Let $K$ be a cyclic cubic number field with conductor $f$ and ring of integers $\mathcal{O}_K$.

Define $K$ to be blue if and only if $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in K$}.$$ Define $K$ to be green if and only if $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in \mathcal{O}_K$}.$$ (So green implies blue).

Question 1: Are all cyclic cubic number fields blue?

Question 2: What is the density of green number fields restricted to blue number fields? That is, defining $$B_N:=\{K:K\text{ is a blue (cyclic cubic) number field of conductor }<N\},$$ $$G_N:=\{K:K\text{ is a green (cyclic cubic) number field of conductor }<N\},$$ what is $$ \lim_{N\to\infty} \frac{\#G_N}{\#B_N}? $$ (and does the limit exist?)

Question 3: Define $$\mathcal{G}:=\{f: K \text{ is green, where $K$ is a cyclic cubic number field of conductor $f$}\}.$$ What is $\mathcal{G}$ explicitly?

Remarks: I wrote some magma code that proved that $K$ is blue for all of the 1822 cubic cyclic number fields given from LMFDB (http://www.lmfdb.org/NumberField/start=0&degree=3&galois_group=C3&count=20). The code also explicitly gives the minimal polynomial of $w$. Here are the first few examples.

\begin{align*} f=7, \quad & t^3 - 2t^2 - t + 1 \\ f=9, \quad & t^3 - 3t + 1 \\ f=13, \quad & t^3 + t^2 - 4t + 1 \\ f=19, \quad & t^3 - 5t^2 + 2t + 1 \\ f=31, \quad & t^3 - (5/2)t^2 - (1/2)t + 1 \end{align*}

The polynomials above prove that $\{7,9,13,19\}\subseteq\mathcal{G}$. Notice that for $f=31$, this polynomial implies $K$ of conductor $31$ is blue, but it may or may not be green.

  • 8
    $\begingroup$ +1 for the colourblind-friendly choice of colours. $\endgroup$ Oct 9, 2018 at 12:46
  • $\begingroup$ After reading GNiklasch's comments, I think for the density question it may make more sense to restrict both $G_N$ and $B_N$ only to number fields in which 2 is inert/Q and then ask the limit of $G_N/B_N$ as $N\to\infty$. $\endgroup$ Oct 9, 2018 at 16:16
  • 1
    $\begingroup$ Nice question. I'm going to edit a bit to improve readability, hope that's okay. Also, I'm sure you're already aware, but there's been lots of work on $w\in\mathcal O_K$ such that $w$ and $1-w$ have norm 1, i.e., they're both units. Then $w$ is called an exceptional unit. So the $w$ in your blue case might be called very exceptional units! $\endgroup$ Oct 9, 2018 at 16:36
  • 1
    $\begingroup$ It almost goes without saying, but $G_N$ and $B_N$ are the cardinalities of the two sets. (Too few characters for me to edit...) $\endgroup$
    – GNiklasch
    Oct 9, 2018 at 17:05
  • $\begingroup$ @GNiklasch I edited the definition of $G_N$ and $B_N$ to make that clearer in the notation. Not sure why you could not edit for that. $\endgroup$
    – KConrad
    Oct 9, 2018 at 19:42

4 Answers 4


The answer to question 1 is yes.

Pick any field element $x_1$ outside the rationals. Let $x_2$ and $x_3$ be its conjugates under the Galois group. Then the cross ratio $$w=\frac{(x_1-1)(x_2-x_3)}{(x_1-x_2)(1-x_3)}$$ does the trick. (This is the same argument as in this answer.)

Note that these $w$ usually won't be algebraic integers, and they cannot be algebraic integers when the prime $2$ splits in $K$ (since $w$ and $1-w$ cannot both map to $1$ in the 2-element residue class field).

The simplest answer to question 3 I know of is that the set is what Daniel Shanks studied in The Simplest Cubic Fields. (This is a little white lie: Shanks had in fact started out by calling all cyclic cubic fields "simplest", and then proceeded to focus on a certain subset where the conductor is prime.) If $w$ is an algebraic unit in a cubic field satisfying the two norm conditions, then the minimal polynomial of $-w$ is necessarily of the Shanks form $t^3-at^2 -(a+3)t-1$ for some rational integer $a$, and passing to $1/w$ if necessary one can assume $a\ge -1$. So you can enumerate all "green" fields by letting $a$ vary (with only a few repetitions, but that's a deep result) and computing conductors (since $w$ won't necessarily generate the full ring of integers: $a=1259$ being the most spectacular case).

I can't answer question 2 offhand. I rather think "green" fields are rare, but I'm insufficiently familiar with fields that don't contain exceptional units to have a good intuition. The possible conductors can of course be enumerated, and a lower bound for $G_N$ follows from the observation that the above family of polynomials yields conductors which must divide $a^2 +3a+9$. It is clear that those $a$ for which this quantity is prime (it will then equal the conductor) contribute a negligible number of "green" fields - there are far too many primes congruent to 1 mod 6. But I don't know how much those $a$ for which the conductor properly divides this quantity will affect the count.

For fun, here's a little table; I've checked this against my archives and it is complete as far as it goes (skipping no "green" fields apart from the gaps indicated by vertical dots). The first four entries cover all cases where multiple $a$ parameters give the same $f$: $$\begin{array}{r|l|l} f & a & \text{reduced polynomial} \\ \hline 7 & \{-1,5,12,1259\} & \\ 9 & \{0,3,54\} & \\ 13 & \{1,66\} & \\ 19 & \{2,2389\} & \\ 31 & \text{---} & \\ 37 & \{4\} & \\ 43 & \text{---} & \\ 61 & \{39\} & \\ 9\cdot 7 & \{6\} & x^3-21x-35 \\ 9\cdot 7 & \text{---} & x^3-21x-28 \\ \ldots & \text{---} & \\ 79 & \{7\} & \\ \ldots & \text{---} & \\ 97 & \{8\} & \\ \ldots & \text{---} & \\ 9\cdot 13 & \{9\} & x^3-39x-91 \\ \ldots & \text{---} & \\ 139 & \{10\} & \\ \vdots & \vdots & \vdots \\ 7\cdot 31 & \{201\} & x^3-x^2-72x+225 \\ 7\cdot 31 & \{13\} & x^3-x^2-72x-209 \\ \ldots & \text{---} & \\ 241 & \{286\} & \\ 13\cdot 19 & \{14\} & x^3-x^2-82x+311 \\ \vdots & \vdots & \vdots \\ 373 & \{1598\} & \\ 379 & \{911\} & \\ 9\cdot 43 & \text{---} & x^3-129x-215 \\ 9\cdot 43 & \{18\} & x^3-129x-559 \end{array}$$

Some further remarks:

Cyclic cubic fields of conductor $f>7$ do not contain any units $w$ of positive norm such that $1-w$ is also a unit. (This is due to Nagell, and elementary.)

When $x$ is a totally positive algebraic integer in a totally real number field, $x+1$ can never be an algebraic unit.

When $x$ is a proper power of a nonzero algebraic integer in a totally real number field, $x+1$ can never be an algebraic unit. (Because the norm of $x+1$ can then be expressed as a resultant which also equals the norm of an obvious non-unit in a cyclotomic field.)

The roots of the Shanks polynomials always generate the subgroup of units of positive norm of the ring generated by these roots. (Due to Thomas 1979, based on geometry of numbers results of Berwick 1932, though Ljunggren may have been aware of this fact by 1942.)

Marie-Nicole Gras conjectured in 1976 that whenever a Galois-invariant subgroup of units of positive norm in a cyclic cubic field contains a unit $x$ such that $x+1 \in \mathcal{O}_K^\times$, the same is true already of a generator (as a Galois module) of this subgroup. In other words, the subgroup itself is then already generated by the roots of a Shanks polynomial. This is wide open, but true in a large range.

There are a lot more congruence obstructions beside the obvious one modulo 2. If $x$ is a unit of positive norm in a cyclic cubic field and $x^\sigma$ one of its conjugates, with common minimal polynomial $x^3-tx^2+s-1$, and $U$ the subgroup of units they generate, necessary conditions for $U+1$ intersecting the set of units of negative norm of $K$ nontrivially include:

  • $\mathrm{Norm}_{K/\mathbb{Q}}(x+1) \equiv -1 \pmod{2f}$
  • (Corollary: $U^{\sigma-1}+1$ meets $\mathcal{O}_K^\times$ only when $f=9$)
  • $\mathrm{gcd}(s+1,t+1) = 1$
  • $\mathrm{gcd}(s,t) \in \{1,3\}$ unless $f=9$
  • $\mathrm{gcd}(s-1,t-1) \in \{1,5\}$
  • $\mathrm{gcd}(s-2,t-2) \in \{1,7\}$
  • $\mathrm{gcd}(s-3,t-3) \in \{1,3,9\}$

(M.-N. Gras 1973, 1976 with a correction by yours truly to the last item where the $9$ had been missing).

Yet more obstructions arise when $U$ maps to $1$ or onto a too-small multiplicative subgroup in some residue class field. A fun example is $f=211$, where one can work modulo 199.

Selected references:

M.-N. Gras: Sur les corps cubiques cycliques dont l'anneau des entiers est monotone. Ann. Sci. Univ. Besançon, $3^e$ sér., Mathém., fasc. 6 (1973), 26pp.

M.-N. Gras: Lien entre le groupe des unités et la monogénéité des corps cubiques cycliques. Semaine. Théories des Nombres Besançon 1975-76. Publ. Math. Fac. Sci. Besançon, fasc. 1 (1976), 19pp.

E. Thomas: Fundamental units for orders in certain cubic number fields. J. reine u. angew. Math. 310 (1979), 33-55.

  • $\begingroup$ Thanks! Perhaps then in the density question, I should consider only those cyclic cubic number fields in which 2 is inert/Q. From the LMFDB, I computed that f is green for at least 89 of the 810 conductors st. 2 is inert in K, cyclic cubic of conductor f. $\endgroup$ Oct 9, 2018 at 15:23
  • $\begingroup$ Regarding your response to question 3, I think you mean $t^3+at^2−(a+3)t+1$. I've basically already done this except backwards; the code I wrote starts with $f$ and then computes $a$ instead of the other way around. If $a$ turns out to be an integer then I know $K$ is green, but if not then I'm not sure I can say whether $K$ is green or not. My code uses ideas from ``On Cyclic Cubic Fields" by Ennola and Turunen. I will continue to think about this. $\endgroup$ Oct 9, 2018 at 16:10
  • $\begingroup$ Fixed the wrong sign (thanks for pointing it out!).- Given a number field by a defining polynomial, it is always possible (at least in principle) to compute the set of all solutions to the unit equation $x+y=1$ in $\mathcal{O}_K^\times$, also known (following Nagell) as the set of exceptional units of $K$; in the cubic case it can be done "the wrong way round" by reducing to a Thue equation (even though computer algebra systems might internally reduce the Thue equation to an ($S$-)unit equation). $\endgroup$
    – GNiklasch
    Oct 9, 2018 at 16:51
  • $\begingroup$ GNiklasch, I'd like to mention you in the acknowledgements of a paper. To whom do I owe credit? $\endgroup$ Nov 4, 2018 at 13:06
  • $\begingroup$ @ChristineMcMeekin Depending on how formal you want it to be, "MO user GNiklasch" or "Gerhard Niklasch" or a reference along the lines of this or this other prior advice on MO meta. I've been out of academia for two decades, so I personally no longer care about citation counts: whichever variant fits the journal style and looks best to you (and your referee)! $\endgroup$
    – GNiklasch
    Nov 4, 2018 at 17:20

One can consider an asymptotic version of question 2. It is well-known, by the work of Delone and Faddeev (though supposedly this result was due to Levi) that cubic orders are parametrized by $\text{GL}_2(\mathbb{Z})$-equivalence classes of integral binary cubic forms. Since maximal, irreducible cubic orders are in one-to-one correspondence with cubic fields, it suffices to consider those cubic forms that correspond to maximal irreducible cubic orders with square discriminant. By this paper of Bhargava and Shnidman (https://projecteuclid.org/euclid.ant/1513730135) we know that all binary cubic forms corresponding to cubic orders of this type take the shape

$$\displaystyle F(x,y) = ax^3 + bx^2 y + (b-3a)xy^2 - ay^3, a, b \in \mathbb{Z}.$$

Moreover, those $F$ with bounded discriminant lie in the ellipse defined by

$$\displaystyle b^2 - 3ab + 9a^2 \leq X.$$

Now, we wish to describe the norm form of the corresponding cubic order $\mathcal{O}_F$ in terms of $F$, or rather, in terms of $a,b$. I can't seem to find a reference (I was told that this was in Bhargava's thesis but it doesn't seem to be in there), but one can write down the norm form for $\mathcal{O}_F$ explicitly. Here is the idea: the norm form for the cubic order is the same as the ideal norm form for the trivial class in the ideal class group, and by Bhargava's Higher Composition Laws II, the class is parametrized by the pair of symmetric matrices

$$\displaystyle (A,B) = \left(\begin{pmatrix} 0 & 0 & 1 \\ 0 & -a & 0 \\ 1 & 0 & 3a - b \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 1 & b & 0 \\ 0 & 0 & -a \end{pmatrix}\right).$$

Using a construction found here https://www.springer.com/gp/book/9781441909985 (the Petitjean chapter), one can construct an auxiliary ternary quadratic form from $A,B$, and using these three quadratic forms, construct a cubic form $G$. One can then compute explicitly that $G$ is proportional to its Hessian, hence it is a decomposable form. I was told that Bhargava proved at some point that $G$ is in fact the norm form for $\mathcal{O}_F$, and when $\mathcal{O}_F$ is a maximal irreducible cubic order, also the norm form for the corresponding cubic field. An explicit form for $G$ is

$$\displaystyle G(x,y,z) = x^3 - a^3 y^3 - a^3 z^3 - a(b-3a)^2y^2 z - 2a^2 by^2 z - ab^2 yz^2 + 2a^2(b-3a)yz^2 + a(b-3a)xy^2 - b(b-3a)xyz - 3a^2 xyz + bx^2 y - abxz^2 - (b-3a)x^2 z.$$

The question is then asking for simultaneous solutions to the equation $G(x,y,z) = -1, G(1-x, -y, -z) = -1$ (the norm form comes from a normalized basis of the shape $\langle 1, \omega, \theta \rangle$). This boils down to first finding an integral point on an affine quadric in $x,y,z$, and whether such a conic contains an integral point should give some information on the existence of $N_{K/\mathbb{Q}}(w) = N_{K/\mathbb{Q}}(1 - w) = -1$.


A supplementary answer which sheds a little more light on question 3, and which extends another result of M.-N. Gras from 1973:

The "green" fields among cyclic cubic fields $K$ are precisely those which contain a Galois-invariant monogenic order $\mathbb{Z}[y] \subseteq \mathcal{O}_K$.

The easy direction: When $-w$ is a root of a Shanks polynomial $t^3-at^2-(a+3)t-1$, then we can take $y=\pm w$, optionally translated by a rational integer.

The other direction is not much harder. When $y^\sigma \in \mathbb{Z}[y]$, say $y^\sigma=a_0+a_1y+a_2y^2$ for rational integer $a_0,a_1,a_2$, then $$y^{\sigma^2}-y^\sigma = a_1(y^\sigma-y) +a_2(y^\sigma-y)^2$$ lies in the principal ideal $(y^\sigma-y)$. Applying the automorphism $\sigma$ repeatedly, we find $$(y^{\sigma^2}-y^\sigma) \subseteq (y^\sigma-y) \subseteq (y-y^{\sigma^2}) \subseteq (y^{\sigma^2}-y^\sigma)$$ so this ideal is invariant under $\sigma$. Therefore the three summands in $$\frac{y^\sigma-y}{y^\sigma-y} + \frac{y^{\sigma^2}-y^\sigma}{y^\sigma-y} + \frac{y-y^{\sigma^2}}{y^\sigma-y} = 0$$ are algebraic units (of positive norm). The second and third summands will thus be roots of Shanks polynomials with integer parameter, and we can pick e.g. $$w=\frac{y^{\sigma^2}-y}{y^\sigma-y}.$$

The maximal order $\mathcal{O}_K$ is always Galois-invariant, but not necessarily monogenic; when it isn't, the field could still contain a non-maximal Galois-invariant order and thus exceptional units. In these cases, the discriminant $(a^2+3a+9)^2$ of the Shanks polynomial will necessarily be a proper multiple of $f^2$.


Thank you @GNiklasch for answering question 1. I had the following realisation towards answering questions 3 and 2...

As in Ennola, Veikko; Turunen, Reino, On cyclic cubic fields, Math. Comput. 45, 585-589 (1985). ZBL0582.12002., we can write the conductor $f$ as

$$ f=\frac{a^2+3b^2}{4} $$ for integers $a$ and $b$ such that

\begin{align*} &a \equiv 6 \bmod 9 \text{ and } b\equiv 3 \text{ or } 6 \bmod 9 \text{ where } b>0 & \text{ if } 3|f,\\ &a \equiv 2 \bmod 3 \text{ and } b\equiv 0 \bmod 3 \text{ where } b>0 & \text{ if } 3\nmid f.\\ \end{align*}

Now define $$ c:= \left\{ \def\arraystretch{1.5} \begin{array}{l l} \frac{-3}{2}\left(1 + \frac{a}{b}\right) & \text{if } 3| f \\ \frac{-3}{2}\left(1 - \frac{a}{b}\right) & \text{if } 3\nmid f. \end{array} \right. $$ If $c\in \mathbb{Z}$ then $K$ is green and the minimal polynomial of $\omega$ is given by $$ m_\omega(x):= x^3 + cx^2 - (c+3)x + 1. $$

  • $\begingroup$ @GNiklasch This is nice because it allows us to start with a conductor $f$ and find $c$ instead of testing all integers $c$ until we get a polynomial with conductor $f$. $\endgroup$ Oct 13, 2018 at 16:32
  • $\begingroup$ But it won't necessarily find a polynomial even when it exists. Consider $f=61$, thus $a=1$, $b=9$, $c=-4/3$. $\endgroup$
    – GNiklasch
    Oct 18, 2018 at 11:41
  • $\begingroup$ You'll find the "missing" examples by applying the formulas to conductors of non-maximal orders. But then you'll have to search for those orders which result in integer $c$... $\endgroup$
    – GNiklasch
    Oct 18, 2018 at 12:27

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