Recall that for polynomials a b n we write a b mod n

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Recall that for polynomials a, b, n , we write a b (mod n ) when n | ( a - b ). For a non-zero polynomial n , and a D , we say that a is a unit modulo n if there exists a 0 D such that aa 0 1 (mod n ), in which case we say that a 0 is a multiplicative inverse of a modulo n . All of the results we proved in Chapter 2 for integer congruences carry over almost identically to polynomials. As such, we do not give proofs of any of the results here. The reader may simply check that the proofs of the corresponding results translate almost directly. Theorem 6.6 An polynomial a is a unit modulo n if and only if a and n are relatively prime. Theorem 6.7 If a is relatively prime to n , then ax ax 0 (mod n ) if and only if x x 0 (mod n ) . More generally, if d = gcd( a, n ) , then ax ax 0 (mod n ) if and only if x x 0 (mod n/d ) . Theorem 6.8 Let n be a non-zero polynomial and let a, b D . If a is relatively prime to n , then the congruence ax b (mod n ) has a solution x ; moreover, any integer x 0 is a solution if and only if x x 0 (mod n ) . Theorem 6.9 Let n be a non-zero polynomial and let a, b D . Let d = gcd( a, n ) . If d | b , then the congruence ax b (mod n ) has a solution x , and any integer x 0 is also a solution if and only if x x 0 (mod n/d ) . If d - b , then the congruence ax b (mod n ) has no solution x . 46
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Theorem 6.10 (Chinese Remainder Theorem) Let k > 0 , and let a 1 , . . . , a k D , and let n 1 , . . . , n k be non-zero polynomials such that gcd( n i , n j ) = 1 for all 1 i < j k . Then there exists a polynomial x such that x a i (mod n i ) ( i = 1 , . . . , k ) . Moreover, any other polynomial x 0 is also a solution of these congruences if and only if x x 0 (mod n ) , where n := Q k i =1 n i . If we set R = D/nD and R i = D/n i D for 1 i k , then in ring-theoretic language, the Chinese Remainder Theorem says the homomorphism from the ring D to the ring R 1 × · · · × R 1 that sends x D to ([ x mod n 1 ] , . . . , [ x mod n k ]) is a surjective homomorphism with kernel nD , and hence R = R 1 × · · · × R k . Let us recall the formula for the solution x (see proof of Theorem 2.6). We have x := k X i =1 z i a i , where z i := n 0 i m i , n 0 i := n/n i , m i n 0 i 1 (mod n i ) ( i = 1 , . . . , k ) . Now, let us consider the special case of the Chinese Remainder Theorem where a i K and and n i = ( T - b i ) with b i K , for 1 i k . The condition that gcd( n i , n j ) = 1 for all i 6 = j is equivalent to the condition that b i 6 = b j for all i 6 = j . Then a polynomial x satisfies the system of congruences if and only if x ( b i ) = a i for 1 i k . Moreover, we have n 0 i = Q j 6 = i ( T - b j ), and m i := 1 / Q j 6 = i ( b i - b j ) is a multiplicative inverse of n 0 i modulo n i . So we get x = k X i =1 a i Q j 6 = i ( T - b j ) Q j 6 = i ( b i - b j ) . The reader will recognize this as the LaGrange Interpolation Formula. Thus, the Chinese Remainder Theorem for polynomials includes LaGrange Interpolation as a special case.
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