Chinese remainder theorem
Over univariate polynomial rings and Euclidean domains

The statement in terms of remainders given in § Theorem statement cannot be generalized to any principal ideal domain, but its generalization to Euclidean domains is straightforward. The univariate polynomials over a field is the typical example of a Euclidean domain which is not the integers. Therefore, we state the theorem for the case of the ring R = K {\displaystyle R=K} for a field K. {\displaystyle K.} For getting the theorem for a general Euclidean domain, it suffices to replace the degree by the Euclidean function of the Euclidean domain.

The Chinese remainder theorem for polynomials is thus: Let P i (X) {\displaystyle P_{i}(X)} (the moduli) be, for i = 1, …, k {\displaystyle i=1,\dots,k}, pairwise coprime polynomials in R = K {\displaystyle R=K}. Let d i = deg P i {\displaystyle d_{i}=\deg P_{i}} be the degree of P i (X) {\displaystyle P_{i}(X)}, and D {\displaystyle D} be the sum of the d i. {\displaystyle d_{i}.} If A i (X), …, A k (X) {\displaystyle A_{i}(X),\ldots,A_{k}(X)} are polynomials such that A i (X) = 0 {\displaystyle A_{i}(X)=0} or deg A i < d i {\displaystyle \deg A_{i}<d_{i}} for every i, then, there is one and only one polynomial P (X) {\displaystyle P(X)}, such that deg P < D {\displaystyle \deg P<D} and the remainder of the Euclidean division of P (X) {\displaystyle P(X)} by P i (X) {\displaystyle P_{i}(X)} is A i (X) {\displaystyle A_{i}(X)} for every i.

The construction of the solution may be done as in § Existence (constructive proof) or § Existence (direct proof). However, the latter construction may be simplified by using, as follows, partial fraction decomposition instead of the extended Euclidean algorithm.

Thus, we want to find a polynomial P (X) {\displaystyle P(X)}, which satisfies the congruences

for i = 1, …, k. {\displaystyle i=1,\ldots,k.}

Consider the polynomials

The partial fraction decomposition of 1 / Q (X) {\displaystyle 1/Q(X)} gives k polynomials S i (X) {\displaystyle S_{i}(X)} with degrees deg S i (X) < d i, {\displaystyle \deg S_{i}(X)<d_{i},} such that

and thus

Then a solution of the simultaneous congruence system is given by the polynomial

In fact, we have

for 1 ≤ i ≤ k. {\displaystyle 1\leq i\leq k.}

This solution may have a degree larger than D = ∑ i = 1 k d i. {\displaystyle D=\sum _{i=1}^{k}d_{i}.} The unique solution of degree less than D {\displaystyle D} may be deduced by considering the remainder B i (X) {\displaystyle B_{i}(X)} of the Euclidean division of A i (X) S i (X) {\displaystyle A_{i}(X)S_{i}(X)} by P i (X). {\displaystyle P_{i}(X).} This solution is
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