Congruence with several variables

A congruence

$$\begin{array}{}\text{(1)}& f(x_1\dots x_n)\equiv 0(\mathrm{mod}m),\end{array}$$

where $f(x_1\dots x_n)$ is a polynomial in $n\ge 2$ variables with integer rational coefficients not all of which are divisible by $m$. The solvability of this congruence modulo $m=p_1^{{\alpha }_1}\dots p_s^{{\alpha }_s}$, where $p_1\dots p_s$ are different prime numbers, is equivalent to the solvability of the congruences

$$\begin{array}{}\text{(2)}& f(x_1\dots x_n)\equiv 0(\mathrm{mod}{p}_i^{{\alpha }_i})\end{array}$$

for all $i=1\dots s$. The number $N$ of solutions of (1) is then equal to the product $N_1\dots N_s$, where $N_i$ is the number of solutions of (2). Thus, when studying congruences of the form (1) it is sufficient to confine oneself to moduli that are powers of prime numbers.

For a congruence

$$\begin{array}{}\text{(3)}& f(x_1\dots x_n)\equiv 0(\mathrm{mod}{p}^{\alpha }),\alpha >1,\end{array}$$

to be solvable, it is necessary that the congruence

$$\begin{array}{}\text{(4)}& f(x_1\dots x_n)\equiv 0(\mathrm{mod}p)\end{array}$$

modulo a prime number $p$ be solvable. In non-degenerate cases, the solvability of (4) is also a sufficient condition for the solvability of (3). More precisely, the following statement is correct: Every solution $x_i\equiv x_i^{(}1)$( $\mathrm{mod}p$) of (4) such that $(df/d{x}_i)(x_1^{(}1)\dots x_n^{(}1))\not\equiv 0$( $\mathrm{mod}p$) for at least one $i=1\dots n$, generates $p^{(\alpha -1)(n-1)}$ solutions $x_i\equiv x_i^{(\alpha )}$( $\mathrm{mod}{p}^{\alpha }$) of (3), whereby $x_i^{(\alpha )}\equiv x_i^{(}1)$( $\mathrm{mod}p$) when $i=1\dots n$.

Thus, in the non-degenerate case, the question of the number of solutions of the congruence (1) modulo a composite number $m$ reduces to the question of the number of solutions of congruences of the form (4) modulo the prime numbers $p$ that divide $m$. If $f(x_1\dots x_n)$ is an absolutely-irreducible polynomial with integer rational coefficients, then for the number $N_p$ of solutions of (4), the estimate

$$|N_p-p^{n-}1|\le C(f)p^{n-}1-1/2$$

holds, where the constant $C(f)$ depends only on $f$ and does not depend on $p$. It follows from this estimate that the congruence (4) is solvable for all prime numbers $p$ that are larger than a certain effectively-calculable constant $C_0(f)$, depending on the given polynomial $f(x_1\dots x_n)$( see also Congruence modulo a prime number). A stronger result in this question has been obtained by P. Deligne [3].

References[edit]

Comments[edit]

See also Congruence equation for more information. A polynomial $f(x_1\dots x_n)$ over $\mathbf{Q}$ is absolutely irreducible if it is still irreducible over any (algebraic) field extension of $\mathbf{Q}$.