Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

‘there is no ontology here’ 393There are solutions a if and only ifr 1 ≡ r 2 (mod GCD(m 1 , m 2 ))where GCD(m 1 , m 2 ) is the greatest common divisor of m 1 and m 2 . And in that case thesolution is unique modulo LCM(m 1 , m 2 ), the least common multiple of m 1 and m 2 .Inaformula, if a and b are both solutions thena ≡ b (mod LCM(m 1 , m 2 ))Note that 2 = GCD(4, 6) and 12 = LCM(4, 6) in the examples above.This theorem supports thinking of integers as functions on the schemeSpec(Z). Compare this familiar fact about the real line R:Theorem. Take any intervals on the line, say [0, 2] ⊆ R and [1, 3] ⊆ R. Then forany continuous functions f 1 and f 2 consider these conditions on an unknown continuousfunction a:a(x) = f 1 (x) for all x ∈ [0, 2] and a(x) = f 2 (x) for all x ∈ [1, 3]There are solutions a if and only if the functions f 1 and f 2 agree on the intersection of theintervals:f 1 (x) = f 2 (x) for all x ∈ [1, 2]And in that case the solution is unique over the union of the intervals: if a and b are bothsolutions thena(x) = b(x) for all x ∈ [0, 3]In the Chinese remainder theorem the given numbers r 1 and r 2 and thesought number a are taken as functions on the line Spec(Z). The equationa ≡ r 1 (mod m 1 )says that a must agree with r 1 not necessarily over the whole line but at leastat all points (p) of the line corresponding to prime factors p of m 1 .²⁰ The otherequationa ≡ r 2 (mod m 2 )says a must agree with r 2 at all points (p) corresponding to prime factors p ofm 2 . The necessary condition is that r 1 and r 2 must agree at all points (p) wherep divides both m 1 and m 2 .Butp divides both m 1 and m 2 if and only if it dividestheir greatest common divisor so this condition can be expressed in an equation:r 1 ≡ r 2 (mod GCD(m 1 , m 2 ))²⁰ For each prime power p n that divides m 1 , the functions a and r 1 must agree on the nth orderinfinitesimal neighborhood of the point (p), orinotherwordsa ≡ r 1 (mod p n ).