Homogenous polynomial over finite field having only trivial zero

Is there a way to construct homogenous polynomials of small degree over a certain finite field having only trivial zero?

For instance, the polynomial $f (a, b, c) = a^3 + b^3 + c^3 - 3abc - 3a^2b - 3b^2c - 3c^2a$ has only a trivial zero over $\mathbb{F}_{13}^3$ (I found this by trial and error). Is there a nice way to show that this polynomial has indeed the forementioned property? Can we find other polynomials over $\mathbb{F}_{13}$ with this property?

• One way of constructing such polynomials would be to consider the norm map $N:\Bbb{F}_{13^3}\to\Bbb{F}_{13}$. When you fix a basis for the field extension, the norm map of a cubic extension becomes a homogeneous cubic polynomial. As the zero element of the extension field is the only one with zero norm, the resulting polynomial will have this property. I don't know whether your polynomial is of that form. A choice of basis will, of course, change the formula of the polynomial, as it amounts to doing an invertible linear substitution of the variables. – Jyrki Lahtonen Mar 5 '14 at 20:50
• @JyrkiLahtonen, I put a method using rational integer matrices as an answer (still needs the matrix appropriate for this problem); it is still norm forms. – Will Jagy Mar 5 '14 at 21:17
• @Will: You talked me into doing this. I took it as a design goal to get away with minimal changes to your $M$ :-) – Jyrki Lahtonen Mar 5 '14 at 21:34
• see mathoverflow.net/questions/127160/… There is a similar example for each entry in the table from the Hudson and Williams article. Your polynomial $f(a,b,c)$ does seem to have the same multiplicative property, represent $3$ and all primes $\pm 1 \pmod {18},$ and is anisotropic for $2$ and all primes $5,7,11,13 \pmod {18}.$ – Will Jagy Mar 6 '14 at 5:45
• see also math.stackexchange.com/questions/329936/… The primes that behave the same way as 13 for that example are all $q = 4 u^2 + 2 u v + 7 v^2,$ which includes 13 with $u=1, v=1.$ – Will Jagy Mar 6 '14 at 6:11

If you can find a matrix $M$ of integers such that $$\det (aI + b M + c M^2)$$ is your polynomial, you have a method. For example, with $$M \; = \; \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) ,$$ we get $$aI + b M + c M^2 = \left( \begin{array}{rrr} a & b & c \\ c & a & b \\ b & c & a \end{array} \right) .$$ In this case $$\det ( aI + b M + c M^2) = a^3 + b^3 + c^3 - 3 a b c.$$ As a result, you can see at a glance that this is singular when $a=b=c,$ but it is also singular when $a+b+c = 0.$ So the determinant is $0$ in those two cases. In your case, if you find an appropriate matrix $M,$ which is the low-budget version of Jyrki's comment, you can do Gauss-Jordan elimination in the field of 13 elements to find out when $aI + b M + c M^2$ is singular, which will (as you say) be only when $$a \equiv b \equiv c \equiv 0 \pmod {13}.$$

I will see if I can come up with a matrix $M$ for your problem. I think Jyrki would know an elegant method for doing that, but I will do trial and error.

EEDDIITTTT: if absolutely necessary, one may use a matrix where the determinant does not come out the same but is equivalent to your polynomial mod 13; so a $-3$ coefficient in your polynomial could become $-16,$ for instance. This variant is limited to one prime at a time...

EEDDDIIIIITT, Thursday, 6 March. It did not turn out as cleanly as I had expected, but it did have an identity matrix and (the negation of) a companion matrix, so everything still holds. In this case we find $$f (a, b, c) = a^3 + b^3 + c^3 - 3abc - 3a^2b - 3b^2c - 3c^2a = \det (a X + b Y + c I),$$ with $$Y \; = \; \left( \begin{array}{rrr} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & -3 & 0 \end{array} \right) ,$$ also $X = I + Y - Y^2,$ so $$X \; = \; \left( \begin{array}{rrr} 1 & -1 & -1 \\ 1 & -2 & -1 \\ 1 & -2 & -2 \end{array} \right) .$$

All we need for the multiplication property to hold is to be able to express $Y^2, XY,YX, X^2$ as sums of $X,Y,I$ with integer coefficients. Note first that $Y^3 = 3 Y + I.$ Then we get $$Y^2 = Y - X + I,$$ $$XY = YX = -Y - X,$$ $$X^2 = - Y - 2 X + I.$$ Done. Everything works. So, there is a middle ground. The identity matrix and a companion matrix, then some polynomial in the companion matrix. Live and learn.

Supplementing Will's answer with the following suggestion for $M$. The key is to observe that the element $3$ is a cubic root of unity in the field $\Bbb{F}_{13}$. But there are no ninth roots of unity in that prime field, as $9\nmid (13-1)$. Therefore the polynomial $x^3-3\in\Bbb{F}_{13}[x]$ is irreducible (all its zeros are clearly of multiplicative order nine). Therefore we can use the companion matrix of $x^3-3$ as $M$ in Will's answer. IOW, pick $$M=\left(\begin{array}{ccc}0&0&3\\1&0&0\\0&1&0\end{array}\right).$$ This gives $$a+b M+ c M^2=\left(\begin{array}{ccc}a&3c&3b\\b&a&3c\\c&b&a\end{array}\right),$$ and also $$\det(a+b M+ c M^2)=a^3 + 3 b^3 - 9 a b c + 9 c^3.$$ As $M$ generates the field $\Bbb{F}_{13^3}$, we know that $a+bM+cM^2$ is invertible for all non-trivial choices $a,b,c\in\Bbb{F}_{13}$.

Another way of looking at the end game is that the matrix $a+bM+cM^2$ is the matrix representing multiplication by $a+bx+cx^2$ in the quotient ring $\Bbb{F}_{13}[x]/\langle x^3-3\rangle$. As that quotient ring is a field by virtue of $x^3-3$ being irreducible, that multiplication is invertible when non-zero.

• Jyrki, I have hidden depths of persuasion. – Will Jagy Mar 5 '14 at 21:36
• Jyrki, very likely that the ternary cubic form given by the OP is the same, under a linear integer change of variables, to the one in the Corollary on the final page of Nowlan (1926), pdf at projecteuclid.org/euclid.bams/1183487069 – Will Jagy Mar 6 '14 at 18:42
• Put in a more general but slower method, lots of changes of variable work as desired. I typed in one with only coefficients $0,\pm 1.$ – Will Jagy Mar 6 '14 at 22:40

It is quite likely that the original cubic polynomial is the same, under a linear invertible change of variables with rational integer coefficients, to the one in the Corollary on the final page of Nowlan 1926.

The behavior as far as represented primes $\pm 1 \pmod 9$ and primes giving only trivial zeroes $2,4,5,7 \pmod 9$ (as your 13: i am calling this "anisotropic") is identical, and the polynomials are quite similar. There are people on this site who could determine this by using the number field in question, I imagine Jyrki could do that, so I left him an additional note. For me, I am doing some trial and error.

EEEEDDDDIIITTT: It worked. The polynomial given by the OP is equivalent, by an invertible linear change of variables with rational integer coefficients, to Nowlan's 1926 polynomial.

As a result, we have Nowlan's Corollary on the final page: for (positive) primes $p \equiv 2,4,5,7 \pmod 9,$ the OP's polynomial can only be divisible by $p$ if all three of his $a,b,c$ are divisible by $p.$ Meanwhile, for prime $q=3$ or $q \equiv \pm 1 \pmod 9,$ we can find rational integers $a,b,c$ such that $f(a,b,c) = p.$ Finally, if the polynomial integrally represents two numbers, it represents their product. So, we can tell exactly what numbers are integrally represented by the polynomial: for every prime factor $p \equiv 2,4,5,7 \pmod 9,$ the exponent must be divisible by $3.$ Those are the only restrictions.

Meanwhile given Nowlan's $$g(u,v,w) = u^3 + 6 u^2 w - 3 u v^2 + 3 uvw +9 u w^2 - v^3 + 3 v w^2 + w^3,$$ we find $$g(a+c, -a-b, -a) = a^3 + b^3 + c^3 -3abc -3 a^2 b - 3 b^2 c - 3 c^2 a = f(a,b,c)$$ We can invert this with $$f(-w, w-v,u+v) = g(u,v,w).$$ So There.

jagy@phobeusjunior:~$gp Reading GPRC: /etc/gprc ...Done. GP/PARI CALCULATOR Version 2.5.0 (released) i686 running linux (ix86/GMP kernel) 32-bit version compiled: Nov 17 2011, gcc-4.6.2 (Ubuntu/Linaro 4.6.2-2ubuntu1) (readline v6.2 enabled, extended help enabled) Copyright (C) 2000-2011 The PARI Group PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER. Type ? for help, \q to quit. Type ?12 for how to get moral (and possibly technical) support. parisize = 4000000, primelimit = 500509 ? ? u = a+c %1 = a + c ? ? v = -a - b %2 = -a - b ? ? w = -a %3 = -a ? ? p = u^3 - v^3 + w^3 + 3 * u * v * w - 3 * u * v^2 + 3 * v * w^2 + 9 * u * w^2 + 6 * u^2 * w %4 = a^3 - 3*b*a^2 + (-3*c^2 - 3*b*c)*a + (c^3 - 3*b^2*c + b^3) ?  • Good job! I was thinking whether they might be cubic form with different value distributions in$\Bbb{Z}_{13}^*$(as an obvious tool of proving that they are not gotten from each other by a linear substitution). Barely possible now that$3\mid(13-1)$. No luck so far. – Jyrki Lahtonen Mar 7 '14 at 6:34 • @JyrkiLahtonen, I'm not entirely sure what it is that you are trying to find; but the ternary cubic polynomial in this question math.stackexchange.com/questions/329936/… is distinct from the ones we have been working on here. I will check again, but i believe the answer did not give a full proof that the cubic is anisotropic for all primes$q = 4 u^2 + 2 u v + 7 v^2,$meaning it is only divisible by such$q$if all three variables are divisible by$q.\$ I would love to know such a proof... – Will Jagy Mar 8 '14 at 23:38
• @JyrkiLahtonen, reading my question again, I did not include a request for a proof of that fact. I did a better job later with mathoverflow.net/questions/127160/… and got a correspondingly complete answer. – Will Jagy Mar 8 '14 at 23:48