A Multivariable Chinese Remainder Theorem

Using an adaptation of Qin Jiushao’s method from the 13th century, it is possible to prove that a system of linear modular equations ai1xi + · · · + ainxn = ~bi mod ~ mi, i = 1, . . . , n has integer solutions if mi > 1 are pairwise relatively prime and in each row, at least one matrix element aij is relatively prime to mi. The Chinese remainder theorem is the special case, where A has only one column. 1. The statement with proof Consider a linear system of equations A~x = ~b mod ~ m, where A is an integer n× n matrix and ~b, ~ m are integer vectors with coefficients mi > 1. Theorem 1.1 (Multivariable CRT). If mi are pairwise relatively prime and in each row, at least one matrix element is relatively prime to mi, then A~x = ~b mod ~ m has solutions for all ~b. There is a solution ~x in an n-dimensional parallelepiped X = ZM/L of volume M = m1 · · ·mn, where L is a lattice in ZM . Proof. The map φ : x → Ax mod ~ m is a group homomorphism from the Abelian group X = Z to the finite Abelian group Y = Zm1 × · · · × Zmn = Y/L, where L = (m1Z)×· · ·×(mnZ) is a lattice subgroup of Y. The kernel of φ is a subgroup LA of X and X = X/LA. The image of φ is a subgroup of Y. By the first isomorphism theorem in group theory, the quotient group X and the image are isomorphic. The kernel LA is a lattice in X spanned by n vectors ~k1, . . . ,~kn. The map φ is injective on X . By the Lagrange theorem in group theory, there exist finitely many vectors ~yi ∈ Y such that ⋃d(A) i=1 A(X ) +~yi = Y. The problem is solvable for all ~b if and only if d(A) = 1. For every ~b, there exists then a unique integer vector ~x in X such that A~x = ~b mod ~ m. As in the usual CRT, we have a solution if each equation has a solution. To construct a solution, pick matrix elements aij(k) such that the i’th row is relatively prime to mi. Let ~ej denote the standard basis in n-dimensional space. Consider a line ~x(t) = t~ej(1) in X, where t is an integer. There exists an integer t1 so that ~x(t) solves the first equation. Now take the line ~x(t) = t1~ej(1) + tm1~ej(2). There is an integer t2 so that ~x(t) solves the second equation. This is possible because m1 is relatively prime to m2. Note that ~x(t) still solves the first equation for all t. We have now a solution to two equations. Continue in the same way until the final solution ~x(t) = ∑ ti(m1 . . .mi)~eij(i) is reached. Date: v1: January 27, 2005, v2: June 22, 2012. 1991 Mathematics Subject Classification. 11Y50,01A25,15A06.