Chinese Remainder Theorem for Rings

This page explains the Chinese remainder theorem for rings (with identity) and is largely based on the Proposition 10 in [Bou89, Chapter I, Section 8].

Objectives

Let $R$ be a ring, $\calI = \set{0, \dots, m - 1}$ be an index set, and $(I_i)_{i \in \calI}$ be a system of pair-wise coprime ideals.

• Then, we have an isomorphism $\eta$ for \[ \frac{R}{\bigcap_{i \in \calI} I_i} \cong \prod_{i \in \calI} \frac{R}{I_i} \] by sending $x \bmod \bigcap_{i \in \calI} I_i$ to $(x \bmod I_i)_{i \in \calI}$.
• If $\bigcap_{i \in \calI} I_i = 0$, we have $\exists e_{\calI} \in R^m, \forall i \in \calI, I_i = (1 - e_i) R$ where $e_{\calI}$ is a system of pair-wise orthogonal central idempotent elements.
• $\eta^{-1}$ is the map $x_{\calI} \mapsto \sum_{i \in \calI} e_i x_i$.
• The existence of $e_{\calI}$ is equivalent to the existence of $I_{\calI}$ with $\bigcap_{i \in \calI} I_i = 0$.

Coprime Ideals

Definitions

• Coprime ideals $I_0$ and $I_1$ of the ring $R$. \[ \exists r_0 \in I_0, \exists r_1 \in I_1, r_0 + r_1 = 1. \] Equivalently, \[ I_0 + I_1 = R. \]
• A system of pair-wise coprime ideals $I_{\calI}$. \[ \forall i, j \in \calI, i \neq j \implies I_i + I_j = R. \]

Case of two coprime ideals

Let $R$ be a ring, and $I_0$ and $I_1$ be ideals of $R$. We say that $I_0$ and $I_1$ are coprime if there are elements $r_0 \in I_0$ and $r_1 \in I_1$ such that $r_0 + r_1 = 1$. This is equivalent to $I_0 + I_1 = R$. The Chinese remainder theorem states that there is an isomorphism $\eta$ from $\frac{R}{I_0 \cap I_1}$ to $\frac{R}{I_0} \times \frac{R}{I_1}$ sending $x \bmod I_0 \cap I_1$ to $(x \bmod I_0, x \bmod I_1)$.

Case of finitely many pair-wise coprime ideals

We can generalize it to ideals that are pair-wise coprime. The ideals $I_0, \dots, I_m$ are called pair-wise coprime if for $i \neq j$, $I_i$ and $I_j$ are coprime. Let $I_{\calI}$ be a system of pair-wise coprime ideals. The Chinese remainder theorem states that there is an isomorphism $\eta$ from $\frac{R}{\bigcap_{i \in \calI} I_i}$ to $\prod_{i \in \calI} I_i$ sending $x \bmod \bigcap_{i \in \calI} I_i$ to $(x \bmod I_i)_{i \in \calI}$.

Examples

TBA

Idempotent Elements

Definitions

• Idempotent element $e \in R$. \[ e^2 = e. \]
• A system of orthogonal idempotent elements $e_{\calI} \in R^m$. \[ \forall i, j \in \calI, e_i e_j = \delta_{i, j} e_i. \]

In a ring $R$, an element $e$ is called idempotent if $e^2 = e$. Furthermore, if $e$ commutes with all the elements of $R$, we call it a central idempotent element. Let $e_1$ and $e_2$ be two distinct idempotent elements in $R$. They are called orthogonal if $e_1 e_2 = e_2 e_1 = 0$. The Chinese remainder theorem for rings is closely related to a system of pair-wise orthogonal central idempotent elements that sums to $1$.

Case of two orthogonal central idempotent elements

Let $e_0, e_1 \in R$ be two orthogonal central idempotent elements with $e_0 + e_1 = 1$ and define $I_0 = (1 - e_0) R$ and $I_1 = (1 - e_1) R$. We claim the following.

• $I_0$ and $I_1$ are coprime.
• $I_0 \cap I_1 = \set{0}$.

For showing $I_0 \cap I_1 = \set{0}$, we will prove the following observations.

• $I_0 I_1 = I_0 \cap I_1$.
• $I_0 I_1 = \set{0}$.

Clearly, if the observations hold, we have $I_0 \cap I_1 = \set{0}$.

Proofs

• Prove that $I_0$ and $I_1$ are coprime.
  • Since $e_2 \in e_2 R = I_0$, $e_1 \in e_1 R = I_1$, and $e_2 + e_1 = 1$, $I_0$ and $I_1$ are coprime as desired.
• Proof for $I_0 I_1 = I_0 \cap I_1$.
  • We first recall that for two coprime ideals $J$ and $K$, $J \cap K = J K + K J$. For the proof, please refer to the supplementary material. Once we show that $I_0 I_1 = I_1 I_0$, we prove the observation. For arbitrary $r_{0, 0}$, $\dots$, $r_{0, c - 1}$, $r_{1, 0}$, $\dots$, $r_{1, c - 1} \in R$, since $\sum_{i = 0}^{c - 1} (1 - e_0) r_{0, i} (1 - e_1) r_{1, i} = \sum_{i = 0}^{c - 1} (1 - e_1) r_{0, i} (1 - e_0) r_{1, i}$, we have $I_0 I_1 = I_1 I_0$ as desired.
• Proof for $I_0 I_1 = \set{0}$.
  • Since for arbitrary $r_0, r_1 \in R$, $(1 - e_0) r_0 (1 - e_1) r_1 = (1 - (e_0 + e_1)) r_0 r_1 = 0$, we have for arbitrary $r_{0, 0}, \dots, r_{0, c - 1}, r_{1, 0}, \dots, r_{1, c - 1} \in R$, $\sum_{i = 0}^{c - 1} (1 - e_0) r_{0, i} (1 - e_1) r_{1, i} = 0$. Therefore, $I_0 I_1 = \set{ \sum_{i = 0}^{c - 1} (1 - e_0) r_{0, i} (1 - e_1) r_{1, i} | r_0, r_1 \in R, c \in \mN^+} = \set{0}$ as desired.

Case of finitely many orthogonal central idempotent elements

We generalize to $e_\calI$ as follows. Let’s write $I_i$ for $(1 - e_i) R$.

• For $i \neq j$, the ideals $I_i$ and $I_j$ are coprime.
• $\bigcap_{i \in \calI} I_i = \set{0}$.

Similarly, for showing $\bigcap_{i \in \calI} I_i = \set{0}$, we will prove the following.

• $\prod_{i \in \calI} I_i = \bigcap_{i \in \calI} I_i$.
• $\prod_{i \in \calI} I_i = \set{0}$.

Proofs

• Prove that for $i \neq j$, the ideals $I_i$ and $I_j$ are coprime.
  • We first observe that $e_i = (1 - e_j) e_i \in I_j $. Now, we choose $1 - e_i \in I_i$ and $e_i \in I_j$ which sums to $1$ as desired.
• Proof for $\prod_{i \in \calI} I_i = \bigcap_{i \in \calI} I_i$.
  • We proceed similarly by first recalling $\bigcap_{i \in \calI} J_i = \sum_{\pi \in S_m} \prod_{i \in \calI} J_{\pi(i)}$ for mutually coprime ideals $J_\calI$. The proof is given as a supplementary material. We now claim that for arbitrary $\pi_0, \pi_1 \in S_m$, $\prod_{i \in \calI} I_{\pi_0(i)} = \prod_{i \in \calI} I_{\pi_1(i)}$. This is immediate since we know that for $i \neq j$, $I_i I_j = I_j I_i$. We conclude that $\bigcap_{i \in \calI} I_i = \sum_{\pi \in S_m} \prod_{i \in \calI} I_{\pi(i)} = \prod_{i \in \calI} I_i$.
• Proof for $\prod_{i \in \calI} I_i = \set{0}$.
  • For arbitrary $r_\calI \in R^m$, we claim that $\prod_{i \in \calI} (1 - e_i) r_i = 0$. Once we show this, we have \[ \prod_{i \in \calI} I_i = \set{ \prod_{i \in \calI} (1 - e_i) r_i | r_\calI \in R^m } = \set{0} \] as desired. We show $\prod_{i \in \calI} (1 - e_i) r_i = 0$ as follows. \[ \prod_{i \in \calI} (1 - e_i) r_i = \left( \prod_{i \in \calI} (1 - e_i) \right) \left( \prod_{i \in \calI} r_i \right) = \left( 1 - \sum_{i \in \calI} e_i \right) \left( \prod_{i \in \calI} r_i \right) = 0. \]

Examples

TBA

Supplementary

• Let $J_\calI$ be pair-wise coprime ideals of $R$. We have \[ \bigcap_{i \in \calI} J_i = \sum_{\pi \in S_m} \prod_{i \in \calI} J_{\pi(i)} \] where $S_m$ is the symmetric group defined on a set of $m$ elements. Since $\sum_{\pi \in S_m} \prod_{i \in \calI} J_{\pi(i)} \subseteq \bigcap_{i \in \calI} J_i$ by the definition of ideals, we only need to show $\bigcap_{i \in \calI} J_i \subseteq \sum_{\pi \in S_m} \prod_{i \in \calI} J_{\pi(i)}$. We prove by induction on $m$ as follows.
  • $m = 2$: since there are elements $e_0 \in J_0$ and $e_1 \in J_1$ with $e_0 + e_1 = 1$, we write an element $x \in I_0 \cap I_1$ as $x = x(e_0 + e_1) = x e_0 + x e_1 \in J_1 J_0 + J_0 J_1$, and therefore $J_0 \cap J_1 \subseteq J_0 J_1 + J_1 J_0$.
  • $m \implies m + 1$: \[ \bigcap_{i \in \set{0, \dots, m}} J_i \subseteq \left( \sum_{\pi \in S_m} \prod_{i \in \set{0, \dots, m - 1}} J_{\pi(i)} \right) J_m + J_m \left( \sum_{\pi \in S_m} \prod_{i \in \set{0, \dots, m - 1}} J_{\pi(i)} \right) \subseteq \sum_{\pi \in S_{m + 1}} \prod_{i \in \set{0, \dots, m}} J_{\pi(i)}. \]

References

• [Bou89] Algebra. Nicolas Bourbaki. 1989.