Congruence with several variables

A congruence

$$ \tag{1 } f( x _ {1} \dots x _ {n} ) \equiv 0 ( \mathop{\rm mod} m), $$

where $ f( x _ {1} \dots x _ {n} ) $ is a polynomial in $ n \geq 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

$$ \tag{2 } f( x _ {1} \dots x _ {n} ) \equiv \ 0 ( \mathop{\rm mod} p _ {i} ^ {\alpha _ {i} } ) $$

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

$$ \tag{3 } f( x _ {1} \dots x _ {n} ) \equiv 0 ( \mathop{\rm mod} p ^ \alpha ),\ \ \alpha > 1, $$

to be solvable, it is necessary that the congruence

$$ \tag{4 } f( x _ {1} \dots x _ {n} ) \equiv 0 ( \mathop{\rm mod} p) $$

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) $( $ \mathop{\rm mod} p $) of (4) such that $ ( df/dx _ {i} )( x _ {1} ^ {(} 1) \dots x _ {n} ^ {(} 1) ) \not\equiv 0 $( $ \mathop{\rm mod} p $) for at least one $ i = 1 \dots n $, generates $ p ^ {( \alpha - 1)( n- 1) } $ solutions $ x _ {i} \equiv x _ {i} ^ {( \alpha ) } $( $ \mathop{\rm mod} p ^ \alpha $) of (3), whereby $ x _ {i} ^ {( \alpha ) } \equiv x _ {i} ^ {(} 1) $( $ \mathop{\rm 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

$$ | N _ {p} - p ^ {n-} 1 | \leq C( f ) p ^ {n-} 1- 1/2 $$

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 $.