# p-adic density of zeroes of a polynomial

I saw the following definition of the p-adic densities of zeroes of a system of polynomial equations:

Definition: Suppose that we have a system of homogeneous polynomials of degree $d$, $f=(f_1,\cdots, f_r)$, where $f_i \in \mathbb Z[x_1, \cdots, x_s]$. Let $p$ be a prime number and let $\ell \in \mathbb Z_{>0}$. Consider the number:

$$\nu_{\ell}(p) := \{ x=(x_1, \cdots, x_s) \in \mathbb Z^s | f(x) \equiv 0 ( mod p^{\ell}) \}$$

which is the number of solutions of the system of the congruences $f(x) \equiv 0 ( mod p^{\ell})$.

Set $\mu_{\ell} = \nu_{\ell} p^{ \ell (r-s) }$. The limit

$$\mu (p) = \lim_{l \to \infty} \mu_{\ell}$$

when it exists is called the p-adic density of zeros of the system $f$.

Because I don't understand it very well I thought to do an example with just one polynomil to see how it works.

Find the p-adic density of zeros of the polynomial $f(x_1, x_2,x_3,x_4)= x_1 +bx_2 +cx_3 +dx_4 \in \mathbb Z [x_1,x_2,x_3,x_4]$.

Can someone help with the above example because I have no idea how to do it.

Thank you in advance!

## 1 Answer

The equation $f(x) \equiv 0 \pmod{p^\ell}$ is easily solvable, since any choice of $x_2, x_3, x_4$ (modulo $p^{\ell}$) gives you a solution precisely with $x_1 = - b x_2 - c x_3 - d x_4$. Thus, in your notation, $\nu_{\ell}(p) = (p^3)^{\ell} = p^{3 \ell}$. Then $\mu_{\ell} = p^{3 \ell} p^{-3 \ell} = 1$, and so the density equals $1$.

• Thank you for your answer! Can you explain me how did you obtain that the number of solutions of the conguence is $(p^3)^{\ell}$ ? – passenger Oct 18 '15 at 17:20