Lecture 7.7: The Chinese remainder theorem Matthew Macauley - PowerPoint PPT Presentation

Lecture 7.7: The Chinese remainder theorem Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 1 / 10

Motivating example Exercise 1 � 2 x ≡ 5 (mod 7) Find all solutions to the system 3 x ≡ 4 (mod 9) M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 2 / 10

Motivating example Exercise 2 � x ≡ 3 (mod 4) Find all solutions to the system x ≡ 0 (mod 6) M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 3 / 10

Number theory version Chinese remainder theorem Let n 1 , . . . , n k ∈ Z + be pairwise co-prime (that is, gcd( n i , n j ) = 1 for i � = j ). For any a 1 , . . . , a k ∈ Z , the system  x ≡ a 1 (mod n 1 )   .  . .   x ≡ a 1 (mod n 1 )  has a solution x ∈ Z . Moreover, all solutions are congruent modulo N = n 1 n 2 · · · n k . This can be generalized. To see how, first recall the following operations on ideals: 1. Intersection : I ∩ J = { r ∈ R | r ∈ I and r ∈ J } . 2. Product : IJ = � ab | a ∈ I , b ∈ J � = { a 1 b 1 + · · · + a k b k | a i ∈ I , b j ∈ J } ⊆ I ∩ J . 3. Sum : I + J = { a + b | a ∈ I , b ∈ J } . Example : R = Z , I = � 9 � = 9 Z , J = � 6 � = 6 Z . 1. Intersection : � 9 � ∩ � 6 � = � 18 � (lcm) 2. Product : � 9 �� 6 � = � 54 � (product) 3. Sum : � 9 � + � 6 � = � 3 � (gcd). M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 4 / 10

Ring theory version Note that gcd( m , n ) = 1 iff am + bn = 1 for some a , b ∈ Z . Or equivalently, � m � + � n � = Z . Definition Two ideals I , J of R are co-prime if I + J = R . Chinese remainder theorem (2 ideals) Let R have 1 and I + J = R . Then for any r 1 , r 2 ∈ R , the system � x ≡ r 1 (mod I ) x ≡ r 2 (mod J ) has a solution r ∈ R . Moreover, any two solutions are congruent modulo I ∩ J . Recall that such a solution r ∈ R satisfies r − r 1 ∈ I and r − r 2 ∈ J . M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 5 / 10

Ring theory version Chinese remainder theorem (2 ideals) Let R have 1 and I + J = R . Then for any r 1 , r 2 ∈ R , the system � x ≡ r 1 (mod I ) x ≡ r 2 (mod J ) has a solution r ∈ R . Moreover, any two solutions are congruent modulo I ∩ J . Proof Write 1 = a + b , with a ∈ I and b ∈ J , and set r = r 2 a + r 1 b . � M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 6 / 10

Ring theory version Chinese remainder theorem Let R have 1 and I 1 , . . . , J n be pairwise co-prime ideals. Then for any r 1 , . . . , r n ∈ R , the system  x ≡ r 1 (mod I 1 )   .  . .   x ≡ r 2 (mod I n )  has a solution r ∈ R . Moreover, any two solutions are congruent modulo I 1 ∩ · · · ∩ I n . Proof n = 1. For j = 2 , . . . , n , write 1 = a j + b j , where a j ∈ I 1 , b j ∈ I j . Then 1 = ( a 2 + b 2 )( a 3 + b 3 ) · · · ( a n + b n ) n � � � � = a 2 ( a 3 + b 3 ) · · · ( a n + b n ) + b 2 ( a 3 + b 3 ) · · · ( a n + b n ) ∈ I 1 + � I j = R . j =2 � x ≡ 1 (mod I 1 ) Now apply the CRT for 2 ideals to the system x ≡ 0 (mod � j � =1 I j ) Let s 1 ∈ R be a solution. M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 7 / 10

Ring theory version Chinese remainder theorem Let R have 1 and I 1 , . . . , J n be pairwise co-prime ideals. Then for any r 1 , . . . , r n ∈ R , the system  x ≡ r 1 (mod I 1 )   .  . .   x ≡ r 2 (mod I n )  has a solution r ∈ R . Moreover, any two solutions are congruent modulo I 1 ∩ · · · ∩ I n . Proof (cont.) n = k . For j = 1 , . . . ✁ ❆ k , . . . , n , write 1 = a j + b j , where a j ∈ I k , b j ∈ I j . Then ❳❳❳ ✘ 1 = ( a 2 + b 2 ) · · · ✘✘✘ � ( a k + b k ) · · · ( a n + b n ) ∈ I k + ❳ I j = R . j � = k � x ≡ 1 (mod I k ) Now apply the CRT for 2 ideals to the system x ≡ 0 (mod � j � =1 I j ) Let s k ∈ R be a solution. M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 8 / 10

Ring theory version Chinese remainder theorem Let R have 1 and I 1 , . . . , J n be pairwise co-prime ideals. Then for any r 1 , . . . , r n ∈ R , the system  x ≡ r 1 (mod I 1 )   .  . .   x ≡ r 2 (mod I n )  has a solution r ∈ R . Moreover, any two solutions are congruent modulo I 1 ∩ · · · ∩ I n . Proof (cont.) By construction, s k ∈ (mod � I j ), and so s k ∈ I j for all j � = k . j � = k We have s k ≡ 1 (mod I k ) and s k ≡ 1 (mod I ) j for j � = k . Set r = r 1 s 1 + · · · + r n s n . It is easy to see that this works. If s ∈ R is another solution, then s ≡ r j ≡ r (mod I j ), for j = 1 , . . . , n , and so n � s ≡ r mod I j . j =1 M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 9 / 10

Applications When is Z n isomorphic to a product? Let R = Z and I j = � m j � , for j = 1 , . . . , n with gcd( m i , m j ) = 1 for i � = j . Then Z m 1 m 2 ··· m n ∼ I 1 ∩ · · · ∩ I n = � m 1 m 2 · · · m n � , and = Z m 1 × · · · × Z m n . Corollary Factor n = p d 1 1 · · · p d n into a product of distinct primes. Then n Z n ∼ = Z p 1 d 1 × · · · × Z p ndn . Remark If R is a Euclidean domain, then the proof of the CRT is constructive . Specifically, we can use the Euclidean algorithm to write � � � � c k m k + d k m j = gcd m k , m j = 1 , where I j = � m j � . j � = k j � = k Then, set s k = d k � m j , and r = r 1 s 1 + · · · + r n s n is the solution. j � = k M. Macauley (Clemson) Lecture 7.7: The Chinese remainder theorem Math 4120, Modern algebra 10 / 10