CRT for Rings

Statement

Let $R$ be a commutative ring and $I_1, \ldots, I_n \subseteq R$ ideals that are pairwise coprime: $I_j + I_k = R$ for all $j \ne k$.

Chinese Remainder Theorem (ring version). There is a ring isomorphism

The classical case: $R = \mathbb{Z}$, $I_k = (m_k)$ with $\gcd(m_j, m_k) = 1$ for $j \ne k$, gives $\mathbb{Z}/(m_1 \cdots m_n) \cong \mathbb{Z}/m_1 \times \cdots \times \mathbb{Z}/m_n$.

Visualization

Example: $\mathbb{Z}/15 \cong \mathbb{Z}/3 \times \mathbb{Z}/5$.

Ideals $(3)$ and $(5)$ in $\mathbb{Z}$ are coprime: $3 \cdot 2 + 5 \cdot (-1) = 1$.

$n \bmod 15$	$n \bmod 3$	$n \bmod 5$	element of $\mathbb{Z}/3 \times \mathbb{Z}/5$
$0$	$0$	$0$	$(0, 0)$
$1$	$1$	$1$	$(1, 1)$
$5$	$2$	$0$	$(2, 0)$
$6$	$0$	$1$	$(0, 1)$
$10$	$1$	$0$	$(1, 0)$
$11$	$2$	$1$	$(2, 1)$
$14$	$2$	$4$	$(2, 4)$

Reconstruction (CRT lift). Given $(2, 4) \in \mathbb{Z}/3 \times \mathbb{Z}/5$, find $n \bmod 15$.

Step 1: Need $n \equiv 2 \pmod{3}$ and $n \equiv 4 \pmod{5}$.

Step 2: Bezout: $2 \cdot 3 + (-1) \cdot 5 = 1$. So $e_1 = 10 \equiv 1 \pmod{3}$, $e_1 \equiv 0 \pmod{5}$ and $e_2 = 6 \equiv 0 \pmod{3}$, $e_2 \equiv 1 \pmod{5}$.

Step 3: $n = 2 \cdot e_1 + 4 \cdot e_2 = 2 \cdot 10 + 4 \cdot 6 = 20 + 24 = 44 \equiv 14 \pmod{15}$.

Proof Sketch

1. Define the map. Let $\phi : R \to \prod_k R/I_k$ by $\phi(r) = (r + I_1, \ldots, r + I_n)$. This is a ring homomorphism.

2. Surjectivity. Given $(a_1, \ldots, a_n) \in \prod R/I_k$, use pairwise coprimality to find idempotents $e_k \in R$ with $e_k \equiv 1 \pmod{I_k}$ and $e_k \equiv 0 \pmod{I_j}$ for $j \ne k$ (via Bezout in $R$). Then $\sum a_k e_k$ maps to $(a_1, \ldots, a_n)$.

3. Kernel is $\bigcap_k I_k$. An element $r$ maps to $0$ iff $r \in I_k$ for all $k$, i.e., $r \in \bigcap_k I_k$. By pairwise coprimality, $\bigcap I_k = I_1 \cdots I_n$ (the product ideal).

4. First Isomorphism Theorem gives $R/\bigcap I_k \cong \prod R/I_k$.

Connections

The number-theoretic Chinese Remainder TheoremChinese Remainder Theorem$$\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$$Simultaneous congruences with coprime moduli have a unique solution mod their productRead more → is the special case $R = \mathbb{Z}$. The ring version generalises it to arbitrary commutative rings and gives Fundamental Theorem of ArithmeticFundamental Theorem of Arithmetic$$n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}$$Every integer greater than 1 has a unique prime factorizationRead more → a structural home: $\mathbb{Z}/n \cong \prod \mathbb{Z}/p_i^{e_i}$ when $n = \prod p_i^{e_i}$.

-- Mathlib: Ideal.quotientInfRingEquivPiQuotient
-- in Mathlib.RingTheory.Ideal.Quotient.Operations

open Ideal in
/-- CRT for rings: if ideals are pairwise coprime, the quotient by their
    intersection is isomorphic to the product of the individual quotients. -/
#check @Ideal.quotientInfRingEquivPiQuotient
-- quotientInfRingEquivPiQuotient :
--   ∀ {R : Type u_1} [inst : CommRing R] {ι : Type u_2} (f : ι → Ideal R),
--     (∀ i j, i ≠ j → f i ⊔ f j = ⊤) →
--     R ⧸ ⨅ i, f i ≃+* ∀ i, R ⧸ f i

/-- Concrete instance: ℤ/15 ≅ ℤ/3 × ℤ/5. -/
example : ZMod 15 ≃+* ZMod 3 × ZMod 5 :=
  (ZMod.chineseRemainder (by norm_num)).toRingEquiv