Weighted enumeration of number fields using Pseudo and Sudo maximal orders

Gaurav Digambar Patil Department of Mathematics, University of Toronto, Bahen Centre, 40 St. George Street, Room 6290, Toronto, Ontario, Canada, M5S 2E4 [email protected]

Abstract.

We establish a fundamental theorem of orders (FTO) which allows us to express all orders uniquely as an intersection of ‘irreducible orders’ along which the index and the conductor distributes multiplicatively. We define a subclass of Irreducible orders named Pseudo maximal orders. We then consider orders (called Sudo maximal orders) whose decomposition under FTO contains only Pseudo maximal orders. These rings can be seen as being “close” to being maximal (ring of integers) and thus there is a limited number of them with bounded index (by X). We give an upper bound for this quantity. We then show that all polynomials which can be sieved using only the Ekedahl sieve correspond to Sudo Maximal Orders. We use this understanding to get a weighted count for the number of number-fields with fixed degree and bounded discriminant using the concept of weakly divisible rings.

1. Introduction

We define the quantities $S(X:n)$ and $N(X:n)$ as follows:

Definition 1.

S(X:n):=\{K:[K:\mathbb{Q}]=n,Gal(\bar{K}/\mathbb{Q})\simeq S_{n},\mathfrak{disc}(K)<X\}

N(X:n):=\#S(X:n)

Malle’s Conjecture (for $S_{n}$ ) in [8] tells us to expect

N(X:n)\simeq c_{n}X.

where $c_{n}$ is a constant dependent on only $n.$ This conjecture is known for small values of $n$ , namely $n\leq 5.$ However, some lower bounds on the quantity $N(X:n)$ which are known are as follows.

•

In [7] (2002), Malle showed that

$N(X:n)\gg X^{\frac{1}{n}}.$
•

In [5](2006) , Akshay Venkatesh and Jordan Ellenberg establish

$N(X:n)\gg X^{\frac{1}{2}+\frac{1}{n^{2}}}.$
•

In [2], Bhargava-Shankar-Wang show

$N(X:n)\gg_{n}X^{\frac{1}{2}+\frac{1}{n}}$
•

In [3], the same authors show

$N(X:n)\gg_{n}X^{\frac{1}{2}+\frac{1}{n-1}}.$
•

In our previous paper [10], we showed that

Theorem 1 (Number-field count).

$N(X:n)\gg_{n}X^{\frac{1}{2}+\frac{1}{n-1}+\frac{(n-3)r_{n}}{(n-1)(n-2)}}$

where $r_{n}=\frac{\eta_{n}}{n^{2}-4n+3-2\eta_{n}(n+\frac{2}{n-2})}$ and where $\eta_{n}$ is $\frac{1}{5n}$ if $n$ is odd and is $\frac{1}{88n^{6}}$ when $n$ is even.

In that paper([10]), we also show that the number of rings with rank $n$ over $\mathbb{Z}$ with bounded discriminant is

\gg_{n}X^{\frac{1}{2}+\frac{1}{n-4/3}}.

Since, for every prime $p$ the ratio of non-maximal rings is $O(1/p^{2})$ , one might expect that $N(X:n)$ is also of the same order. In this paper, we show that that is true up to weighted enumeration and an $\epsilon$ in the order. Thus, showing that the liminf of $N(X:n)$ is bounded below by $X^{\frac{1}{2}+\frac{1}{n-4/3}}.$

Each of the above bounds uses different classifications of rings (see Introduction of our previous paper titled “Weakly Divisible Rings” [10].).

There are two main obstructions (there are three but we will only look at two) on to getting better and better lower bounds for $N(X:n)$ . One side of the problem requires more general parametrizations of rings(as discussed in [10]). The first three results given above use monic polynomials parametrizing monogenic rings. Heuristically, the third bound best possible using monogenic rings. This optimality is a result of solving a sieve theoretic problem. The fourth bound is based on binary rings and their parametrization via binary forms. Heuristically, this is the best possible result using just binary rings and a result of solving the second problem which is sieve theoretic in nature. In [10] we introduce a different parametrization, namely weakly divisible rings, to get the fifth result. Our other computations heuristically suggests that the lower bound can be improved to $\gg X^{\frac{1}{2}+\frac{1}{n-4/3}},$ provided we can optimize the seive that sieves for maximal (weakly divisible) rings among all (weakly divisible) rings.

To get lower bounds, one has to restrict the given parametrization to certain local conditions that force the ring under consideration to be the whole ring of integers.

Ekedahl sieve allows us to sieve out points in a parameter space that are lifts of points in a codimension two (or more) Variety over $\mathbb{F}_{p}.$ We thus refer to such a condition as Ekedahlian. In general local conditions are a combination of $\bmod p$ , $\bmod p^{2}$ , etc conditions. A specific type of local condition that is not easily handled by the Ekedahl sieve (in its base form) is often the biggest deterrent to our ability to sieve for maximal rings of integers. In this paper, we show that in some such cases, using the Ekedahl sieve only we can instead get a weighted lower bound of the appropriate order. We do this by carefully looking at what it means for a polynomial to be weakly divisible in terms of the ring associated to it. The concept of weakly divisible polynomials is defined in [3] and [2].

We define

Definition 2.

N^{*}(X:n):=\sum_{\mathbb{K}\in S(X,n)}\frac{1}{\sqrt{\mathfrak{disc}(\mathbb{K})}}.

Malle’s conjecture for $N(X:n)$ is clearly equivalent (via discrete integration by parts computation) to

N^{*}(X:n)\simeq 2c_{n}X^{\frac{1}{2}}.

In our previous paper we showed that, the number of “weakly divisible” rings of rank $n$ over $\mathbb{Z}$ with discriminant less than or equal to $X$ is

\gg_{n}X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}}.

one would expect that the number of number fields one gets sieving these rings would also be

\gg_{n}X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}}.

This is because the product of the local probabilities of a ring under consideration having a index co-prime to $p$ for any given prime $p$ is (in all the currently known parametrizations) $\prod(1-O(p^{-2})),$ which is bounded below by an absolute constant dependent on only $n.$

In this paper, we will show (using only the Ekedahl Sieve in its base form)

Theorem 2.

N^{*}(X:n)\gg X^{\frac{1}{n-\frac{4}{3}}-\epsilon}.

As a corollary, this implies

\limsup_{X\longrightarrow\infty}\frac{N(X:n)}{X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}-\epsilon}}=\infty.

If we get some good upper-bound on the size of the space of parametrization used (weakly divisible polynomials) or some sort of equi-distribution, it would follow that

N(X:n)\gg_{n}X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}-\epsilon}

improving the result of our previous paper.

In section 2 and section 3 we discuss a Fundamental Theorem of Orders, which describes Orders being described as an intersection of irreducible orders in a unique way. For this decomposition of an order, the conductor of the given order distributes multiplicatively along the conductors of the the irreducible orders in its decomposition. The index of an order in its ring of integers also distributes (multiplicatively) along the decomposition of the order into irreducible orders. The above theorem and understanding give us a slightly more nuanced way to understand conductor ideals and give a nice proof of Furtẅangler’s condition for a conductor ideal of order in a given number field.

We will apply the context of FTO to binary rings and as a consequence show a Dedekind-Kummer Type Theorem for Binary Rings and factorization of associated binary form. This was previously shown by [4], which we found in the references of [11], after writing this paper.

We discuss small prime splitting restrictions for Binary Rings, a restriction similar to the one the small prime splitting restrictions for Monogenic Rings.

We then show a structure theorem of all ‘large’ irreducible orders. We call these orders as Pseudo Maximal Orders. We will give a more general structure theorem for irreducible orders in a future paper.

We then define Sudo Maximal Orders as orders whose decomposition in the Fundamental Theorem of Orders consists of only‘large’ irreducible orders (i.e. Pseudo Maximal orders). We have a structure theorem for such orders. We bound the number of such orders in a given number field of bounded index. This quantity can be bound in terms of only the bound on the index and the degree of the number field. Notably, this bound is independent of the discriminant of the number field. This bound can be seen as an upper bound for all orders in a number field with squarefree conductors. This is evidence that indicates that the theorem shown in [6] about orders in number fields with degree less than or equal to five, is generalizable to higher degrees. In an upcoming paper, we will do this.

We use the Ekedahl sieve on the space of polynomials and get polynomials which we refer to as Ultra-Weakly divisible polynomials. As a consequence of the Dedekind-Kummer Theorem for Binary Rings and the Ultra-Weakly Divisible condition on Binary Rings associated to Ultra-Weakly Divisible polynomials it follows that the Binary Rings (and Weakly Divisible Rings) associated to Ultra-Weakly Divisible Polynomials are all Sudo Maximal Rings.

Combining with our bound on Sudo Maximal Orders in a number field we show the result in theorem 2.

2. Dedekind Domains in number-fields

Notation 3.

Let $\mathbb{K}$ denote a number-field, $\mathcal{O}_{\mathbb{K}}$ its ring of integers. Let $M(\mathbb{K})$ denote the set of non-Archimedean places of $\mathbb{K}$ i.e. the set of non-zero prime ideals in $\mathcal{O}_{\mathbb{K}}$ (or maximal ideals). Let $\mathbb{P}_{\mathbb{Z}}$ denote the set of prime numbers in integers.

For $R\subseteq\mathbb{K}$ be a ring (not necessarily of finite rank over $\mathbb{Z}$ ) such that $Frac(R)=\mathbb{K},$ let $\bar{R}$ denote its integral closure.

Lemma 1.

$\forall p\in\mathbb{P}_{\mathbb{Z}}:|\nicefrac{{R}}{{pR}}|\leq p^{[\mathbb{K}:\mathbb{Q}]}.$

Proof.

We see that $\nicefrac{{R}}{{pR}}$ as a $\nicefrac{{\mathbb{Z}}}{{p\mathbb{Z}}}=\mathbb{F}_{p}$ -module.
For $x\in R$ let $x^{*}$ denote its image under the canonical map from $R$ to $\nicefrac{{R}}{{pR}}.$ We follow our previous observation with another,

if $x_{1}^{*},x_{2}^{*},\cdots x_{k}^{*}$ are linearly independent over $\mathbb{F}_{p},$ then $x_{1},x_{2},\cdots,x_{n}$ are linearly independent over $\mathbb{Z}$ (or equivalently, over $\mathbb{Q}$ ).

This is is easy to see since if $x_{i}$ ’s satisfy a $\mathbb{Z}$ linear relationship we may re write the relationship with coefficients such that the gcd of all the coefficients is $1$ i.e. we may find $a_{i}$ such that $\sum_{i=1}^{k}a_{i}x_{i}=0$ and $(a_{1},a_{2},\cdots a_{k})=1$ which immediately allows us to see $\sum_{i=1}^{k}a_{i}^{*}x_{i}^{*}=0$ where not all $a_{i}^{*}=0$ as $p\nmid(a_{1},a_{2},\cdots a_{k})=1.$ ∎

Lemma 2.

$\forall p\in\mathbb{P}_{\mathbb{Z}}:|\nicefrac{{R}}{{p^{k}R}}|=|\nicefrac{{R}}{{pR}}|^{k}.$

Proof.

Multiplication by $p^{m}$ from $R\longrightarrow p^{m}R$ and $pR\longrightarrow p^{m+1}R$ are bijections and thus the canonical induced map, $\nicefrac{{R}}{{pR}}\longrightarrow\nicefrac{{p^{m}R}}{{p^{m+1}R}}$ is also a bijection, giving us $|\nicefrac{{p^{m}R}}{{p^{m+1}R}}|=|\frac{R}{pR}|.$ The lemma follows. ∎

It follows that $\nicefrac{{R}}{{nR}}$ is a finite module for all $n.$

If $x\in R$ then one can see that the numerator of $\mathfrak{Nm}(x)$ (say $t$ ) can be expressed as an integral polynomial of $x$ and thus $\nicefrac{{R}}{{xR}}\subseteq\nicefrac{{R}}{{tR}}$ which is a finite set. Implying that every ideal in $R$ is finitely generated and thus no-etherian. We also note that for every non zero prime ideal in $R,$ $\nicefrac{{R}}{{\rho}}$ is a finite integral domain and hence a field i.e. $\rho$ is a maximal ideal. Thus, $\bar{R}$ is Noetherian, integrally closed domain with Krull dimension equal to one. It follows that $\bar{R}$ is a Dedekind Domain.

We naturally have $\mathcal{O}_{\mathbb{K}}\subseteq\bar{R}$ for every such ring. (Since, integral closure of $\mathbb{Z}$ in $\mathbb{K}$ (or $\mathcal{O}_{\mathbb{K}}$ ) will sit in the integral closure of $R$ in $\mathbb{K}.$ )

Every non-zero prime ( $\rho$ ) in $\bar{R}$ thus naturally corresponds to a prime ( $\rho\cap\mathcal{O}_{\mathbb{K}}$ ) in $\mathcal{O}_{\mathbb{K}}$ .
Furthermore, since $\bar{R}$ is a Dedekind domain $\bar{R}_{\rho}$ ( $\bar{R}$ localized at $\rho$ ) is a DVR and hence, a prime $\rho$ in $\bar{R}$ will naturally induce a non-Archimedean metric on $\mathbb{K}$ . Ostrowski’s theorem tells us that $M(\mathbb{K})$ simultaneously denotes the set of all non zero primes in $\mathcal{O}_{\mathbb{K}}$ and all possible non-Archimedean valuations on $\mathbb{K}$ . If $v=\rho\cap(\mathcal{O}_{\mathbb{K}})\in M(\mathbb{K})$ , it follows that $\bar{R}_{\rho}=(\mathcal{O}_{\mathbb{K}})_{v}.$ This allows us to see non-zero prime ideals in $\bar{R}$ as a subset of $M(\mathbb{K}).$ On the other hand, given any nonempty subset ( $S$ ) of $M(\mathbb{K})$ one can obtain a Dedekind domain from this set by defining

Notation 4.

R_{S}:=\bigcap_{v\in S}(\mathcal{O}_{\mathbb{K}})_{v}.

We thus get the following correspondence,

Theorem 3.

\bigg{\{}\textit{Dedekind Domains in }\mathbb{K}\bigg{\}}\leftrightarrow\bigg{\{}\textit{ Non-Empty subsets of }M(\mathbb{K})\bigg{\}}

Notation 5.

When $R$ is a Dedekind Domain in $\mathbb{K}$ , we use the notation

M(R):=\{v\in M(\mathbb{K}):R\subseteq(\mathcal{O}_{\mathbb{K}})_{v}\}.

We can write the correspondence as

	$\displaystyle R\leftrightarrow M(R)$
	$\displaystyle R_{S}\leftrightarrow S.$

Remark 4.

We can let $g$ denote the class number of $\mathbb{K}$ and observe that $v^{g}$ can be seen as a principal ideal $(t_{v}).$ For $S\subseteq M(\mathbb{K})$ we may define

A(S):=\{\prod_{v\in S}^{*}t_{v}^{n_{v}}:n_{v}\in\mathbb{Z}_{\geq 0}\}

where $\prod^{*}$ denotes finitely supported product. Then, the above correspondence from set to domain can be written as

S\longrightarrow A(S)^{-1}\cdot\mathcal{O}_{\mathbb{K}}.

In particular, we notice that for any ring $R$ ,

\bar{R}=R\mathcal{O}_{\mathbb{K}}.

3. Orders and Separation Lemmata

Notation 6.

For $R\subseteq\mathbb{K}$ a ring with $Frac(R)=\mathbb{K}$ , we define $M(R)$ to be the set of non-zero prime ideals (same as maximal ideals in this context) of $R.$

When $R$ is integrally closed (and thus a Dedekind Domain) this matches up with the understanding that $M(R)$ is the set of valuations corresponding to non-zero prime ideals of $R$ from above.

We note that for every ring $R$ satisfying $Frac(R)=\mathbb{K},$ $M(\bar{R})$ can be seen as a cover of $M(R).$

To see this, we note that for $\rho,\rho^{\prime}\in M(R)$ we clearly have $\rho+\rho^{\prime}=R$ since they are distinct maximal ideals. Thus, valuations over $\rho$ and $\rho^{\prime}$ have to be disjoint.

Notation 7.

For $\rho\in R,$ we define

\bar{\rho}:=\{v\in M(\bar{R}):R_{\rho}\subseteq(\mathcal{O}_{\mathbb{K}})_{v}\}.

In other words, $\{\bar{\rho}:\rho\in M(R)\}$ is a partition of $M(\bar{R}).$

We also note that

R=\bigcap_{\rho\in R}R_{\rho}.

We also note that there will only exist finitely many $\rho\in M(R)$ such that $R_{\rho}\subsetneq(\mathcal{O}_{\mathbb{K}})_{v}.$ See, remark 5.

Notation 8.

We set

S(R):=\{\rho\in M(R):R_{\rho}\subsetneq(\mathcal{O}_{\mathbb{K}})_{v}\}.

For all $\rho\notin S(R),$ we have $|\bar{\rho}|=1.$

Definition 9.

We say an ring $R$ with $Frac(R)=\mathbb{K}$ is irreducible if $|S(R)|=1.$

In other words, $R$ is irreducible if and only if $R\neq\bar{R}$ and $\exists!\rho$ , a maximal ideal in $R$ such that $\forall\rho^{\prime}\neq\rho$ in $M(R),$ we have

R_{\rho^{\prime}}=(\mathcal{O}_{\mathbb{K}})_{v}

for some $v\in M(\bar{R}).$

We call $R$ an irreducible order, when $R$ is an order that is irreducible.

We thus have every ring $R$ with $Frac(R)=\mathbb{K},$ we have

R=\bigcap_{\rho\in S(R)}(R_{\rho}\cap\bar{R}).

(1)

Notation 10.

For an order $\mathcal{O},$ we let $c_{\mathcal{O}}$ denote the conductor of the order $\mathcal{O}.$ In other words,

c_{\mathcal{O}}:=(\mathcal{O}:\mathcal{O}_{\mathbb{K}})_{\mathbb{K}}=\{x\in\mathbb{K}:x\mathcal{O}_{\mathbb{K}}\subseteq\mathcal{O}\}

Remark 5.

We recall some properties of the conductor ideal of an order.

•

$c_{\mathcal{O}}$ is simultaneously an ideal of $\mathcal{O}$ and $\mathcal{O}_{\mathbb{K}}.$
•

$c_{\mathcal{O}}$ is the largest ideal of $\mathcal{O}_{\mathbb{K}}$ sitting inside $\mathcal{O}$ as a subset.
•

$\mathcal{O}_{\rho}$ is not a DVR (i.e. is not equal to $(\mathcal{O}_{\mathbb{K}})_{v}$ for some valuation $v\in M(\mathbb{K})$ ) if and only if $c_{\mathcal{O}}\subseteq\rho.$

Thus, the radical of $c_{\mathcal{O}}$ in $\mathbb{K}$ will be the exactly equal to the product of valuations in $\sqcup_{\rho\in S(\mathcal{O})}\bar{\rho}.$

Lemma 3 (Separation Lemma 1).

Let $\mathcal{O}$ is an order in $\mathcal{O}_{\mathbb{K}},$ and

\mathcal{O}=\bigcap_{\rho\in S(\mathcal{O})}(\mathcal{O}_{\rho}\cap\mathcal{O}_{\mathbb{K}})

from eq. 1.

Let $c_{\rho}$ denote the conductor of $\mathcal{O}_{\rho}\cap\mathcal{O}_{\mathbb{K}}.$ We have

•

[\mathcal{O}_{\mathbb{K}}:\mathcal{O}]=\prod_{\rho\in S(\mathcal{O})}[\mathcal{O}_{\mathbb{K}}:(\mathcal{O}_{\rho}\cap\mathcal{O}_{\mathbb{K}})].

•

For $\rho,\rho^{\prime}\in S(\mathcal{O})$

$c_{\rho}+c_{\rho^{\prime}}=\mathcal{O}_{\mathbb{K}}$

and

$c_{\mathcal{O}}=\prod_{\rho\in S(\mathcal{O})}c_{\rho}$

Proof.

Since, for distinct primes $\rho,\rho^{\prime}\in M(\mathcal{O})$ we have $\rho+\rho^{\prime}=\mathcal{O}.$ This implies that, for any primary ideals $\partial,\partial^{\prime}$ satisfying $\sqrt{\partial}=\rho$ and $\sqrt{\partial^{\prime}}=\rho^{\prime},$ we have $\partial+\partial^{\prime}=\mathcal{O}.$ In other words,

\nicefrac{{\mathcal{O}}}{{\partial\partial^{\prime}}}\simeq\nicefrac{{\mathcal{O}}}{{\partial}}\times\nicefrac{{\mathcal{O}}}{{\partial^{\prime}}}.

It follows that

\mathcal{O}\otimes\mathbb{Z}_{p}=\prod_{\rho\in S(\mathcal{O})}(\mathcal{O}_{\rho}\cap\mathcal{O}_{\mathbb{K}})\otimes\mathbb{Z}_{p}=\prod_{\rho\in S(\mathcal{O})}\mathcal{O}_{\rho}\otimes\mathbb{Z}_{p}

where $\mathcal{O}_{\rho}\otimes\mathbb{Z}_{p}\subseteq\prod_{v\in\bar{\rho}}(\mathcal{O}_{\mathbb{K}})_{v}.$

We can look at all of these as $\mathbb{Z}_{p}$ modules and compare the $\mathbb{Z}_{p}$ indexes. We get

v_{p}([\mathcal{O}_{\mathbb{K}}:\mathcal{O}])=\sum_{\rho\in S(\mathcal{O})}v_{p}([\mathcal{O}_{\mathbb{K}}:(\mathcal{O}_{\rho}\cap\mathcal{O}_{\mathbb{K}})]).

The result follows.

The other part follows naturally from Remark remark 5 as

\sqrt{c_{\rho}}=\prod_{v\in\bar{\rho}}v

and the fact that for any distinct primes $\rho,\rho^{\prime}\in M(R),$

\bar{\rho}\cap\bar{\rho}^{\prime}=\phi.

∎

Lemma 4 (Separation Lemma 2).

If $\mathcal{O}=\mathcal{O}_{1}\cap\mathcal{O}_{2}$ such that $c_{\mathcal{O}_{1}}+c_{\mathcal{O}_{2}}=\mathcal{O}_{\mathbb{K}},$ then $S(\mathcal{O}_{1})$ and $S(\mathcal{O}_{2})$ can be naturally identified with disjoint subsets of $S(\mathcal{O})$ such that

•

$S(\mathcal{O}_{1})\sqcup S(\mathcal{O}_{2})=S(\mathcal{O})$
•

$\rho\in S(\mathcal{O}_{1})\subseteq S(\mathcal{O})\iff\mathcal{O}_{\rho}=(\mathcal{O}_{1})_{\rho}$
•

$\rho\in S(\mathcal{O}_{2})\subseteq S(\mathcal{O})\iff\mathcal{O}_{\rho}=(\mathcal{O}_{2})_{\rho}.$

Proof.

Since, $c_{\mathcal{O}_{1}}+c_{\mathcal{O}_{2}}=\mathcal{O}_{\mathbb{K}}$ let $r_{1}+r_{2}=1$ with $r_{1}\in c_{\mathcal{O}_{1}}$ and $r_{2}\in c_{\mathcal{O}_{2}}.$

Fix a prime ideal $\rho\in S(\mathcal{O}_{1})$ (that is, $c_{\mathcal{O}_{1}}\subseteq\rho)$ . Let $\tau=\rho\cap\mathcal{O}.$

Clearly, $\tau$ is a prime ideal in $\mathcal{O}$ and $\mathcal{O}_{\tau}\subseteq(\mathcal{O}_{1})_{\rho}.$

If $s\in(\mathcal{O}_{1})_{\rho}$ then $s=q/t$ for some $q\in\mathcal{O}_{1}$ and $t\in\mathcal{O}_{1}\backslash\rho$ . We simply note that since $tr_{2}\in c_{\mathcal{O}_{2}}\subseteq\mathcal{O}_{2}$ and $tr_{2}=t-tr_{1}\in\mathcal{O}_{1}$ we get $tr_{2}\in\mathcal{O}.$ We further note that $tr_{2}\notin\tau$ as $tr_{2}\in\tau\subseteq\rho$ would imply $r_{2}\in\rho$ and since $r_{1}\in c_{\mathcal{O}_{1}}$ we get $1=r_{1}+r_{2}\in\rho.$ ) Similarly, $qr_{2}\in c_{\mathcal{O}_{2}}\subseteq\mathcal{O}_{2}$ and $qr_{2}=q-qr_{1}\in\mathcal{O}_{1}$ we get $qr_{2}\in\mathcal{O}.$

We get $s=q/t=(qr_{2})/(tr_{2})\in\mathcal{O}_{\tau}$ implying

\rho\in S(\mathcal{O}_{1})\Rightarrow(\mathcal{O}_{1})_{\rho}=\mathcal{O}_{\rho\cap\mathcal{O}}

If $\rho,\rho^{\prime}\in S(\mathcal{O}_{1})$ such that $\rho\cap\mathcal{O}=\rho^{\prime}\cap\mathcal{O},$ then $(\mathcal{O}_{1})_{\rho}=(\mathcal{O}_{2})_{\rho^{\prime}}$ and thus $\rho=\rho^{\prime}.$

Thus, the canonical map $S(\mathcal{O}_{1})\longrightarrow S(\mathcal{O})$ given by $\rho\longrightarrow\rho\cap\mathcal{O}$ is injective. Similarly we can identify $S(\mathcal{O}_{2})$ with a subset of $S(\mathcal{O}).$ To show that $S(\mathcal{O}_{1})$ and $S(\mathcal{O}_{2})$ are disjoint, we only need to note that if some $\rho\in S(\mathcal{O})$ lies in both $S(\mathcal{O}_{1})$ and $S(\mathcal{O}_{2}),$ we would immediately have $r_{1}\in c_{\mathcal{O}_{1}}\subseteq\rho$ and $r_{2}\in c_{\mathcal{O}_{2}}\subseteq\rho.$ Forcing $1=r_{1}+r_{2}\in\rho.$ Contradicting $\rho$ is a prime ideal. ∎

Remark 6.

Note that the above lemma tells us that irreducible orders mimic Euclid’s property defining prime numbers, that is

If $\mathcal{O}$ is an irreducible order, then $\mathcal{O}_{A},\mathcal{O}_{B}$ are orders with $c_{\mathcal{O}_{A}}+c_{\mathcal{O}_{B}}=\mathcal{O}_{\mathbb{K}}$ , then

\mathcal{O}_{A}\cap\mathcal{O}_{B}\subseteq\mathcal{O}\Rightarrow\mathcal{O}_{A}\subseteq\mathcal{O}\textbf{ or }\mathcal{O}_{B}\subseteq\mathcal{O}.

Theorem 7 (Fundamental Theorem of Orders).

Every order $\mathcal{O}$ can be written as an intersection of irreducible orders in a unique way such that the conductors of the irreducible orders are pairwise co-prime. Furthermore, we get that the index (and conductor) of $\mathcal{O}$ in $\mathcal{O}_{\mathbb{K}}$ will be the product of the indices (and conductors) of the irreducible orders in $\mathcal{O}_{\mathbb{K}}$ in the given decomposition.

Proof.

Clearly follows from lemma 3 and lemma 4. ∎

Notation 11.

For any valuation $v\in M(\mathbb{K})$ , we denote the ramification index and the inertial degree of $v$ by $e_{v}$ and $f_{v},$ respectively. That is, if $v\cap\mathbb{Z}=(p)$ for some prime in $\mathbb{Z},$ then $e_{v}$ and $f_{v}$ are defined by

p\mathcal{O}_{\mathbb{K}}=\prod_{p\in v}v^{e_{v}}

and

\mathfrak{Nm}_{\mathcal{O}_{\mathbb{K}}}(v)=p^{f_{v}}.

Definition 12.

For any order $\mathcal{O}$ and any non-zero prime ideal of $\rho\in M(\mathcal{O}),$ we define

ef(\rho):=\sum_{v\in\bar{\rho}}e_{v}\cdot f_{v}.

Definition 13.

If $\mathcal{O}$ is an irreducible order with $S(\mathcal{O})=\{\rho\},$ we define

ef(\mathcal{O}):=ef(\rho).

4. Dedekind-Kummer type Theorems and the Fundamental theorem of Orders.

We recall the traditional versions of Dedekind-Kummer and Dedekind Criterion theorems surrounding polynomials.

4.1. Dedekind-Kummer Theorem

Theorem 8 (Dedekind-Kummer Theorem).

Suppose $\alpha\in\overline{\mathbb{Z}}$ satisfies $\mathbb{Q}[\alpha]=\mathbb{K},$ and suppose that $f(x)$ is the minimal polynomial of $\alpha$ (will be monic) over $\mathbb{K}$ .

If $p\nmid[\mathcal{O}_{\mathbb{K}}:\mathbb{Z}[\alpha]]$ and

f(x)\equiv\prod_{i}(f_{i})^{e_{i}}\bmod p,

then

p\mathcal{O}_{\mathbb{K}}=\prod_{i}(\beta_{i})^{e_{i}}

where $\beta_{i}$ are prime ideals in $\mathcal{O}_{\mathbb{K}}$ sitting over $p,$ with the understanding that

\beta_{i}=p\mathcal{O}_{\mathbb{K}}+f_{i}(\alpha)\mathcal{O}_{\mathbb{K}}.

Notation 14.

Let $\overline{g}$ denote the image of $g\in\mathbb{Z}[x]$ in $\mathbb{F}_{p}[x].$

Dedekind’s Criterion (see Lemma 3.1 in [1]) for index adds

Theorem 9 (Dedekind’s Criterion).

We have

p|[\mathcal{O}_{\mathbb{K}}:\mathbb{Z}[\alpha]]\iff\exists i:1\leq i\leq k,\>\overline{f_{i}}|\overline{\frac{f-\prod_{i}(f_{i})^{e_{i}}}{p}}.

Equivalently, we can say

p|[\mathcal{O}_{\mathbb{K}}:\mathbb{Z}[\alpha]]\iff\exists g\in\mathbb{Z}[x]:f\in(p^{2},pg,g^{2})\subseteq\mathbb{Z}[x]

We will unpack the standard proof of Dedekind-Kummer theorem with the Dedekind’s Criterion by talking about the ring $\mathbb{Z}[\alpha]$ rather than $\mathcal{O}_{\mathbb{K}}$ as follows.

Theorem 10.

If $\mathcal{O}=\mathbb{Z}[\alpha]$ with $\alpha\in\overline{\mathbb{Z}}$ and if $f(x)$ the minimal polynomial of $f$ shows the following decomposition modulo $p,$

f(x)\equiv\prod_{i}(f_{i})^{e_{i}}\bmod p,

then, we get

(\mathbb{Z}\backslash p\mathbb{Z})^{-1}\mathcal{O}=\bigcap_{i}\mathcal{O}_{\rho_{i}}

and

p\mathcal{O}=\prod_{i}\beta_{i}

where $\beta_{i}$ are primary ideals in $\mathcal{O}$ sitting over $p$ such that

\sqrt{\beta_{i}}=\rho_{i}

with the understanding that

\rho_{i}=p\mathcal{O}+f_{i}(\alpha)\mathcal{O}.

are all distinct prime ideals and all prime ideals in $\mathcal{O}$ sitting over $p$ and

\sum_{i}ef(\rho_{i})=[\mathbb{K}:\mathbb{Q}].

Remark 11.

The theorem 10 says that, “The factorization of a monic polynomial modulo $p$ informs us about the local ring decomposition of $\mathbb{Z}[\alpha]$ over $p$ ” and the theorem 9 is about when a particular factor (corresponding to a particular prime ideal) does or does not correspond to a DVR.

4.2. Dedekind-Kummer-Dedekind System for Binary Rings and Primitive Binary Forms

Theorem 12.

If $\mathcal{O}=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}]$ with $\delta\in\overline{\mathbb{Q}}$ and if $f(x,y)$ the minimal primitive binary form such that $f(\delta,1)=0$ and shows the following decomposition/factorization modulo $p,$

f(x,y)\equiv\prod_{i}(f_{i})^{e_{i}}\bmod p,

then, we get

(\mathbb{Z}\backslash p\mathbb{Z})^{-1}\mathcal{O}=\bigcap_{i}\mathcal{O}_{\rho_{i}}

and

p\mathcal{O}=\prod_{i}\beta_{i}

where $\beta_{i}$ are primary ideals in $\mathcal{O}$ sitting over $p$ such that

\sqrt{\beta_{i}}=\rho_{i}

with the understanding that

•

if $f_{i}\neq Y,$ then

$\rho_{i}=(p\mathbb{Z}[\delta]+f_{i}(\delta,1)\mathbb{Z}[\delta])\cap\mathbb{Z}[\delta^{-1}]$
•

and if $f_{i}\neq X,$ then

$\rho_{i}=(p\mathbb{Z}[\delta^{-1}]+f_{i}(1,\delta^{-1})\mathbb{Z}[\delta^{-1}])\cap\mathbb{Z}[\delta]$

such that $\rho_{i}$ are all distinct prime ideals and all prime ideals in $\mathcal{O}$ sitting over $p$ and

\sum_{i}ef(\rho_{i})=[\mathbb{K}:\mathbb{Q}].

Theorem 13 (Dedekind’s Criterion Binary Rings).

We have

p|[\mathcal{O}_{\mathbb{K}}:(\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}])]\iff\exists i:1\leq i\leq k,\>\overline{f_{i}}|\overline{\frac{f-\prod_{i}(f_{i})^{e_{i}}}{p}}.

Equivalently, we can say

p|[\mathcal{O}_{\mathbb{K}}:(\mathbb{Z}[\delta]\cap\mathbb{Z}[\alpha^{-1}])]\iff\exists g\in\mathbb{Z}[x,y]_{m}:f\in(p^{2},pg,g^{2})\subseteq\mathbb{Z}[x,y]_{n}

Before jumping into the proof we will review some properties of binary rings associated to primitive forms.

4.3. Localizations of Binary Rings associated to primitive forms

Let’s review Binary rings: All of the following results can be found in [11].

Let

f(X,Y):=a_{n}X^{n}+a_{n-1}X^{n-1}Y+\cdots+a_{0}Y^{n}

denote a binary form of degree $n$ . Let $a_{n}\neq 0$ and suppose that $\delta$ denotes the image of $X$ in the algebra $\mathbb{Q}[X]/f(X,1).$

Definition 15.

When $a_{n}\neq 0,$

R_{f}:=\mathbb{Z}\langle 1,\>a_{n}\delta,\>a_{n}\delta^{2}+a_{n-1}\delta,\>\cdots,\>\sum_{i=0}^{k-1}a_{n-i}\delta^{k-i},\>\cdots,\>\sum_{i=0}^{n-2}a_{n-i}\delta^{n-1-i}\rangle.

(2)

We set

$\displaystyle\langle B_{0},B_{1},\cdots,B_{n-1}\rangle:=$	$\displaystyle\langle 1,\>a_{n}\delta,\>a_{n}\delta^{2}+a_{n-1}\delta,\>\cdots,\>\sum_{i=0}^{k-1}a_{n-i}\delta^{k-i},\>\cdots,\>\sum_{i=0}^{n-2}a_{n-i}\delta^{n-1-i}\rangle$	(3)
that is,
$\displaystyle B_{0}:=$	$\displaystyle 1$	(4)
$\displaystyle B_{1}:=$	$\displaystyle a_{n}\delta$	(5)
	$\displaystyle\cdots$
	$\displaystyle\cdots$
	$\displaystyle\cdots$
$\displaystyle B_{k}:=$	$\displaystyle a_{n}\delta^{k}+a_{n-1}\delta^{k-1}+\cdots+a_{n-k+1}\delta$	(6)
	$\displaystyle\cdots$
$\displaystyle B_{n-1}:=$	$\displaystyle a_{n}\delta^{n-1}+a_{n-1}\delta^{n-2}+\cdots+a_{2}\delta$	(7)

Definition 16.

$I_{f}$ denotes the (fractional) ideal class generated by $(1,\delta)$ in $R_{f}$

Theorem 14.

$R_{f}$ is a ring of rank $n.$

Theorem 15.

Properties of $R_{f}$ :

(1)

$\mathfrak{disc}_{\mathbb{Z}}(R_{f})=\mathfrak{disc}(f)$ (8)
(2)

If $\delta$ is invertible, and $f$ is primitive, then

$R_{f}=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}].$ (9)
(3)

If $f$ is primitive, $I_{f}$ is invertible in $R_{f}.$
(4)

$R_{f}$ and $I_{f}$ invariant a under the natural $GL_{2}(\mathbb{Z})$ action on binary forms of degree $n.$

(5)

We also have

R_{f}+R_{f}\delta=\mathbb{Z}\langle 1,\delta,B_{2},\cdots,B_{n-1}\rangle

(10)

and

R_{f}\cap R_{f}\delta^{-1}=\mathbb{Z}\langle a_{n},B_{1},B_{2},\cdots,B_{n-1}\rangle

(11)

and

(R_{f}+R_{f}\delta)\cdot(R_{f}\cap R_{f}\delta^{-1})=\mathbb{Z}\langle gcd(a_{n},a_{n-1},\cdots,a_{0}),B_{1}-a_{1},B_{2}-a_{2},\cdots,B_{n-1}-a_{n-1}\rangle.

(12)

Remark 16.

We make note of two facts. One is that the product (of modules) in eq. 12 is in fact equal to $R_{f}$ when $gcd(a_{n},a_{n-1},\cdots,a_{0})=1$ (that is, $f$ is primitive). The second is

\nicefrac{{R_{f}}}{{R_{f}\cap R_{f}\delta^{-1}}}\simeq\nicefrac{{\mathbb{Z}}}{{a_{n}\mathbb{Z}}}.

(13)

This implies that the prime ideals in $R_{f}$ containing $R_{f}\cap R_{f}\delta^{-1}$ are of the form $pR_{f}+(R_{f}\cap R_{f}\delta^{-1})$ where $p\mid a_{n}.$ We can conclude that for each prime number dividing $a_{n}$ there exists exactly one prime in $R_{f}$ containing $p$ and $R_{f}\cap R_{f}\delta^{-1}$ , and for $p\nmid a_{n}$ there is no prime ideal in $R_{f}$ containing $p.$ Furthermore, when $p\mid a_{n}$ the prime ideal in $R_{f}$ containing $p$ and $R_{f}\cap R_{f}\delta^{-1}$ clearly has norm $p.$

Notation 17.

Let $\mathcal{O}=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}].$ Let $f(x,y)$ denote the minimal primitive binary form satisfying $f(\delta,1)=0.$

Thus, $R_{f}\simeq\mathcal{O}.$

Lemma 5.

(\mathcal{O}\cap\mathcal{O}\delta^{-1})\mathbb{Z}[\delta]=\mathbb{Z}[\delta]

Proof.

We know that from item 5,

\mathcal{O}\cap\mathcal{O}\delta^{-1}=\mathbb{Z}\langle a_{n},B_{1},B_{2},\cdots,B_{n-1}\rangle.

Thus, $(\mathcal{O}\cap\mathcal{O}\delta^{-1})\mathbb{Z}[\delta]$ is spanned by the set organized in the table,

$a_{n}$	$B_{1}$	$B_{2}$	$\cdots$	$B_{n-1}$
$a_{n}\delta$	$B_{1}\delta$	$B_{2}\delta$	$\cdots$	$B_{n-1}\delta$
$a_{n}\delta^{2}$	$B_{1}\delta^{2}$	$B_{2}\delta^{2}$	$\cdots$	$B_{n-1}\delta^{2}$
$a_{n}\delta^{3}$	$B_{1}\delta^{3}$	$B_{2}\delta^{3}$	$\cdots$	$B_{n-1}\delta^{3}$
$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$

We note that $B_{i+1}-B_{i}\delta=a_{n-i}$ is in the $\mathbb{Z}$ -span of the above set for all $n-1\geq i\geq 1$ . We can also see that $a_{0}=-B_{n-1}\delta$ is also in the span. Thus,

1=\gcd(a_{0},a_{1},a_{2},\cdots,a_{n})\in(\mathcal{O}\cap\mathcal{O}\delta^{-1})\mathbb{Z}[\delta].

Result follows. ∎

We note that if $\rho\in M(\mathcal{O})$ (prime ideal in $\mathcal{O}$ ) such that

(\mathcal{O}\cap\mathcal{O}\cdot\delta^{-1})\not\subseteq\rho

i.e. the denominator (proper) ideal of $\delta$ in $\mathcal{O}$ is not contained in $\rho,$ then $\exists t\in\mathcal{O}\cap\mathcal{O}\cdot\delta^{-1}$ such that $t^{-1}\in\mathcal{O}_{\rho}.$ Since, $t\delta\in\mathcal{O}$ it follows that $\mathbb{Z}[\delta]\subseteq\mathcal{O}_{\rho}.$ For such primes $\rho$ (primes not containing $\mathcal{O}\cap\mathcal{O}\cdot\delta^{-1}$ ), we note that $\nicefrac{{\mathbb{Z}[\delta]}}{{\rho\mathbb{Z}[\delta]}}$ contains $\nicefrac{{\mathcal{O}}}{{\rho}}$ and is contained in $\nicefrac{{\mathcal{O}_{\rho}}}{{\rho\mathcal{O}_{\rho}}}\simeq\nicefrac{{\mathcal{O}}}{{\rho}}$ and hence must be equal to $\nicefrac{{\mathcal{O}}}{{\rho}}$ forcing $\rho\mathbb{Z}[\delta]$ to be a prime ideal in $\mathbb{Z}[\delta]$ . This identifies $\rho\in M(\mathcal{O})$ with a prime in $\mathbb{Z}[\delta]$ in a localization preserving way. That is to say,

\mathbb{Z}[\delta]_{\rho\mathbb{Z}[\rho]}=\mathcal{O}_{\rho}.

Combining with lemma 5, we argue that the only prime ideals in $\mathcal{O}$ which will not be mapped to any prime ideal in $\mathbb{Z}[\delta]$ will be exactly those prime ideals containing $\mathcal{O}\cap\mathcal{O}\cdot\delta^{-1}$

So, if we look at the map $M(\mathbb{Z}[\delta])\longrightarrow M(\mathcal{O})$ given by $\rho\longrightarrow\rho\cap\mathcal{O}$ , then this map is clearly the inverse of the localization preserving map above.

4.4. Proof of Dedekind Kummer in Binary/One-fine Rings or theorem 12.

This is easily observed using the previous understanding of prime ideals in $\mathcal{O}=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}]$ and their relationship to prime ideals in $\mathbb{Z}[\delta]$ and the identification

\mathbb{Z}[\delta]\simeq\nicefrac{{\mathbb{Z}[x]}}{{(f(x,1))}}.

Suppose, $f(x,y)$ has the following decomposition/factorization modulo $p.$

f(x,y)\equiv y^{k}\prod_{i}(f_{i})^{e_{i}}\bmod p,

where $f_{i}$ are irreducible binary forms $\bmod p$ such that $f_{i}(x,1)$ has the same degree in $X$ as the binary form $f_{i}$ in $(x,y),$ (note that this will not happen if and only if $f_{i}=y$ ). We set

\rho_{i}=(p\mathbb{Z}[\delta]+f_{i}(\delta,1)^{e_{i}}\mathbb{Z}[\delta]).

It follows that

\nicefrac{{\mathbb{Z}[\delta]}}{{\rho_{i}}}\simeq\nicefrac{{\mathbb{Z}[X]}}{{(p,f_{i}(x,1)^{e_{i}},f(x,1)}}\simeq\nicefrac{{\mathbb{Z}[X]}}{{(p,f_{i}(x,1))}}\simeq\nicefrac{{\mathbb{F}_{p}[x]}}{{(f_{i}(x)^{e_{i}})}}.

Thus, $\rho_{i}$ are primary ideals in $\mathbb{Z}[\delta]$ with

\sqrt{\rho_{i}}=(p,f_{i}(\delta)).

On the other hand, it is easy to see that

	$\displaystyle(p,f(x,1))=\prod_{i}(p,f_{i}(x,1)^{e_{i}})$
	$\displaystyle\textit{where }i\neq j\implies(p,f_{i}(x,1)^{e_{i}})+(p,f_{j}(x,1)^{e_{j}})=\mathbb{Z}[x],$

since $f_{i}(x,1)\bmod p$ and $f_{j}(x,1)\bmod p$ are irreducible distinct polynomials in $\mathbb{F}_{p}[x].$ Thus,

\nicefrac{{\mathbb{Z}[\delta]}}{{(p)}}\simeq\prod_{i}\nicefrac{{\mathbb{F}_{p}[x]}}{{(f_{i}(x,1)^{e_{i}})}}.

We can conclude that $\sqrt{\rho_{i}}$ are all distinct prime ideals and all prime ideals in $\mathbb{Z}[\delta]$ containing $p$ . This concludes the proof of the Dedekind Kummer theorem for Binary rings associated to primitive forms.

4.5. Proof of Dedekind’s Criterion for Primitive binary forms or theorem 13

The proof is very much the same as we can look at $\mathbb{Z}[\delta]$ and $\mathbb{Z}[\delta^{-1}]$ separately.

Suppose that $f(x,y)$ is the minimal primitive binary form satisfying $f(\delta,1)=0.$

Notation 18.

Given a polynomial $h(x)\in\mathbb{Z}[x]$ we let its image in $\mathbb{F}_{p}[x]$ be denoted by $\bar{h}(x).$

Thus, if $f(x,y)$ shows the following decomposition/factorization modulo $p,$

f(x,y)\equiv y^{k}\prod_{i}(f_{i})^{e_{i}}\bmod p,

where $f_{i}$ are irreducible binary forms $\bmod p$ such that $f_{i}(x,1)$ has the same degree in $x$ as the binary form $f_{i}$ in $(x,y),$ (we note that this will not happen if and only if $f_{i}=y$ ) then we may set

\beta_{i}=(p\mathbb{Z}[\delta]+f_{i}(\delta,1)\mathbb{Z}[\delta])

as the prime ideal associated to $f_{i}.$

It follows that

\beta_{i}^{2}=(p^{2}\mathbb{Z}[\delta]+pf_{i}(\delta,1)\mathbb{Z}[\delta]+f_{i}(\delta,1)^{2}\mathbb{Z}[\delta])

Based on the structure of the prime ideal, we conclude a random element

t=m(\delta)\in\beta_{i}\textit{ if and only if }\bar{f_{i}}(x,1)|\bar{m}(x).

(14)

We will consider two cases:

$f_{i}(\delta,1)\in\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}$ and $f_{i}(\delta,1)\not\in\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}.$

•

Case: $f_{i}(\delta,1)\in\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}.$

	$\displaystyle f_{i}(\delta,1)\in\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}\implies$	$\displaystyle f_{i}(\delta,1)=p^{2}A+pf_{i}(\delta,1)B+f_{i}(\delta,1)^{2}C$
	$\displaystyle\iff$	$\displaystyle f_{i}(\delta,1)(1-pB-f_{i}(\delta,1)C)=p^{2}A$

Since, $1-pB-f_{i}(\delta,1)C\notin\beta_{i},$ it is a unit this means that $p^{2}R(\delta)=f_{i}(\delta,1)S(\delta)$ where $S(\delta)$ is a unit in $\mathbb{Z}[\delta]_{\beta_{i}}.$

In this case, $\mathbb{Z}[\delta]_{\beta_{i}}$ is a DVR where the uniformizer can be taken to be $p$ as it generates the maximal ideal in $\mathbb{Z}[\delta]_{\beta_{i}}.$ Furthermore, since the uniformizer of $\mathbb{Z}[\delta]_{\beta_{i}}\otimes\mathbb{Z}_{p}$ is the same as the uniformizer of $\mathbb{Z}_{p}$ , $\mathbb{Z}[\delta]_{\beta_{i}}\otimes\mathbb{Z}_{p}$ is an un-ramified extension of $\mathbb{Z}_{p}.$ This forces $e_{i}=1,$ which implies $f(x,y)\notin(p^{2},pf_{i},(f_{i})^{2}).$

•

Case: $f_{i}(\delta,1)\notin\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}.$

In this case, $f_{i}(\delta,1)\in\beta_{i}\mathbb{Z}[\delta]_{\beta_{i}}\backslash\beta_{i}^{2}\mathbb{Z}[\delta]_{\beta_{i}}.$ Thus, if $\mathbb{Z}[\delta]_{\beta_{i}}$ is a DVR then $f_{i}(\delta,1)$ may be chosen as a uniformizer.

We see $\mathbb{Z}[\delta_{i}]\otimes\mathbb{Z}_{p}$ as the ring of integers of a totally ramified extension of a totally un-ramified extension of $\mathbb{Z}_{p}.$ Since, $\nicefrac{{\mathbb{Z}[\delta]}}{{\beta_{i}}}$ is a $deg(f_{i})$ extension of $\mathbb{F}_{p}.$ Thus, the ramification degree of $\mathbb{Z}[\delta_{i}]\otimes\mathbb{Z}_{p}$ is $deg(f_{i})$ and the ramification index is $e_{i}$ . This will imply that

$f_{i}(\delta,1)^{e_{i}}=pu$

for some unit $u$ .

Now substituting $(\delta,1)$ in $f(x,y)=y^{k}\prod_{j}f_{j}(x,y)^{e_{i}}+ph(x,y)$ we get

$\prod_{j}f_{j}(\delta,1)^{e_{j}}=-ph(\delta,1)$

Since for $j\neq i,$ $f_{j}(\delta,1)$ are all units in $\mathbb{Z}[\delta]_{\beta_{i}},$ as they, by definition, are distinct irreducible polynomials over $\mathbb{F}_{p}.$

This means that $h(\delta,1)$ has to be a unit in $\mathbb{Z}[\delta]_{\beta_{i}}$ i.e. $\bar{f}_{i}(x,1)\nmid\bar{h}(x,1)$ (using eq. 14) which is equivalent to $h(x,y)\notin(p,f_{i}(x,y))\iff f(x,y)\notin(p^{2},pf_{i},f_{i}^{2}).$

To show the other direction, we note that, if $e_{i}=1$ then clearly $f(x,y)\notin(p^{2},pf_{i},f_{i}^{2})$ . At the same time,

\prod_{j}f_{j}(\delta,1)^{e_{j}}=-ph(\delta,1)

and for $j\neq i,$ $f_{j}(\delta,1)$ are all units in $\mathbb{Z}[\delta]_{\beta_{i}}$ which means that $\beta_{i}\mathbb{Z}[\delta]_{\beta_{i}}=p\mathbb{Z}[\delta]_{\beta_{i}}$ making it a principal ideal and thus $\mathbb{Z}[\delta]_{\beta_{i}}$ a DVR.

If $e_{i}>1$ and $f\notin(p^{2},pf_{i},f_{i}^{2})$ then $h(x,y)\notin(p,f_{i}(x,y)).$ Subbing in $\delta$ we see that this means that $h(\delta,1)$ is a unit in $\mathbb{Z}[\delta]_{\beta_{i}}.$ At the same time,

\prod_{j}f_{j}(\delta,1)^{e_{j}}=-ph(\delta,1)

and for $j\neq i,$ $f_{j}(\delta,1)$ are all units in $\mathbb{Z}[\delta]_{\beta_{i}}$ which means that $\beta_{i}\mathbb{Z}[\delta]_{\beta_{i}}=f_{i}(\delta,1)\mathbb{Z}[\delta]_{\beta_{i}}$ making it a principal ideal and thus $\mathbb{Z}[\delta]_{\beta_{i}}$ a DVR. This completes the proof.

5. Small prime splitting restrictions.

Just as a small prime splitting poses an obstruction for a ring to be monogenic, a small prime splitting poses a restriction for a ring to be binary. In this section we will quatify this restriction.

Let $\mathbb{K}$ denote a finite extension of $\mathbb{Q}$ . Let $M_{\mathbb{K}}$ denote the set of non-archimedean places of $K.$ We fix a prime $p$ in $\mathbb{Z}$ for this section. Let $\mathcal{O}$ denote an order in $\mathcal{O}_{\mathbb{K}}$ Let $p\mathcal{O}=\delta_{1}\cdot\delta_{2}\cdots\delta_{r}$ denote the unique m-primary ideal factorization of the ideal $p\mathcal{O}$ in $\mathcal{O}$ where $\delta_{i}+\delta_{j}=\mathcal{O}.$ Let $\tau_{i}$ denote the radical of the ideal $\delta_{i}$ in $R.$ These will be all the maximal ideals of $\mathcal{O}$ containing $p$ . For any non zero proper ideal $I$ of $\mathcal{O}$ we define

Norm_{\mathcal{O}}(I)=|\nicefrac{{\mathcal{O}}}{{I}}|.

And let

T_{\mathcal{O}}(p,f):=\{\tau_{i}:Norm(\tau_{i})=p^{f}\}\qquad H(p,f):=\frac{1}{f}\sum_{d|f}p^{d}\mu(\frac{f}{d})

Remark 17.

$H(p,f)$ is the number of prime ideals in $\mathbb{F}_{p}[x]$ with the norm $p^{f}.$

Theorem 18.

If $\mathcal{O}$ is binary ring, then

	$\displaystyle H(p,1)+1\geq\|T_{\mathcal{O}}(p,f)\|$
	$\displaystyle f\geq 2\implies H(p,f)\geq\|T_{\mathcal{O}}(p,f)\|.$

Proof.

Let $\mathcal{O}=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}]$ . From Dedekind Kummer Theorem for Binary forms (theorem 12), we have a neat (norm preserving) correspondence from all prime ideals of $\mathbb{Z}[\delta]$ containing $p$ to either all or all but one prime ideal(that of norm $p$ ) of $\mathcal{O}$ . We identify $\mathbb{Z}[\delta]$ with $\nicefrac{{\mathbb{Z}[X]}}{{g(x)}},$ where $g(x)=f(x,1).$ Let $A=|T_{\mathcal{O}}(p,f)|.$

Consider the following diagram of canonical quotient maps,

Clearly each arrow is surjective. Consider the lower row of maps culminating in the following surjective ring homo-morphism

\mathbb{F}_{p}[x]\longrightarrow(\mathbb{F}_{p^{f}})^{|A|}

In particular, there must exist at least $|A|$ distinct prime ideals in $\mathbb{F}_{p}[x]$ with norm $p^{f}.$ This corresponds to surjective maps from $\mathbb{F}_{p}[x]$ to $\mathbb{F}_{p^{f}}$ up-to $Gal(\mathbb{F}_{p^{f}}/\mathbb{F}_{p})$ action. In other words, the number of degree $f$ elements in $\mathbb{F}_{p^{f}}$ up-to $Gal(\mathbb{F}_{p^{f}}/\mathbb{F}_{p})$ action. This is $H(p,f).$ Thus, $H(p,f)\geq|A|$ . Combining with Dedekind Kummer Theorem for Binary rings, we see that,

	$\displaystyle f\geq 2\implies$	$\displaystyle H(p,f)\geq\|T_{\mathcal{O}}(p,f)\|$
		$\displaystyle H(p,1)+1\geq\|T_{\mathcal{O}}(p,f)\|.$

∎

Remark 19.

\mathbb{P}(p,f):=\#\{\textit{ prime ideals in }\mathbb{P}(\overline{\mathbb{F}}_{p})\textit{ with norm }p^{f}\},

then,

\mathbb{P}(p,1)=H(p,1)+1

and

f\geq 2\implies\mathbb{P}(p,f)=H(p,f).

Remark 20.

Binary rings are best viewed as a subset of orders which are locally monogenic, more specifically there are locally monogenic rings which are almost globally monogenic. Monogenic rings are globally monogenic (simultaneously across all primes). In case of $n=2$ the concept of freely locally monogenic is equivalent to the version of almost globally monogenic used here. We compare this to, “Every ideal class in a rank two ring occurs in binary quadratic forms.” When $n=3$ , this concept of locally monogenic is the same as almost globally monogenic used here, but not freely locally monogenic. We compare this to, “Every ring of integers occurs as a binary ring not every ideal class in ring of integers occurs as a binary ring and associated invertible ideal.” We generalize the concept in our thesis and in upcoming papers and relate it to other classification of ring theorems.

Remark 21.

We compare this condition with the small prime splitting condition in monogenic rings, which is

f\geq 1\implies H(p,f)\geq|T_{\mathcal{O}}(p,f)|.

In-spite of a very minimal weakening on this condition we note that the rings captured as monogenic rings are expected to be proportion $0$ in binary rings (almost globally monogenic rings).

6. Pseudo Maximal Orders and Sudo Maximal orders.

Remark 22.

In general, if we have an irreducible order $\mathcal{O}$ with $S(\mathcal{O})=\{\rho\}$ such that $\overline{\rho}=\{v_{1},v_{2},\cdots,v_{k}\}$ and if we let the conductor of $\mathcal{O}$ to be $c_{\mathcal{O}},$ an ideal in $\mathcal{O}_{\mathbb{K}},$ then we know that

c_{\mathcal{O}}=(v_{1})^{a_{v_{1}}}(v_{2})^{a_{v_{2}}}\cdots(v_{k})^{a_{v_{k}}}

where $a_{i}\geq 1.$

Furthermore, the since $c_{\mathcal{O}}\subseteq\mathcal{O}\subseteq\mathcal{O}_{\mathbb{K}}$ we see that $\mathcal{O}$ is completely determined by $\nicefrac{{\mathcal{O}}}{{c_{\mathcal{O}}}}$ a sub-ring of $\nicefrac{{\mathcal{O}_{\mathbb{K}}}}{{c_{\mathcal{O}}}}.$ Hence, we can search for sub-rings of the appropriate complete local finite extension of $p$ -adic integers.

From here it is easy to get the Furtwrangler’s condition for a conductor ideals by just looking at the ideal $(p^{m},c_{\mathcal{O}})$ in a reduced order where $\mathcal{O}$ where $p^{m+1}\in c_{\mathcal{O}}$ and $p^{m}\notin c_{\mathcal{O}}$ and observing when it will be an ideal $\mathcal{O}_{\mathbb{K}}$ . This ideal can be an ideal of $\mathcal{O}_{\mathbb{K}}$ if and only if $c_{\mathcal{O}}$ cannot be a conductor ideal. In fact, combining with Fundamental Theorem of Orders or theorem 7 it gives a more nuanced answer about orders and associated conductors and what can exist. For example, it tells us that there cannot exist an order $\mathcal{O}$ with conductor $vw$ with $f_{v}=e_{v}=f_{w}=e_{w}=1$ and $\nicefrac{{\mathcal{O}}}{{vw}}\simeq\mathbb{Z}_{p}^{2}.$ This one is obvious, by things we already know.

6.1. Classification of Irreducible orders $\mathcal{O}$ with $ef(\mathcal{O})=1$ and $ef(\mathcal{O})=2$

6.1.1. $ef(\mathcal{O})=1$

There exist no irreducible order $\mathcal{O}$ in $\mathcal{O}_{K}$ with $ef(\mathcal{O})=1$ as this would be akin to finding proper unital sub-rings of the p-adic integers $\mathbb{Z}_{p}$ .

6.1.2. $ef(\mathcal{O})=2$

Lemma 6.

If $A=\mathbb{Z}_{p}+\mathbb{Z}_{p}\cdot t$ is a algebra of rank $2$ over $\mathbb{Z}_{p}$ then all sub-rings of $A$ which have rank $2$ over $\mathbb{Z}_{p}$ and containing $\mathbb{Z}_{p}$ are given by $\mathbb{Z}_{p}+p^{k}(A)$ for some $k\in\mathbb{Z}^{\geq 0}.$ The index of $\mathbb{Z}_{p}+p^{k}(A)$ in $A$ is $p^{k}.$

Proof.

If $M$ is a sub-ring of $A$ which is rank $2$ over $\mathbb{Z}_{p}$ may be seen as $\mathbb{Z}_{p}+\mathbb{Z}_{p}r$ where $r=a+bt$ with $a,b\in\mathbb{Z}_{p}$ . We further write $b=p^{k}u$ where $u\in\mathbb{Z}_{p}$ is a unit, i.e. $u\mathbb{Z}_{p}=\mathbb{Z}_{p}$ Clearly,

\mathbb{Z}_{p}+\mathbb{Z}_{p}r=\mathbb{Z}_{p}+\mathbb{Z}_{p}bt=\mathbb{Z}_{p}+\mathbb{Z}_{p}+\mathbb{Z}_{p}up^{k}t=\mathbb{Z}_{p}+\mathbb{Z}_{p}(p^{k}t)=\mathbb{Z}_{p}+p^{k}A.

∎

Remark 23.

There is no irreducible(non $\mathcal{O}_{\mathbb{K}}$ ) order $\mathcal{O}$ with $ef(\mathcal{O})=1.$ Consequence of the Furtwangler condition for an ideal to be a conductor ideal. We give a structure theorem for irreducible orders with $ef(\mathcal{O})=2$ . Similar Structure theorems can be given easily using corresponding ring classification theories for irreducible orders with $ef(\mathcal{O})\leq 4.$ With some difficulty, the authors also expect such theorems possible for $ef(\mathcal{O})=5$ by extending Bhargava’s work for quintic rings to $\mathbb{Z}_{p}.$

Definition 19.

A Pseudo Maximal Order in $\mathbb{K}$ is an irreducible order $\mathcal{O}$ satisfying $ef(\mathcal{O})=2.$ See definition 13 for the definition of $ef.$

If $\mathcal{O}$ is an irreducible order with $S(\mathcal{O})=\{\rho\}$ , then its conductor is made up of prime ideals (in $\mathbb{K}$ ) in $\overline{\rho}$ . Thus, when

Lemma 7.

$\mathcal{O}$ is a Pseudo Maximal Order, then one of the following must be true:

	$\displaystyle\overline{\rho}=$	$\displaystyle\{v,w\}\textit{ with }f_{v}=f_{w}=e_{v}=e_{w}=1$
	$\displaystyle=$	$\displaystyle\{v\}\textit{ with }f_{v}=2=2e_{v}$
	$\displaystyle=$	$\displaystyle\{w\}\textit{ with }2f_{v}=2=e_{v}.$

Theorem 24 (A).

If $\{v,w\}\subseteq M(\mathbb{K})\textit{ with }f_{v}=f_{w}=e_{v}=e_{w}=1,$ then the irreducible orders with conductor $v^{a}w^{b}$ are given by

•

if $a=b=r$ then there is a unique irreducible order $\mathcal{O}$ with conductor $(vw)^{r}$ given by $\mathbb{Z}+(vw)^{r}.$ This order will have index in $\mathcal{O}_{\mathbb{K}}$ given by $[\mathcal{O}_{\mathbb{K}}:\mathbb{Z}+(vw)^{r}]=p^{r}.$
•

if $a\neq b$ then there is no irreducible (or otherwise) order with conductor $v^{a}w^{b}.$

Proof.

Clearly, we are looking for irreducible orders in $\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}$ which is treated an extension of $\mathbb{Z}_{p}$ via diagonal embedding. So, we may write it as $\mathbb{Z}_{p}+\mathbb{Z}_{p}t$ where $t=(0,1)$ . Thus every order here must be of the form $\mathbb{Z}_{p}+p^{k}(\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}).$ Coordinate wise $p=(p,p)$ due to diagonal embedding. Thus, the sub-ring is given by $\mathbb{Z}_{p}+((p,1)^{k}\cdot(1,p)^{k})(\mathbb{Z}_{p}\oplus\mathbb{Z}_{p})$ . Intersecting with $\mathcal{O}_{\mathbb{K}}$ we get the order is given by $\mathbb{Z}+(vw)^{k}$ whose conductor is easily seen as $(vw)^{k}.$ ∎

Theorem 25 (B).

If $\{v\}\subseteq M(\mathbb{K})\textit{ with }f_{v}=2=2e_{v},$ then there is a unique irreducible order with conductor $v^{a}$ which is given by $\mathbb{Z}+v^{a}$ with $[\mathcal{O}_{\mathbb{K}}:(\mathbb{Z}+v^{a})]=p^{a}$

Proof.

Clearly, we are looking for irreducible orders in the ring of integers of the unique totally un-ramified extension of degree 2 over $\mathbb{Z}_{p}.$ Thus, $p$ may be seen as the uniformizer for this completion. So, we may write it as $O_{v}:=\mathbb{Z}_{p}+\mathbb{Z}_{p}t$ . Thus every order here must be of the form $\mathbb{Z}_{p}+p^{k}(\mathcal{O}_{v}).$ Thus, the sub-ring is given by $\mathbb{Z}_{p}+p^{k}(O_{v})$ . Intersecting with $\mathcal{O}_{\mathbb{K}}$ we get the order is given by $\mathbb{Z}+v^{k}$ whose conductor is easily seen as $v^{k}.$ ∎

Theorem 26 (C).

If $\{v\}\subseteq M(\mathbb{K})\textit{ with }2f_{v}=2=e_{v},$ then

•

if $a\geq 2$ and $a$ is even then there is a unique irreducible order with conductor $v^{a}$ which is given by $\mathbb{Z}+v^{a}$ with $[\mathcal{O}_{\mathbb{K}}:(\mathbb{Z}+v^{a})]=p^{a/2}.$
•

if $a\geq 1$ and $a$ is odd then there is no irreducible (or otherwise) order with conductor $v^{a}.$

Proof.

Clearly, we are looking for irreducible orders in the ring of integers of some totally ramified extension of degree 2 over $\mathbb{Z}_{p}.$ Thus, $p$ may be seen as the square of the uniformizer for this completion. So, we may write it as $O_{v}:=\mathbb{Z}_{p}+\mathbb{Z}_{p}t$ where $t$ is the uniformizer(since it is totally ramified. Thus, every order here must be of the form $\mathbb{Z}_{p}+p^{k}(\mathcal{O}_{v}).$ Thus, the sub-ring is given by $\mathbb{Z}_{p}+p^{k}(O_{v})$ . Intersecting with $\mathcal{O}_{\mathbb{K}}$ we get the order is given by $\mathbb{Z}+v^{2k}$ (as $(p)=v^{2}$ , as it is a totally ramified extension) whose conductor is easily seen as $v^{2k}.$ ∎

A corollary we will use for counting purposes is,

Corollary 1.

Given a set of valuations $S\subseteq M(\mathbb{K})$ such that $\sum_{v\in S}e_{v}f_{v}=2$ then given $r\geq 1$ there exists a unique irreducible order $\mathcal{O}$ with $S(\mathcal{O})=\{\rho\}$ such that $\overline{\rho}=S$ and $[\mathcal{O}_{\mathbb{K}}:\mathcal{O}]=p^{r}.$

Definition 20.

A Sudo Maximal Order in $\mathbb{K}$ is an order $\mathcal{O}$ such that

\displaystyle\forall\rho\in S(\mathcal{O}):ef(\rho)=2.

We say that this is Restricted Sudo Maximal Order if for every prime $p\in\mathbb{Z}$ we have at most one prime ideal in $S(\mathcal{O})$ which contains $p.$

Theorem 27.

The number of Restricted Sudo Maximal Orders in $\mathbb{K}$ with index $\leq X$ is

\leq A_{n}X\log(X)^{\binom{n}{2}-1}

where $A_{n}$ is a constant only dependent on $n.$

Proof.

We note that if $C_{\mathbb{K},p^{k}}$ denote the number of Restricted Sudo Maximal Order in $\mathbb{K}$ with index $p^{k},$ we may look at

S_{\mathbb{K},p}(s):=\sum_{k=0}^{\infty}C_{\mathbb{K},p^{k}}p^{-ks}

and look at the L-function

L_{\mathbb{K}}(s):=\prod_{p\in\mathbb{P}}S_{\mathbb{K},p}

Since we know from the above section that

	$\displaystyle C_{\mathbb{K},p^{k}}=$	$\displaystyle\|\{(v,w):v,w\in M(\mathbb{K}),f_{v}=f_{w}=1,v\neq w,v(p)=w(p)=1\}\|$
		$\displaystyle+\|\{v\in M(\mathbb{K}):f_{v}=2v(p)=1\}\|+\|\{v\in M(\mathbb{K}):f_{v}=1v(p)=2\}\|.$

Thus, $C_{\mathbb{K},p^{k}}\leq\binom{n}{2}.$

It follows that the Dirichlet coefficients of $L_{\mathbb{K}}(s)$ are all between $0$ and the Dirichlet coefficients of $\zeta(s)^{\binom{n}{2}}$ .

Sum of Dirichlet coefficients up to $X$ of $\zeta(s)^{\binom{n}{2}}$ grows asymptotically like

\frac{X\log(X)^{\binom{n}{2}-1}}{(\binom{n}{2}-1)!}.

Thus, the number of orders in $\mathbb{K}$ with index $\leq X$ is

\leq A_{n}\frac{X\log(X)^{\binom{n}{2}-1}}{(\binom{n}{2}-1)!}

for some constant $A_{n}$ only dependant on $n.$ ∎

Remark 28.

We note that this gives a strong indication that the number of orders in $\mathbb{K}$ with index bounded by $X$ should be of the order of $X.$

6.2. Strongly divisible-ness and Ultra-weakly divisible polynomials

Remark 29.

Recall that Bhargava-Shankar-Wang define $f$ is strongly divisible by $p$ (in [3] and [2]) if $p^{2}|\mathfrak{disc}(f+p\cdot g)$ for any choice of $g$ such that $deg(f)\geq deg(g)$ .

Recall the theorem.

Theorem 30.

$p^{2}|\mathfrak{disc}(f)$ if and only if one of the following holds:

•
$f$ is strongly divisible by $p$ if and only if one of the following is true.
1. (1)
  
  $f\bmod p$ has a triple root in $\mathbb{P}(\mathbb{F}_{p})$
2. (2)
  
  $f\bmod p$ has two double roots in $\mathbb{P}(\overline{\mathbb{F}}_{p}).$
•
If $f$ is not strongly divisible by $p$ , then one of the following holds
1. (1)
  
  $f$ is weakly divisible by $p.$ This is the case where the linear double root in $\mathbb{P}(\mathbb{F}_{p})$ is in $\mathbb{F}_{p}$ (seen as a base affine component) and not at the point at infinity.
2. (2)
  
  The lead coefficient is divisible by $p^{2}$ and the second lead is divisible by $p$ . In other words, the palindromic reverse of $f$ is weakly divisible by $p$ at $0.$ This is the case where the linear double root in $\mathbb{P}(\mathbb{F}_{p})$ is at the point at infinity. (We will try and ignore this case by making sure that the lead coefficient of our polynomial is squarefree or making sure that the leading two coefficients are co-prime)

Proof.

•
We start by showing that if
- –
  
  $f\bmod p$ has a triple root in $\mathbb{P}(\mathbb{F}_{p})$ or
- –
  
  $f\bmod p$ has two double roots in $\mathbb{P}(\overline{\mathbb{F}}_{p})$
then $p^{2}|\mathfrak{disc}(f).$ Note that this automatically implies that $f$ is strongly divisible by $p.$
Since, we just proved the Dedekind-Kummer Theorem for Binary Rings (theorem 12), let us use it here.

If $f(x,y)$ shows the following decomposition/factorization modulo $p,$

$f(x,y)\equiv\prod_{i}(f_{i})^{e_{i}}\bmod p,$

then we know $(f_{i})^{e_{i}}$ corresponds to either a DVR or it corresponds to a proper subring in associated DVR.

Now if some $(f_{i})^{e_{i}}$ corresponds to a proper subring of a DVR, then $p$ will divide the index of corresponding local ring in DVR. This will imply $p^{2}|\mathfrak{disc}(f).$

On the other hand, if it does not correspond to an irreducible order, then $R_{f}$ localized at that prime is a DVR, which corresponds to a degree $e_{i}$ totally ramified extension of the totally un-ramified extension of degree $f_{i}$ of the local field $\mathbb{Q}_{p}.$ This means that $p^{e_{i}-1}$ divided the discriminant of the corresponding local field extension and thus $p^{\sum_{i}e_{i}-1}$ divides the discriminant of $\mathbb{K}_{f}$ . In this case, however, the power of $p$ dividing the discriminant of $\mathbb{K}_{f}$ and $f$ is the same.

Thus, $p^{2}|\mathfrak{disc}(f+pg)$ for any choice of $g$ provided that for some $i,$ $deg(f_{i})\geq 2\And e_{i}\geq 2.$

Similarly, $p^{2}|\mathfrak{disc}(f+pg)$ for any choice of $g$ provided that for some $i,$ $deg(f_{i})\geq 1\And e_{i}\geq 3.$

•

If $p^{2}|\mathfrak{disc}(f),$ then $p|\mathfrak{disc}{f}.$ Thus, $f$ has a double root in $\mathbb{P}(\overline{\mathbb{F}}_{p}).$ If this root is non linear then we fall in the case discussed above which implies $f$ is strongly divisible by $p.$ Similarly, if the root is linear and the multiplicity is greater than $3$ we again fall in the above case.

Thus, since $f$ is not strongly divisible by $p$ , we may assume that $f$ has only one double root in $\mathbb{P}^{1}(\bar{\mathbb{F}}_{p}).$ Now if this linear double root is in $\mathbb{F}_{p},$ we will translate appropriately so that the double root is at zero. If the linear root is a point at infinity we take the palindromic inverse or the reciprocal polynomial to shift its double root to $0.$ Then, showing initial polynomial is weakly divisible by $p$ as in definition 32 at the root will be same as showing new $f$ is weakly divisible by $p$ at $0.$

We note that, using lemma 15, the discriminant of $f$ can be written as

\mathfrak{disc}(f)=2f(0)(\frac{f^{(2)}(0)}{2!})^{3}\cdot\Delta_{1}+f^{\prime}(0)^{2}\cdot\Delta_{2}+f(0)f^{\prime}(0)\cdot\Delta_{3}+f(0)^{2}\Delta_{4}

where

\Delta_{1}=\mathfrak{disc}(\frac{f-f(0)Y^{n}-f^{\prime}(0)XY^{n-1}}{XY}).

Since, $f$ is not strongly divisible by $p,$ we see that $p\nmid\Delta_{1}$ (no two double roots) and $p\nmid\frac{f^{(2)}(0)}{2!}$ (no triple linear root).

Since, $p|f(0)$ and $p|f^{\prime}(0)$ and $p^{2}|\mathfrak{disc}{f}$ and $p\nmid\Delta_{1}$ and $p\nmid\frac{f^{(2)}(0)}{2!}$ it follows that $p^{2}|f(0).$

∎

See appendix.

Definition 21.

We say a $f$ ultra weakly divisible if

p^{2}|\mathfrak{disc}(f)\Rightarrow f\textit{ is weakly divisible by }p.

If we treat $\mathfrak{disc}(f)$ as a function of its coefficients (say $a_{i}$ by abuse of notation), then this condition corresponds to $p^{2}|\mathfrak{disc}(f)$ and $p|\frac{\partial(\mathfrak{disc})}{\partial a_{i}}(f)$ for all $0\leq i\leq n$ ( $f\bmod p$ is in the singular locus of $\mathfrak{disc}(f)=0\bmod p$ ).

Remark 31.

While it may seem that the singular locus has a very large co-dimension, it has a component of co-dimension 2.

By definition, if $f$ is ultra-weakly divisible, then $f$ does not have non linear double root or a linear triple root modulo any prime $p.$

Theorem 32.

If $f$ is ultra-weakly divisible, then there exist a maximal $m_{f}$ such that $f$ is weakly divisible by $m_{f}$ (at some $l_{f}$ ) and $R^{\prime}_{(f,m_{f},l_{f})}$ is maximal i.e. $R^{\prime}_{(f,m_{f},l_{f})}$ is THE ring of integers for $\mathbb{K}_{f}\simeq\mathbb{Q}[x]/(f).$ In fact, if $\mathfrak{disc}(f)=st^{2}$ where $s$ is squarefree, then $m_{f}=t.$

Proof.

Let $p$ denote a prime such that $p^{2}|\mathfrak{disc}(f).$ And let $0\leq l_{0}<p$ denote the linear double root in $\mathbb{F}_{p}$ of $f.$ Then, we note that if

f(x+l)=f(l)+f^{\prime}(l)\cdot x+\frac{f^{(2)}(l)}{2!}\cdot x^{2}+\cdots+a_{0}x^{n}

then using lemma 15 we see that,

\mathfrak{disc}(f)=f(l)\cdot\Delta_{1}+f^{\prime}(l)^{2}\cdot\Delta_{2}+f(l)f^{\prime}(l)\cdot\Delta_{3}.

Furthermore, if $l\equiv l_{0}\bmod p$ then we can realize $\Delta_{1}$ as

\Delta_{1}\equiv 4(\frac{f^{(2)}(l)}{2!})^{3}\mathfrak{disc}(f^{*})\bmod p

where $f^{*}=\frac{\overline{f(x)}}{\overline{(x-l_{0})^{2}}}\in\mathbb{F}_{p}[x]$ .

Since $f$ is ultra-weakly divisible by $p,$ we have $f$ is not strongly divisible by $p.$ This means that if $l\equiv l_{0}\bmod p$ then $p\nmid\frac{f^{2}(l)}{2}$ as this will force $f$ to not have triple linear root at $l_{0}$ in $\mathbb{F}_{p}.$ We can further say that that $p\nmid\Delta_{1}.$ If $p\mid\Delta_{1}$ then it is easy to see that $f$ will have at least two double roots in $\bar{\mathbb{F}}_{p}.$

Now, by Hensels lemma for any $k$ we can find $l_{k}$ such that $p^{k}|f^{\prime}(l_{k}).$

Let $D=v_{p}(\mathfrak{disc}(f))$ and $l\equiv l_{k}\bmod p^{k}$ with $0\leq l<p^{k}$ where $k=[\frac{D}{2}].$

Then, we note that $p^{r}|f(l)$ and $r<2[\frac{D}{2}].$ It follows that $\mathfrak{disc}(f)\equiv p^{r}\cdot\Delta_{1}\bmod p^{r+[D/2]}$ which contradicts the fact that $p^{D}||\mathfrak{disc}(f).$

Thus, $p^{2\cdot[D/2]}|f(l)$ and $p^{[D/2]}|f^{\prime}(l)$ or $f$ is weakly divisible by $p^{[D/2]}.$ Thus, $f$ is weakly divisible by $m_{f}$ where $\mathfrak{disc}(f)=s(m_{f})^{2}$ and $s$ is squarefree. This tells us that $R^{\prime}_{(f,m_{f},l_{f})}$ is $p^{D-2\cdot[D/2]}$ which is at most one. This makes $R^{\prime}_{(f,m_{f},l_{f})}$ the ring of integers of $\mathbb{K}_{f}$ as its discriminant is squarefree. ∎

7. Polynomials that correspond to distinct rings

Definition 22.

W(s:t):=\Big{\{}\sum_{i=0}^{n}a_{i}x^{n-i}\in\mathbb{Z}[x]:1\leq a_{1}<s/2<a_{0}\leq s,(a_{0},a_{1})=1,|a_{i}|\leq st^{i}\text{ for all }2\leq i\leq n\Big{\}}

Remark 33.

The above set is a subset of a fundamental domain for action of integers by translation on polynomials. That is, for any polynomial $f$ there exists at most one integer $l$ such that $f(x+l)\in W(s:t).$

Definition 23.

W(s:t:m):=\{(f,l):f\in W(s:t),0\leq l<m,f\textit{ is weakly divisible by }m\textit{ at }l\}

with the understanding that each element $(f,l)\in W(s:t:m)$ where $f$ is Reduced- $m$ -Polynomial will give distinct rings, see Theorem theorem 50.

Lemma 8.

Given $a_{0},a_{1},\cdots,a_{n-2},l,m$ there is a unique choice of $a_{n-1}$ and $a_{n}$ satisfying

•

$B+1\leq a_{n-1}\leq B+m$
•

$C+1\leq a_{n-1}\leq C+m^{2}$
•

$a_{0}x^{n}+a_{1}x^{n-1}+\cdots+a_{n}\textit{ is weakly divisible by $m$ at $l.$}$

Proof.

Follows from the definition directly. ∎

Discussion 24.

Let $B_{\epsilon}$ denote a compact set in

\{f\in\mathbb{R}[x]:H(f)\leq 1,\mathfrak{disc}_{\mathbb{Z}}(f)\neq 0\}

such that

Vol(B_{\epsilon})\geq(1-\epsilon)Vol(\{f:H(f)\leq 1\}).

Then for this region we can construct a $\rho_{B}$ such that all polynomials $f$ with height $\geq\rho_{B}$ satisfying

H(f)^{n}f(\frac{X}{H(f)})\in B_{\epsilon}

are Reduced- $1$ -Polynomials.

From theorem 50, we can say that polynomials of this type with height

\geq m^{1/(n-2)}\rho_{B}

are Reduced- $m$ -Polynomials.

Furthermore, we may consider $B_{\epsilon}$ to be a finite union of disjoint boxes $B_{i}$ (depending on epsilon).

Thus, correspondingly, if $d_{i,r}$ are the dimensions of the box $B_{i},$ we define $W(s:t:m)_{B_{i}}$ to be those polynomials that satisfy

\frac{H(f)^{n}}{s}f(\frac{X}{H(f)})\in B_{i}

and

\frac{f^{(2)}(0)}{2!}\geq s\rho_{B}^{n-2}m.

The latter condition $a_{n-2}\geq s\rho_{B}^{(n-2)}m$ ensures that polynomials we count are of ‘large enough’ (see lemma 14) height to be a Reduced- $m$ -Polynomial.

Definition 25.

W(s:t:m)^{Red}:=\{(f,l)\in W(s,t,m):f\textit{ is a Reduced-}m\textit{-Polynomial.}\}

7.1. Ekedahlian Sieve- Quantitative version

Let $V_{n}$ denote the variety in $\mathbb{A}^{n+1}$ which describes the Singular locus of the curve given by $G_{n}=0$ where

G_{n}:=\mathfrak{disc}_{X}(X_{0}X^{n}+X_{1}X^{n-1}+\cdots X_{n}).

(15)

We describe the structure of the discriminant polynomial as a polynomial of $X_{n}.$

Lemma 9.

	$\displaystyle\mathfrak{disc}_{X}(X_{0}X^{n}+X_{1}X^{n-1}$	$\displaystyle+\cdots+X_{n})=$
		$\displaystyle n^{n}(X_{0}X_{n})^{n-1}$
		$\displaystyle+\sum_{i=1}^{n-2}X_{n}^{i}\Delta_{i}$
		$\displaystyle+X_{n-1}^{2}\mathfrak{disc}_{X}(X_{0}X^{n-1}+X_{1}X^{n-2}+\cdots+X_{n-1})$

Proof.

Obvious from the the understanding that $X_{0}\mathfrak{disc}_{X}(X_{0}X^{n}+X_{1}X^{n-1}+\cdots+X_{n})$ is the resultant of $X_{0}X^{n}+X_{1}X^{n-1}+\cdots+X_{n}$ and $\frac{\partial}{\partial X}(X_{0}X^{n}+X_{1}X^{n-1}+\cdots+X_{n})$ can the consideration of this resultant as the determinant of the Sylvester matrix. See eq. 28 and lemma 15. ∎

Let

F_{n}:=\mathfrak{disc}_{X_{n}}(\mathfrak{disc}_{X}(X_{0}X^{n}+X_{1}X^{n-1}+\cdots+X_{n}))

(16)

Thus, for any prime $p,$ and integer tuple $(a_{0},a_{1},\cdots,a_{n})$ ,

(a_{0},a_{1},\cdots,a_{n})\in V_{n}(\mathbb{F}_{p})\Rightarrow p|(G_{n}(a_{0},a_{1},\cdots,a_{n}),F_{n}(a_{0},a_{1},\cdots,a_{n-1)}).

(17)

This follows from the fact that $G_{n},F_{n}\in I(V_{n}).$

We note that

Remark 34.

A polynomial $f$ is UWD or ultra-weakly divisible if and only if

\forall p\in\mathbb{P}:f\bmod p\notin V_{n}(\mathbb{F}_{p}).

We will now look for UWD polynomials in $W(s:t:m).$ We will then look for Reduced- $m$ -polynomials in this set which will allow us to count UWD Rings. We note that UWD polynomials are easily sieve-able via the Ekedahl sieve.

Discussion 26.

Given $f=\sum_{i=0}^{n}a_{i}x^{n-i}$ such that $f(x)$ is weakly divisible by $m$ at $0$ and $(a_{0},a_{1})=1$ ,

\frac{\mathfrak{disc}(f)}{m^{2}}\equiv 2(a_{n-2})^{3}\mathfrak{disc}(f^{*})\bmod m

where $f^{*}=\sum_{i=0}^{n-2}a_{i}x^{n-i-2}.$ This follows from the structure of the discriminant polynomial in lemma 15.

Thus, the only way $f$ can be strongly divisible by $p$ for some $p|m$ if

•

either $p|a_{n-2}$ (this will force a triple root at $0$ )
•

or $p|\mathfrak{disc}(f^{*})$ (this will force some other double root in $\bar{\mathbb{F}}_{p}$ ).

Furthermore, applying the structure of the discriminant (lemma 15) to $f^{*}$ we can say that for every $a_{0},a_{1},\cdots,a_{n-3}$ and $p|m$ there are at most $n+1$ solutions $\bmod p$ for $a_{n-2}$ such that

2(a_{n-2})^{3}\mathfrak{disc}(\sum_{i=0}^{n-2}a_{i}x^{n-i-2})\equiv 0\bmod p.

We will use the traditional form of the Ekedahl sieve. This will serve us better. Our parameter of $m$ which interferes with the sieving in the main term. And since we wish to maximize $m$ to be used, we wish to use the fact that $a_{n-2}$ freely moves in a much larger range than other coefficients and based on the structure of $G_{n}$ can be used to give a usable lower bound for $f\notin V_{n}(\mathbb{F}_{p})$ for $p|m$ for much larger values of $m.$

Theorem 35.

If $s\gg\frac{st^{n}}{m^{2}},$ then

|\{(f,l)\in W(s:t:m):\exists p\in\mathbb{P}_{\mathbb{Z}},p>M,p\nmid m,f\in V_{n}(\mathbb{F}_{p})\}|

=O(\frac{s^{n}t^{\frac{n(n+1)}{2}-1}}{m^{2}})\cdot(\frac{1}{M\log M}+\frac{m^{2}\log(st^{n})}{st^{n}})

Proof.

As usual we put this set into the union of 3 sets:

•

$S_{1}:=\{(f,l)\in W(s:t:m):F_{n}(f)=0\}$

For any choice of $(a_{0},a_{1},\cdots,a_{n-2},l)$ such that $F_{n}$ is non-degenerate as a polynomial in $a_{n-1},$ then we can have at most $n(n-1)$ possible choices for $a_{n-1}.$

Inductively, the number of choices for $(a_{0},a_{1},\cdots,a_{n-2})$ for which the polynomial $F_{n}$ is degenerate as a polynomial in $a_{n-1}$ would be space cut out by all coefficient polynomials in $\mathbb{A}^{n-1}.$ Thus, the number of elements in $S_{1}$ is

$O_{n}(\frac{s^{n}t^{\frac{n(n+1)}{2}-1}}{m^{2}})\cdot(\frac{1}{s}+\frac{m}{st^{n-1}}).$

•

S_{2}:=\{(f,l)\in W(s:t:m):\exists p\in\mathbb{P}_{\mathbb{Z}},\frac{st^{n}}{m^{2}}>p>M,p\nmid m,f\in V_{n}(\mathbb{F}_{p})\}

For every such prime $p$ the number of solutions ( $\bmod p$ ) to

G_{n}(f)\equiv F_{n}(f)\equiv 0\bmod p

is $O(p^{n-1})$ . Thus, the total number of solutions possible $f\in W(s:t:m)$ is

	$\displaystyle O(p^{n-1})\cdot m\cdot(\frac{s}{p}+O(1))^{2}\cdot(\frac{2st^{2}}{p}+O(1))\cdots(\frac{2st^{k}}{p}+O(1)\cdots(\frac{2st^{n-2}}{p}+O(1))$
	$\displaystyle\cdot(\frac{2st^{n-1}}{pm}+O(1))\cdot(\frac{2st^{n}}{pm^{2}}+O(1))$
	$\displaystyle=O(\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}})(\frac{1}{p^{2}}).$

Summing this value over the given range for primes, we get that $|S_{2}|$ is

O(\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}M\log M})

•

S_{3}:=\{(f,l)\in W(s:t:m):F_{n}(f)\neq 0,\exists p\in\mathbb{P}_{\mathbb{Z}},\frac{st^{n}}{m^{2}}<p,p\nmid m,f\in V_{n}(\mathbb{F}_{p})\}

In this case, we first pick any $a_{0},a_{1},\cdots,a_{n-1}$ . $F_{n}$ is not zero. We then pick a prime $p$ such that $p|F_{n}(a_{0},a_{1},\cdots,a_{n-1})$ and $p>\frac{st^{n}}{m^{2}}$ . The size of $F_{n}$ is less than $(st^{n})^{k_{n}}$ for some $k_{n}$ . Thus, the number of distinct prime divisors of $F_{n}$ is

\ll k_{n}\log(st^{n}).

Since, if $p\nmid a_{0},$ then we can deduce from lemma 9 that $G_{n}$ is a polynomial in $a_{n}$ of degree $n-1.$ If $p|a_{0},$ then $p\nmid a_{1}$ (as $(a_{0},a_{1})=1$ ) and thus

G_{n}(a_{0},\cdots,a_{n})\equiv a_{1}^{2}G_{n-1}(a_{1},a_{2},\cdots,a_{n})\bmod p.

This follows directly from applying lemma 9 to the palindromic inverse of $f$ . We basically do not have to worry about $G_{n}\bmod p$ being degenerate as a polynomial in $a_{n}$ . Thus, there will be at most $n-1$ possibilities for $a_{n}$ as the range it moves through is of size less than $p.$ Therefore, $|S_{3}|$ will be

O(\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}})\cdot\frac{m^{2}}{st^{n}}\cdot\log(st^{n}).

∎

Definition 27.

Let $W_{n}$ denote the variety which is union of $V_{n}$ with the variety given by $X_{0}=X_{1}=0.$ Let

c_{p}:=1-\frac{|W_{n}(\mathbb{F}_{p})|}{p^{n+1}}.

Since, for every $p$ we have polynomials with discriminant co-prime to $p,$ $c_{p}$ is never zero. Furthermore, since $V_{n}$ is a variety contained in $V(F_{n},G_{n})$ of co-dimension $2$ and $X_{0}=X_{1}=0$ is also co-dimension 2 we can conclude that $c_{p}\geq(1-\frac{r_{n}}{p^{2}})$ for some constant $r_{n}.$ For $p\nmid m,$ $c_{p}$ will serve as the measure of how many polynomials mod p are not strongly divisible. The addition of this second variety will save us some trouble as the condition $(a_{0},a_{1})=1$ will be taken care of internally.

7.2. Actual sieve

Definition 28.

W^{*}(s:t:m)=\{(f,l)\in W(s,t,m):\forall p|m,f\notin V_{n}(\mathbb{F}_{p})\}

Before moving forward, we refer to 24. Let $B_{i}$ denote boxes as there, and let

Definition 29.

W^{*}(s:t:m)_{B_{i}}:=W(s:t:m)_{B_{i}}\cap W^{*}(s:t:m)

and

W^{*}(s:t:m)_{B_{i}}^{UWD}:=\{(f,l)\in W^{*}(s:t:m)_{B_{i}}:\forall p\in\mathbb{P},f\notin V_{n}(\mathbb{F}_{p})\}

and

W^{*}(s:t:m:M)_{B_{i}}:=\{(f,l)\in W(s:t:m)^{*}_{B_{i}}:\forall p<M,f\notin V_{n}(\mathbb{F}_{p})\}.

We begin with defining $M^{\prime}=\prod_{\begin{subarray}{c}p<M\\ p\nmid m\end{subarray}}p.$ We will search for solutions to $\forall p<M,f\notin V_{n}(\mathbb{F}_{p})$ by looking at the system modulo $M^{\prime}$ . So we let

C_{M^{\prime}}=\prod_{\begin{subarray}{c}p<M\\ p\nmid m\end{subarray}}c_{p}.

We will restrict ourselves to squarefree $m$ . For $p|m,$ we will look at $a_{n-2}$ only and note that for any choice of $a_{0},a_{1},\cdots,a_{n-3}$ there are at most $n^{\omega(m)}$ possible solutions modulo $m$ for $a_{n-2}$ for $f$ to be strongly divisible by $m.$ See 26.

Thus, making boxes of size $M^{\prime},$ we see that

	$\displaystyle\|$	$\displaystyle W^{*}(s:t:m:M)_{B_{i}}\|$
		$\displaystyle=C_{M}\cdot m\cdot(\frac{s}{2M^{\prime}}+O(1))^{2}\cdot(\frac{2st^{2}}{M^{\prime}}+O(1))\cdots(\frac{2st^{k}}{M^{\prime}}+O(1)\cdots(\frac{2st^{n-2}}{M^{\prime}}+O(n^{\omega(m)})+O(s\rho_{B}m))$
		$\displaystyle\cdot(\frac{2st^{n-1}}{M^{\prime}m}+O(1))\cdot(\frac{2st^{n}}{M^{\prime}m^{2}}+O(1))\cdot(M^{\prime})^{n+1}.$
		$\displaystyle\textit{ If we have }M^{\prime}\leq\min\{s,\frac{st^{n}}{m^{2}},\frac{t^{n-2}}{m}\},\textit{ we get,}$
		$\displaystyle=2^{n-3}Vol(B_{i})C_{M}\cdot\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}}(1+O(\frac{m^{2}M^{\prime}}{st^{n}})+O(\frac{M^{\prime}m}{t^{n-2}})+O(\frac{M^{\prime}}{s})).$

This will not be a problem as we will be taking $M^{\prime}$ to be small.

We quickly write $|W^{*}(s:t:m)_{B_{i}}^{UWD}|$ as

|W^{*}(s:t:m:M)_{B_{i}}|+O(|\{(f,l)\in W(s:t:m):\exists p\in\mathbb{P}_{\mathbb{Z}},p>M,p\nmid m,f\in V_{n}(\mathbb{F}_{p})\}|)

Let $C=\lim_{M\longrightarrow\infty}C_{M}.$ Using theorem 35, we get

Theorem 36.

If $s\gg\frac{st^{n}}{m^{2}},$ then

	$\displaystyle\|W^{*}(s:t:m)_{B_{i}}^{UWD}\|\gg$	$\displaystyle 2^{n-3}Vol(B_{i})C\cdot\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}}(1+O(\frac{m^{2}e^{M}}{st^{n}})+O(\frac{me^{M}}{t^{n-2}}))$
		$\displaystyle+O(\frac{s^{n}t^{\frac{n(n+1)}{2}-1}}{m^{2}})\cdot(\frac{1}{M\log M}+\frac{m^{2}\log(st^{n})}{st^{n}})$

Setting $e^{M}=(st^{n})^{\epsilon/2},$ we get the following corollary.

Corollary 2.

If $(st^{n})^{\epsilon}\ll s$ then

|W^{*}(s:t:m)_{B_{i}}^{UWD}|\gg 2^{n-3}Vol(B_{i})C\cdot\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}}(1+O_{\epsilon}(\frac{1}{\log(st^{n})})+O(\frac{m^{2}(st^{n})^{\epsilon}}{st^{n}})+O(\frac{m(st^{n})^{\epsilon}}{t^{n-2}})).

Summing over all boxes $B_{i}$ we get the following corollary.

Corollary 3.

If $(st^{n})^{\epsilon}\ll s$ then

|W^{*}(s:t:m)|\gg(1-\epsilon)2^{n-3}C\frac{s^{n+1}t^{\frac{n(n+1)}{2}-1}}{m^{2}}(1+O_{\epsilon}(\frac{1}{\log(st^{n})})+O(\frac{m^{2}(st^{n})^{\epsilon}}{st^{n}})+O(\frac{m(st^{n})^{\epsilon}}{t^{n-2}})).

We set $X=\frac{s^{2n-2}t^{n(n-1)}}{m^{2}}$ .

Let $S(X,s,m):=W^{*}(s:t:m)$ .

Note that this set counts distinct rings.

It follows that

Theorem 37.

For $n\geq 5,$

S(X,s,m)\gg_{n}\frac{X^{\frac{1}{2}+\frac{1}{n}}\cdot s^{\frac{2}{n}}}{m^{1-\frac{2}{n}}}

(18)

provided that

(1)

$s\gg_{\epsilon}(st^{n})^{\epsilon}$
(2)

$X^{1-\epsilon}\gg_{\epsilon}s^{2n-2}\cdot m^{n-4-\frac{1}{n-2}}$ (Condition corresponding to $t^{n-2}\gg_{\epsilon}m^{1+\epsilon}$ - required for injectivity).
(3)

$X^{1-\epsilon}\gg_{\epsilon}s^{n-1}\cdot m^{2n-4}$ (Condition corresponding to $(st^{n})^{1-\epsilon}\gg_{\epsilon}m^{2}$ -required so the final coefficient has enough space to very to have polynomials which are weakly divisible by $m$ )

To count ultra-weakly divisible rings which are weakly divisible by $m$ rings, we wish to choose $s$ appropriately. Note that within the constraints for a fixed $X$ , the rings counted will all be distinct for any choice of $m$ and then any choice of $s.$

We can chose $s$ such that $st^{n}\simeq t^{2n-4}.$ This will make the condition on $m$ a singular condition. May not be optimal.

st^{n}\simeq t^{2n-4}\iff s\simeq(m^{2}X)^{\frac{n-4}{(n-1)(3n-8)}}

Both conditions combine to

X^{1-\epsilon}\geq(m^{2}X)^{\frac{n-4}{3n-8}}m^{2n-4}\iff X^{\frac{n-2}{3n^{2}-13n+12}-\epsilon}\geq m

Thus, if we let $S(X:m)$ denote the set of all UWD Rings, weakly divisible by $m,$ then we have

S(X:m)\gg\frac{X^{\frac{1}{2}+\frac{1}{n}+\frac{2(n-4)}{n(n-1)(3n-8)}}m^{\frac{2}{n}+\frac{4(n-4)}{n(n-1)(3n-8)}}}{m}

	$\displaystyle T(X):=\sum_{m\leq X^{\frac{n-2}{3n^{2}-13n+12}-\epsilon}}S(X,m)$	$\displaystyle\gg\sum_{m\leq X^{\frac{n-2}{3n^{2}-13n+12}-\epsilon}}\frac{X^{\frac{1}{2}+\frac{1}{n}+\frac{2(n-4)}{n(n-1)(3n-8)}}m^{\frac{2}{n}+\frac{4(n-4)}{n(n-1)(3n-8)}}}{m}$
		$\displaystyle\gg_{n}X^{\frac{1}{2}+\frac{1}{n}+\frac{2(n-4)}{n(n-1)(3n-8)}}(X^{\frac{n-2}{3n^{2}-13n+12}-\epsilon})^{\frac{2}{n}+\frac{4(n-4)}{n(n-1)(3n-8)}}$
		$\displaystyle\sim_{\epsilon}X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}-\epsilon}$

Remark 38.

We are summing over squarefree $m$ .

7.3. Number-field Counting Strategy

The strategy now comes into play by noting that

Theorem 39.

A UWD ring is a Restricted Sudo maximal Ring.

Proof.

Using Dedekind Kummer for Binary Rings i.e.theorem 12, we see that factorization of the corresponding UWD polynomial modulo $p$ captures the irreducible decomposition of the associated binary ring. We see that if $f$ is Ultra-weakly divisible then at every prime $p$ it can never have 2 double roots or a triple root. This means that $f$ can at most have one double root modulo $p.$ The corresponding factor will correspond to either a maximal ring at $p$ or an irreducible order with $ef(\mathcal{O})=2.$ Thus, $R_{f}$ is restricted Sudo maximal Ring. Recall that Sudo maximal rings were Rings whose theorem 7 decomposition only consist of irreducible orders $\mathcal{O},$ satisfying $ef(\mathcal{O})=2.$ Our description of Restricted Sudo Maximal Rings and Pseudo maximal rings immediately tells us that any intermediate Ring that sits in between a Restricted Sudo maximal Ring and its integral Closure(the ring of integers it sits in) is also a Restricted Sudo maximal Ring. ∎

Let $A(X)$ denote the set of all Restricted Sudo maximal Rings with discriminant $\leq X.$ Let $a(X):=|A(X)|$ . Thus, as a corollary we have

Corollary 4.

$a(X)\geq T(X).$

Let

B(X):=\{frac(R):R\in A(X)\}.

and $b(X):=|B(X)|.$

On the other hand, we may bound $a(X)$ above by using theorem 27 as follows.

	$\displaystyle a(X)$	$\displaystyle\leq\sum_{\mathbb{K}\in B(X)}S_{\mathbb{K}}(X)\leq\sum_{\mathbb{K}\in B(X)}\sqrt{\frac{X}{D_{\mathbb{K}}}}\log(\frac{X}{D_{\mathbb{K}}})^{c_{n}}$
		$\displaystyle\leq\sqrt{X}\sum_{\mathbb{K}\in B(X)}\frac{1}{\sqrt{D_{\mathbb{K}}}}\log(X)^{c_{n}}.$

Combine with the fact that

a(X)\gg T(X)\gg X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}-\epsilon}

We get

N^{*}(X:n)\geq\sum_{\mathbb{K}\in B(X)}\sqrt{\frac{1}{D_{\mathbb{K}}}}\gg X^{\frac{1}{n-\frac{4}{3}}-\epsilon}

Corollary 5.

\limsup_{X\longrightarrow\infty}\frac{N(X:n)}{X^{\frac{1}{2}+\frac{1}{n-\frac{4}{3}}-\epsilon}}=\infty.

Remark 40.

We note that if the number of UWD polynomial tuples $(f,m)$ with discriminant $<X$ is of the expected order then we can easily convert this to be an appropriate lower bound for $N(X:n).$

Appendix A Defining Weakly Divisible Rings

Most of this section is from our previous paper on Weakly Divisible Rings.

We recall the following theorem from [11] defining Binary Rings.

Let

f(X,Y):=a_{n}X^{n}+a_{n-1}X^{n-1}Y+\cdots+a_{0}Y^{n}

denote a binary form of degree $n$ . Let $a_{n}\neq 0$ and suppose that $\delta$ denotes the image of $X$ in the algebra $\mathbb{Q}[X]/(f(X,1)).$

We recall eq. 3.

Remark 41.

We refer to the basis given for $R_{f}$ in eq. 3 as the canonical basis attached to $f.$

Theorem 42.

When $f$ is integral(i.e. $f$ is a binary form of degree $n$ with integer coefficients), $R_{f}$ is a ring of rank $n$ over $\mathbb{Z}.$

Definition 30.

We define $I_{f}$ as the (fractional) ideal class generated by $(1,\delta)$ over $R_{f},$ when $f$ is integral.

Definition 31.

When $f$ is integral, $R_{f}$ is known as the binary ring associated to the binary form $f.$

These are a few properties of binary rings.

Proposition 1.

Properties of $R_{f}$ (when $f$ is integral):

(1)

$\mathfrak{disc}_{\mathbb{Z}}(R_{f})=\mathfrak{disc}(f).$ (19)
(2)

If $\delta$ is invertible, and $f$ is primitive, then

$R_{f}:=\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}].$ (20)
(3)

If $f$ is primitive, $I_{f}$ is invertible in $R_{f}.$
(4)

Both $R_{f}$ and $I_{f}$ are invariant under the natural $GL_{2}(\mathbb{Z})$ action on binary forms of degree $n.$ In particular, for $\delta\in\overline{\mathbb{Q}}\backslash\mathbb{Q}$ this means that if $\lambda=\frac{a\delta+b}{c\delta+d}$ with $a,b,c,d\in\mathbb{Z}$ and $ad-bc=\pm 1$ then

$\mathbb{Z}[\delta]\cap\mathbb{Z}[\delta^{-1}]=\mathbb{Z}[\lambda]\cap\mathbb{Z}[\lambda^{-1}].$

You can find the proof of all of these statements in [11].

Remark 43.

Note that one can write down the multiplication table for the defining basis of $R_{f}$ in terms of coefficients of $f$ explicitly. Using this table to define binary rings, one can give a definition for $R_{f}$ without the need for the condition “ $a_{n}\neq 0$ ”.

We will note the following part of the multiplication table of the canonical basis of $R_{f}$ (eq. 3), which is easily verified.

Lemma 10.

Let $f(X,Y):=a_{n}X^{n}+a_{n-1}X^{n-1}Y+\cdots+a_{0}Y^{n}$ denote a binary form of degree $n,$ and let $\langle B_{0},B_{1},\cdots,B_{n-1}\rangle$ denote the canonical basis for $R_{f}$ associated to $f$ as in eq. 3. Then, we have

\displaystyle B_{n-1}\cdot B_{n-i}=-a_{0}\cdot B_{n-i-1}+a_{i}\cdot B_{n-1}

(21)

Proof.

We make note of the fact, $B_{n-1}=-a_{0}\delta^{-1}.$
Thus,

	$\displaystyle B_{n-1}\cdot B_{n-i}=-a_{0}\delta^{-1}\cdot(\sum_{j=0}^{n-i}a_{n-j}\delta^{j})$
	$\displaystyle=-a_{0}\cdot\sum_{j=0}^{n-i}a_{n-j}\delta^{j-1}$
	$\displaystyle=-a_{0}\cdot(B_{n-i-1}+\frac{a_{i}}{\delta})$
	$\displaystyle=-a_{0}\cdot B_{n-i-1}+a_{i}\cdot(\frac{-a_{0}}{\delta})$
	$\displaystyle=-a_{0}\cdot B_{n-i-1}+a_{i}\cdot B_{n-1}.$

∎

Definition 32.

We say a binary form $f(x,y)\in\mathbb{Z}[x,y]$ of degree $n$ is weakly divisible by $m\in\mathbb{Z}_{>0}$ if there exists an $\ell\in\mathbb{Z}$ such that

	$\displaystyle f(l,1)$	$\displaystyle\equiv 0\bmod{m^{2}}$
	$\displaystyle\frac{\partial f}{\partial x}(l,1)$	$\displaystyle\equiv 0\bmod m$

When appropriate we say $f$ is weakly divisible by $m$ at $l$ .

Remark 44.

We use “weakly divisible” as these polynomials are defined in [3] and [2] and the main parts of these papers is about establishing an upper bound for the number of polynomials which are weakly divisible by $m$ and bounded height, for all large $m.$

Definition 33.

Given a binary form $f$ we set $f_{l}=f_{l}(x,y):=f(x+ly,y).$

The condition that “ $f$ is weakly divisible by $m$ at $l$ ” is equivalent to having integers $a_{0},a_{1},\cdots,a_{n}$ such that

f_{l}=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{2}x^{2}y^{n-2}+ma_{1}xy^{n-1}+m^{2}a_{0}y^{n}.

Definition 34.

We define,

R^{\prime}_{(f,m)}:=\mathbb{Z}\langle B_{0},B_{1},\cdots,B_{n-2},\frac{B_{n-1}}{m}\rangle.

Theorem 45.

If $f$ is weakly divisible by $m$ at $l,$ then $R^{\prime}_{(f_{l},m)}$ is a ring.

Proof.

Let $\langle B_{0},B_{1},\cdots,B_{n-1}\rangle$ denote the canonical (old) basis of $R_{f_{l}}$ associated to $f_{l}.$ Since, $f$ is weakly divisible by $m$ at $l,$ we can write

f_{l}=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{2}x^{2}y^{n-2}+ma_{1}xy^{n-1}+m^{2}a_{0}y^{n},

where $a_{i},m\in\mathbb{Z}.$

We will show that $R^{\prime}_{(f_{l},m)}$ is a ring, by showing that product of any two elements in the (new) basis given by $\langle B_{0},B_{1},\cdots,B_{n-2},\frac{B_{n-1}}{m}\rangle$ (basis for $R^{\prime}_{(f_{l},m)})$ is in the $\mathbb{Z}$ -span of itself. We will achieve this by comparing the product of every two elements in the old basis with the product of the corresponding two elements in the new basis.)

We note that $B_{i}\cdot B_{j}$ for $i,j\neq n-1$ can be written as a $\mathbb{Z}$ -linear combination of $\langle B_{i}\rangle$ . This follows directly from the fact that the $\mathbb{Z}$ -span of $\langle B_{i}\rangle$ forms a ring (the prodigal binary ring) (we can also refer to the multiplication tables given in [9] or section 2.1 in [11]).

It immediately follows that $B_{i}\cdot B_{j}$ for $i,j\neq n-1$ can also be written as a $\mathbb{Z}$ -linear combination of our new basis $\langle B_{0},B_{1},\cdots,B_{n-2},\frac{B_{n-1}}{m}\rangle$ (which only differs from the original basis at the index $n-1$ ) by simply replacing $B_{n-1}$ with $m\cdot\frac{B_{n-1}}{m}$ in the multiplication table of the old basis.

On the other hand, lemma 10 immediately tells us that, when $i\neq 1$

\displaystyle\frac{B_{n-1}}{m}\cdot B_{n-i}=-ma_{0}\cdot B_{n-i-1}+a_{i}\cdot\frac{B_{n-1}}{m}

and

\displaystyle\frac{B_{n-1}}{m}\cdot\frac{B_{n-1}}{m}=-a_{0}\cdot B_{n-2}+a_{1}\cdot\frac{B_{n-1}}{m}.

Thus, the products of elements in this new basis are $\mathbb{Z}$ -linear combinations of the same new basis. It follows that $R^{\prime}_{(f_{l},m)}$ is in-fact a ring. ∎

Definition 35.

We say $R^{\prime}_{(f_{l},m)}$ is the weakly divisible ring (at $l$ with respect to $m$ ) associated to $f$ , when $f$ is weakly divisible by $m$ at $l$ . When appropriate we will also represent this ring as $R^{\prime}_{(f,m,l)}.$

Remark 46.

Every binary ring is a weakly divisible ring at every value with respect to 1.

Remark 47.

We note that weakly divisible rings may also defined by multiplication tables to avoid dependence on a condition like “ $a_{n}\neq 0$ or $a_{0}\neq 0$ ”.

Appendix B Effective injectivity of the map: $(f,m,l)\longrightarrow R^{\prime}_{(f_{l},m)}$

B.1. Matrix of transformation

For convenience sake, we define $\binom{n}{a}:=0$ if $n\geq 0$ and $a<0.$

We mention a useful matrix of transformation for base change from the canonical basis of $R_{f}$ associated to $f(X,Y)$ to the canonical basis of $R_{f_{l}}$ associated to $f_{l}.$

Theorem 48.

We let $f(X,Y)=A_{0}X^{n}+A_{1}X^{n-1}Y+\cdots+A_{n}Y^{n}$ and $\langle B_{0},B_{1},\cdots,B_{n-1}\rangle$ be the canonical basis of $R_{f}$ associated to $f(X,Y)$ and $\langle C_{0},C_{1},\cdots,C_{n-1}\rangle$ be the canonical basis of $R_{f_{l}}$ associated to $f_{l},$ then we have

\displaystyle\begin{matrix}\langle C_{0},C_{1},\cdots,C_{n-1}\rangle=\\ \\ \\ \\ \\ \\ \\ \\ \\ \end{matrix}\begin{matrix}\langle B_{0},B_{1},\cdots,B_{n-1}\rangle\\ \\ \\ \\ \\ \\ \\ \\ \\ \end{matrix}\begin{bmatrix}1&\binom{n-1}{1}lA_{0}&\binom{n-1}{2}l^{2}A_{0}&\cdots&\binom{n-1}{k-1}l^{k-1}A_{0}&\cdots&\binom{n-1}{n-1}l^{n-1}A_{0}\\ 0&1&\binom{n-2}{1}l&\cdots&\binom{n-2}{k-2}l^{2}&\cdots&\binom{n-2}{n-2}l^{n-2}\\ 0&0&1&\cdots&\binom{n-3}{k-3}l^{k-3}&\cdots&\binom{n-3}{n-3}l^{n-3}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&\binom{n-k^{\prime}}{k-k^{\prime}}l^{k-k^{\prime}}&\cdots&\binom{n-k^{\prime}}{n-k^{\prime}}l^{n-k^{\prime}}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&\binom{1}{k-n+1}l^{k-n+1}&\cdots&l\\ 0&0&0&\cdots&\binom{0}{k-n}l^{k-n}&\cdots&1\end{bmatrix}

B.2. Proof

\begin{split}&\text{ Let }F(x,y)=\sum_{i=0}^{n}a_{i}x^{n-i}y^{i}\text{ denote an irreducible integral binary form.}\\ &\text{Let }\delta\text{ denote the root of }f(x):=F(x,1).\\ &\text{Let }F_{l}(x,y):=f(x+ly,y)=\sum_{i=0}^{n}b_{i}x^{n-i}y^{i}.\\ &\text{Note that }\delta-l\text{ is the root of }f_{l}(x):=F_{l}(x,1).\\ &\text{Let }s_{i}(x):=\sum_{j=0}^{i-1}a_{j}x^{i-j}=a_{0}x^{i}+a_{1}x^{i-1}+...+a_{i-1}x.\\ &\text{Let }m_{i}(x):=\sum_{j=0}^{i-1}b_{j}x^{i-j}=b_{0}x^{i}+b_{1}x^{i-1}+...+b_{i-1}x.\end{split}

(22)

Thus the basis for $R_{F_{l}}$ will be given by

\displaystyle\mathbb{Z}\langle 1,\>m_{1}(\delta-l)+b_{1},\>m_{2}(\delta-l)+b_{2},\>\cdots,\>m_{k}(\delta-l)+b_{k},\>\cdots,\>m_{n-1}(\delta-l)+b_{n-1}\rangle

and that of $R_{f}$ will be given by

\displaystyle\mathbb{Z}\langle 1,\>s_{1}(\delta)+a_{1},\>s_{2}(\delta)+a_{2},\>\cdots,\>s_{k}(\delta)+a_{k},\>\cdots,\>s_{n-1}(\delta)+a_{n-1}\rangle

Lemma 11.

b_{k}=\sum_{i=0}^{k}{{n-i}\choose{n-k}}a_{i}\\

(23)

Proof.

f(x+l)=\sum_{i=0}^{n}a_{i}(x+l)^{n-i}=\sum_{i=0}^{n}a_{i}\sum_{j=0}^{n-i}{n-i\choose j}x^{j}l^{n-i-j}

=\sum_{i=0}^{n}\sum_{j=0}^{n-i}a_{i}{n-i\choose j}x^{j}l^{n-i-j}=\sum_{j=0}^{n}x^{j}\sum_{i=0}^{n-j}a_{i}{n-i\choose j}l^{n-i-j}

Thus,

b_{k}=\sum_{i=0}^{k}a_{i}{n-i\choose n-k}l^{k-i}

∎

Lemma 12.

	$\displaystyle m_{k}(x)+b_{k}$	$\displaystyle=x(m_{k-1}(x)+b_{k-1})+b_{k}$
	$\displaystyle s_{k}(x)+a_{k}$	$\displaystyle=x(s_{k-1}(x)+a_{k-1})+a_{k}$

Proof.

By definition. ∎

Proposition 2.

m_{k}(x-l)+b_{k}=\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j}+{n-1\choose k}l^{k}a_{0}\\

(24)

Proof.

We proceed by induction on $k$ ,
Base Case.
$m_{1}(x-l)=b_{0}(x-l)+b_{1}=(a_{0})(x-l)+a_{1}+na_{0}l$
$=(a_{0}x+a_{1})+{n-1\choose 1}a_{0}l$
$={n-2\choose 0}(s_{1}(x)+a_{1})+{n-1\choose 1}a_{0}$

Induction Hypothesis:

m_{k}(x-l)+b_{k}=\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j}+{n-1\choose k}l^{k}a_{0}

We have from lemma 12,

m_{k+1}(x-l)=(x-l)(m_{k}(x-l)+b_{k}).

Substituting in our induction hypothesis, we get

	$\displaystyle m_{k+1}(x-l)$
	$\displaystyle=(x-l)(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j}+{n-1\choose k}l^{k}a_{0})$
	$\displaystyle=\sum_{j=0}^{k-1}{n-k+j-1\choose j}((s_{k-j}(x)+a_{k-j})l^{j}x-(s_{k-j}(x)+a_{k-j})l^{j+1})+{n-1\choose k}a_{0}l^{k}x-{n-1\choose k}l^{k+1}a_{0}$
	$\displaystyle=(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)x+a_{k-j}x)l^{j}+{n-1\choose k}a_{0}xl^{k})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j+1}+{n-1\choose k}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j+1}(x))l^{j}+{n-1\choose k}s_{1}(x)l^{k})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j+1}+{n-1\choose k}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k+j-1\choose j}(s_{k-j+1}(x))l^{j}-(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j+1}+{n-1\choose k}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k+j-1\choose j}(s_{k-j+1}(x)+a_{k-j+1})l^{j}-(\sum_{j=0}^{k-1}{n-k+j-1\choose j}(s_{k-j}(x)+a_{k-j})l^{j+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-{n-1\choose k}a_{0}l^{k+1}-\sum_{j=0}^{k-1}{n-k+j-1\choose j}a_{k-j+1}l^{j}$
	$\displaystyle=\sum_{j=0}^{k}({n-k+j-1\choose j}-{n-k+j-2\choose j-1})(s_{k+1-j}+a_{k+1-j})l^{j})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-{n-1\choose k}a_{0}l^{k+1}-\sum_{j=0}^{k-1}{n-k+j-1\choose j}a_{k-j+1}l^{j}$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j})-{n-1\choose k}a_{0}l^{k+1}-\sum_{j=0}^{k}{n-k+j-1\choose j}a_{k-j+1}l^{j}$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-{n-1\choose k}a_{0}l^{k+1}-(\sum_{j=0}^{k}{n-k+j-1\choose j}a_{k-j+1}l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-(\sum_{j=0}^{k}{n-k+j-1\choose j}a_{k-j+1}l^{j}+({n-1\choose k+1}+{n-1\choose k})a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-(\sum_{j=0}^{k}{n-k+j-1\choose j}a_{k-j+1}l^{j}+{n\choose k+1}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-(\sum_{j=0}^{k}{n-k+j-1\choose n-k-1}a_{k-j+1}l^{j}+{n\choose k+1}a_{0}l^{k+1})$
	$\displaystyle=(\sum_{j=0}^{k}{n-k-2+j\choose j}(s_{k+1-j}+a_{k+1-j})l^{j}+{n-1\choose k+1}a_{0}l^{k+1})$
	$\displaystyle\text{\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt\hskip 28.45274pt}-b_{k+1}$

∎

Substituting $x=\delta$ in the above Proposition we get the result.

B.3.

Given $f$ a binary form, we let $\langle B_{0},B_{1},\cdots,B_{n-1}\rangle$ be the canonical basis of $R_{f}$ associated to $f$ as in eq. 3. Let $\langle v_{1},v_{2},\cdots,v_{n}\rangle$ denote a real basis for $\mathbb{R}^{n}.$ We perform a Gram-Schmidt reduction on the basis using some canonical distance form. We write

\langle v_{1},v_{2},\cdots,v_{n}\rangle=\langle B\rangle M[(t_{1},t_{2},...,t_{n})]

where $B$ is an orthonormal ordered basis of vectors, all of which are of the same size. $M$ is an upper triangular uni-potent matrix. And $[(t_{1},t_{2},..,t_{n})]$ denotes a diagonal matrix with $t_{i}$ placed in the $(i,i)^{th}$ place.

Let $v_{i}^{\prime}$ denote the projection of $v_{i}$ to the space orthogonal to the space spanned by $\{v_{1},v_{2},\cdots,v_{i-1}\}$ . Then, Gram-Schmidt process forces

|t_{i}|=||v_{i}^{\prime}||.

(25)

We note that for each $1\leq i\leq n$ , and any choice of coefficients $a_{1},\cdots,a_{i}$ we have

||a_{i}v_{i}+\sum_{j=1}^{i-1}a_{j}v_{j}||\ \geq\ |a_{i}||t_{i}|.

This follows by looking at the component of the given vector (on RHS) along $v^{\prime}_{i}.$ Moreover, for each $1\leq i\leq n$ , there exist $b_{1},\cdots,b_{i-1}\in\mathbb{Z}$ so that

||v_{i}+\sum_{j=1}^{i-1}b_{j}v_{j}||\leq|t_{i}|+|t_{i-1}|+\cdots+|t_{1}|.

One can simply choose $b_{j}$ such that the $\sum_{j=1}^{i-1}b_{j}v_{j}$ approximates the vector $v_{i}-v^{\prime}_{i}$ which clearly lies in the space spanned by $\langle v_{1},v_{2},\cdots,v_{i-1}\rangle.$

Now if $t_{i+1}/{t_{i}}>2$ for all $i$ , one can see that inductively the $i^{th}$ vector in the Minkowski reduced basis of the lattice spanned by $\{v_{1},v_{2}\cdots,v_{n}\}$ will have the form $\pm v_{i}+\sum_{j=1}^{i-1}b_{j}v_{j}.$

In fact, having

t_{i}/{t_{i-1}}>\sqrt{1+\frac{1}{i}}\textit{ for all }1\leq i\leq n

is sufficient to conclude this.

Furthermore, if we just know that,

\frac{t_{i}}{t_{2}}\geq 2\textit{ for all }3\leq i\leq n\textit{ and }\frac{t_{2}}{t_{1}}\geq 2

(26)

then, the first two vectors (which will be unique up to sign) in the Minkowski reduced basis for the lattice $\langle v_{1},v_{2},\cdots,v_{n}\rangle$ will have the above form. That is

\{\pm v_{1},\pm(v_{2}+a\cdot v_{1})\}

will be the first two elements in the Minkowski reduced basis for this lattice spanned by $\langle v_{1},v_{2},\cdots,v_{n}\rangle.$

Definition 36.

If $\langle v_{1},v_{2},\cdots,v_{n}\rangle$ is a basis for $\mathbb{R}^{n}$ and $v_{i}^{\prime}$ denotes the projection of $v_{i}$ to the space orthogonal to the space spanned by $\{v_{1},v_{2},\cdots,v_{i-1}\}$ and $t_{i}=||v_{i}^{\prime}||,$ then we say $\langle v_{1},v_{2},\cdots,v_{n}\rangle$ is Normally Minkowski Reduced if $t_{i}$ satisfy

\frac{t_{i}}{t_{2}}\geq 2\textit{ for all }3\leq i\leq n\textit{ and }\frac{t_{2}}{t_{1}}\geq 2

(27)

Our discussion above gives us the following lemma:

Lemma 13.

If $\langle B_{1},B_{2},\cdots,B_{n}\rangle$ is Normally Minkowski Reduced, then

(1)

$B_{1}$ is the unique smallest vector in $L$ up to sign.
(2)

$\exists a\in\mathbb{Z}$ such that, $B_{2}+a\cdot B_{1}$ is the smallest vector in $L$ which is not in $\mathbb{Z}\cdot B_{1}$ up to sign.

B.4. Reduced-m-Polynomials

Definition 37.

We say a tuple $(f,m,l)$ is a Reduced- $m$ -Polynomial (at $l$ ) if

•

The canonical basis for $R^{\prime}_{(f_{l},m)}$ is Normally Minkowski Reduced when seen in $R^{\prime}_{(f_{l},m)}\otimes\mathbb{R}\simeq\mathbb{R}^{r}\oplus\mathbb{C}^{s}$ under the canonical norm.

Remark 49.

We will ignore the $l$ part as $(f,m,l)$ is a Reduced- $m$ -polynomial (at $l$ ) $\iff$ $(f,m,r)$ is a Reduced- $m$ -Polynomial (at $r$ ) for any real value $r.$ This is easy to see from the fact that the matrix in theorem 48 is upper triangular and hence will leave relative sizes of normal components unchanged.

Theorem 50.

Given $f,g$ polynomials of degree $n\geq 4$ such that $f$ is a Reduced- $m$ -polynomial and $f$ is weakly divisible by $m$ at $e$ and $g$ is a Reduced- $m^{\prime}$ -polynomial and $g$ is weakly divisible by $m^{\prime}$ at $d$ ( $m,m^{\prime}\geq 1$ ) with $R^{\prime}_{(f_{e},m)}=R^{\prime}_{(g_{d},m)}$ then

m=m^{\prime}\textit{ and }\exists r\in\mathbb{Z}:g(x)=f(x+mr-d+e).

Proof.

Let $\beta$ and $\alpha$ denote roots of $f$ and $g$ respectively ( $\in\mathbb{R}^{r}\oplus\mathbb{C}^{s}$ ). Let $a_{0}$ and $b_{0}$ denote the positive leading coefficients of $f$ and $g$ and $a_{1}$ and $b_{1}$ denote the second leading coefficient of $f$ and $g$ respectively.

We know that the smallest two elements of any Normally Minkowski Reduced basis are unique. Since $R^{\prime}_{(f_{e},m)}=R^{\prime}_{(g_{d},m^{\prime})},$ comparing the smallest two elements elements in these, we get

\langle 1,a_{0}\cdot\beta\rangle\begin{bmatrix}a&b\\ 0&c\end{bmatrix}=\langle 1,b_{0}\cdot\alpha\rangle

where $a=\pm 1$ and $c=\pm 1.$ Clearly, $a=1$ and thus it follows that

\alpha=\frac{\pm a_{0}}{b_{0}}\cdot\beta+\frac{b}{b_{0}}.

\frac{f(\frac{\pm a_{0}}{b_{0}}\cdot x+\frac{b}{b_{0}})}{a_{0}}=(\pm\frac{a_{0}}{b_{0}})^{n}\frac{g(x)}{b_{0}}.

This immediately tells us that the matrix of transfer from $\langle 1,a_{0}\beta,a_{0}\beta^{2}+a_{1}\beta\rangle$ to $\langle 1,b_{0}\alpha,b_{0}\alpha^{2}+b_{1}\alpha\rangle$ is

\begin{bmatrix}1&b&*\\ 0&c&*\\ 0&0&\frac{a_{0}}{b_{0}}\end{bmatrix}.

Since $n\geq 4$ the above matrix must also be integral and invertible. It follows that $a_{0}=b_{0}$ and thus $\mathfrak{disc}(f)=\mathfrak{disc}(g)$ .

Since $R^{\prime}_{(f_{e},m)}=R^{\prime}_{(g_{d},m^{\prime})}$ we also have

\frac{\mathfrak{disc}(f)}{m^{2}}=\frac{\mathfrak{disc}(g)}{m^{\prime 2}},

and we immediately get $m=m^{\prime}.$

Now, without loss of generality, we may assume $c=1$ , for if $c=-1$ we may change $g(x)$ to $(-1)^{n}g(-x).$ We let $l=\frac{b}{b_{0}}\in\mathbb{Q}.$ Thus, $f(x+l)=g(x).$

We make note of the fact that $f$ and $g$ are weakly divisible by the same value $m$ .

Observing the matrix of transformation from $R_{f_{e}}$ to $R_{g_{d}}$ given using LABEL:l_matrix_of_tranformation, we notice that the entry in the $(n-1)^{th}$ row and $n^{th}$ column of the matrix of transfer for the canonical bases of $R_{f_{e}}$ to $R_{g_{d}}$ is $d-e+l$ .

Thus, the entry in the $(n-1)^{th}$ row and $n^{th}$ column of the matrix of transfer for the canonical bases of $R^{\prime}_{(f_{e},m)}$ to $R^{\prime}_{(g_{d},m)}$ is $\frac{d-e+l}{m}$ . Thus $r=\frac{d-e+l}{m}$ must be an integer. It follows that $\exists r\in\mathbb{Z}:g(x)=f(x+mr-d+e).$ ∎

Lemma 14.

If $f$ is a real monic polynomial, then there exists a $\rho_{f}\in\mathbb{R}$ (continuously varying with $f$ ) such that $\lambda\rho^{n}f(\frac{x}{\rho})$ is a Reduced- $1$ -polynomial for all $\rho\geq\rho_{f}$ and $\lambda\geq 1$ .

If $\rho\geq m^{1/(n-2)}\rho_{f}$ and $\lambda\geq 1$ the polynomial $\lambda\rho^{n}f(\frac{x}{\rho})$ is a Reduced- $m$ -polynomial.

Proof.

We follow the argument of Bhargava-Shankar-Wang proving in Lemma 5.2 in[2].

Set

f(x)=x^{n}+a_{n-1}x^{n-1}+\cdots+a_{k}x^{n-k}+\cdots+a_{0}

and set $a_{n}=1.$

We perform a Gram-Schmidt reduction of the basis for $R_{f}$ using the canonical distance form on $R_{f}\bigotimes\mathbb{R}\simeq\mathbb{R}^{r}\bigoplus\mathbb{C}^{s}$ ( $r+2s=n$ ). We write

\langle 1,\>\delta,\>\delta^{2}+a_{n-1}\delta,\>\cdots,\>\sum_{i=0}^{k-1}a_{n-i}\delta^{k-i},\>\cdots,\>\sum_{i=0}^{n-2}a_{n-i}\delta^{k-i}\rangle=\langle B\rangle M[(t_{1},t_{2},...,t_{n})]

where $B$ is an orthonormal ordered basis of vectors, $M$ is an upper triangular unipotent matrix, and $[(t_{1},t_{2},..,t_{n})]$ denotes a diagonal matrix with those entries along the diagonal. Now we note that if

	$\displaystyle\langle 1,\>\delta,\>$	$\displaystyle\delta^{2}+a_{n-1}\delta,\>\cdots,\>\sum_{i=0}^{k-1}a_{n-i}\delta^{k-i},\>\cdots,\>\sum_{i=0}^{n-2}a_{n-i}\delta^{n-1-i}\rangle$
		$\displaystyle=\langle B\rangle M[(t_{1},t_{2},\cdots,t_{k},\cdots,t_{n})]$

then, using eq. 25, we see that

	$\displaystyle\langle 1,\>\lambda(\rho\delta),\>$	$\displaystyle\lambda((\rho\delta)^{2}+a_{n-1}\rho\delta),\>\cdots,\>\lambda(\sum_{i=0}^{k-1}a_{n-i}(\rho\delta)^{k-i}),\>\cdots,\>\lambda(\sum_{i=0}^{n-2}a_{n-i}(\rho\delta)^{n-1-i}\rangle)$
		$\displaystyle=\langle B\rangle M_{\rho}[(t_{1},\lambda\rho t_{2},\cdots,\lambda\rho^{k-1}t_{k},\dots,\lambda\rho^{n-1}t_{n})].$

Thus, for each polynomial $f$ one may find $\rho_{f}$ which is a continuous function of $f$ such that $\lambda\rho^{n}f(\frac{x}{\rho})$ is Minkowski reduced for all $\lambda\geq 1$ and $\rho\geq\rho_{f}$ . One can simply take

\rho_{f}:=\max\{\max_{i\geq 3}\{(\frac{2t_{2}}{t_{i}})^{1/(i-2)}\},\frac{2t_{1}}{t_{2}}\}.

Furthermore, we note that translating the polynomial changes the canonical basis by an upper triangular matrix.

Thus, if $g(x)=\lambda\rho^{n}f(\frac{X}{\rho}),$ then the canonical basis of $R^{\prime}_{(g_{l},m)}$ after Gram-Schmidt process will look like

\langle B\rangle M_{\rho,m,l}[(t_{1},\lambda\rho t_{2},\cdots,\lambda\rho^{k-1}t_{k},\cdots,\lambda\rho^{n-2}t_{n-1},\frac{\lambda\rho^{n-1}}{m}t_{n})]

where $M_{\rho,m,l}$ is a uni-potent upper triangular matrix.

It follows that for $\rho\geq m^{1/(n-2)}\rho_{f}$ and $\lambda\geq 1,$ $\lambda\rho^{n}f(\frac{X}{\rho})$ is Reduced- $m$ -polynomial. ∎

Appendix C The structure of the discriminant polynomial

Let

f(X,Y):=a_{0}X^{n}+a_{1}X^{n-1}Y+\cdots+a_{n}Y^{n}

denote a binary form of degree $n$ . Let $a_{n}\neq 0$ . Let

\mathfrak{disc}(a_{0},a_{1},\cdots,a_{n}):=\mathfrak{disc}(f)

denote the discriminant as a polynomial in the coefficients of $f.$ Then, we know that

a_{0}\mathfrak{disc}(f)=Res(f,\frac{\partial}{\partial X}f).

Since, the resultant can be seen as the determinant of the Sylvester matrix. The Sylvester matrix for these polynomials is

\setcounter{MaxMatrixCols}{11}\begin{bmatrix}a_{0}&a_{1}&a_{2}&\cdots&a_{n-1}&a_{n}&0&\cdots&0&0&0\\ 0&a_{0}&a_{1}&\cdots&a_{n-2}&a_{n-1}&a_{n}&\cdots&0&0&0\\ 0&0&a_{0}&\cdots&a_{n-3}&a_{n-2}&a_{n-1}&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&a_{3}&a_{4}&a_{5}&\cdots&a_{n}&0&0\\ 0&0&0&\cdots&a_{2}&a_{3}&a_{4}&\cdots&a_{n-1}&a_{n}&0\\ 0&0&0&\cdots&a_{1}&a_{2}&a_{3}&\cdots&a_{n-2}&a_{n-1}&a_{n}\\ na_{0}&(n-1)a_{1}&(n-2)a_{2}&\cdots&a_{n-1}&0&0&\cdots&0&0&0\\ 0&na_{0}&(n-1)a_{1}&\cdots&2a_{n-2}&a_{n-1}&0&\cdots&0&0&0\\ 0&0&na_{0}&\cdots&3a_{n-3}&2a_{n-2}&a_{n-1}&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&(n-2)a_{2}&(n-3)a_{3}&\cdots&\cdots&a_{n-1}&0&0\\ 0&0&0&\cdots&(n-1)a_{1}&(n-2)a_{2}&(n-3)a_{3}&\cdots&2a_{n-2}&a_{n-1}&0\\ 0&0&0&\cdots&na_{0}&(n-1)a_{1}&(n-2)a_{2}&\cdots&3a_{n-3}&2a_{n-2}&a_{n-1}\end{bmatrix}.

(28)

The following lemmata are easy to observe from the following:

Lemma 15.

\mathfrak{disc}(f)=a_{n}\Delta_{1}+a_{n}a_{n-1}\Delta_{2}+a_{n-1}^{2}\Delta_{3}+a_{n}^{2}\Delta_{4}.

where

\Delta_{1}=4a_{n-2}^{3}\cdot\mathfrak{disc}(a_{0}X^{n-2}+a_{1}X^{n-3}Y+\cdots+a_{n-2}Y^{n-2})

and

\Delta_{3}=a_{n-1}^{2}\mathfrak{disc}(a_{0}X^{n-1}+a_{1}X^{n-2}Y+\cdots+a_{n-2}XY^{n-2}+a_{n-1}Y^{n-1}).

Proof.

We carefully expand the determinant of the Sylvester Matrix along the last two columns. We can clearly see we can write the determinant as

2a_{n}a_{n-2}\Delta_{1}+a_{n-1}^{2}\Delta_{2}+a_{n}a_{n-1}\Delta_{3}+a_{n}^{2}\Delta_{4}.

(29)

We will pick $\Delta_{1}\in\mathbb{Z}[a_{0},a_{1},\cdots,a_{n-2}]$ and as the components of $\Delta_{1}$ containing $a_{n-1}$ and $a_{n}$ can be pushed inside $\Delta_{2},\Delta_{3},\Delta_{4}$ . One can then choose $\Delta_{3}\in\mathbb{Z}[a_{0},a_{1},\cdots,a_{n-1}]$ as components of $\Delta_{2}$ containing $a_{n}$ can be pushed inside $\Delta_{2}$ and $\Delta_{4}.$

Then, $\Delta_{1}$ can be seen as the determinant of (i.e. the minor corresponding to $a_{n}\cdot(2a_{n-2})$ where $a_{n}$ and $a_{n-1}$ are set to zero looks as follows)

\begin{bmatrix}a_{0}&a_{1}&a_{2}&\cdots&0&0&0&\cdots&0\\ 0&a_{0}&a_{1}&\cdots&a_{n-2}&0&0&\cdots&0\\ 0&0&a_{0}&\cdots&a_{n-3}&a_{n-2}&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&a_{3}&a_{4}&a_{5}&\cdots&0\\ 0&0&0&\cdots&a_{2}&a_{3}&a_{4}&\cdots&0\\ na_{0}&(n-1)a_{1}&(n-2)a_{2}&\cdots&0&0&0&\cdots&0\\ 0&na_{0}&(n-1)a_{1}&\cdots&2a_{n-2}&0&0&\cdots&0\\ 0&0&na_{0}&\cdots&3a_{n-3}&2a_{n-2}&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&(n-2)a_{2}&(n-3)a_{3}&\cdots&\cdots&0\\ 0&0&0&\cdots&(n-1)a_{1}&(n-2)a_{2}&(n-3)a_{3}&\cdots&2a_{n-2}\end{bmatrix}.

Now doing operations $R_{n}-R_{1},R_{n+1}-R_{2},R_{n+1}-R_{3},\cdots,R_{2n-2}-R_{n-2}$ we get the following matrix with the same determinant.

\begin{bmatrix}a_{0}&a_{1}&a_{2}&\cdots&0&0&0&\cdots&0\\ 0&a_{0}&a_{1}&\cdots&a_{n-2}&0&0&\cdots&0\\ 0&0&a_{0}&\cdots&a_{n-3}&a_{n-2}&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&a_{3}&a_{4}&a_{5}&\cdots&0\\ 0&0&0&\cdots&a_{2}&a_{3}&a_{4}&\cdots&0\\ (n-1)a_{0}&(n-2)a_{1}&(n-3)a_{2}&\cdots&0&0&0&\cdots&0\\ 0&(n-1)a_{0}&(n-2)a_{1}&\cdots&a_{n-2}&0&0&\cdots&0\\ 0&0&(n-1)a_{0}&\cdots&2a_{n-3}&a_{n-2}&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&(n-3)a_{2}&(n-4)a_{3}&\cdots&\cdots&0\\ 0&0&0&\cdots&(n-1)a_{1}&(n-2)a_{2}&(n-3)a_{3}&\cdots&2a_{n-2}\end{bmatrix}.

We note that the above matrix is very close to Sylvester matrix for computing the discriminant of

\frac{f(x,y)-a_{n}Y^{n}-a_{n-1}XY^{n-1}}{Y}=X(a_{0}X^{n-2}+a_{1}X^{n-3}Y+\cdots+a_{n-2}Y^{n-2}).

More particularly, the determinant of the above matrix is twice of the discriminant of the above form. This can be easily observed by expanding the determinant of the above matrix and that of the Sylvester Matrix along the final column.

Clearly, since $a_{n-2}$ is the Resultant of $X$ and $a_{0}X^{n-2}+a_{1}X^{n-3}Y+\cdots+a_{n-2}Y^{n-2},$ we get

\mathfrak{disc}(\frac{f(x,y)-a_{n}Y^{n}-a_{n-1}XY^{n-1}}{Y})=a_{n-2}^{2}\cdot\mathfrak{disc}(a_{0}X^{n-2}+a_{1}X^{n-3}Y+\cdots+a_{n-2}Y^{n-2}).

Substituting back into eq. 29 we get the appropriate structure for $\Delta_{1}$ .

To get the structure of $\Delta_{3}$ we see that

a_{n-1}^{2}\Delta_{3}=\mathfrak{disc}(f(X,Y)-a_{n}Y^{n})

by definition. ∎

References

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[2] M. Bhargava, A. Shankar, and X. Wang. Squarefree values of polynomial discriminants I. Invent. Math., 228(3):1037–1073, 2022.
[3] M. Bhargava, A. Shankar, and X. Wang. Squarefree values of polynomial discriminants II, 2022.
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[5] J. S. Ellenberg and A. Venkatesh. The number of extensions of a number field with fixed degree and bounded discriminant. Annals of Mathematics, 163:723–741, 2003.
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Weighted enumeration of number fields using Pseudo and Sudo maximal orders

Abstract.

1. Introduction

Definition 1.

Theorem 1 (Number-field count).

Definition 2.

Theorem 2.

2. Dedekind Domains in number-fields

Notation 3.

Lemma 1.

Proof.

Lemma 2.

Proof.

Notation 4.

Theorem 3.

Notation 5.

Remark 4.

3. Orders and Separation Lemmata

Notation 6.

Notation 7.

Notation 8.

Definition 9.

Notation 10.

Remark 5.

Lemma 3 (Separation Lemma 1).

Proof.

Lemma 4 (Separation Lemma 2).

Proof.

Remark 6.

Theorem 7 (Fundamental Theorem of Orders).

Proof.

Notation 11.

Definition 12.

Definition 13.

4. Dedekind-Kummer type Theorems and the Fundamental theorem of Orders.

4.1. Dedekind-Kummer Theorem

Theorem 8 (Dedekind-Kummer Theorem).

Notation 14.

Theorem 9 (Dedekind’s Criterion).

Theorem 10.

Remark 11.

4.2. Dedekind-Kummer-Dedekind System for Binary Rings and Primitive Binary Forms

Theorem 12.

Theorem 13 (Dedekind’s Criterion Binary Rings).

4.3. Localizations of Binary Rings associated to primitive forms

Definition 15.

Definition 16.

Theorem 14.

Theorem 15.

Remark 16.

Notation 17.

Lemma 5.

Proof.

4.4. Proof of Dedekind Kummer in Binary/One-fine Rings or theorem 12.

4.5. Proof of Dedekind’s Criterion for Primitive binary forms or theorem 13

Notation 18.

5. Small prime splitting restrictions.

Remark 17.

Theorem 18.

Proof.

Remark 19.

Remark 20.

Remark 21.

6. Pseudo Maximal Orders and Sudo Maximal orders.

Remark 22.

6.1. Classification of Irreducible orders 𝒪\mathcal{O} with e​f​(𝒪)=1ef(\mathcal{O})=1 and e​f​(𝒪)=2ef(\mathcal{O})=2

6.1.1. e​f​(𝒪)=1ef(\mathcal{O})=1

6.1.2. e​f​(𝒪)=2ef(\mathcal{O})=2

Lemma 6.

Proof.

Remark 23.

Definition 19.

Lemma 7.

Theorem 24 (A).

Proof.

Theorem 25 (B).

Proof.

Theorem 26 (C).

Proof.

Corollary 1.

6.1. Classification of Irreducible orders $\mathcal{O}$ with $ef(\mathcal{O})=1$ and $ef(\mathcal{O})=2$

6.1.1. $ef(\mathcal{O})=1$

6.1.2. $ef(\mathcal{O})=2$

Appendix B Effective injectivity of the map: $(f,m,l)\longrightarrow R^{\prime}_{(f_{l},m)}$