Chinese Remainder Theorem (Commutative Algebra)

Theorem

Let $A$ be a commutative ring.

Let $I_1, \ldots, I_k$ for some $k \ge 1$ be pairwise coprime ideals in $A$, that is:

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

Then there is an isomorphism of rings:

$A / \left({I_1 \cap \ldots \cap I_k}\right) \to A / I_1 \times \cdots \times A / I_k$

which is induced by the ring homomorphism $\phi: A \to A / I_1 \times \cdots \times A / I_k$ defined as:

$\phi \left({x}\right) = \left({x + I_1, \ldots, x + I_k}\right)$

which passes through the quotient.

Proof

The mapping $\phi$ is indeed a ring homomorphism, because each $A \to A / I_i$ is a ring homomorphism.

The kernel of $\phi$ is given by:

$\displaystyle \ker \phi = \left\{{x \in A: \forall i, 1 \le i \le k : x \in I_i}\right\} = \bigcap_{1 \mathop \le i \mathop \le k} I_i$

So $\phi$ defines an injective homomorphism by passing through the quotient.

It remains then to prove that:

$\tilde \phi \equiv \phi \circ \pi$

where $\pi$ is the canonical quotient-map, is surjective.

That is:

$\forall x_i \in A, 1 \le i \le k: \exists x \in A: x - x_i \in I_i, 1 \le i \le k$

Then it follows that:

$\displaystyle \left({x_1 + I_1, \ldots, x_k + I_k}\right) = \tilde \phi \left({x + \bigcap I_i}\right)$

Note that:

$\left({x_1 + I_1, \ldots, x_k + I_k}\right) = \left({x_1 + I_1}\right) e_1 + \cdots + \left({x_k + I_k}\right) e_k$

where the unit element lies at the $i$-th component:

$e_i = \left({0, \ldots, 0, 1_{A / I_i}, 0, \ldots, 0}\right)$

This implies that it is enough to find $a_i \in A, 1 \le i \le k$, such that:

$\tilde \phi (a_i + \bigcap I_i) = e_i$

since:

\(\displaystyle \tilde \phi \left({x_1 a_1 + \cdots + x_k a_k}\right)\)

\(=\)

\(\displaystyle \tilde \phi \left({x_1}\right) \tilde \phi \left({a_1}\right) + \cdots + \tilde \phi \left({x_k}\right) \tilde \phi \left({a_k}\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle \left({\tilde \phi \left({x_1}\right), \ldots, \tilde \phi \left({x_k}\right)}\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle \left({x_1 + I_1, \ldots, x_n + I_n}\right)\)

We show that $e_1$ is in the image of $\tilde \phi$.

The other instances of $e_i$ then follow in a similar manner.

We then need to find an $a \in A$ such that $a -1 \in I_1, a \in I_2, \ldots, a \in I_k$.

Since then $I_1$ is coprime with the other ideals, we have that $I_1 + I_j = A, 2 \le j \le k$.

So there exists a $b_j \in I_1$, $c_j \in I_j$ such that $b_j + c_j = 1$.

Note that this implies that $c_j - 1 \in I_1, c_j \in I_j$.

Define now $a = c_2 c_3 \cdots c_k \in A$.

Then for $2 \le j \le k$:

$a = c_j \left({c_2 \cdots c_{j - 1} c_{j + 1} \cdots c_k}\right) \in I_j$

and:

\(\displaystyle a + I_1\)

\(=\)

\(\displaystyle \left({c_2 + I_1}\right) \cdots \left({c_k + I_1}\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle \left({1 + I_1}\right) \cdots \left({1 + I_1}\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle 1 + I_1\)

Hence:

$a - 1 \in I_1$

$\blacksquare$