Chinese remainder theorem for rings

Let be a commutative ring and suppose are pairwise Relatively prime ideals. Then the homomorphism #m/thm/ring

is surjective and induces an isomorphism

Proof

Since by ^P1 we have , it suffices to show surjectivity of . We argue by induction on . For , there is nothing to show. For assume the statement holds for fewer ideals, so all we need to prove is that the homomorphism

is surjective. By ^P2, , so we are reduced to the case of two ideals.

If , then there exist for such that . We need to verify that for there exists an such that . We get this from , for which

and by the same token , as required.

It should be clear that the classical Chinese remainder theorem is a special case.

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