Chinese Remainder Theorem (Commutative Algebra)

Theorem
Let $A$ be a commutative and unitary ring.

Let $I_1, \ldots, I_n$ for some $n \ge 1$ be ideals of $A$.

Then the ring homomorphism $\phi: A \to A / I_1 \times \cdots \times A / I_n$ defined as:


 * $\map \phi x = \tuple {x + I_1, \ldots, x + I_n}$

has the kernel $\ds I := \bigcap_{i \mathop = 1}^n I_i$, and is surjective the ideals are pairwise coprime, that is:
 * $\forall i \ne j: I_i + I_j = A$

Hence in that case, it induces an ring isomorphism:


 * $A / I \to A / I_1 \times \cdots \times A / I_n$

through the First Isomorphism Theorem.

Proof
The mapping $\phi$ is indeed a ring homomorphism, because each canonical projection $\phi_i: A \to A / I_i$ is a ring homomorphism.

The kernel of $\phi$ is given by:
 * $\ds \ker \phi = \set {x \in A: \forall i, 1 \le i \le n : x \in I_i} = \bigcap_{1 \mathop \le i \mathop \le n} I_i =: I$

It remains then to be proved that $\phi$ is surjective the ideals are pairwise coprime.

Stated explicitly, we will show that the statement:
 * $\forall x_i \in A, 1 \le i \le n: \exists x \in A: x - x_i \in I_i, 1 \le i \le n$

holds :
 * $\forall i \ne j: I_i + I_j = A$

To reach this goal, we now define $e_i \in A / I_1 \times \cdots \times A / I_n$ so that a unity lies at the $i$th coordinate:
 * $e_i := \tuple {0, \ldots, 0, 1_{A / I_i}, 0, \ldots, 0}$

Sufficient Condition
We will start by showing the condition is sufficient for surjectivity.

So suppose $\phi$ is surjective.

Then in particular, for each $i$, there is $a_i \in A$ such that $\map \phi {a_i} = e_i$.

Clearly, $\map {\phi_j} {a_i} = 0$ for $j \ne i$ while $\map {\phi_i} {1 - a_i} = \map {\phi_i} 1 - \map {\phi_i} {a_i} = 1 - 1 = 0$.

Hence for all $i \ne j$, we find:
 * $1 = a_i + \paren {1 - a_i} \in I_j + I_i$

Since Sum of Ideals is Ideal, we can conclude $r \cdot 1 \in I_j + I_i$ for all $r \in R$.

This completes the proof that $I_i + I_j = R$.

Necessary Condition
We will now show the converse that the ideals being coprime is necessary.

Note that each $x \in A / I_1 \times \cdots \times A / I_n$ may then be written as:
 * $x = \tuple {x_1 + I_1, \ldots, x_n + I_n} = \paren {x_1 + I_1} e_1 + \cdots + \paren {x_n + I_n} e_n$

for some choice of $x_i \in A$.

This implies that it is enough to find $a_i \in A, 1 \le i \le n$, such that:
 * $\map \phi {a_i} = e_i$

since then:

To construct the $a_i \in A$, we need that $a_i - 1 \in I_i$, but $a_i \in I_j$ for all $j \ne i$.

Since $I_i$ is coprime with the other ideals, we have that:
 * $I_i + I_j = A, i \ne j$

In particular there exist $u_{i j} \in I_i$, $v_{i j} \in I_j$ for each pair $\tuple {i, j}$ with $i \ne j$ such that $u_{i j} + v_{i j} = 1$.

Define now $\ds a_i = \prod_{k \mathop \ne i} v_{i k}$.

Then for $k \ne i$:


 * $a_i = v_{i k} \tuple {v_{i 2} \cdots v_{i \paren {k - 1} } v_{i \paren {k + 1} } \cdots v_{i n} } \in I_k$

and:

Hence:
 * $a_i - 1 \in I_i$

which verifies that $\map {\phi} {a_i} = e_i$.

This concludes the proof of the necessity.

Now to conclude, let $\phi$ be surjective again.

Let $\tilde \phi$ be the injective homomorphism obtained by the First Isomorphism Theorem:
 * $\map {\tilde \phi} {x + I} = \tuple {x_1 + I_1, \ldots, x_n + I_n}$

for all $x \in A$.

It immediately follows that $\tilde \phi$ is also surjective, and hence constitutes an isomorphism.