

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

$P(x,y,z)$ is a polynomial function on an algebraic surface $S$ in $F_{q}^{3}$. Suppose that the derivative of $P$ along any tangent vector of $S$ is zero. Can we say that $P$ is constant on $S$?

Here $q$ is a prime, and we assume the degrees of $P$ and $S$ are significantly smaller than $q$.

share|cite|improve this question
Here is a convoluted argument that seems to indicate the answer is "yes": if not, then by taking a sequence of counterexamples with P,S of bounded degree and q going off to infinity and forming an ultraproduct (or using the compactness theorem in logic), one can create a counterexample in char 0, which by the Lefschetz principle gives a counterexample over ${\bf C}$, which can be ruled out by differential geometry. But presumably there is a direct algebraic proof... – Terry Tao Jul 8 '13 at 22:59
from the answers given so far I see that one also has to assume that $S$ is connected, since otherwise things don't work even over ${\bf C}$. (But connectedness is preserved via ultraproducts; I wrote a proof for this for irreducibility in… but one can do something similar for connectedness, e.g. by splitting into irreducible components and seeing whether they overlap.) – Terry Tao Jul 9 '13 at 17:59
@Terry Tao: I do not really think this has to do with connectedness, cf. my second example. My feeling is this has much more to do with the fact that even low degree polynomials over finite fields may have few rational points. – Jason Starr Jul 9 '13 at 23:06
@TerryTao Your ultraproduct+compactness 1-2-punch precludes me using the same formula and just changing q. My guess is in several variables (and on surfaces) there are many counter examples like $x^q - x$ which may differ wildly as you change the prime q. – john mangual Jul 10 '13 at 12:55

Personally, I think this problem is ill-posed. What geometric properties would the OP like to assume about $S$? What does "constant" mean -- constant on the set of rational points, or truly a constant polynomial? Also, what precisely does $d$ "significantly smaller" than $q$ mean? There are "surfaces" in $\mathbb{F}_q^3$ that have few rational points, e.g., the vanishing set of $x^2-x$ is a surface with only $q^2$ points. Worse yet, a polynomial like $P(x,y,z) = [(x-1)^2-1]^2 = x^2(x-2)^2$ will vanish on all tangent vectors at these $q^2$ points. Of course $S$ is reducible, but the OP says nothing about irreducibility. I suspect that there are similar irreducible examples: the crucial point is that the (few) rational points (mostly) lie on a small number of curves cut out by low degree equations.

Edit. Vivek complains that my example $S$ above is disconnected. However, one can use my second suggestion to easily produce connected counterexamples. Start with a disconnected set of small size compared to $q$, e.g., $\{(0,0,0),(1,0,0)\}$, and a small number $d$ of low degree defining equations, e.g., $x^2-x=y=z=0$ and $d=3$. Now take a normic form $g$ over $\mathbb{F}_q$ of degree $d$ in $d$ variables: these always exist over finite fields (cf. Lang's thesis). Now plug in the defining equations of your disconnected set for the variables of the normic form to get a new polynomial $h$, e.g., $h(x,y,z) = g(x^2-x,y,z)$. Now the only rational points of the zero set of $h$ will be the points in the original disconnected set, e.g., $\{(0,0,0),(1,0,0)\}$. Now you can use a polynomial $P$ such as my polynomial above. Really this has NOTHING to do with $S$, and only to do with the (incredibly small) set of rational points of $S$.

share|cite|improve this answer
S is worse than reducible, it's not connected; presumably this is meant to be excluded. In particular your objection also works over $\mathbb{C}$... – Vivek Shende Jul 9 '13 at 11:05
@Vivek: For a smooth variety, such as mine, reducible is equivalent to connected. If you know precisely what conditions the OP has in mind, I ask that you edit the statement of the question so that the problem is well-posed. – Jason Starr Jul 9 '13 at 22:06
I think it is very frustrating to try to answer vague and ambiguous questions when the OP disappears and it's not worth trying. – Felipe Voloch Jul 9 '13 at 23:53
I agree with Felipe about over-interpreting a vague question. I do agree that it is reasonable to at least assume that $S$ is irreducible (although I wish the OP would have stated that). However, the OP never says anything about varying the field with the polynomials held fixed. I agree that, if one allows the field to grow with the degrees of the polynomials, then Lang-Weil will take over. But, in fact, I don't think you even need something as precise as Lang-Weil. There are infinitely many closed points: I think that is all you need. – Jason Starr Jul 10 '13 at 1:57
Thanks for all the answers and I apologize for the vagueness and my absence yesterday. I do want the polynomial defining $S$ to be irreducible and we assume there are at least $q^{2}$ points on $S$. All polynomials are defined over $F_{q}$, not any extension of it. – Ben Jul 11 '13 at 18:50

In one variable, $x^q - x \equiv 0 $ for all $x \in \mathbb{F}_q$.
This function would not be constant over an algebraic extension $\mathbb{F}_{q^2}$, which can always be constructed

In 3 variables, $(x^q+y^q+z^q) - (x+y+z) \equiv 0 $ .

The notions of "algebraic surface" and "tangent vector" are shaky in finite characteristic. I guess you could formally write the partial derivatives:

$$ a \partial_x P + b \partial_y P + c \partial_z P = 0 $$

Still not sure what it "tangent to $S$" means in finite characteristic.

There may need to be some condition on $P,S$ even if the degrees are small.

In two variables, $S = \{ xy = 1: x,y \in \mathbb{F}_q\}$ and then $P(x,y) = x^k - y^{(q-1)-k}$ is identically zero. We can check:

$$ x^k - y^{(q-1)-k} = x^k - x^{k-(q-1)} = x^k - x^k = 0$$

for any $x \in \mathbb{F}_q$. The partial derivatives are $\partial_x P = k \,x^{k-1}$ and $\partial_yP = [(q-1)-k] \;y^{(q-2)-k}$ . That... disproves the converse.

In one variable try $\fbox{$ p(x) = \frac{1}{q+1}x^{q+1} - \frac{1}{2} x^2$}$

Then $p'(x) = x^q - x \equiv 0$, the derivative is identically zero.

This function is not constant $p(0)=0$ and $p(1) = 1 - \frac{1}{2} \neq 0$.

I've been thinking about what it means to be "tangent" to a surface $f(x,y,z)=0$ in finite characteristic. In differential geometry we could consider a function $(x(t), y(t), z(t))$ and take the derivative of $P(x,y,z)$ in the "direction" of the line tangent going through the curve.

\[ \frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} + \frac{df}{dz} \frac{dz}{dt} =
a \frac{df}{dx} + b\frac{df}{dy} + c\frac{df}{dz} = 0 \]

This is a linear equation to define which "directions" are "tangent". In these directions we could define the derivative of our polynomial

\[ a \frac{dP}{dx} + b\frac{dP}{dy} + c\frac{dP}{dz} \]

For varieties and schemes, there is a notion of Zariski Tangent Space when more intuitive geometric notions over $\mathbb{C}$ no longer hold.

We are really looking at residue classes of polynomials modulo functions that vanish identically on that surface. $\mathbb{F}_q[x,y,z] / (f)$.

The prime ideals $\mathfrak{p}$ in this ring are "points" on your surface. You define a maximal ideal of "O(1)" functions $\mathfrak{m} = \mathfrak{p}A_{\mathfrak{p}}$. Your tangent space is (quite succinctly) $\mathfrak{m}/\mathfrak{m}^2$.

share|cite|improve this answer
Notice his polynomials are of degree smaller than the characteristic. – Mariano Suárez-Alvarez Jul 9 '13 at 4:53
In several variables, it would be possible to work around it. I keep trying when I have a moment. – john mangual Jul 9 '13 at 10:32
I do want to assume that the degree of $P$ and $S$ is small enough, say $log(q)$ – Ben Jul 11 '13 at 18:56

Your Answer


