In the Green-Tao method appear quite often expressions such as \mathbb{E}\prod_{i}1_{d_i\mid\psi_i(n)} for some system of affine linear forms \psi_i:\mathbb{Z}^d\rightarrow\mathbb{Z} where the d_i are some positive integers. The most basic example would be \mathbb{E}_{n\leq N}1_{d\mid n}=\frac{1}{N}\lfloor \frac{N}{d}\rfloor=\frac{1}{d}+O(1/N)=\mathbb{E}_{n\in\mathbb{Z}/d\mathbb{Z}}1_{n=0}+O(1/N) where the last equality is here only to insist heavily on the form a general statement will have to take. A bit more generally, one easily sees that if f:\mathbb{Z}\rightarrow \mathbb{Z} is d-periodic, then its average until N is basically its average until d+O(d/N). Following the ideas from Green-Tao, in fact we can show that for our general problem, the asymptotic should be analogously \mathbb{E}_{n\in (\mathbb{Z}/m\mathbb{Z})^d}1_{\forall i,d_i\mid\psi_i(n)} where m is the gcd of the d’s.

Thus, applying this method, one has frequently to deal with such « local densities », the intersection of the zero sets of affine forms. More generally again, one may have to deal with things like \alpha_P(p^2):=\mathbb{P}_{n\in(\mathbb{Z}/p^2\mathbb{Z})^d}(P(n)=0) for some polynomial P\in\mathbb{Z}[X_1,\dots,X_d]. Say P=\prod_{i=1}^t\phi_i where the \phi_i are affine linear forms, so that P is of degree t. Suppose also that no two of them are affinely dependant, even when reduced modulo p. Then to compute \alpha_P(p^2), notice that in order for n to satisfy P(n)\equiv 0[p^2], either two of the forms must be divisible by p in n, or one must be divisible by p^2 there.

To analyse the first possibility, reduce the forms modulo p, so they are as seen affine forms on the field \mathbb{F}_p. They are non-zero form, so the image of each linear part is of dimension 1 and the kernels of dimension d-1; so the zero locus of each of our affine forms is a hyperplane, and the hypothesis of no affine dependance ensures that the intersection of any two of these hyperplanes is of dimension d-2. So there are only p^{d-2} common zero for each pair of forms. But each point of \mathbb{F}_p^d gives rise to p^d antecedents in (\mathbb{Z}/p^2\mathbb{Z})^d under the map of reduction modulo p. Finally, this possibility provides a summand of at most \binom{t}{2}p^{-2} to \alpha_P(p^2).

To compute now \mathbb{P}_{n\in (\mathbb{Z}/p^2\mathbb{Z})^d}(\phi_i(n)=0), notice that any such 0 must arise from a 0 mod p. Now there are p^{d-1} 0 mod p, because the set of zeros is a hyperplane. Then we can use a form of Hensel’s lemma in several variables. To obtain such a statement, we rely on the following simple identy for any Q\in\mathbb{Z}[X_1,\dots,X_d]:

Q(x+p\overrightarrow{k})\equiv Q(x)+p\,\overrightarrow{\mathrm{grad}}(Q)\cdot \overrightarrow{k}\text{ mod }p^2

which shows that if Q(x)=0\text{ mod }p, so Q(x)=mp\text{ mod }p^2 and latex \text{grad}(Q)\neq 0\text{ mod }p$ then the vector \overrightarrow{k} is constrained to live in a hyperplane if Q(x+p\overrightarrow{k}) is to vanish. So there are only p{d-1} elements y\in\mathbb{Z}/p^2\mathbb{Z} with x\equiv y [p] and Q(y) \equiv 0[p^2]. Of course if Q is affine then the gradient is simply the linear part. Thus the number of roots modulo p^2 is p^{2d-2}. By the union bound, we see that:

\mathbb{P}_{n\in (\mathbb{Z}/p^2\mathbb{Z})^d}(\exists i:\phi_i(n)=0)\leq tp^{-2}. In the end, \alpha_P(p^2)\ll_t p^{-2}.

To find more about this calculation and similar ones and the reason why I am interested in them, look at this article here.

 

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