanced (multi)set of tuples. So by Proposition 4.1,
$(a_{1},\dots,a_{n})$ is a Smyth tuple. ∎

######  Remark 5.2.

By setting $N=d=\max_{i}\deg(a_{i})$ and $n=3$ in Proposition 5.1, we see that
if $(a_{1},a_{2},a_{3})$ is a Smyth triple, then $T_{d}$ is a 1-factor.

##  6\. Degree of Galois Conjugates in Linear Galois Relations

In the previous section, we produced balanced multisets for all Smyth tuples,
i.e. all tuples for which it is possible to find a balanced multiset. We
relied on Proposition 4.1 to turn the balanced multisets into linear Galois
relations. In this section, we consider the question of how close these linear
Galois relations are to “as small as possible.” As a first attempt to
formulate this question, we might ask for a uniform lower bound $\ell_{n}(D)$
on the degree of all linear Galois relations whose coefficients are an
$n$-tuple of height $D$. However, this is quickly seen to be an uninteresting
question, as there exist Smyth tuples of arbitrarily large height that are
coefficients of a bounded degree linear Galois relation. (For instance, if
$\sum_{i=1}^{n}\epsilon_{i}a_{i}=0$ where $\epsilon_{i}\in\\{1,-1\\}$, then
there is a balanced multiset with respect to $(a_{1},...,a_{n})$ of size $2$.)
So instead, we ask for a lower bound $L_{n}(D)$ on the “worst” Smyth $n$-tuple
of height $D$.

######  Definition 6.1.

Let $L^{K}_{n}(D)$ be the minimal $L$ so that there is some
$(a_{1},...,a_{n})\in K^{n}$ of height $D$ such that every linear Galois
relation with coefficients $a_{1},...,a_{n}$ has degree at least $L$

In this section the underlying field $K$ will always be $\mathbb{F}_{q}(t)$,
and so we will omit it from the notation. In Section 9 we will record
analogous results about $L^{\mathbb{Q}}_{3}(D)$.

In Corollary 6.4, we prove that there exist Smyth triples
$(a_{1},a_{2},a_{3})$ of height $D$ for which one cannot find a balanced
multiset of size less than $|V_{D}\setminus\\{0\\}|=q^{D}-1$. In other words,
besides removing zero from the multiset, our construction in Section 4 cannot
be uniformly improved upon.

In Proposition 6.2, we (non-constructively) show that a $3$-term linear Galois
relation of degree $d$ with coefficients $(a_{1},a_{2},a_{3})$ implies the
existence of a balanced multiset with respect to $(a_{1},a_{2},a_{3})$ of size
$d$. Thus we obtain in Corollary 6.5 the lower bound $q^{D}-1\leq L_{3}(D)$.

Note that this does not imply that the linear Galois relations arising from
our balanced multisets in the $n=3$ case have degree as small as possible; the
linear Galois relations arising from a balanced multiset of size $q^{D}-1$ may
have degree as large as $(q^{D}-1)!$ (see Theorem 7.2).

###  6.1. A Map of Representations

We will use the framework of representation theory to prove Proposition 6.2.
We now introduce in a general setting the representations we will be working
with.

Let $K$ be a field and let $L/K$ be a Galois extension with Galois group $G$.
Let $\gamma\in L$ be any element and let $V_{GalConj}$ be the $K$-vector space
spanned by the conjugates of $\gamma$. Then $V_{GalConj}$ is a $G$-invariant
subspace of $L$, and so the action of $G$ on $L$ as a $K$-vector space gives
an action of $G$ on $V_{GalConj}$. We call this the Galois conjugate
representation.

Suppose that $\gamma$ has degree $d$ over $K$. Let
$\gamma_{1},\dots,\gamma_{d}$ be the Galois conjugates of $\gamma$. Let
$V_{Perm}$ be a $d$-dimensional vector space over $K$ with basis
$\alpha_{1},...,\alpha_{d}$ and define an action of $G$ on $V_{Perm}$ by
$g\cdot\alpha_{i}=\alpha_{j}$ whenever $g(\gamma_{i})=\gamma_{j}$. This is the
permutation representation. We will write $(V_{Perm},\rho_{Perm})$ and
$(V_{GalConj},\rho_{GalConj})$ to denote the permutation and Galois conjugate
representations of $G$, respectively.

Notice that specializing the formal variables $\alpha_{i}$ to the Galois
conjugates $\gamma_{i}$ gives a surjective map $\phi$ of $G$-representations
from $(V_{Perm},\rho_{Perm})$ to $(V_{GalConj},\rho_{GalConj})$.

######  Proposition 6.2.

Let $K={\mathbb{F}}_{q}(t)$. Suppose that
$(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ is a coprime tuple and that
$\gamma=\gamma_{1},...,\gamma_{n}$ are Galois conjugates over $K$ such that
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$. Let $m:=[K(\gamma):K]$. Then there exists a
balanced multiset of tuples with respect to $(a_{1},...,a_{n})$ of size $m$.

###### Proof.

Let $E/K$ be the Galois closure of $K(\gamma_{1},...,\gamma_{n})$, and write
$G:=\operatorname{Gal}(E/K)$. Let $\gamma=\gamma_{1}$ and choose
$g_{1}=1,...,g_{n}\in G$ so that $g_{i}\gamma=\gamma_{i}$. Notice that
$\sum_{i=1}^{n}a_{i}g_{i}$ gives a linear transformation on both $V_{Perm}$
and $V_{Gal}$, which we’ll call $\psi_{Perm}$ and $\psi_{Gal}$ respectively.
Moreover, they commute with the map of $G$-representations $\phi$, i.e. the
following diagram commutes.

${V_{Perm}}$${V_{GalConj}}$${V_{Perm}}$${V_{GalConj}}$$\scriptstyle{\psi_{Perm}}$$\scriptstyle{\phi}$$\scriptstyle{\psi_{GalConj}}$$\scriptstyle{\phi}$

By assumption there is a nonzero $\gamma\in\ker\psi_{GalConj}$ and a simple
diagram-chasing argument shows that $\psi_{Perm}$ has nontrivial kernel too.
Therefore $\det(\sum_{i=1}^{n}a_{i}g_{i})=0$ on $V_{Perm}$, and the
construction from Proposition 4.1 yields a balanced multiset of tuples of size
$m$.

∎

Smyth also produced balanced multisets of tuples from linear Galois relations
in [4, Thm. 2]. Given a linear Galois relation
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$, Smyth’s argument yielded balanced multisets
of size the degree of the smallest normal extension ${\mathbb{Q}}(\beta)$ of
${\mathbb{Q}}$ containing $\gamma$ so that
$[{\mathbb{Q}}(\beta):{\mathbb{Q}}(\gamma)]\geq n$. Our argument shows the
existence of a smaller balanced set, of size
$[{\mathbb{Q}}(\gamma):{\mathbb{Q}}]$.

######  Proposition 6.3.

Let $(a,b,c)\in\mathbb{F}_{q}[t]^{3}$ be a coprime triple. Any nonzero
balanced set with respect to $(a,b,c)$ (if one exists) has size as least the
order of $-a/b$ in $(\mathbb{F}_{q}[t]/c)^{*}$.

###### Proof.

Let $T=\\{(t_{1j},t_{2j},t_{3j})\\}_{j=1}^{m}$ be any balanced multiset of
triples with respect to $(a,b,c)$. Without loss of generality, suppose that
the $t_{1j}$ ($j=1,2,...,m$) are coprime polynomials. Choose $k$ so that
$t_{1k}$ is not divisible by $c$.

For a polynomial $x\in\mathbb{F}_{q}[t]$, we will use the notation
$\overline{x}$ to denote the image of $x$ under the mod-c quotient map
$\mathbb{F}_{q}[t]\to\mathbb{F}_{q}[t]/c$. From the equation
$at_{1k}+bt_{2k}+ct_{3k}=0$, we have
$\overline{t_{2k}}=\overline{(-a/b)t_{1k}}$. By balancedness, this shows that
there is some $j$ so that $\overline{t_{1j}}=\overline{(-a/b)t_{1k}}$.

Iterating this argument, we see that
$\\{\overline{(-a/b)^{n}t_{1k}}:n\in\mathbb{Z}^{+}\\}\subset\\{\overline{t_{1j}}\\}_{j=1}^{m}$,
and the result follows. ∎

In the case where $c$ is irreducible and $-a/b$ is a generator for
$({\mathbb{F}}_{q}[t]/c)^{*}$, we get the following.

######  Corollary 6.4.

Let $(a,b,c)\in{\mathbb{F}}_{q}[t]^{3}$ be a Smyth triple. Suppose that $c$ is
irreducible with $\deg(c)=d\geq\deg(b),\deg(a)$ and also that $(-a/b)$ is a
generator for $({\mathbb{F}}_{q}[t]/c)^{*}$. Then the smallest balanced
multiset of triples with respect to $(a,b,c)$ has size $q^{d}-1$.

###### Proof.

From Proposition 5.1, setting $N=d$ and $n=3$, and removing the triples
$(0,0,0)$, we obtain a balanced multiset (in fact set) of triples of size
$q^{d}-1$.

On the other hand, Proposition 6.3 shows that this there cannot be a smaller
balanced multiset. ∎

######  Corollary 6.5.

$L_{3}(D)\geq q^{D}-1$.

###### Proof.

By Proposition 6.2, it suffices to show that for all $D$, there is some triple
$(a,b,c)$ with respect to which every balanced multiset of tuples has size at
least $q^{D}-1$. In other words, we are reduced to constructing a triple
satisfying the hypotheses of Corollary 6.4 with $d=D$.

Let $b,c$ be distinct irreducible polynomials of degree $D$. Let
$g\in(\mathbb{F}_{q}[t]/c)^{*}$ be any multiplicative generator and let $a$ be
the unique polynomial of degree $\leq D-1$ such that $a=-gb$ (mod c). Then
$a,b,c$ are pairwise coprime and their maximum degree is achieved twice, so
they are a Smyth triple by Theorem 4.2. Thus they satisfy the hypotheses of
Corollary 6.4. ∎

##  7\. Galois Group of Constructed Linear Galois Relations

In this section we show that the conjugates in our construction for Smyth
triples from Section 5 can be chosen to have the full symmetric group as their
Galois group.

We begin with a lemma.

######  Lemma 7.1.

Let $L$ be an $S_{d}$-extension of $K:=\mathbb{F}_{q}(t)$. There exist
conjugates $\alpha_{1},\dots,\alpha_{d}$ with $\sum_{i=1}^{d}\alpha_{i}\neq 0$
such that $L=K(\alpha_{1},\dots,\alpha_{d})$.

###### Proof.

Choose $\alpha\in L$ of degree $d$ with conjugates
$\alpha=\alpha_{1},...,\alpha_{d}$ such that $L=K(\alpha_{1},...,\alpha_{d})$.
We will show that $\sum_{i=1}^{d}\alpha_{i}\neq 0$ for some $1\leq k\leq d$.
Let $p_{k}=\sum_{i=1}^{d}x_{i}^{k}$ be the $k^{th}$ power sum polynomial.
Suppose for contradiction that $p_{k}(\alpha_{1},\dots,\alpha_{d})=0$ for all
$k\geq 1$. Let $e_{i}$ be the $i^{th}$ elementary symmetric polynomial and let
$f(x)=\sum_{i=0}^{d}a_{i}x^{i}$ be the minimal polynomial of $\alpha_{1}$. We
have that $e_{k}(\alpha_{1},\dots,\alpha_{d})=(-1)^{k}a_{d-k}$ for $1\leq
k\leq d$.

Since we assume $p_{i}(\alpha_{1},\dots,\alpha_{d})=0$ for all $1\leq i\leq
d$, the Newton identities give

$(-1)^{k-1}ke_{k}(\alpha_{1},\dots,\alpha_{d})=0$

So, $e_{k}(\alpha_{1},\dots,\alpha_{d})=0$ for all $p\nmid k$. Thus, the
minimal polynomial of $\alpha_{1}$ is inseparable, contradicting the fact that
$\alpha$ generates a separable extension. So, there is some $k$ for which
$\sum_{i=1}^{d}\alpha_{i}^{k}\neq 0$. If $\sum_{i=1}^{d}\alpha_{i}=0$, then
$k>1$ and the $\alpha_{i}^{k}$ are not guaranteed to generate $L$. But we can
find some nonzero $c\in K$ for which $\alpha_{1}+c\alpha_{1}^{k}$ and its
conjugates generate $L$. Indeed, for all but finitely many $c\in K$, the
elements $\alpha_{i}+c\alpha_{i}^{k}$ are pairwise distinct, in which case no
nontrivial $K$-automorphisms of $L$ fix the field
$K(\alpha_{i}+c\alpha_{i}^{k})_{i=1}^{d}$. Since
$\sum_{i=1}^{d}(\alpha_{1}+c\alpha_{1}^{k})=c\sum_{i=1}^{d}\alpha_{1}^{k}\neq
0$ for nonzero $c$, we are done.

∎

######  Theorem 7.2.

Let $K=\mathbb{F}_{q}(t)$ and fix an integer $n\geq 3$. Let
$(a_{1},\dots,a_{n})\in\mathbb{F}_{q}[t]^{n}$ be a coprime tuple. Suppose that
there is a 1-factor $T=\\{(x_{i_{1}},...,x_{i_{n}})\in K^{n}\\}_{i=1}^{m}$
with respect to $(a_{1},...,a_{n})$ of size $m$. Then there exist Galois
conjugates $\gamma_{1},\dots,\gamma_{d}$ such that
$\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ and
$\operatorname{Gal}(K(\gamma_{1})/K)=S_{m}$.

###### Proof.

By Lemma 7.1, we obtain $\alpha_{1},\dots,\alpha_{d}$ such that
$K(\alpha_{1},\dots,\alpha_{d})$ is an $S_{d}$-extension of $K$ and
$\sum_{i=1}^{m}\alpha_{i}\neq 0$. Define
$\gamma_{1}:=\sum_{j=1}^{m}v_{1,j}\alpha_{j}$. Note that since $S$ is a
1-factor, the entries $v_{1,j}$ $(j=1,2,...,m)$ are distinct. We will write
$v_{j}$ for $v_{1,j}$. If $K(\gamma_{1})$ is a proper subextension of $L$,
then it must be fixed by a nontrivial element $\sigma$ of $S_{m}$. So suppose
that

$\sum_{j=1}^{m}v_{j}\alpha_{j}=\gamma_{1}=\sigma(\gamma_{1})=\sum_{j=1}^{m}v_{\sigma^{-1}(j)}\alpha_{j}.$

Then $\sum_{j=1}^{m}(v_{j}-v_{\sigma^{-1}(j)})\alpha_{j}=0$. By [4, Thm
13.4.2], which is presented as a lemma about ${\mathbb{Q}}$ but is valid for
$K$ as well, the coefficients $v_{j}-v_{\sigma^{-1}(j)}$ must all be equal.
Then for $c:=v_{1}-v_{\sigma(1)}$, we have $c\sum_{j=1}^{m}\alpha_{j}=0$. By
assumption, $\sum_{j=1}^{m}\alpha_{j}\neq 0$, so $c=0$. Therefore $\sigma=1$,
and we have proved that $\gamma_{1}$ is not contained in a proper subextension
of $K(\gamma_{1},...,\gamma_{d})$.

∎

##  8\. A Heuristic for the Existence of a Linear Galois Relation with
Prescribed Galois Group

We just saw in Section 7 that the conjugates appearing in the linear Galois
relations we constructed for Smyth triples $(a,b,c)\in\mathbb{F}_{q}[t]^{3}$
can be chosen to generate an $S_{q^{d}}$-extension of $\mathbb{F}_{q}(t)$
where $d$ is the height of $(a,b,c)$. In this section, we consider which
Galois groups we expect to arise in this context.

######  Definition 8.1.

If $(a_{1},...,a_{n})$ is a Smyth tuple, $G\subset S_{m}$ is a finite group,
and there exist Galois conjugates $\gamma_{1},...,\gamma_{n}$ over $K$ such
that $\sum_{i=1}^{n}a_{i}\gamma_{i}=0$ and
$\operatorname{Gal}(K(\gamma_{1})/K)\cong G$, then we say $(a_{1},...,a_{n})$
is a Smyth $G$-tuple.

######  Question 8.2.

Suppose that $(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ is a Smyth tuple. For
which $m>0$ and transitive subgroups $G\subset S_{m}$ is $(a_{1},...,a_{n})$ a
Smyth $G$-tuple?

Heuristic 8.3, which depends only on $|G|$ and $m$, will correctly predict
that Smyth tuples $(a_{1},...,a_{n})$ are always $S_{q^{N}}$-tuples for
sufficiently large $N$. (Recall that Remark 5.2 and Theorem 7.2 together show
that every Smyth triple is an $S_{q^{N}}$-triple for sufficiently large $N$.)

By Proposition 4.1, a coprime $n$-tuple
$(a_{1},...,a_{n})\in\mathbb{F}_{q}[t]^{n}$ of height $d$ is a Smyth tuple if
and only if there exist permutation matrices $X_{1},...,X_{n}$ such that
$\det(\sum_{i=1}^{n}a_{i}X_{i})=0$. Our approach so far has been to look for
balanced multisets, which is essentially asking, given $(a_{1},...,a_{n})$,
whether or not there exists a vector $v\in\mathbb{F}_{q}[t]^{n}$ and
permutation matrices $X_{1},..X_{n}$ such that
$(\sum_{i=1}^{n}a_{i}X_{i})v=0$. We now shift our perspective slightly by
fixing both the tuple $(a_{1},...,a_{n})$ and a vector $v$, and asking whether
or not there exist such permutation matrices in $G$. Since $v$ will have no
repeated entries, the heuristic is suited only to detect 1-factors, as opposed
to a general balanced multiset.

In particular, we will fix a vector $v_{N}\in\mathbb{F}_{q}[t]^{q^{N}}$ whose
coordinates are all the elements of $V_{N}$ in some order, each occurring
exactly once. We have already seen that for any Smyth triple
$(a_{1},a_{2},a_{3})$, there do exist permuation matrices
$X_{1},X_{2},X_{3}\in S_{q^{N}}$ such that
$(a_{1}X_{1}+a_{2}X_{2}+a_{3}X_{3})v_{N}=0$ (Remark 5.2). Our heuristic, in
the case of triples, suggests an answer to the question of whether this
remains true if we insist on choosing $X_{1},X_{2},X_{3}$ from some subgroup
$G\subset S_{q^{N}}$.

######  Heuristic 8.3.

Fix $N$ and enumerate the set of polynomials over $\mathbb{F}_{q}$ of degree
$<N$: $V_{N}=\\{f_{1},f_{2},...,f_{q^{N}}\\}$, and fix a vector
$v=(f_{i})_{i=1}^{q^{N}}\in\mathbb{F}_{q}[t]^{q^{N}}$ whose coordinates are
all the elements of $V_{N}$, each occurring exactly once. Let $X_{n}=I$ be the
$q^{N}\times q^{N}$ identity matrix. Then choose random permutations
$X_{1},...,X_{n-1}\in G\subset S_{q^{N}}$ and assume that for each $j:1\leq
j\leq q^{N}$, the sum $\sum_{i=1}^{n}a_{i}v_{X_{i}^{-1}(j)}$ takes values in
$V_{N+d}$ uniformly and independently at random.

######  Remark 8.4.

A shortcoming of Heuristic 8.3 is that it doesn’t see the (necessary) “local”
conditions – it would predict that any $n$-tuple $(a_{1},...,a_{n})$, even
those which are not Smyth tuples, is a Smyth $G$-tuple for large enough $G$.

######  Remark 8.5.

The reason for fixing $X_{n}=I$ is that if $\sum_{i=1}^{n}a_{i}X_{i}v=0$, then
also $\sum_{i=1}^{n}a_{i}(XX_{i})v=0$ for every $X\in G$. This means that we
are really concerned with $G$-orbits of $n$-tuples of $G$, and we identify
these with $(n-1)$-tuples of elements of $G$ by fixing the last coordinate.

The result of this assumption is that we have $\sum_{i=1}^{n}(a_{i}X_{i})v=0$
with probability $(q^{-(N+d)})^{q^{N}}$. On the other hand, we have
$|G|^{n-1}$ choices for $X_{1},...,X_{n-1}$, so, according to this model, the
probability of failing to find $X_{1},...,X_{n}$ such that
$\sum_{i=1}^{n}(a_{i}X_{i})v=0$ is $p_{N}:=(1-q^{-(N+d)q^{N}})^{|G|^{(n-1)}}$.

This probability goes to $0$ as $N\to\infty$ for
$|G|=\omega_{q,d}(q^{(N+d)q^{N}/(n-1)})$ (Proposition 8.6). (We are using the
standard Landau asymptotic notation here, in which $f(n)=\omega(g(n))$ means
that $\lim_{n\to\infty}\frac{g(n)}{f(n)}=0$.)

Notably, if we plug in $n=2$, we see that Heuristic 8.3 correctly does  not
predict that a random pair $(a_{1},a_{2})$ is a Smyth pair; indeed, one can
check that $(a_{1},a_{2})\in\mathbb{Z}^{2}$ is a Smyth pair if and only if
$a_{1}=\pm a_{2}$.

######  Proposition 8.6.

Write $|G_{N}|=c_{N}(q^{(N+d)q^{N}})^{1/(n-1)}$ and suppose that
$\lim_{N\to\infty}c_{N}=\infty$. Then $\lim_{N\to\infty}p_{N}=0$.

By Stirling’s formula, Heuristic 8.3 suggests an affirmative answer to
Question 8.2 for $G=S_{q^{N}}$ when $n\geq 3$, in accordance with the fact
that we found linear Galois relations among triples of conjugates which
generate $S_{q^{N}}$-extensions. It also predicts an affirmative answer for
the alternating groups $A_{q^{N}}$, but predicts negative answers for the
cyclic group of order $q^{N}$ and the dihedral group of order $2q^{N}$.

##  9\. Smyth’s Conjecture Over $\mathbb{Q}$

In this brief section we note that some of our results from Section 6 can be
easily translated from $\mathbb{F}_{q}(t)$ over to $\mathbb{Q}$. In
particular, we have the following analogues of Proposition 6.3, Corollary 6.4,
and Corollary 6.5.

######  Proposition 9.1.

If $(a,b,c)\in{\mathbb{Z}}^{3}$ then any balanced multiset of triples with
respect to $(a,b,c)$ (if one exists) has size at least the order of $-a/b$ in
$(\mathbb{Z}/c\mathbb{Z})^{*}$.

The proof of this proposition is virtually identical to the proof of
Proposition 6.3. We also obtain analogous corollaries.

######  Corollary 9.2.

Let $(a,b,c)\in\mathbb{Z}^{3}$ be a coprime triple. Suppose that $c$ is prime
and that $-a/b$ is a generator for $(\mathbb{Z}/c\mathbb{Z})^{*}$. Then the
smallest balanced multiset of triples with respect to $(a,b,c)$ (if one
exists) has size at least $|(\mathbb{Z}/c\mathbb{Z})^{*}|=c-1$.

######  Corollary 9.3.

Let $p_{D}$ denote the largest prime at most $e^{D}$. Assuming Conjecture 2.1,
we have $L^{\mathbb{Q}}_{3}(D)\geq p_{D}-1$ for $D\geq 5$, and consequently,
$L_{3}^{\mathbb{Q}}(D)=\Omega(e^{D})$.

###### Proof.

Similar to Corollary 6.5, we just have to construct a fitting Smyth triple.
Assuming the Smyth Conjecture, for $(a,b,c)\in{\mathbb{Z}}^{3}$ to be a Smyth
triple is the same as for $a,b,c$ to be pairwise coprime and satisfy the
triangle inequalities ($a+b\geq c$, $a+c\geq b$, $b+c\geq a$).

Assume that $D\geq 5$ and let $p_{D}$ be the largest prime at most $e^{D}$ and
let $g\in({\mathbb{Z}}/p_{D}{\mathbb{Z}})^{*}$ be any multiplicative generator
other than $-1\mod p_{D}$.

Take $n$ to be the representative of $\frac{-1}{g+1}\mod p_{D}$ between $0$
and $p_{D}-1$. Then, $-\frac{n+1}{n}\equiv g\mod p_{D}$ and if $n+(n+1)\geq
p_{D}$, then $(n,n+1,p_{D})$ satisfy the absolute value criteria over
$\mathbb{Q}$. (Since we are assuming Conjecture 2.1, this is equivalent to
being a Smyth triple.) On the other hand, if $n+(n+1)<p_{D}$, then
$(p_{D}-(n+1),p_{D}-n,p_{D})$ satisfies the absolute value criteria. In either
case, the inequality now follows from Proposition 9.1.

By the prime number theorem, for any $k<1$, we have $p_{N}>kN$ for large
enough $N$. Therefore $p_{D}-1=\Omega(e^{D})$.

∎

##  10\. Smyth’s Conjecture Over Number Fields

Recall that in any field, the absolute value criteria are necessary conditions
for being a Smyth tuple (Lemma 4.3). We showed in Theorem 4.2 that these
criteria are sufficient for being a Smyth tuple over $\mathbb{F}_{q}(t)$, and
Smyth conjectured the same over $\mathbb{Q}$ (Conjecture 2.1).

However, an example presented by David Speyer in a MathOverflow post shows
that the absolute value criteria are not sufficient for being a Smyth tuple in
a general number field [1]. In particular, the triple
$(1,1,\frac{1+\sqrt{-15}}{2})$ satisfies the absolute value criteria, but is
not a Smyth triple. Note that this triple achieves equality in the archimedean
absolute value inequalities.

Speyer showed in the same post that for triples of the form $(1,1,a_{3})$, if
one amends the absolute value criteria to be strict inequalities for the
archimedean absolute values, then they become a sufficient condition for being
a Smyth triple. On the other hand, examples such as $(2,3,-5)$ show that we
cannot simply amend the archimedean absolute value criteria to be strict
inequalities, as $(2,3,-5)$ trivially is a Smyth triple. Instead, if some
analogue of Smyth’s Conjecture is true in number fields, it must be a little
more sensitive to the cases in which there is equality in one of the
archimedean absolute value criteria.

In order to formulate what we think the right conjecture is, we define the
strong absolute value criteria over $K$ as follows.

  * (1”)

For any archimedean absolute value $|\cdot|$ of $K$, we have
$|a_{i}|<\sum_{j\neq i}|a_{j}|$ for all $i$.

  * (2”)

For any nonarchimedean absolute value $|\cdot|$ of $K$, we have
$|a_{i}|\leq\max_{j\neq i}|a_{j}|$ for all $i$.

The strong absolute value criteria are obtained from the absolute value
criteria by making the archimedean inequalities strict.

We are now ready to formulate our generalization of Conjecture 2.1.

######  Conjecture 10.1.

Let $K$ be a number field and ${\mathcal{O}}_{K}$ its ring of integers.
$(a_{1},...,a_{n})\in{\mathcal{O}}_{K}^{n}$ is a Smyth tuple if and only if
$(a_{1},...,a_{n})$ satisfy the strong absolute value criteria over $K$ or
there exist roots of unity $\omega_{1},...,\omega_{n}$ in some extension of
$K$ such that $\sum_{i=1}^{n}a_{i}\omega_{i}=0$.

######  Remark 10.2.

The $K=\mathbb{Q}$ case of Conjecture 10.1 is equivalent to Conjecture 2.1.

We will show in Proposition 10.6 that Conjecture 10.1 correctly deals with the
tuples in which equality is achieved in one of the archimedean absolute value
criteria. In particular, if $(a_{1},...,a_{n})$ is a tuple such that equality
in one of the archimedean absolute value criteria, then any tuple in a
balanced multiset with respect to $(a_{1},...,a_{n})$ (if one exists) is a
scalar multiple of a tuple of roots of unity.

But first we need two lemmas, the first of which shows that the property of
being a Smyth tuple is preserved by multiplying the coordinates by (possibly
different) roots of unity.

######  Lemma 10.3.

Let $(a_{1},...,a_{n})\in{\mathcal{O}}_{K}^{n}$. If
$\omega_{1},...,\omega_{n}$ are roots of unity in some extension of $K$ and
$(a_{1},...,a_{n})$ is a Smyth tuple, then
$(\omega_{1}a_{1},...,\omega_{n}a_{n})$ is a Smyth tuple in
${\mathcal{O}}_{K(\omega_{1},\dots,\omega_{n})}^{n}$.

###### Proof.

Without loss of generality we may assume that $\omega_{2}=...=\omega_{n}=1$,
as we can make the following argument about each coordinate in turn. Denote
$\omega:=\omega_{1}$ and $L:=K(\omega)$. Suppose that $\omega^{m}=1$.

Let $\\{(x_{i_{1}},...,x_{i_{n}})\in K^{n}\\}_{i=1}^{N}$ be a balanced
multiset with respect to $(a_{1},...,a_{n})$. Then
$\bigcup_{k=0}^{m-1}\\{(\omega^{k-1}x_{i_{1}},\omega^{k}x_{i_{2}},...,\omega^{k}x_{i_{n}})\in
L^{n}\\}_{i=1}^{N}$ is a balanced multiset with respect to $(\omega
a_{1},a_{2},...,a_{n})$. ∎

######  Remark 10.4.

In particular, Lemma 10.3 shows that if there are roots of unity
$\omega_{1},...,\omega_{n}$ such that $\sum_{i=1}^{n}a_{i}\omega_{i}=0$, then
$(a_{1},...,a_{n})$ is a Smyth tuple. Linear relations among roots of unity
are a well-studied topic, going back at least to the 1960s. There are several
results constraining the prevalence of such relations, indicating that such
coefficients represent quite a small subset of Smyth tuples. A survey of some
of these results is given in [5]. For instance, when $a_{1},...,a_{n}$ are
rational, a result of Mann gives an explicit upper bound depending only on $n$
for the order of the roots of unity $\omega_{i}$ occurring in a minimal
relation $\sum_{i=1}^{n}a_{i}\omega_{i}=0$ [3]. (Here minimality means that no
nonempty proper sub-sum vanishes, and that the equation is normalized so that
$\omega_{1}=1$.)

######  Lemma 10.5.

Let $(a_{1},...,a_{n})\in{\mathcal{O}}_{K}^{n}$. Suppose that there exists an
archimedean absolute value $|\cdot|_{\nu}$ of $K$ and some $i$ for which
$|a_{i}|_{\nu}=\sum_{j\neq i}|a_{j}|_{\nu}$. If there exists a balanced
multiset with respect to $(a_{1},...,a_{n})$, then there exists a balanced
multiset $\\{(y_{i1},...,y_{in})\in K^{n}\\}_{i=1}^{N}$ with respect to
$(a_{1},...,a_{n})$ whose coordinates $y_{ij}$ all satisfy $|y_{ij}|_{\nu}=1$.

###### Proof.

Without loss of generality assume that
$|a_{1}|_{\nu}=\sum_{j>1}|a_{j}|_{\nu}$. Let $S=\\{(x_{i1},...,x_{in})\in
K^{n}\\}_{i=1}^{N}$ be a balanced multiset with respect to
$(a_{1},...,a_{n})$. Let $X=\\{x_{ij}:1\leq i\leq N,1\leq j\leq n\\}$ be the
set of all coordinates appearing in $S$. Write $M=\max_{x\in X}|x|_{\nu}$. Any
reference to “absolute value” in this proof refers to $|\cdot|_{\nu}$.

We claim that if a tuple in $S$ has a coordinate of absolute value $M$, then
all coordinates of that tuple have absolute value $M$. To see this, first
suppose that $|x_{i_{0}1}|=M$ for some $i_{0}$. Along with the assumptions
that $\sum_{j=1}^{n}a_{j}x_{i_{0}j}=0$ and
$|a_{1}|_{\nu}=\sum_{j>2}|a_{j}|_{\nu}$, this implies that
$|x_{i_{0}j}|_{\nu}=M$ for all $j=1,...,n$. What we’ve shown so far is that if
the first coordinate in a tuple in $S$ has absolute value $M$, then all
coordinates in that tuple do.

But $S$ is balanced, which means that the multiset of first coordinates is the
same as the multiset of $j^{th}$ coordinates for every $j=1,2,...,n$. In
particular, each of these multisets has the same number of elements of
absolute value $M$, with the same multiplicities. Therefore coordinates of
absolute value $M$ can only occur in tuples whose first coordinate has
absolute value $M$, and the claim is proved.

Thus the tuples whose coordinates have absolute value $M$ form a balanced sub-
multiset of $S$, and dividing all of these coordinates by an element of $K$ of
absolute value $M$, we obtain the desired balanced multiset. ∎

######  Proposition 10.6.

Let $(a_{1},...,a_{n})\in{\mathcal{O}}_{K}^{n}$. Suppose that there exists an
archimedean absolute value $|\cdot|_{\nu}$ of $K$ and some $i$ for which
$|a_{i}|_{\nu}=\sum_{j\neq i}|a_{j}|_{\nu}$. Then $(a_{1},...,a_{n})$ is a
Smyth tuple if and only if there exist roots of unity
$\omega_{1},...,\omega_{n}$ (not necessarily in $K$) such that
$\sum_{i=1}^{n}a_{i}\omega_{i}=0$.

###### Proof.

($\Leftarrow$): By assumption, $(\omega_{1}a_{1},...,\omega_{n}a_{n})$ is a
Smyth tuple. The result now follows from Lemma 10.3.

($\Rightarrow$): Let $\phi:K\hookrightarrow\mathbb{C}$ be an embedding
corresponding to the archimedean absolute value $|\cdot|_{\nu}$ and let
$\psi:\overline{\mathbb{Q}}\hookrightarrow\mathbb{C}$ be an embedding of the
algebraic closure of $\mathbb{Q}$ which extends $\phi$. We will write
$|\cdot|$ for the standard absolute value of complex numbers. Without loss of
generality assume that $|\phi(a_{1})|=\sum_{j>1}|\phi(a_{j})|$.

By Lemma 10.5, there exists a balanced multiset $S=\\{(x_{i1},...,x_{in})\in
K^{n}\\}_{i=1}^{N}$ with respect to $(a_{1},...,a_{n})$ such that all
$|x_{ij}|_{\nu}=1$. By definition of balanced multiset, we have

(10.1) $\sum_{j=1}^{n}a_{j}x_{ij}=0$

Now (10.1) along with $|\phi(a_{1})|=\sum_{j>1}|\phi(a_{j})|$ and the
assumption that $|\phi(x_{ij})|=1$ implies that

(10.2) $\arg\phi(a_{j}x_{ij})=\pi+\arg\phi(a_{1}x_{i1})(\text{mod
}2\pi),\text{ for all }i,j\text{ with }j>1.$

In words, (10.2) is saying that given a fixed $i$, the $\phi(a_{j}x_{ij})$ all
“point in the same direction” for $j>1$, and $\phi(a_{1}x_{i1})$ “points in
the opposite direction.”

The rest of the argument is most easily expressed in polar coordinates. For
all $j$, let $\phi(a_{j})=r_{j}\theta_{j}$ where $r_{j}\in\mathbb{R}^{\geq 0}$
and $|\theta_{j}|=1$. Fix any $i_{0}\in\\{1,2,..,N\\}$ and any
$j\in\\{2,...,n\\}$. Then by (10.2) and the fact that all $|\phi(x_{ij})|=1$,
we have $\phi(x_{{i_{0}}j})=-\frac{\theta_{1}}{\theta_{j}}\phi(x_{i_{0}1})$.

By balancedness, there is some $i_{1}$ so that $x_{i_{1}1}=x_{i_{0}j}$, so
repeating the above argument, we get
$\phi(x_{i_{1}j})=-\frac{\theta_{1}}{\theta_{j}}\phi(x_{i_{1}1})=-\frac{\theta_{1}}{\theta_{j}}\phi(x_{i_{0}j})=(-\frac{\theta_{1}}{\theta_{j}})^{2}\phi(x_{i_{0}1})$.
Iterating, this argument shows that
$(-\frac{\theta_{1}}{\theta_{j}})^{m}\phi(x_{i_{0}1})\in\\{\phi(x):x\in X\\}$
for all $m\in\mathbb{Z}$, implying that $-\frac{\theta_{1}}{\theta_{j}}$ is a
root of unity.

Now let $\omega_{1}=1$ and $\omega_{j}=-\frac{\theta_{1}}{\theta_{j}}$ for
$j>1$. Dividing the equation (10.1) with $i=i_{0}$ by $x_{i_{0}1}$ and
applying $\phi$ to both sides, we have
$\sum_{i=1}^{n}\phi(a_{i})\omega_{i}=0$. Finally, letting
$\rho_{i}=\psi^{-1}(\omega_{i})$, we see that
$\psi(\sum_{i=1}^{n}a_{i}\rho_{i})=\sum_{i=1}^{n}\phi(a_{i})\omega_{i}=0$, and
hence $\sum_{i=1}^{n}a_{i}\rho_{i}=0$. ∎

The above work, along with Lemma 4.3, reduces Conjecture 10.1 to the
following.

######  Conjecture 10.7.

Let $K$ be a number field and ${\mathcal{O}}_{K}$ its ring of integers. If
$(a_{1},...,a_{n})\in{\mathcal{O}}_{K}^{n}$ satisfies the strong absolute
value criteria, then $(a_{1},...,a_{n})$ is a Smyth tuple.

Speyer [1] gives a proof of Conjecture 10.7 in the case where $n=3$ and
$a_{1}=a_{2}$.

Speyer’s argument works for general $n$ and $a_{1}=...=a_{n-1}$ with minimal
modification; this result is our final proposition.

######  Proposition 10.8.

Let $n\geq 3$ be an integer. Let $K$ be a number field and ${\mathcal{O}}_{K}$
its ring of integers. Let $\alpha\in{\mathcal{O}}_{K}$ so that every
archimedean absolute value of $\alpha$ is less than $n-1$. Then
$(1,1,\dots,1,\alpha)\in{\mathcal{O}}_{K}^{n}$ is a Smyth tuple.

###### Proof.

By Lemma 10.3, it suffices to show $(1,1,\dots,1,-\alpha)$ is a Smyth tuple.
By Proposition 4.1, it suffices for us to show that there are permutation
matrices $X_{i}$ so that $\sum_{i=1}^{n-1}X_{i}$ has $\alpha$ as an
eigenvalue.

We will follow the argument from [1], starting with a slight generalization of
Speyer’s Step 1, which we write out in full for the sake of clarity.

Step 1: There is a nonnegative integer matrix $C$ whose rows sum to $n-1$ with
eigenvalue $\alpha$.

Consider the lattice $A={\mathbb{Z}}[\alpha]$ and the vector space
$V=A\otimes_{\mathbb{Z}}{\mathbb{R}}$. Since $\alpha$ is an algebraic integer,
$A$ is a discrete full sublattice of $V$. We take the norm
$\sum_{\nu}|x|_{\nu}^{2}$, where the sum runs over all archimedean places. Let
$c=\max_{\nu}|\alpha|_{\nu}$. By hypothesis, $c<n-1$. Denote by $B_{R}$ the
closed ball of radius $R$ around $0$.

Let $M$ be large enough so that any ball of radius $M$ around any point in $V$
contains a point in $A$. Take $R$ large enough so that
$\frac{c}{n-1}R+(n-2)M<R$. Now, for any $z\in A\cap B_{R}$, let $z_{1}\in
A\cap B_{R}$ be the closest point to $\frac{\alpha z}{n-1}$. Let $z_{2}=\alpha
z-(n-2)z_{1}$. Now,

$|z_{1}|\leq|z_{1}-\frac{\alpha z}{n-1}|+|\frac{\alpha z}{n-1}|\leq
M+\frac{c}{n-1}R<R$

Similarly,

$|z_{2}|=|\alpha z-(n-2)z_{1}|\leq|\alpha z-\frac{n-2}{n-1}\alpha
z|+(n-2)|\frac{\alpha z}{n-1}-z_{1}|\leq\frac{c}{n-1}R+(n-2)M<R$

Thus, for any $z\in A\cap B_{R}$, we can find $z_{1},z_{2}\in A\cap B_{R}$ so
that $(n-2)z_{1}+z_{2}=z$. Enumerate the elements of $A\cap B_{R}$ as
$z_{1},z_{2},\dots,z_{l}$. Then, we can form an $l\times l$ matrix $C$ with
the following entries. For the $i$-th row, consider $z_{i}$. As before, we may
write $(n-2)z_{j}+z_{k}=z_{i}$ for some $1\leq j,k\leq l$. In the $i$-th row,
put $n-2$ in the $j$-th column and $1$ in the $k$-th column if $j\neq k$; if
$j=k$, put an $n-1$ in the $j^{th}$ column. This matrix $C$ has all row sums
$n-1$. By construction, it has $\alpha$ as an eigenvalue with right
eigenvector $(z_{1},z_{2},\dots,z_{l})^{T}$.

The rest of Speyer’s argument can now be applied with virtually no
modification; using the Perron-Frobenius theorem, one obtains a matrix $D$
from $C$ which is the sum of $n-1$ permutation matrices and still has $\alpha$
as an eigenvalue. ∎

## References

  * [1] David E Speyer (https://mathoverflow.net/users/297/david-e-speyer)  “Tell me an algebraic integer that isn’t an eigenvalue of the sum of two permutations” URL:https://mathoverflow.net/q/264035 (version: 2017-03-09), MathOverflow  URL: https://mathoverflow.net/q/264035
  * [2] N. Berry et al.  “The conjugate dimension of algebraic numbers”  In _The Quarterly Journal of Mathematics_ 55.3, 2004, pp. 237–252  DOI: 10.1093/qmath/hah003
  * [3] H.B. Mann  “On linear relations between roots of unity”  In _Mathematika_ 12, 1965, pp. 107–117
  * [4] C.J. Smyth  “Additive and Multiplicative Relations Connecting Conjugate Algebraic Numbers”  In _Journal of Number Theory_ , 1986, pp. 243–254  DOI: https://doi.org/10.1016/0022-314X(86)90094-6
  * [5] U. Zannier  “Vanishing sums of roots of unity”  In _Rend. Sem. Mat. Univ. Pol. Torino_ 53, 4, 1995, pp. 487–495


# On the existence of twin prime in an interval

Shaon Sahoo  Indian Institute of Technology, Tirupati 517506, India
<EMAIL_ADDRESS>
###### Abstract.

Let $S_{(x,y]}=\left\\{\frac{p_{n}}{p_{n+1}-2}:~{}n\in I\right\\}$, where
$I=\left\\{n:~{}x<p_{n}\leq y\right\\}$, $p_{n}$ is the $n$-th prime and
$x,y\in\mathbb{R}_{>0}$. If $M_{\alpha}(x,y)$ denotes the $\alpha$-power mean
of the elements of $S_{(x,y]}$, it is shown that the existence of a twin prime
pair in $(x,y]$ is implied if
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)>1-2/y+O(y^{-2})$
for a sufficiently large $y$. For a special choice of $y$, we also find a
lower bound for the mean:
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,x^{\beta})>1-c/x^{\beta}+O(x^{-\beta}\log^{-1}x)$,
where the constant $c>0$ and $\beta=1+c/\log^{2}x$ or equivalently,
$x^{\beta}=x+cx/\log x+O(x/\log^{2}x)$. With $c<2$, the lower bound for
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,x^{\beta})$ satisfies
the inequality on the existence of a twin prime in the interval
$(x,x^{\beta}]$.

###### Key words and phrases:

Twin primes, Power mean

###### 2010 Mathematics Subject Classification:

11N05

##  1\. Introduction

Let $p_{n}$ be the $n$-th prime and $x,y\in\mathbb{R}_{>0}$ with $x<y$. Let
$S_{(x,y]}=\\{r_{n}:~{}n\in I\\}$, where $r_{n}=\frac{p_{n}}{p_{n+1}-2}$ and
$I=\\{n:~{}x<p_{n}\leq y\\}$. The power mean or Hölder mean of the elements of
$S_{(x,y]}$ is given by $M_{\alpha}(x,y)=(\frac{1}{\pi(x,y)}{\sum_{n\in
I}}r_{n}^{\alpha})^{\frac{1}{\alpha}}$, where $\alpha\in\mathbb{R}$ and
$\pi(x,y)$ denotes the number of primes in $(x,y]$. By construction,
$r_{n}\leq 1$ for all $n\in I$, and $r_{n}=1$ only when $p_{n+1}-p_{n}=2$. By
Theorem 3, $\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)$
converges to the largest element in $S_{(x,y]}$. Therefore,
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)=1$ implies that
there exists a twin prime in $(x,y]$. In fact, as stated in Theorem 2, the
existence of a twin prime pair in the interval $(x,y]$ is implied if
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)>1-2/y+O(y^{-2})$
for a sufficiently large $y$.

In this work we show that, for a special $x$-dependent choice of $y$, a lower
bound for the mean
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)$ is actually
larger than $1-2/y+O(y^{-2})$ for a sufficiently large $y$ (or $x$). The
result on the lower bound is stated as the following theorem.

######  Theorem 1.

Let $S_{(x,y]}=\left\\{\frac{p_{n}}{p_{n+1}-2}:~{}n\in I\right\\}$ for
$\displaystyle I=\left\\{n:~{}x<p_{n}\leq y\right\\}$, where $p_{n}$ is the
$n$-th prime. With $r_{n}=\frac{p_{n}}{p_{n+1}-2}$, the power mean of the
elements of $S_{(x,y]}$ is given by $\displaystyle
M_{\alpha}(x,y)=\left(\frac{1}{\pi(x,y)}\sum_{n\in
I}r_{n}^{\alpha}\right)^{\frac{1}{\alpha}}$, where $\alpha$ is a real number
and $\pi(x,y)$ denotes the number of primes in the interval $(x,y]$. We have

(1.1)
$\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,x^{\beta})>1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log
x}),$

where $c$ is a fixed positive constant ($c>0$) and $\beta=1+c/\log^{2}x$ or
equivalently, $x^{\beta}=x+cx/\log x+O(x/\log^{2}x)$.

The proof of Theorem 1 is given in Section 4. If we take $c<2$ in Theorem 1,
the inequality in Theorem 2 is satisfied for the interval $(x,x^{\beta}]$:
${\displaystyle\lim_{\alpha\rightarrow\infty}}M_{\alpha}(x,x^{\beta})>1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log
x})>1-\frac{2}{x^{\beta}}+O(\frac{1}{x^{2\beta}})$. This shows that there
exists a twin prime in the interval $(x,x+\frac{cx}{\log
x}+O(\frac{x}{\log^{2}x})]$ for sufficiently large $x$ and $0<c<2$. If a twin
prime exists in an interval, it also exists in an extension of the interval.
We, accordingly, conclude that a twin prime exists in the aforementioned
interval for any fixed $c>0$.

##  2\. Background

It is conjectured that there are infinite number of primes $p$, for each of
which $p+2$ is also a prime. Many important contributions have been made in
the efforts to prove the conjecture. In this regard we may note the Chen’s
work [1] and more recent work by Zhang [2]. Chen showed that there are
infinite number of primes $p$ so that $p+2$ has at most two prime factors. On
the other hand, Zhang showed that there are infinite number of prime pairs
with bounded gaps. More specifically, he showed that
$\liminf_{n\rightarrow\infty}(p_{n+1}-p_{n})<7\times 10^{7}$, where $p_{n}$ is
the $n$-th prime number. Subsequent works by Tao (through Polymath projects)
and Maynard brought down the upper bound of 70 million to 246 [3, 4]. It is
still an open problem whether the upper bound of the limit can be brought down
to the conjectured value of 2.

We take here a different approach to study the question on the twin primes.

##  3\. Some useful results

######  Theorem 2.

Let $S_{(x,y]}$ and $M_{\alpha}(x,y)$ respectively be the set and the power
mean of the elements of the set, as defined in Theorem 1. The existence of a
twin prime pair in the interval $(x,y]$ is implied if
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)>1-\frac{2}{y}+O(\frac{1}{y^{2}})$
for a sufficiently large $y$.

###### Proof.

Taking $g_{n}=p_{n+1}-p_{n}$, we rewrite the elements of $S_{(x,y]}$ as
$r_{n}=\frac{1}{1+(g_{n}-2)/p_{n}}$. We note that $r_{n}\leq 1$; the equality
sign holds only when $g_{n}=2$, i.e., when $p_{n}$ represents (the first
member of) a twin prime. We, therefore, see that 1 is the largest possible
element of $S_{(x,y]}$ and $1\in S_{(x,y]}$ implies the existence of twin
prime in $(x,y]$.

Let $R$ be the largest element in $S_{(x,y]}$ which is not 1. A non-trivial
upper bound of $R$ corresponds to $g_{n}=4$ and the largest prime in $(x,y]$.
If $P$ is the largest prime in $(x,y]$, we have
$R\leq\frac{1}{1+(4-2)/P}\leq\frac{1}{1+2/y}=1-2/y+O(y^{-2})$ for sufficiently
large $y$. By Theorem 3,
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)$ converges to the
largest element in $S_{(x,y]}$. Therefore, the inequality
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)>1-2/y+O(y^{-2})$
implies that there is an element in $S_{(x,y]}$ which is larger than $R$. But
1 can be the only element which is larger than $R$. Hence
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,y)>1-2/y+O(y^{-2})$
actually confirms that $1\in S_{(x,y]}$. The statement of the lemma now
follows. ∎

######  Theorem 3.

Let $A=\\{a_{1},a_{2},\cdots,a_{N}\\}$ be a set of $N$ positive real numbers.
Let $\max\\{A\\}$ and $\min\\{A\\}$ denote respectively the largest and the
smallest elements in $A$. The $\alpha$-power mean of the elements of $A$ is
given by
$M_{\alpha}(A)=(\frac{1}{N}\sum_{i=1}^{N}a_{i}^{\alpha})^{\frac{1}{\alpha}}$,
where $\alpha\in\mathbb{R}$. We have

  1. (1)

$M_{\alpha}(A)\geq M_{\alpha^{\prime}}(A)$ for $\alpha>\alpha^{\prime}$; the
equality sign holds iff all the elements in $A$ are equal.

  2. (2)

${\displaystyle\lim_{\alpha\rightarrow
0}M_{\alpha}(A)}=(\prod_{i=1}^{N}a_{i})^{\frac{1}{N}}$.

  3. (3)

$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(A)=\max\\{A\\}$ and
$\displaystyle\lim_{\alpha\rightarrow-\infty}M_{\alpha}(A)=\min\\{A\\}$.

These are standard results; the proof can be found elsewhere (eg. [8]).

######  Theorem 4.

We have,

$\sum_{p\leq x}\frac{1}{p}=\log\log x+M+O(\exp(-\sqrt[14]{\log x})),$

where $M$ ($\approx$ 0.261) is the Meissel-Mertens constant.

This result (Theorem 4) is due to Landau [9]. Results with improved error term
are known but are not necessary for our work here. A corollary of this result
is the following.

######  Corollary 1.

We have,

$\sum_{p\leq x}\frac{1}{p}=\log\log x+M+O(1/(\log x)^{A}),$

for any constant $A>0$.

For our work we need the above result with $A>1$; for definiteness, we will
take $A=2$.

######  Lemma 1.

We have,

$(1-\frac{1}{2})\prod_{2<p\leq
x}(1-\frac{2}{p})=\frac{e^{-D}}{\log^{2}x}\left(1+O(\frac{1}{\log^{2}x})\right),$

where $D$ ($\approx$ 0.877) is a constant.

###### Proof.

Let ${\displaystyle G(x)=\prod_{2<p\leq x}(1-\frac{2}{p})}$. Therefore,
${\displaystyle\log G_{x}=\sum_{2<p\leq x}\ln(1-\frac{2}{p})=\sum_{2<p\leq
x}\left[-\frac{2}{p}-\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}\right]}$.
Since
$\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}\leq\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{2}{p}\right)^{k}=\frac{2}{p^{2}(1-2/p)}\leq\frac{6}{p^{2}}$,
the infinite sum
${\displaystyle\sum_{p>2}\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}}$
converges absolutely to a constant, say, $C$. By numerical calculation we
estimate that ${\displaystyle
C=-\sum_{p>2}\left[\log(1-\frac{2}{p})+\frac{2}{p}\right]\approx 0.660413}$.
Moreover, $\sum_{2<p\leq
x}\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}=\sum_{p>2}\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}-\sum_{p>x}\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}$.
Here the first term is the constant $C$ and the second term
$\sum_{p>x}\sum_{k=2}^{\infty}\frac{1}{k}\left(\frac{2}{p}\right)^{k}\leq\sum_{p>x}\frac{6}{p^{2}}=O(\frac{1}{x})$.
With this, we have
${\displaystyle\log G(x)=-2\sum_{2<p\leq x}\frac{1}{p}-C+O(\frac{1}{x})}$.
Using now the Corollary 1 (with $A=2$), we get
${\displaystyle\log G(x)=-2\log\log
x-2M+1-C+O(\frac{1}{\log^{2}x})+O(\frac{1}{x})}$, or
${\displaystyle\log G(x)=-2\log\log x-D^{\prime}+O(\frac{1}{\log^{2}x})}$,
where $D^{\prime}=2M+C-1\approx 0.183407.$
This gives us ${\displaystyle
G(x)=\frac{e^{-D^{\prime}}}{\log^{2}x}(1+O(\frac{1}{\log^{2}x}))}$.
Finally we get ${\displaystyle(1-\frac{1}{2})\prod_{2<p\leq
x}(1-\frac{2}{p})=\frac{1}{2}G_{x}=\frac{e^{-D}}{\log^{2}x}(1+O(\frac{1}{\log^{2}x}))}$,
where $D=D^{\prime}+\log 2\approx 0.876554$. ∎

######  Lemma 2.

Let $\displaystyle T(x,x^{\beta})=\prod_{n\in I}\frac{p_{n}}{p_{n+1}-2}$,
where $p_{n}$ is the $n$-th prime and $I=\left\\{n:~{}x<p_{n}\leq
x^{\beta}\right\\}$. Here $\beta>1$ and additionally, let $\beta$ also depend
on $x$ in the following way: $\beta=O(1)$. We have
$T(x,x^{\beta})=\frac{\beta^{2}}{x^{\beta-1}}\left(1+O(\frac{1}{\log^{2}x})\right).$

###### Proof.

Let $p_{s}$ be the smallest prime greater than $x$ and similarly, $p_{e}$ be
the smallest prime greater than $x^{\beta}$.

Now, $\displaystyle T(x,x^{\beta})=\prod_{n\in
I}\frac{p_{n}}{p_{n+1}-2}=\left(\frac{p_{s}-2}{p_{e}-2}\right)\frac{(1-\frac{1}{2})\prod_{2<p\leq
x}(1-\frac{2}{p})}{(1-\frac{1}{2})\prod_{2<p\leq x^{\beta}}(1-\frac{2}{p})}$.
Now using Lemma 1, we get,

(3.1)
$\begin{split}T(x,x^{\beta})=&\left(\frac{p_{s}-2}{p_{e}-2}\right)\frac{e^{-D}\log^{-2}x(1+O(\log^{-2}x))}{e^{-D}\log^{-2}x^{\beta}(1+O(\log^{-2}x^{\beta}))}\\\
=&\beta^{2}\frac{p_{s}-2}{p_{e}-2}\left(1+O(\frac{1}{\log^{2}x})\right).\end{split}$

We know from the prime number theorem that the average gap between the
consecutive primes upto $x$ is $(1+o(1))\log x$. In fact it is conjectured, by
Cramér [5] and later with some refinement by Granville [6], that, if $G(x)$ is
the largest gap between two consecutive primes upto $x$, then
$G(x)=O(\log^{2}x)$. Therefore, one would expect that $p_{s}=x+O(\log^{t}x)$
and $p_{e}=x^{\beta}+O(\log^{t}x)$, for some $t\geq 1$.

In this regard, currently the best unconditional result is due to Baker,
Harman and Pintz [7]. They showed that $G(x)=O(x^{\theta})$, where
$\theta=0.525$. This result is sufficient to prove the lemma here. Taking
$p_{s}=x+O(x^{\theta})$ and $p_{e}=x^{\beta}+O(x^{\beta\theta})$ in Equation
3.1, we get,

$\begin{split}T(x,x^{\beta})=&\beta^{2}\frac{x(1+O(\frac{1}{x^{1-\theta}}))}{x^{\beta}(1+O(\frac{1}{x^{\beta-\beta\theta}}))}\left(1+O(\frac{1}{\log^{2}x})\right)\\\
=&\frac{\beta^{2}}{x^{\beta-1}}\left(1+O(\frac{1}{\log^{2}x})\right).\end{split}$

∎

##  4\. Proof of Theorem 1

###### Proof.

By Theorem 3, $M_{\alpha}(x,x^{\beta})\geq M_{\alpha^{\prime}}(x,x^{\beta})$
for $\alpha>\alpha^{\prime}$. For the time being, we take $\beta>1$; its
appropriate form will be mentioned soon. We are here, in particular,
interested in the inequality $M_{\infty}(x,x^{\beta})\geq M_{0}(x,x^{\beta})$,
where $M_{\infty}(x,x^{\beta})$ and $M_{0}(x,x^{\beta})$ respectively denote
$\displaystyle\lim_{\alpha\rightarrow\infty}M_{\alpha}(x,x^{\beta})$ and
$\displaystyle\lim_{\alpha\rightarrow 0}M_{\alpha}(x,x^{\beta})$. In the
following we first argue that a strict inequality holds between the two means,
i.e., $M_{\infty}(x,x^{\beta})>M_{0}(x,x^{\beta})$, and then we show that
$M_{0}(x,x^{\beta})=1-c/x^{\beta}+O(x^{-\beta}\log^{-1}x)$, where $c>0$ and
$\beta=1+c/\log^{2}x$ or equivalently, $x^{\beta}=x+cx/\log x+O(x/\log^{2}x)$.

For the first part, we now note that $p_{n}\leq p_{n+1}-2$ and, by Bertrand-
Chebyshev theorem, $2p_{n}>p_{n+1}>p_{n+1}-2$. Therefore, $p_{n}\leq
p_{n+1}-2<2p_{n}$. This implies that $gcd(p_{n},p_{n+1}-2)=1$, except when
$p_{n}=p_{n+1}-2$. This observation helps us to determine if two elements of
the set $S_{(x,x^{\beta}]}$ are equal or not. Since the elements of
$S_{(x,x^{\beta}]}$ are in the form $\frac{p_{n}}{p_{n+1}-2}$ ($=r_{n}$), the
numerators of all the elements are different, except when the elements
correspond to twin primes. Equivalently, we have $r_{n}\neq r_{m}$ whenever
$n\neq m$, with an exception where both $p_{n}$ and $p_{m}$ are (first members
of) twin primes. Since all the consecutive primes do not form twin prime
pairs, we conclude that all elements in $S_{(x,x^{\beta}]}$ are not equal, and
hence by Theorem 3, $M_{\infty}(x,x^{\beta})>M_{0}(x,x^{\beta})$.

Next we analyze the mean $M_{0}(x,x^{\beta})$. Let $\pi(x,x^{\beta})$ denote
the number of primes in $(x,x^{\beta}]$. If $\pi(x)$ denotes the number of
primes upto $x$, we have $\pi(x)=\frac{x}{\log x}(1+O(\frac{1}{\log x}))$.
Therefore,
$\pi(x,x^{\beta})=\pi(x^{\beta})-\pi(x)=\frac{1}{\beta}\frac{x^{\beta}}{\log
x}(1+O(\frac{1}{\log x}))$. Using Lemma 2, we now have

(4.1)
$\begin{split}M_{0}(x,x^{\beta})&=T(x,x^{\beta})^{\frac{1}{\pi(x,x^{\beta})}}\\\
&=\left(\frac{\beta^{2}}{x^{\beta-1}}\right)^{\frac{1}{\pi(x,x^{\beta})}}\left(1+O\left(\frac{1}{\pi(x,x^{\beta})\log^{2}x}\right)\right)\\\
&=\left(\frac{\beta^{2}}{x^{\beta-1}}\right)^{\frac{1}{\pi(x,x^{\beta})}}\left(1+O(x^{-\beta}\log^{-1}x)\right).\end{split}$

Now to determine the main term, let
$z=\left(\frac{\beta^{2}}{x^{\beta-1}}\right)^{\frac{1}{\pi(x,x^{\beta})}}$.
Taking logarithm on both sides, we have

(4.2) $\begin{split}\log z&=\frac{2\log\beta-(\beta-1)\log
x}{\pi(x,x^{\beta})}\\\ &=\frac{2\beta\log\beta\log
x-\beta(\beta-1)\log^{2}x}{x^{\beta}}\left(1+O(\frac{1}{\log
x})\right).\end{split}$

At this stage, we take $\beta=1+\frac{c}{\log^{2}x}$, for a fixed $c>0$.
Accordingly, we have
$\beta\log\beta=\frac{c}{\log^{2}x}+O(\frac{1}{\log^{4}x})$,
$\beta(\beta-1)=\frac{c}{\log^{2}x}+O(\frac{1}{\log^{4}x})$ and
$x^{\beta}=x(x)^{c/\log^{2}x}=x(1+\frac{c}{\log x}+O(\frac{1}{\log^{2}x}))$.
We now get from Equation 4.2,

(4.3) $\begin{split}\log z&=\left(\frac{2c}{x^{\beta}\log
x}-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log^{2}x})\right)\left(1+O(\frac{1}{\log
x})\right)\\\ &=-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log x}).\end{split}$

This gives us $z=1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log x})$. Plugging
this value of $z$ in Equation 4.1, we get
$M_{0}(x,x^{\beta})=\left(1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log
x})\right)\left(1+O(\frac{1}{x^{\beta}\log
x})\right)=1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log x})$.

Finally we get
$M_{\infty}(x,x^{\beta})>M_{0}(x,x^{\beta})=1-\frac{c}{x^{\beta}}+O(\frac{1}{x^{\beta}\log
x})$. ∎

## References

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  * [2] Y. Zhang, Bounded gaps between primes, Ann. of Math. 179 (2014), 1121–1174.
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