Chinese Remainder Theorem

The Chinese Remainder Theorem

Let $R$ be a ring with unity and $I_{1}, \dots, I_{k} \subseteq R$ be two-sided ideals in $R$ . Let $$\pi:R\to R/I_1\times \cdots \times R/I_k$$ be the ring morphism induced by the projection morphisms $π_{i} : R \to R / I_{i}$ . In other words, $π$ is the map given by $r \mapsto (r + I_{1}, \dots, r + I_{k})$ . Then:

$\ker (π) = I_{1} \cap \dots \cap I_{k}$ ;
If the ideals are pairwise coprime, then $π$ is surjective. In this case, the First Isomorphism Theorem gives an isomorphism

R / I_{1} \dots I_{k} ≃ R / I_{1} \times \dots \times R / I_{k} .

If $R$ is commutative, then we also have $I_{1} \cap \dots \cap I_{k} = I_{1} \dots I_{k}$ .

Examples

Quotients of polynomial rings

Consider the ideals $I_{1} = (x - 1)$ and $I_{2} = (x + 1)$ in the polynomial ring $Q [x]$ . Upon noting that $1 = - \frac{1}{2} (x - 1) + \frac{1}{2} (x + 1) \in I_{1} + I_{2}$ it follows that $I_{1} + I_{2} = Q [x]$ and so the ideals $I_{1}$ and $I_{2}$ are coprime. The Chinese Remainder Theorem thus gives a ring isomorphism

Q [x] / ⟨ (x - 1) (x + 1) ⟩ \tilde{\to} Q [x] / ⟨ x - 1 ⟩ \times Q [x] / ⟨ x + 1 ⟩ .

(Here we are using the angled-bracket notation for ideals, to avoid having too many nested parentheses.)

Explicitly, this isomorphism maps each coset $p (x) + ⟨ (x - 1) (x + 1) ⟩$ to the pair of cosets $(p (x) + ⟨ x - 1 ⟩, p (x) + ⟨ x + 1 ⟩)$ .