zeros of a homogeneous polynomial

Hi All,

Let $F$ be a finite field, $\lambda\in F$, and $$p_\lambda (x,y,z)=\left|\begin{array}{ccc}x & y & z \\ y & z & x +\lambda z \\ z& x+\lambda z & y+\lambda x+\lambda ^2z \end{array}\right|.$$ Can one choose $\lambda$ so that $(0,0,0)$ is the only zero in $F^3$ for $p_\lambda (x,y,z)$? Also, I'm wondering if the matrix that defines $p_\lambda$ looks familiar to anyone. Thanks!

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I tested $F=\mathbb{F}_2$, $F=\mathbb{F}_3$, $F=\mathbb{F}_5$, where it is true (thank you, Zack). –  Dietrich Burde May 30 '13 at 21:09

Let $t$ be a root of $t^3 + \lambda t + 1 = 0$. Suppose $\lambda$ can be chosen so this polynomial is irreducible. Then the cubic decomposes into three lines, one of which is given by $x = ty + z/t$ (and the other two are the Galois conjugates). Now it's easy to see this line has no $\mathbb{F}_p$-rational points (else $t$ would satisfy an equation of degree less than $3$). Similarly the others don't have rational points either. So all you have to do is ensure that this poly is irreducible, and also that $4\lambda^3 + 27$ is nonzero, so you don't have a singular point over the base field.

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... and that has to happen once $p$ is large enough: the discriminant $-(4\lambda^3+27)$ must be a nonzero square, which happens for $p/2+O(p^{1/2})$ choices of $\lambda$, and then the polynomial must not factor completely, which happens for $p/6+O(p^{1/2})$ choices (corresponding to points on the elliptic curve $t_1+t_2+t_3=0$, $t_1 t_2 t_3 = -1$ where the $t_i$ are the roots, up to permutation and minus the few points where the $t_i$ are not pairwise distinct). –  Noam D. Elkies May 30 '13 at 22:35
The number of $(t,\lambda)$ such that $t^3+\lambda t+1=0$ is at most $q-1$, hence there must exist $\lambda_0 \in \mathbf{F}_q$ such that the polynomial is irreducible. –  François Brunault May 30 '13 at 22:50
(Then $4\lambda_0^3+27$ is automatically nonzero, since it is the discriminant of an irreducible cubic polynomial.) –  François Brunault May 30 '13 at 22:55

The existence of $\lambda$ has by now been proved by A.Kumar's answer and the ensuing comments, but this still doesn't explain the behavior of the determinant $p_\lambda(x,y,z)$: usually the zero-locus of $3 \times 3$ determinant of linear forms is an irreducible curve, not the union of three lines. A.Kumar's answer does strongly hint at the reason, though. Here's an explanation, as well as a formula for the number of $\lambda \bmod p$ for which $p_\lambda$ has no nontrivial zeros.

Let $k$ be any field, $R$ the ring $k[T] \left/ (T^3 - \lambda T^2 - 1) \right.$, and $r \in R$ the element $y+xT+zT^2$. Then multiplication by $r$ is a $k$-linear transformation of $R$ whose matrix with respect to the basis $(T,1,T^2)$ is none other than $$\left( \begin{array}{ccc} x & y & z \\ y & z & x + \lambda z \\ z & x + \lambda z & y + \lambda x + \lambda ^2 z \end{array} \right)\phantom0.$$ Thus $p_\lambda(x,y,z)$ is nonzero iff $r$ is invertible in $R$, and this is true for all nonzero $r \in R$ iff $R$ is a field iff the cubic $T^3 - \lambda T^2 - 1$ is irreducible. Taking $T = 1/t$ then recovers A.Kumar's polynomial $t^3 + \lambda t + 1$.

Now as F.Brunault observes, if $k$ is finite then the existence of a $\lambda$ that makes $t^3 + \lambda t + 1$ irreducible is an elementary counting argument: in degree $3$, irreduicble still means no rational root; if $t^3 + \lambda t + 1 = 0$ then $t \neq 0$ and $\lambda = -(t^2 + t^{-1})$, so each nonzero $t$ arises exactly once as a root, and $t=0$ does not arise at all. Since there are $\left|k\right|$ choices of $\lambda$, at least one of them must make $t^3 + \lambda t + 1$ irreducible.

To enumerate such $\lambda$ we need only count the polynomials $t^3 + \lambda t + 1$ that split completely. These are parametrized by solutions of $t_1 + t_2 + t_3 = 0$, $t_1 t_2 t_3 = -1$ up to permutation. These equations give an elliptic curve isogenous with the Fermat cubic (if $a^3+b^3+c^3 = 0$ then $$(t_1^{\phantom.}, t_2^{\phantom.}, t_3^{\phantom.}) = \frac{-1}{abc} (a^3, b^3, c^3)$$ satisfies $t_1 + t_2 + t_3 = 0$ and $t_1 t_2 t_3 = -1$; this gives the isogeny in one direction). The number of rational points on this curve is known, and eventually one finds (if I did this right) that the count is $(p+1)/3$ if $p \equiv -1 \bmod 3$, and $(p+1-a)/3$ if $p \equiv +1 \bmod 3$, where in the latter case $a \equiv -1 \bmod 3$ is determined uniquely by $4p = a^2 + 27 b^2$. In particular, for large $p$ the count is $p/3 + O(p^{1/2})$, as predicted by Čebotarev (since $1/3$ is the proportion of $3$-cycles in the symmetric group $S_3$).

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The responses to my question have been extremely helpful. Many thanks for the additional explanation. –  user34548 Jun 1 '13 at 21:18

I did some calculations with maple (see below) which suggest that for a fixed finite field $F$ the probability that $p_\lambda(x,y,z)=0$ has only $(x,y,z)=(0,0,0)$ as a solution, is high:

Consider first $p_\lambda$ first as a polynomial in the ring $\mathbb{Q}(\lambda)[x,y,z]$, i.e., consider $\lambda$ as a transcendental element of the ground field. Maple claims that this polynomial is irreducible. However, the curve in $\mathbb{P}^2_{\mathbb{Q}(\lambda)}$ defined by this polynomial has genus -2 and three singular points, which means that there is finite extension $L$ of $\mathbb{Q}(\lambda)$ over which this polynomial factors in three linear polynomials.

Fix now a finite field $F$ and an element of $\lambda$ of $F$. If $p_\lambda$ is reducible then it contains a linear factor and this factor will yield a nontrivial solution, has you want to discard these $\lambda$.

Suppose now $p_\lambda$ that is irreducible and $p_\lambda=0$ has a nontrivial solution. There is a finite (Galois) extension $K$ of $F$ over which $p_\lambda$ is a product of a linear polynomials. Over $K$ the curve $p_\lambda=0$ in $\mathbb{P}^2$ is the union of three lines. The Galois group permutes the three lines, but fixes the rational point. Hence this point is on each of the three lines. In this case the curve $p_\lambda=0$ has a unique singular point. Maple claims that there are (at least) three distinct singular point if $4\lambda^3+27$ is nonzero in $K$. Hence if $p_\lambda$ is irreducible and $4\lambda^3+27$ is nonzero then $p_\lambda=0$ has only $(0,0,0)$ as a solution. (It turns out that if $4\lambda^3+27=0$ then $p_\lambda$ is reducible anyway.)

Maple claims that we can take $\mathbb{Q}(\lambda)(\alpha)$ for $L$, where the minimal polynomial of $\alpha$ is $X^3+2\lambda X^2+\lambda^2X-1$. This means that $\mathbb{Q}$ is algebraically closed in $L$ and that for most" choices of $\lambda$, $p_\lambda$ is irreducible.

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Over the field of three elements, you can take $\lambda=-1$. Over the field of five or seven elements, you can take $\lambda=1$. Over the field of eleven or thirteen elements, you can take $\lambda=4$.

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Over a field of $5$ elements, you can take just $\lambda=1$ or $\lambda=2$, if I am not mistaken. –  Dietrich Burde May 30 '13 at 21:12
Dietrich: Yes, I agree with this. –  Steven Landsburg May 30 '13 at 21:14
More data: Here are all allowable $\lambda$ for small fields. For $F_3$, $\lambda=2$. For $F_5$, $\lambda=1,2$. For $F_7$, $\lambda=1,2,4$. For $F_{11}$, $\lambda=4,5,7,8$. For $F_{13}$, $\lambda=4,10,12$. –  Steven Landsburg May 30 '13 at 22:07