Chinese remainder theorem for rings
Let
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.