I have a question regarding the finite field method to compute the characteristic polynomial of a hyperplane arrangement. I am using the book "Enumerative Combinatorics, Volume 1" second edition by Richard P. Stanley. This method may be found in chapter 3.11.4.
I will begin by stating a problem I am interested solving using the finite field method. Then I will state the theorem and tell you might thoughts on this. In the bottom of this post, you may find my questions.
I'm interested in solving 3.114 (b) in the book by Stanley:
For an arrangement $\mathcal{A}$ below (in $\mathbb{R}^n$) show that the characteristic polynomials are as indicated. $x_i=x_j$ for $1\leq i<j\leq n$ and $\sum_{i=1}^n x_i=0$. Then $$\chi_{\mathcal{A}}(x)=(x-1)^2(x-2)(x-3)\cdots (x-n+1).$$
If I am allowed to use the finite field method, I have a solution to this problem that gives me the correct polynomial. However, by reading the theorem (I will state it down below), I'm not sure why I would be allowed to use it.
The theorem only says something about $\mathbb{Q}^n$ and not $\mathbb{R}^n$.
Let us now state the theorem.
Theorem:
Let $\mathcal{A}$ be an arrangement in $\mathbb{Q}^n$, and suppose that $L(\mathcal{A})\cong L(\mathcal{A}_q)$ for some prime power $q$. Then $$ \chi_{\mathcal{A}}(q)=\#\Big ( \mathbb{F}_q - \bigcup_{H\in\mathcal{A}_q} H \Big )=q^n-\#\bigcup_{H\in\mathcal{A}_q} H. $$ To me, it really seems as we need to work in the vector space $\mathbb{Q}^n$ to apply this theorem. This theorem does not seem to tell us that we can do this in the vector space $\mathbb{R}^n$.
I still have some hope that I might be able to apply the above theorem for two reasons.
(1) Just to quote Stanley in the first paragraph when he starts to talk about the finite field method
In this subsection we will describe a method based on finite fields for computing the characteristic polynomial of an arrangement defined over $\mathbb{Q}$.
It seems to me that he wants to tell the reader that the finite field method works if your hyperplanes are defined over $\mathbb{Q}$. After doing some research what "defined over $\mathbb{Q}$" means, it seems as this means the hyperplanes have rational coefficients. Surely, in the problem I want to solve the hyperplanes does, indeed, have rational coefficients.
(2) In example 3.11.11 in the book, Stanley uses the finite field method to find the characteristic polynomial of the braid arrangement $\mathcal{B}_n$. This is an arrangement in $\mathbb{K}^n$ with hyperplanes $x_i-x_j=0$ for $1\leq i<j\leq n$.
He hasn't said anything about which field $\mathbb{K}$ we consider. Thus, I guess, we could for instance let $\mathbb{K}=\mathbb{R}$? Notice the polynomials are defined over $\mathbb{Q}$.
It is very likely that I am misinterpreting something. Perhaps I am misinterpreting the theorem or the example?
I would be really happy if someone could tell me why we can apply the theorem when we have a hyperplane arrangement in $\mathbb{R}^n$ (if it is possible) and if not perhaps tell me what's happening in example 3.11.11.
Thanks for taking your time!