## Solving simultaneous polynomial congruences

Let be polynomials in , all of degree . Let be pairwise relatively prime integers. Use the Chinese Remainder Theorem to prove that there exists a polynomial of degree such that for each – that is, the coefficients of and agree mod . Show also that if the are all monic, then may be chosen monic.

We begin with some lemmas.

Lemma 1: Let be a ring, and let be ideals. Then . Proof: is clear, since . Let . Then .

Lemma 2: Let be a ring, and let be ideals. Then . Proof: Certainly . Now if , then and , so that .

Note that the ideals are pairwise comaximal. By Lemma 1, the ideals are pairwise comaximal in . By the Chinese Remainder Theorem, the map given by is a surjective ring homomorphism. Thus, given , there exists such that .

Suppose that each of the are monic- that is, the leading coefficient of is congruent to 1 mod . By the previous exercise, there is an integer which is congruent to 1 mod for all . The solution above is unique only modulo , and thus we can choose the leading coefficient of to be 1.

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