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Let $h_1, \ldots, h_r$ be linear forms in variables $x_1, \ldots, x_n$ with integer coefficients. Let $\mathbb F_q$ denote the finite field with $q = p^e$ elements. I am asked to prove that except in a finite number of characteristics $p$, the number of vectors $v \in \mathbb F_q^n$ such that $h_i(v) = 0$ for all $i$ is given for all $q$ by a polynomial $\chi(q)$ in $q$ with integer coefficients, and that $(-1)^n \chi(-1)$ is equal to the number of connected regions into which $\mathbb R^n$ is separated by the removal of all the hyperplanes $h_i = 0$. My confusion is with this part (see next paragraph for more context; see below for question).

This is an exercise from a set of notes on Lie algebras. The upshot of this result is meant to be the following: if we let the $h_i$ be the root hyperplanes of a finite root system, then $\chi(q)$ has integer roots $e_1, \ldots, e_n$; call these the exponents of the root system. The hyperplanes $h_i$ divide $\mathbb R^n$ into Weyl chambers, so in fact the order of the Weyl group is $$(-1)^n \chi(-1) = (-1)^n \prod_{i=1}^n (-1-e_i) = \prod_{i=1}^n (1+e_i)$$

My question is the following: is it not the case that a linear system over $\mathbb F_q$ will always have $q^k$ elements where $k$ is the dimension of the null space of coefficient matrix of the $h_i$? For all but finitely many characteristics, we can perform row reduction over the rationals and all of the values by which we multiply will be elements of $\mathbb F_q$ (e.g. if we multiply by, say, $\frac3{10}$, $\frac{5}{22}$, and $\frac{2}{15}$ in the row reduction process, then any $p$ larger than $11$ will do). So the rank, and hence the nullity, is some fixed constant for all but finitely many characteristics. This would imply that $\chi(q) = q^k$, but then $(-1)^n \chi(-1) \in \{\pm 1\}$, which (except possibly the $h_i$ are all zero) is not the number of connected regions of $\mathbb R^n \setminus \{h_i\}$.

Am I missing something here? I haven't been able to find any reference on the exponents of a root system as defined above (I've found plenty on the exponents of a Weyl/Coxeter group which satisfy the product relation above); perhaps there is some modification to the definition of exponents that makes this work out.

Note: this is homework; just a mild push in the right direction would be ideal.

Haydn Gwyn
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1 Answers1

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Your doubts are very convincing. I have a strong feeling that in the set-up, it should actually be

the number of vectors $v \in \mathbb F_q^n$ such that $h_i(v) \color{red}{\neq} 0$ for all $i$

Namely, if we consider that, then I get for all but finitely many $p$ that $$\chi(q)= q-1$$ for the root system $A_1$, and $$\chi(q)=(q-1)(q-2)$$ for the root system $A_2$, which seems right so far. So this might be the "push in the right direction" you need. Good luck with the general case.