CRT in General

Recall that all rings in this book are commutative with unity.

Definition 5.1.1 (Coprime)   Ideals $I$ and $J$ are coprime if $I+J=(1)$.

If $I$ and $J$ are nonzero ideals in the ring of integers of a number field, then they are coprime precisely when the prime ideals that appear in their two (unique) factorizations are disjoint.

Lemma 5.1.2   If $I$ and $J$ are coprime ideals in a ring $R$, then $I\cap{}J = IJ$.

Proof. Choose $x\in I$ and $y\in J$ such that $x+y=1$. If $c\in{} I\cap{} J$ then

$$c=c\cdot 1=c\cdot (x+y) = cx + cy \in IJ + IJ = IJ,$$

so $I\cap{} J\subset IJ$. The other inclusion is obvious by definition of ideal. $\qedsymbol$

Lemma 5.1.3   Suppose $I_1,\ldots, I_s$ are pairwise coprime ideals. Then $I_1$ is coprime to the product $I_2\cdots I_s$.

Proof. It suffices to prove the lemma in the case $s=3$, since the general case then follows from induction. By assumption, there are $x_1 \in I_1, y_2 \in I_2$ and $a_1 \in I_1, b_3 \in I_3$ such

$$x_1 + y_2 = 1$$   and$$\qquad a_1 + b_3 = 1.$$

Multiplying these two relations yields

$$x_1 a_1 + x_1 b_3 + y_2 a_1 + y_2 b_3 = 1 \cdot 1 = 1.$$

The first three terms are in $I_1$ and the last term is in $I_2 I_3 = I_2 \cap I_3$ (by Lemma 5.1.2), so $I_1$ is coprime to $I_2 I_3$. $\qedsymbol$

Next we prove the general Chinese Remainder Theorem. We will apply this result with $R=\O_K$ in the rest of this chapter.

Theorem 5.1.4 (Chinese Remainder Theorem)   Suppose $I_1,\ldots, I_r$ are nonzero ideals of a ring $R$ such $I_m$ and $I_n$ are coprime for any $m\neq n$. Then the natural homomorphism $R \to \bigoplus_{n=1}^r R/I_n$ induces an isomorphism

$$\psi: R/\prod_{n=1}^r I_n \to \bigoplus_{n=1}^r R/I_n.$$

Thus given any $a_n \in R$, for $n=1,\ldots,r$, there exists some $a\in R$ such that $a\equiv a_n\pmod{I_n}$ for $n=1,\ldots,r$; moreover, $a$ is unique modulo $\prod_{n=1}^r I_n$.

Proof. Let $\varphi :R \to \bigoplus_{n=1}^r R/I_n$ be the natural map induced by reduction modulo the $I_n$. An inductive application of Lemma 5.1.2 implies that the kernel $\cap_{n=1}^r I_n$ of $\varphi$ is equal to $\prod_{n=1}^r I_n$, so the map $\psi$ of the theorem is injective.

Each projection $R\to R/I_n$ is surjective, so to prove that $\psi$ is surjective, it suffices to show that $(1,0,\ldots,0)$ is in the image of $\varphi$, and similarly for the other factors. By Lemma 5.1.3, $J=\prod_{n=2}^rI_n$ is coprime to $I_1$, so there exists $x\in I_1$ and $y\in J$ such that $x+y=1$. Then $y = 1-x$ maps to $1$ in $R/I_1$ and to 0 in $R/J$, hence to 0 in $R/I_n$ for each $n\geq 2$, since $J\subset I_n$. $\qedsymbol$