Chinese Remainder Theorem (Commutative Algebra)

Theorem
Let $A$ be a commutative ring. Let $I_1, ..., I_k$ for some $k \geq 1$ be pairwise coprime ideals in $A$, i.e. $I_i + I_j = A$ for $i \neq j$. Then there is an isomorphism of rings $ A / (I_1 \cap ... \cap I_k) \rightarrow A/I_1 \times ... \times A / I_k $ which is induced by the ring-homomorphism $$ \left\{ \begin{array}{lr} A \rightarrow A / I_1 \times ... \times A / I_k \\ x \mapsto (x + I_1, ..., x + I_k) \end{array} \right. $$ which passes through the quotient.

Proof
The homomorphism $A \rightarrow A / I_1 \times ... \times A / I_k$ is indeed one, because each $A \rightarrow A / I_i$ is a ring morphism.

The kernel of that homomorphism - say: $f$ - is $\ker f = \{ x \in A : \forall i, 1 \leq i \leq k : x \in I_i \} = \bigcap_{1 \leq i \leq k} I_i$, so $f$ defines an injective homomorphism by passing through the quotient. It remains then to prove that $\tilde{f} \equiv f \circ \pi$ (where $\pi$ is the canonical quotient-map) is surjective, that is: $\forall x_i \in A, 1 \leq i \leq k$, there exists an $x \in A$ such that $x - x_i \in I_i$, $1 \leq i \leq k$ - then it follows that $(x_1 + I_1, ..., x_k + I_k) = \tilde{f}(x + \bigcap I_i)$.

Note that $(x_1 + I_1, ..., x_k + I_k) = (x_1 + I_1)e_1 + ... + (x_k + I_k)e_k$ where $e_i = (0, ..., 0, 1_{A/I_i}, 0, ..., 0)$ (the unit element lies at the $i$-th component). This implies that it is enough to find $a_i \in A, 1 \leq i \leq k$, such that $\tilde{f}(a_i + \bigcap I_i) = e_i$ since $$\tilde{f}(x_1 a_1 + ... + x_k a_k) = \tilde{f}(x_1) \tilde{f}(a_1) + ... + \tilde{f}(x_k) \tilde{f}(a_k) = (\tilde{f}(x_1), ..., \tilde{f}(x_k)) = (x_1 + I_1, ..., x_n + I_n)$$

We show that $e_1$ is in the image of $\tilde{f}$ (the other $e_i$'s are similar). We then need to find an $a \in A$ such that $a -1 \in I_1, a \in I_2, ..., a \in I_k$. Since then $I_1$ is coprime with the other ideals, we have that $I_1 + I_j = A$, $2 \leq j \leq k$. 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 $a = c_2 c_3 ... c_k \in A$. Then for $2 \leq j \leq k$, $a = c_j(c_2 ... c_{j - 1} c_{j + 1} ... c_k) \in I_j$ and $a + I_1 = (c_2 + I_1) ... (c_k + I_1) = (1 + I_1) ... (1 + I_1) = 1 + I_1$. Hence $a - 1 \in I_1$. $\square$