primes represented integrally by a homogeneous cubic form

Expired by this question Show determinant of matrix is non-zero I am moved to ask:

Given integers $a,b,c,$ and cubic form $$f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|,$$ what primes $p$ can be integrally represented as $$p = f(a,b,c)?$$

I think it is $3,$ all primes $p \equiv 2 \pmod 3,$ and all $p = u^2 + 27 v^2$ in integers, but not any $q = 4 u^2 + 2 u v + 7 v^2.$ I checked for $p < 10000.$

Note that, if $-p$ is represented, so is $p.$

Although it does not finish things, note that if $f$ integrally represents both $m,n$ then it represents $mn.$ That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where $$X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}.$$ Then $X^3 = 2 I$ and $X^4 = 2 X.$

I once asked a guy at MSRI about pretty much the same problem, only instead of the important polynomial being $\lambda^3 - 2$ it was $\lambda^3 - \lambda^2 - \lambda - 1.$ The phrase norm forms came up, and he laughed at me.

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   p           a           b           c
2           0           1           0
3          -1           0           1
5           1           0           1
11          -1           1           1
17          -1          -1           2
23           1           1           2
29          -3          10          -6
31          -1          26         -20
41           1          -2           2
43           1          -1           2
47          -1           4          -2
53           1          -4           3
59           1           3          -2
71          -1           2           2
83           3           1           3
89           1           2           3
101           3          -7           4
107          -1           0           3
109           1         -12           9
113           1           4           2
127          -1          16         -12
131           3           3          -1
137          -3           1           3
149           1           4          -1
157          -1           5          -2
167          -3           3           2
173          -3           7          -3
179           1         -31          24
191          -1          -2           4
197           5           2          -1
223           1           5           2
227           3          -2           3
229          -1          -1           4
233           1           5          -3
239           1           3           4
251          -1          -4           5
257           1           0           4
263           3           4          -3
269          -1           9          -6
277           1           5          -1
281          -1           1           4
283          -1           8          -5
293           1          -9           7
307          -1           4           3
311           3           3           5
317          -3           5           1
347           3         -12           8
353           3          -1           4
359          -5          23         -15
383          -5          28         -19
389          -3           2           4
397           1           7          -5
401           1          -5           5
419           3           6           5
431           1          -7           6
433          -1          -5           6
439           3           1           5
443           3          -4           4
449           1           8          -6
457           1           2           5
461           5           4          -2
467          -1          -1           5
479          -1           4           4
491           3          18         -16
499          -1           0           5
503           5           3           6
509           1           4           5
521           5           5          -1
557          -1          89         -70
563           3           6          -1
569          -1           7          -2
587           3           4           6
593          -7           2           5
599           1           7          -4
601           1         -22          17
617          -5         -59          50
641           3          23         -20
643          -1           3           5
647          -1          14         -10
653           1          16         -13
659           3         -10           7
677          -1         -11          10
683          -5           5           3
691           3          -2           5
701          -1          -3           6
719           5           5          -4
727           3          -5           5
733           3           9          -8
739           1           7          -2
743           3          -3           5
761           1         -14          11
773           5          -1           5
797          -3           3           5
809           1           2           6
811           1           3           6
821           3           7           6
827          -1          11          -7
839          -3           5           4
857          -9           5           4
863          -1           0           6
881           1          -4           6
887           7           3           7
911          -1          -5           7
919          -1           7           3
929           9           2          -2
941           9           3          -1
947          -3           1           6
953          -7          26         -16
971          -1           8          -1
977           1           7           5
983           3           7          -3
997           3         -11           8


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• Expired or inspired? :) – Amzoti Mar 14 '13 at 2:23
• @Amzoti, I think expired gives the more accurate feel. – Will Jagy Mar 14 '13 at 2:23
• I can't tell you how many times I've been there! :) + 1 for a very interesting question! – Amzoti Mar 14 '13 at 2:24
• Is there a path to go from $p|f(a,b,c)$ for some $a,b,c$ to $p = f(a,b,c)$ for some $a,b,c$? – user27126 Mar 14 '13 at 4:50
• @Sanchez, always possible. The situations I know where that argument works are quadratic forms rather than cubic. – Will Jagy Mar 14 '13 at 5:02

1 Answer

The discriminant of $\Bbb Q \subset \Bbb Q(\sqrt[3]2)$ is $-108$, and the Minkowski bound for this extension is $\frac {3!}{3^3} \frac 4 \pi \sqrt {108} \approx 2.940$. So to prove that this number field has class number $1$ we only need to find a way to represent $2$, and $2$ is indeed represented by $(0,1,0)$. Thus $p$ is represented by this norm form if and only if the ideal $(p)$ has an ideal factor of norm $p$, which happens if and only if $2$ is a cube modulo $p$.

If $p \equiv 2 \pmod 3$ then any nonzero element of $\Bbb F_p$ has one cube root in $\Bbb F_p$ and two cube roots in $\Bbb F_p^2$, so $2$ is a cube modulo $p$.

If $p \equiv 1 \pmod 3$ then $(p)$ splits in $\Bbb Q(\zeta_3)$, and $2$ is a cube if and only it further splits in $\Bbb Q(\zeta_3,\sqrt[3]2)$. Since $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$ is an abelian extension, it is a ray class field for some modulus $\mathfrak m$ of $\Bbb Q(\zeta_3)$.

Working modulo $6$, we have $(a+b\zeta_3)^3 = (a^3+b^3) - 3ab^2+3ab(a-b)\zeta_3 = a^3 + b^3-3ab^2 \in \Bbb Z/6 \Bbb Z$, and thus for any $a,b,c \in \Bbb Z[\zeta_3]$, $a^3+2b^3+4c^3-6abc = a^3+2b^3+4c^3 \in \Bbb Z/6\Bbb Z$. So, norms that are coprime to $6$ are units ($\pm 1$) modulo $6$. So $\Bbb Q(\zeta_3,\sqrt[3]2)$ is an extension of the ray class field of modulus $(6)$ for $\Bbb Q(\zeta_3)$.

On the other hand, $G = (\Bbb Z[\zeta_3]/(6))^*/\langle \overline{\zeta_6} \rangle$ is isomorphic to $\Bbb Z/3 \Bbb Z$, which is the Galois group of the extension $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$, so $\mathfrak m = (6)$, and $2$ is a cube modulo $p$ if and onlt if $p \equiv 2 \mod 3$ or $p = a^2-ab+b^2$ where $a+\zeta_3 b$ is congruent modulo $6$ to one of $\{1,1+\zeta_3,\zeta_3,-1,-1- \zeta_3,- \zeta_3\}$.

Each element of $G$ (modulo complex conjugation) corresponds to a class of primitive binary quadratic forms of discriminant $-108$, or a corresponding lattice class (modulo multiplication by a unit and complex conjugation) whose endomorphism ring is $\Bbb Z[3\sqrt{-3}]$:

$\Lambda = \langle 1, 3\sqrt{-3} \rangle$ is a lattice corresponding to the neutral element of $G$ : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the neutral class.
while $\Lambda = \langle 2, \frac {1+3\sqrt{-3}}2 \rangle$ corresponds to the other two classes : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the other two classes.

So if $p \equiv 1 \pmod 3$, then $p$ is represented either as $a^2 + 27b^2$ (when $2$ is a cube) or as $4u^2 \pm 2uv + 7v^2$ (when $2$ is not a cube), and never both at the same time.

• Minor comment: the criterion in the first paragraph, that $p$ is a norm iff it is not prime in the cubic field, requires noting that $-1$ is itself a norm so that there is no need to distinguish between $-p$ or $p$ being a norm; they both are or both aren't. If $-1$ is not a norm in a quadratic field, for instance, then $-p$ may be a norm of some algebraic integer while $p$ is not. – KCd Mar 20 '13 at 12:44
• @KCd and mercio, Thank you. I just posted the original question as math.stackexchange.com/questions/336191/… but his time I put the source (Hudson and Williams) I should have posted with this question. The area where I would enjoy additional tutoring is "Thus $p$ is represented by this norm form if and only if the ideal $(p)$ is not prime in this extension" – Will Jagy Mar 20 '13 at 19:51
• @WillJagy : it goes along these lines : there is an element of norm $p$, iff. there is a principal ideal of norm $p$, iff. (class number $1$) there is an ideal of norm $p$ (which necessarily contains the ideal $(p)$ of norm $p^3$), iff. there is a factorisation $(p) = (\mathfrak P_1)(\mathfrak P_2)$ with $N(\mathfrak P_1) = p$, iff. $X^3-2$ has a linear factor modulo $p$, iff. $2$ is a cube modulo $p$ – mercio Mar 20 '13 at 20:23
• @mercio, thank you. I simply never studied algebraic number theory, but in the natural course of things made up some problems where that is the method of choice, some still open. – Will Jagy Mar 20 '13 at 20:46