---
abstract: 'We classify all quadratic imaginary number fields that have a Euclidean ideal class. There are seven of them, they are of class number at most two, and in each case the unique class that generates the class-group is moreover norm-Euclidean.'
author:
- Hester Graves
- Nick Ramsey
title: Euclidean Ideals in Quadratic Imaginary Fields
---

Introduction
============

In [@lenstra], Lenstra generalized the notion of a Euclidean domain to that of a Dedekind domain $R$ with a Euclidean ideal $C\subseteq R$. He proved that if $C$ is a Euclidean ideal then the class-group of $R$ is cyclic and generated by $C$. Moreover, if $C=R$, then his notion reduces to that of a Euclidean domain and the above result reduces to the familiar fact that a Euclidean domain is a principal ideal domain. Building on work of Weinberger ([@weinberger]), Lenstra ([@lenstra]) showed (conditional of the generalized Riemann hypothesis) that any generator of the class group of the ring of integers in a number field with infinite unit group is Euclidean. As the only number fields with finite unit group aside from ${{\mathbb{Q}}}$, it is natural to inquire about the situation for quadratic imaginary fields.

It is known that, among the nine quadratic imaginary fields of class number one, exactly five have Euclidean integer rings and in each case the norm serves as a Euclidean algorithm (see [@samuel]). The purpose of this paper is to extend this result to the setting of Euclidean ideal classes by determining all quadratic imaginary fields that have a Euclidean ideal. We record them in the following theorem.

\[thm1\] The quadratic imaginary fields with a Euclidean ideal are as follows.

   [class number]{}                          [fields]{}
  ------------------ ----------------------------------------------------------
          1           ${{\mathbb{Q}}}(\sqrt{-D})$ for $D\in\{1, 2, 3, 7, 11\}$
          2                ${{\mathbb{Q}}}(\sqrt{-D})$ for $D\in\{5,15\}$

In each case the unique class that generates the class-group is moreover norm-Euclidean.

If one is only interested in *norm*-Euclidean ideals, then this result is contained in Proposition 2.4 of [@lenstra]. Of course, as the results of Weinberger and Lenstra mentioned above conditionally demonstrate and examples of Clark ([@clark]) and Harper ([@harper]) unconditionally demonstrate (in the class number one case), there are ideals in integer rings that are are Euclidean but not with respect to the norm.

In the Euclidean ring setting, a construction of Motzkin ([@motzkin]) has proven to be a fruitful tool in the study of Euclidean rings that are not norm-Euclidean. In her thesis, the first author adapts this construction to the Euclidean ideal setting. Her techniques are the main tool used to prove Theorem \[thm1\].

\
[**Convention**]{} - In this paper all Euclidean algorithms are taken to be ${{\mathbb{N}}}$-valued and ${{\mathbb{N}}}$ is taken to include $0$.

A Motzkin-type construction for Euclidean ideals
================================================

Let $R$ be a Dedekind domain with fraction field $K$. We denote by $E$ the set of fractional ideals of $R$ in $K$ that contain $R$ itself. Recall from [@lenstra] that $C$ is called *Euclidean* if there exists a function $\psi:E\longrightarrow {{\mathbb{N}}}$ such that for all $I\in E$ and all $x\in IC\setminus C$, there exists $y\in C$ such that $$\psi((x+y)^{-1}IC)<\psi(I)$$ In this case $\psi$ is called a *Euclidean algorithm* for $C$. If $C$ is Euclidean then it generates the class-group of $R$. Also, if $\psi$ is a Euclidean algorithm for $C$ then it is also a Euclidean algorithm for any ideal in the same class as $C$ and no ideal in a different class than $C$. These facts are all elementary and can be found in [@lenstra].

The following definition, given in [@gravesthesis], is an adaptation of Motzkin’s construction to the Euclidean ideal setting.

\[hesterconstruction\] Let $C$ be a non-zero ideal in $R$. We define a nested sequence of subsets of $E$ as follows. Set $A_{C,0} = \{R\}$ and for $i>0$ we set $$A_{C,i} = A_{C,i-1}\cup \left\{ I\in E \left| \begin{array}{c}
    \forall x\in IC\setminus C\ \ \exists y\in C
    \\ \mbox{such\ that}\ (x-y)^{-1}IC \in
    A_{C,i-1}\end{array}\right.\right\}$$ Finally, set $A_C = \cup_i
  A_{C,i}$.

When the ideal $C$ is fixed or otherwise clear from the context, we will often omit it from the notation and simply use $A_i$ and $A=\cup_i A_i$. The significance of this construction is the following lemma of the first author (see [@gravesthesis]).

\[lem1\] The ideal $C$ is Euclidean if and only if $A=E$.

In fact, one can say more. Namely, if $A=E$, then the function $\psi:
E\longrightarrow {{\mathbb{N}}}$ defined by $\psi(I) = i$ if $I\in A_i\setminus
A_{i-1}$ is a Euclidean algorithm for $C$ and is minimal with respect to this property.

The following two lemmas furnish constraints on the sets $A_{C,i}$ that will be useful in what follows. The first is general in nature and highlights the role of cyclicity of the class-group.

\[lem2\] If $I\in A_i\setminus A_{i-1}$ then $[I] = [C^{-i}]$.

This is an immediate inductive consequence of Definition \[hesterconstruction\].

By definition, any $I\in A_i\setminus A_{i-1}$ has the property that for all $x\in IC\setminus C$ there exists $y\in C$ such that $(x+y)^{-1}IC\in A_{i-1}$. However, using the previous lemma and ideal class considerations, one can often cut down the set “$A_{i-1}$” in this statement. When $R^\times$ is finite, this observation is particularly useful because one can use it to efficiently bound the norm of a new element of $A_i$, as the following lemma demonstrates.

\[lem3\] Suppose that $R^\times$ is finite, and suppose that $S\subseteq
  A_{i-1}$ is a subset with the property that, if $I\in A_i\setminus
  A_{i-1}$ then for all $x\in IC\setminus C$ there exists $y\in C$ such that $$(x-y)^{-1}IC\in S$$ Then all $I\in
  A_i\setminus A_{i-1}$ have the property that $${{\mathrm{Nm}}}(I^{-1})\leq
  |R^\times||S|+1.$$

For $x\in IC\setminus C$, the condition that there exists $y\in C$ such that $(x-y)^{-1}IC$ is a particular ideal depends only on the class of $x$ in $IC/C$. Fix an ideal $I\in A_i\setminus A_{i-1}$. For each non-zero class in $IC/C$ choose a representative $x\in IC\setminus C$ and a $y\in C$ such that $(x-y)^{-1}IC\in S\subseteq A_{i-1}$. This collection of choices amounts to a (decidedly non-canonical) function $$(IC\setminus C)/C\longrightarrow S$$ Suppose that two classes in $(IC\setminus C)/C$ map to the same ideal and let $x_1$ and $x_2$ be their chosen representatives. Then there exist $y_1,y_2\in C$ such that $$(x_1-y_1)^{-1}IC =
  (x_2-y_2)^{-1}IC.$$ It follows that there exists a unit $u\in {{\EuScript{O}}}_K^\times$ such that $x_1-y_1=u(x_2-y_2)$, and hence $x_1-ux_2 = y_1-uy_2\in C$. That is, the classes of $x_1$ and $x_2$ in $IC/C$ differ (multiplicatively) by a unit. The upshot is that the set of nonzero classes in $IC/C$ modulo the multiplicative action of $R^\times$ injects into $S$, and hence $$|IC/C| -1 \leq
  |R^\times||S|$$ But since $R$ is Dedekind, the left side is simply ${{\mathrm{Nm}}}(I^{-1})-1$, so this is the desired inequality.

Application to quadratic imaginary fields
=========================================

For the remainder of the paper, $K$ will denote a quadratic imaginary field and ${{\EuScript{O}}}_K$ its ring of integers. One approach to classifying the Euclidean ${{\EuScript{O}}}_K$ is to break into cases according to the factorizations of small rational primes in ${{\EuScript{O}}}_K$ and use Lemmas \[lem2\] and \[lem3\] of the previous section to glean consequences about the sets $A_i$. If one uses the crutch of known lists of quadratic imaginary fields of small class number, then this approach *nearly* yields Theorem \[thm1\]. Indeed, aside from the known norm-Euclidean cases detailed in this theorem, one finds in nearly all cases that the sequence of sets $A_i$ stabilizes very quickly (one needn’t ever consider ideals with prime factors of norm larger than 7). The one vexing exception is the field $K={{\mathbb{Q}}}(\sqrt{-23})$. A bit of computation with SAGE ([@sage]) reveals that the $A_i$ in this case contain at least the inverses of every ideal of norm up to $47$. Lacking the patience to continue this computation to its end (and indeed the confidence that it had one), we decided to switch perspective.

It is convenient to first dispense with the cases where ${{\EuScript{O}}}_K^\times$ is unusually large, namely $K={{\mathbb{Q}}}(\sqrt{-1})$ and $K={{\mathbb{Q}}}(\sqrt{-3})$. These two fields are well-known to have norm-Euclidean rings of integers, and for any other $K$ we have ${{\EuScript{O}}}_K^\times = \{\pm 1\}$. From this point on we assume that $K$ is among the latter fields. It then follows from Lemma \[lem3\] that any $I\in A_1\setminus A_0$ has ${{\mathrm{Nm}}}(I^{-1})\leq 3$. As a result, by Lemma \[lem2\], a Euclidean ideal class in $K$ is represented by a residue degree one prime lying over $2$ or $3$.

Fix an embedding of $K$ into ${{\mathbb{C}}}$. We will freely identify $K$ with its image in ${{\mathbb{C}}}$ in what follows. Under this embedding, the field norm corresponds to the square of the complex absolute value. Note that a nonzero fractional ideal $C$ of $K$ is identified with a lattice in ${{\mathbb{C}}}$. Consider the union of the open disks of radius $\sqrt{{{\mathrm{Nm}}}(C)}$ centered about these lattice points. It is a simple consequence of the definition and the above comments that $C$ is norm-Euclidean if these disks cover all of ${{\mathbb{C}}}$ (see also [@lenstra]). The moral of the following result is that, if this covering fails too badly, then $C$ cannot possibly be Euclidean for *any* choice of algorithm.

\[prop1\] Let $K$ and $C$ be as above, and let $U$ denote the union of the open disks of radius $\sqrt{{{\mathrm{Nm}}}(C)}$ centered at the elements of $C$. If the complement of $U$ in ${{\mathbb{C}}}$ contains a nonempty open set, then $C$ is not a Euclidean ideal.

Before proceeding with the proof, we need the following lemma, which effectively states that inverses of fractional ideals of increasingly large norm are increasingly dense in $K$.

\[lemdense\] Let $K$ be a quadratic imaginary field and let $\varepsilon>0$ be any positive real number. There exists a number $M$ such that, for all $z\in K$ and all fractional ideals $I$ with ${{\mathrm{Nm}}}(I)>M$, there exists an element $x\in I^{-1}$ such that ${{\mathrm{Nm}}}(x-z)<\epsilon$.

Let $I_1, I_2, \dots, I_h$ be a set of representatives of the ideal class group of $K$. Viewing each fractional ideal $I_i^{-1}$ as a lattice in ${{\mathbb{C}}}$, we see that disks of sufficiently large radius centered at the elements of $C$ will cover ${{\mathbb{C}}}$. Thus, for each $i$ there exists a positive number $M_i$ such that, for each $z'\in K$ there exists $x'\in I_i^{-1}$ such that $${{\mathrm{Nm}}}(x'-z')=|x'-z'|^2<M_i$$ Now choose $M$ so that $M>\max_i (M_i{{\mathrm{Nm}}}(I_i)/\varepsilon)$. Let $z\in K$ and let $I$ be a fractional ideal with ${{\mathrm{Nm}}}(I)>M$. Choose $i$ so that $I=gI_i$ for some $g\in K^\times$ and pick $x'\in
  I_i^{-1}$ such that ${{\mathrm{Nm}}}(x'-gz)< M_i$. Then $$\begin{aligned}
  {{\mathrm{Nm}}}(g^{-1}x'-z) &= & {{\mathrm{Nm}}}(g^{-1}){{\mathrm{Nm}}}(x'-gz) \\ &<&
  \frac{{{\mathrm{Nm}}}(I_i)}{{{\mathrm{Nm}}}(I)}M_i <\varepsilon\end{aligned}$$ so that $x=g^{-1}x'\in (I_iI^{-1})I_i^{-1} = I^{-1}$ is the desired element.

(of Proposition \[prop1\]) Suppose that the complement of $U$ in ${{\mathbb{C}}}$ contains a nonempty open set. Arguing by contradiction, let us suppose that $C$ is Euclidean for the algorithm $\psi:E\longrightarrow {{\mathbb{N}}}$, so by Lemma \[lem1\], $A=\cup A_i = E$. Since $K$ is dense in ${{\mathbb{C}}}$ under its embedding, the complement of $U$ contains an $\sqrt{\varepsilon}$-neighborhood of an element $z\in K$ for some $\varepsilon>0$. Let $M$ be as in Lemma \[lemdense\] for this $K$ and $\varepsilon$.

Suppose that $I_0\in E$ and ${{\mathrm{Nm}}}(I_0^{-1}C^{-1})>M$. By Lemma \[lemdense\], there exists $x\in I_0C$ such that $$|x-z| =
  ({{\mathrm{Nm}}}(x-z))^{1/2}<\sqrt{\varepsilon}$$ It follows that $x$ lies in the complement of $U$. Since $x\in I_0C\setminus C$ and $I_0\in
  E=A$, there exists $y\in C$ such that $\psi((x+y)^{-1}I_0C)<\psi(I_0)$. Define $I_1 = (x+y)^{-1}I_0C$ and note that $I_1\in E$ and $\psi(I_1)<\psi(I_0)$. By the above, we also have $${{\mathrm{Nm}}}(I_1) = {{\mathrm{Nm}}}(I_0){{\mathrm{Nm}}}(C)/{{\mathrm{Nm}}}(x+y) =
  {{\mathrm{Nm}}}(I_0){{\mathrm{Nm}}}(C)/|x+y|^2 \leq {{\mathrm{Nm}}}(I_0)$$ since $x$ lies in the complement of $U$.

Because of this norm inequality, we again have ${{\mathrm{Nm}}}(I_1^{-1}/C)>M$ and can repeat the argument with $I_0$ replaced by $I_1$ to obtain a fractional ideal $I_2\in E$ with $\psi(I_2)<\psi(I_1)$ and ${{\mathrm{Nm}}}(I_2)\leq {{\mathrm{Nm}}}(I_1)$. Proceeding in this fashion, we obtain a sequence of ideals $I_0, I_1, \dots$ in $E$ with $${{\mathrm{Nm}}}(I_0)\geq {{\mathrm{Nm}}}(I_1)\geq {{\mathrm{Nm}}}(I_2)\geq \cdots$$ and $$\psi(I_0)>\psi(I_1)>\psi(I_2)>\cdots$$ But the latter is clearly impossible, as ${{\mathbb{N}}}$ is well-ordered. We conclude that $C$ could not have been Euclidean to begin with.

With the running restrictions on $K$, we know that any Euclidean ideal class is represented by a prime of norm $2$ or $3$. We are led by the above to examine the union $U$ as above for $C$ a degree one prime dividing $2$ or $3$. In determining the extent to which $U$ covers ${{\mathbb{C}}}$, it is clear that one need only consider a particular fundamental domain for $C$ in ${{\mathbb{C}}}$. Let $K = {{\mathbb{Q}}}(\sqrt{-D})$ for a square-free positive integer $D$. In each of the following cases, we identify which $D$ correspond to the case, and for such $D$ we draw a fundamental domain and the covering circles comprising $U$ that meet this fundamental domain. The pictures below are were generated with SAGE ([@sage]).

As we will see, as $D$ increases, the fundamental domains we choose below get too tall to be covered entirely by these disks. In each case, we illustrate the fundamental domain and the disks comprising $U$ that meet it. We do this for the following $D$ in each class: those for which $U$ covers all of ${{\mathbb{C}}}$ and the first $D$ for which it does not (keeping in mind that we are only interested in square-free $D$). For the latter $D$, as we will see, it always happens that the complement of $U$ moreover contains an open set (as opposed to having a nonempty but discrete complement), so Proposition \[prop1\] implies that $C$ is not Euclidean.

We note that in each case below, the given *ideal* generators of $C$ are also generators of $C$ as an Abelian group, as is easy to check. Thus the parallelogram that they span forms the boundary of a fundamental domain, which is the one that we consider in each case.

Since there is a degree one prime dividing $2$, $2$ either ramifies or splits in $K$, corresponding to the conditions $D\equiv 1,2\pmod{4}$ and $D\equiv 7\pmod{8}$, respectively. We consider the various sub-cases separately.

Here ${{\EuScript{O}}}_K$ is generated as an algebra over ${{\mathbb{Z}}}$ by $\sqrt{-D}$, so a defining polynomial of ${{\EuScript{O}}}_K$ is $x^2+D$. Modulo $2$, this is congruent to $(x+1)^2$, so the unique prime above $(2)$ is $(2,\sqrt{-D}+1)$. Thus we use $2$ and $\sqrt{-D}+1$ to span a parallelogram bounding a fundamental domain, and obtain the following pictures for increasing $D$.

![image](2_1_d1.png) ![image](2_1_d5.png) ![image](2_1_d13.png)

$D=1$ $D=5$ $D=13$

We conclude that the only $D$ under consideration (recall that $D=1$ was treated separately because of additional units) in this class for which a degree one prime over $2$ is Euclidean is $D=5$.

Again the defining polynomial of ${{\EuScript{O}}}_K$ is $x^2+D$. Modulo $2$, this is simply $x^2$, so the prime above $(2)$ is $(2,\sqrt{-D})$, and working as above we obtain the pictures.

![image](2_2_d2.png) ![image](2_2_d6.png)

$D=2$ $D=6$

We conclude that the only $D$ in this class for which a degree one prime over $2$ is Euclidean is $D=2$

Here ${{\EuScript{O}}}_K$ is generated as an algebra over ${{\mathbb{Z}}}$ by $\frac{1+\sqrt{-D}}{2}$. The defining polynomial is then $x^2-x+\frac{1+D}{4}$, which is congruent modulo $2$ to $x(x-1)$. The primes above $(2)$ are $(2,\frac{1+\sqrt{-D}}{2})$ and $(2,\frac{-1+\sqrt{-D}}{2})$. As these are Galois-conjugate, we need only examine the first, which gives the following pictures.

![image](2_3_d7.png) ![image](2_3_d15.png) ![image](2_3_d23.png)

$D=7$ $D=15$ $D=23$

We conclude that the only $D$ in this class for which a degree one prime over $2$ is Euclidean are $D=7$ and $D=15$. It is worth mentioning that the complement $U$ for $D=23$ is very small. This explains the atypical behavior of the sets $A_i$ for ${{\mathbb{Q}}}(\sqrt{-23})$ in that they do not stabilize quickly.

\

Next, we examine the case of a degree one primes dividing $3$. Again, this means that either $3$ ramifies or splits in $K$, but in order associate a congruence condition on $D$, we must also take into account the residue of $D$ mod $4$ since this effects the nature of the ring of integers ${{\EuScript{O}}}_K$.

These conditions are equivalent to $D\equiv 6, 9\pmod{12}$. Here ${{\EuScript{O}}}_K$ is generated over ${{\mathbb{Z}}}$ by $\sqrt{-D}$, so a defining polynomial is $x^2+D$. Modulo $3$ this is $x^2$, so the prime above $(3)$ is $(3,\sqrt{-D})$. Already for $D=6$, we see that the complement of $U$ contains a nonempty open set.

![image](3_1_d6.png)

$D=6$

We conclude that there are no $D$ in this class for which a degree one prime over $3$ is Euclidean.

These conditions amount to $D\equiv 2, 5\pmod{12}$. Again, $x^2+D$ is a defining polynomial, which is congruent modulo $3$ to $(x-1)(x+1)$. Thus the primes above $(3)$ are $(3,\sqrt{-D}+1)$ and $(3,\sqrt{-D}-1)$. As these are Galois-conjugate, we need only consider the first, which gives the following pictures.

![image](3_2_d2.png) ![image](3_2_d5.png) ![image](3_2_d14.png)

$D=2$ $D=5$ $D=14$

We conclude that the only $D$ in this class for which a degree one prime over $3$ is Euclidean are $D=2$ and $D=5$.

This amounts to $D\equiv 3\pmod{12}$, and in this case $\frac{1+\sqrt{-D}}{2}$ generates ${{\EuScript{O}}}_K$, and $x^2-x+\frac{1+D}{4}$ is a defining polynomial. Modulo $3$, this is congruent to $(x+1)^2$, so the prime above $(3)$ is $(3,\frac{3+\sqrt{-D}}{2})$.

![image](3_3_d3.png) ![image](3_3_d15.png) ![image](3_3_d39.png)

$D=3$ $D=15$ $D=39$

We conclude that the only $D$ under consideration (recall that $D=3$ was treated separately because of extra units) in this class for which a degree one prime over $3$ is Euclidean is $D=15$.

This amounts to $D\equiv 11 \pmod{12}$. Again, ${{\EuScript{O}}}_K$ is generated by $\frac{1+\sqrt{-D}}{2}$ and $x^2-x+\frac{1+D}{4}$ is a defining polynomial. Modulo $3$, this is $x(x-1)$, so the primes above $(3)$ are $(3,\frac{1+\sqrt{-D}}{2})$ and $(3,\frac{-1+\sqrt{-D}}{2})$. As these are Galois-conjugate, we need only consider the first, which gives the following pictures.

![image](3_4_d11.png) ![image](3_4_d23.png)

$D=11$ $D=23$

We conclude that the only $D$ in this class for which a degree one prime over $3$ is Euclidean is $D=11$.

\

The upshot of this enumeration is that the only $D$ for which ${{\mathbb{Q}}}(\sqrt{-D})$ has a Euclidean ideal class are $D\in \{1, 2, 3, 5, 7,
11, 15\}$, and each the unique generator of the class-group is in fact norm-Euclidean, establishing Theorem \[thm1\].

[1]{}

– <http://www.sagemath.org/>.

– [[“]{}A quadratic field which is [E]{}uclidean but not norm-[E]{}uclidean[”]{}]{}, *Manuscripta Math.* **83** (1994), no. 3-4, p. 327–330.

– [[“]{}On euclidean ideal classes[”]{}]{}, *University of Michigan Thesis* (2009).

– [[“]{}[$\Bbb Z[\sqrt{14}]$]{} is [E]{}uclidean[”]{}]{}, *Canad. J. Math.* **56** (2004), no. 1, p. 55–70.

– [[“]{}Euclidean ideal classes[”]{}]{}, in *Journées [A]{}rithmétiques de [L]{}uminy ([C]{}olloq. [I]{}nternat. [CNRS]{}, [C]{}entre [U]{}niv. [L]{}uminy, [L]{}uminy, 1978)*, Astérisque, vol. 61, Soc. Math. France, Paris, 1979, p. 121–131.

– [[“]{}The [E]{}uclidean algorithm[”]{}]{}, *Bull. Amer. Math. Soc.* **55** (1949), p. 1142–1146.

– [[“]{}About [E]{}uclidean rings[”]{}]{}, *J. Algebra* **19** (1971), p. 282–301.

– [[“]{}On [E]{}uclidean rings of algebraic integers[”]{}]{}, in *Analytic number theory ([P]{}roc. [S]{}ympos. [P]{}ure [M]{}ath., [V]{}ol. [XXIV]{}, [S]{}t. [L]{}ouis [U]{}niv., [S]{}t. [L]{}ouis, [M]{}o., 1972)*, Amer. Math. Soc., Providence, R. I., 1973, p. 321–332.
