Chinese remainder theorem for rings

Let $𝑅$ be a commutative ring and suppose $𝔞 1, \dots, 𝔞 𝑛 ⊴ 𝑅$ are pairwise Relatively prime ideals. Then the homomorphism ring

𝜑 = (𝜋 1, \dots, 𝜋 𝑛) : 𝑅 ↠ 𝑛 ⨁ 𝑖 = 1 𝑅 𝔞 𝑖

is surjective and induces an isomorphism

˜ 𝜑 : 𝑅 𝔞 1 \dots 𝔞 𝑛 \to 𝑛 ⨁ 𝑖 = 1 𝑅 𝔞 𝑖 .

Proof

Since by ^P1 we have $k e r 𝜑 = 𝔞 1 \dots 𝔞 𝑛$ , it suffices to show surjectivity of $𝜑$ . We argue by induction on $𝑛$ . For $𝑛 = 1$ , there is nothing to show. For $𝑛 > 1$ assume the statement holds for fewer ideals, so all we need to prove is that the homomorphism
$𝑅 \to 𝑅 𝔞 1 \dots 𝔞 𝑛 - 1 \oplus 𝑅 𝔞 𝑛$
is surjective. By ^P2, $𝔞 1 \dots 𝔞 𝑛 - 1 + 𝔞 𝑛 = ⟨ 1 ⟩$ , so we are reduced to the case of two ideals.

If $𝔞 1 + 𝔞 2 = ⟨ 1 ⟩$ , then there exist $𝑎 𝑖 \in 𝔞 𝑖$ for $𝑖 = 1, 2$ such that $𝑎 1 + 𝑎 2 = 1$ . We need to verify that for $𝑟 1, 𝑟 2 \in 𝑅$ there exists an $𝑟 \in 𝑅$ such that $𝑟 \equiv 𝑟 𝑖 (m o d 𝔞 𝑖)$ . We get this from $𝑟 = 𝑎 1 𝑟 2 + 𝑎 2 𝑟 1$ , for which
$𝑟 = 𝑎 1 𝑟 2 + (1 - 𝑎 1) 𝑟 1 = 𝑎 1 (𝑟 2 - 𝑟 1) + 𝑟 1 \equiv 𝑟 1 (m o d 𝔞 1)$
and by the same token $𝑟 \equiv 𝑟 2 (m o d 𝔞 2)$ , as required.

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

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Chinese remainder theorem for rings

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