Chinese Reminder Theorem

First I’ll go over some definitions that are relevant to the Chinese Reminder Theorem and content that will be covered in the future.

The fracional field of an integral domain R is defined as $Frac(R) = \{(\frac{a}{b} : a, b \in R \backslash \{0\}\}$ where $(\frac{a}{b}) = (\frac{p}{q})$ if and only if there is some $x \in R \backslash \{0\}$ such that $xaq = xpb$.

If $(R_j)_{j \in J}$ is a collection of rings, $\prod_j R_j \{(a_j)_{j \in J}, a_k \in R_j\}$ can be made into a ring by considering coordinate wise addition and multiplication: $(a_i)_j +(b_j)_j = (a_j+b_j)_{j \in J}$ (addition) $(a_j)_j(b_j)_j = (a_j)\cdot b_{j_{j \in J}}$ (multiplication).

Let $(I_j)_{j \in J}$ be a collection of ideals. 1) the $I_j$ are said to be comaximal/coprime if $(\bigcup_j I_j) = R$ ($(\bigcup_j I_j)$ is the ideal generated by the union of the $I_j$). 2) the $I_j$ are said to be pairwise comaximal if for all $i \neq j, (I_i \cup I_j) = R$. For example, $n\mathbb{Z}, m\mathbb{Z}$ are comaximal if and only if gcd(m,n) = 1.

Chinese reminder theorem: let R be a commutative unitary ring ($1 \neq 0$) and $I_1,...,I_k$ be pairwise comaximal ideals, then $R/\cap_j I_j \cong \prod_j R/I_j$

Proof: we proceed by induction on k. For k = 2, Let $\phi: R \rightarrow R/I_1 \times R/I_2$ be a map defined by $x \mapsto (\pi_1(x), \pi_2(x))$ where $\pi_i : R \rightarrow R/I_i$. $\phi(x+y) = (\pi_1(x+y), \pi_2(x+y)) =$ $(\pi_1(x)+\pi_1(y), \pi_2(x)+\pi_2(y)) =$ $(\pi_1(x), \pi_2(x)) + (\pi_1(y), \pi_2(y)) =$ $\phi(x) + \phi(y)$. Since $\phi(1) = (\pi_1(1), \pi_2(1)) = (1,1)$ it follows that $\phi$ is a unitary ring homomorphism. ker$(\pi) = ker(\pi_1) \cap ker(\pi_2) = I_1 \cap I_2$. $I_1, I_2$ are comaximal, by definition $(I_1 \cup I_2) = R$ $\{ \sum a_i r_i : a_i \in I_1 \cup I_2, r_i \in R\} =$ $\{a_1+a_2: a_1 \in I_1, a_2 \in I_2\}$. Then there exists $a_i \in I_i$ such that $a_1+a_2 = 1$. $a_1 \mod I_2 = 1$, $a_1 = 1 \mod I_2$, $a_2 = 1 \mod I_1$. $\phi(a_1) = (0,1), \phi(a_2) = (1,0)$. Let $x, y \in R$ then $\phi(ya_1+xa_2) =$ $\phi(y)\phi(a_1)+\phi(x)\phi(a_2) =$ $(\pi_1(y), \pi_2(y))(0,1)+(\pi_1(x),\pi_2(x))(1,0) =$ $(0, \pi_2(y))+(\pi_1(x),0) =$ $(\pi_1(x), \pi_2(y))$ hence $\phi$ is surjective. By the 1st isomorphism theorem $K/I_1 \cap I_2 \cong R/I_1 \times R/I_2$, $x \mod I_1 \cap I_2 \mapsto (x \mod I_i)_{i=1,2}$ Assume case k. For k+1: we have $I_1,...,I_k,I_{k+1}$ pairwise comaximal ideals. Then for all $i \leq k, I_i$ and $I_{k+1}$ are comaximal, which means there exists an $x_i \in I_i$ such that $x_i+y_i=1$ where $y_i \in I_{k+1}$. Note that $x_i+y_i \mod I_{k+1} = x_i \mod I_{k+1}$ so $1 = \prod x_i+y_i = \prod x_i (\in \bigcap_{i \leq k}) \mod I_{k+1}$ hence $1 \in I_{k+1} + \bigcap_{i \leq k}I_i$ so they are comaximal. It follows that $R/\bigcap_{i \leq k+1}I_i \cong R/\bigcap_{i \leq k} I_i \times R/I_{k+1} \cong \prod_{i \leq k} R/I_i \times R/I_{k+1}$, $x \mapsto (x \mod \cap I_i, x \mod I_{k+1}) \mapsto (x \mod I_i) i \leq k+1$.

Remark: $\bigcap_{i \leq k} I_i = \prod_{i \leq k}I_i = \{\sum_{j}\prod_{i} a_{ij} : a_{ij} \in I_i \}$