Chinese Remainder Theorem (Commutative Algebra)

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Theorem

Let $A$ be a commutative and unitary ring.


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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 if and only if 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.


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Proof 1

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 if and only if 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 if and only if:

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

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:

$e_i := \tuple {0, \ldots, 0, 1_{A / I_i}, 0, \ldots, 0}$


Necessary Condition

We will start by showing the condition is necessary 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 $I_i + I_j = R$.

$\Box$


Sufficient Condition

We will now show the converse that the ideals being coprime is sufficient.

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:

$\map \phi {a_i} = e_i$

because then:

\(\ds \map \phi {x_1 a_1 + \cdots + x_n a_n}\) \(=\) \(\ds \map \phi {x_1} \map \phi {a_1} + \cdots + \map \phi {x_n} \map \phi {a_n}\)
\(\ds \) \(=\) \(\ds \tuple {\map \phi {x_1}, \ldots, \map \phi {x_n} }\)
\(\ds \) \(=\) \(\ds \tuple {x_1 + I_1, \ldots, x_n + I_n}\)
\(\ds \) \(=\) \(\ds x\)

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:

\(\ds \map {\phi_i} {a_i}\) \(=\) \(\ds \map {\phi_i} {\prod_{k \mathop \ne i} v_{i k} }\)
\(\ds \) \(=\) \(\ds \map {\phi_i} {\prod_{k \mathop \ne i} \paren {1 - u_{i k} } }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop \ne i} \map {\phi_i} {1 - u_{i k} }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop \ne i} \paren {1 - \map {\phi_i} {u_{i k} } }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop \ne i} 1\)
\(\ds \) \(=\) \(\ds 1\)

Hence:

$a_i - 1 \in I_i$

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

This concludes the proof of the sufficiency.

$\Box$


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.

$\blacksquare$

Proof 2

Consider $\pi$ only as a homomorphism of groups.

Then Chinese Remainder Theorem (Groups) is applicable as Subgroup of Abelian Group is Normal.

We only need to demonstrate that the condition $I_i + I_j = R$ for all $i \neq j$ assumed here is equivalent to:

$\ds \forall k \le n - 1: I_{k + 1} + \bigcap_{i \mathop = 1}^k I_i = R$

The implication from the latter condition is immediate.

For the converse, let $i$ be arbitrary. The result follows from:

\(\ds R\) \(=\) \(\ds \prod_{j \neq i} (I_i + I_j)\) $R$ has a unit and $I_i + I_j = R$ for $j \neq i$
\(\ds \) \(\subseteq\) \(\ds I_i + \prod_{j \neq i} I_j\) distribution, absorption property of ideals
\(\ds \) \(\subseteq\) \(\ds I_i + \bigcap_{j \neq i}^n I_j\) Intersection of Ideals of Ring contains Product


$\blacksquare$

Also see

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