The Chinese Remainder Theorem

Definition. Let $R$ be a commutative ring with identity. If $I$ and $J$ are two ideals of $R$ such that
$$
I+J=R,
$$
then $I$ and $J$ are said to be coprime.

Proposition. Let $I$, $J$, and $K$ be ideals of $R$. Then:

1. If $IJ\subseteq K$ and $J$ is coprime to $K$, then
   $$
   I\subseteq K.
   $$

2. If $I$ is coprime to both $J$ and $K$, then $I$ is coprime to $JK$.

3. If $I$ and $J$ are coprime, then for every natural number $n$,
   $$
   I^n \text{ and } J^n \text{ are coprime}.
   $$

4. If $I$ and $J$ are coprime, then
   $$
   IJ=I\cap J.
   $$

Proof: We first prove part (1). Since $J$ and $K$ are coprime, we have
$$
J+K=R.
$$
Therefore,
$$
IJ+IK=I(J+K)=IR=I.
$$
Here $IR=I$ because $R$ has an identity element: clearly $IR\subseteq I$, and for every $x\in I$ we have
$$
x=x\cdot 1\in IR.
$$
So indeed $IR=I$.

Now, since
$$
IJ\subseteq K,\qquad IK\subseteq K,
$$
and $K$ is an ideal, hence closed under addition, we obtain
$$
I=IJ+IK\subseteq K.
$$

We now prove part (2). Since $I$ is coprime to both $J$ and $K$, we have
$$
I+J=R,
$$
and
$$
I+K=R.
$$
Hence
$$
(I+J)(I+K)=RR=R.
$$
On the other hand,
$$
\begin{aligned}
(I+J)(I+K)
&=II+JI+IK+JK \\\
&=II+IJ+IK+JK \\\
&=II+I(J+K)+JK \\\
&=II+IR+JK \\\
&=II+I+JK \\\
&=I+JK.
\end{aligned}
$$
The equality $JI=IJ$ holds because $R$ is commutative, and the last equality follows from the fact that
$$
II\subseteq I.
$$
Therefore,
$$
I+JK=R.
$$
So $I$ and $JK$ are coprime.

Part (3) now follows from repeated application of part (2). Taking $K=J$ in part (2), we see that if $I$ and $J$ are coprime, then $I$ and $J^2$ are coprime. Repeating this argument, we obtain that $I$ and $J^n$ are coprime for every $n\in\mathbb{N}$. Similarly, we then obtain that $I^2$ and $J^n$ are coprime, and continuing in this way yields that
$$
I^n \text{ and } J^n \text{ are coprime}.
$$

Finally, we prove part (4). We already know that
$$
IJ\subseteq I\cap J.
$$
Now let $x\in I\cap J$. Since $I+J=R$, there exist $a\in I$ and $b\in J$ such that
$$
a+b=1.
$$
Therefore,
$$
x=x(a+b)=xa+xb=ax+xb.
$$
Since $x\in J$ and $a\in I$, we have
$$
ax\in IJ.
$$
Since $x\in I$ and $b\in J$, we have
$$
xb\in IJ.
$$
Thus
$$
x\in IJ.
$$
So
$$
I\cap J\subseteq IJ.
$$
Combining the two inclusions, we conclude that
$$
IJ=I\cap J.
$$

Theorem (Chinese Remainder Theorem). Let $R$ be a commutative ring with identity, and let $I_i$ $(1\le i\le n)$ be ideals of $R$ which are pairwise coprime. Then for any given elements $x_i\in R$ $(1\le i\le n)$, there exists an element $x\in R$ such that
$$
\begin{cases}
x\equiv x_1 \pmod{I_1}, \\\
x\equiv x_2 \pmod{I_2}, \\\
\cdots \\\
x\equiv x_n \pmod{I_n}.
\end{cases}
$$
Here the notation
$$
x\equiv y \pmod{I}
$$
means that
$$
x-y\in I.
$$

Proof: Since $I_1$ is coprime to each of $I_2,\dots,I_n$, by the proposition above it follows that $I_1$ is coprime to the product
$$
I_2I_3\cdots I_n.
$$
Thus
$$
I_1+I_2I_3\cdots I_n=R.
$$
Therefore, there exist
$$
y_1\in I_2I_3\cdots I_n\subseteq I_i \quad (i\neq 1),
$$
and
$$
y_1’\in I_1
$$
such that
$$
y_1+y_1’=1.
$$
Hence
$$
y_1-1=-y_1’\in I_1,
$$
and
$$
y_1-0=y_1\in I_i \quad (i\neq 1).
$$
Therefore,
$$
\begin{cases}
y_1\equiv 1 \pmod{I_1}, \\\
y_1\equiv 0 \pmod{I_i}, \quad i\neq 1.
\end{cases}
$$

Similarly, for each $j=2,3,\dots,n$, there exists $y_j\in R$ such that
$$
\begin{cases}
y_j\equiv 1 \pmod{I_j}, \\\
y_j\equiv 0 \pmod{I_i}, \quad i\neq j.
\end{cases}
$$

Now define
$$
x=x_1y_1+x_2y_2+\cdots+x_ny_n.
$$
Then for each $j$, all terms except $x_jy_j$ lie in $I_j$, while
$$
x_jy_j\equiv x_j \pmod{I_j}.
$$
Hence
$$
x\equiv x_j \pmod{I_j}
$$
for every $j=1,2,\dots,n$. This is exactly what we wanted.

The cover image of this article was taken in Irkutsk Oblast, Russia.