[proofplan] We verify each property of $\Theta$ in turn. Well-definedness is the observation that each $m_i$ divides $M$, so the residue class modulo $m_i$ depends only on the class modulo $M$. The ring homomorphism axioms reduce to the componentwise definition of addition and multiplication on the product ring. Finally, bijectivity is exactly the content of the classical Chinese Remainder Theorem: surjectivity is the existence of simultaneous solutions, and injectivity is their uniqueness modulo $M$. Since the domain and codomain are finite sets of the same cardinality $M$, we may deduce bijectivity from either surjectivity or injectivity alone. [/proofplan] [step:Verify that $\Theta$ is well-defined] Suppose $a + M\mathbb{Z} = a' + M\mathbb{Z}$, i.e., $M \mid (a - a')$. For each $i \in \{1, \ldots, k\}$, since $m_i \mid M$ we have $m_i \mid (a - a')$, so $a + m_i\mathbb{Z} = a' + m_i\mathbb{Z}$. Therefore the tuple \begin{align*} (a + m_1\mathbb{Z},\, \ldots,\, a + m_k\mathbb{Z}) \end{align*} does not depend on the representative $a$ chosen for the class $a + M\mathbb{Z}$. This shows that the map \begin{align*} \Theta: \mathbb{Z}/M\mathbb{Z} &\to \mathbb{Z}/m_1\mathbb{Z} \times \cdots \times \mathbb{Z}/m_k\mathbb{Z} \\ a + M\mathbb{Z} &\mapsto (a + m_1\mathbb{Z},\, \ldots,\, a + m_k\mathbb{Z}) \end{align*} is well-defined. [/step] [step:Verify that $\Theta$ is a ring homomorphism] The addition and multiplication on the product ring are defined componentwise: for tuples $(x_1, \ldots, x_k)$ and $(y_1, \ldots, y_k)$, \begin{align*} (x_1, \ldots, x_k) + (y_1, \ldots, y_k) &= (x_1 + y_1, \ldots, x_k + y_k), \\ (x_1, \ldots, x_k) \cdot (y_1, \ldots, y_k) &= (x_1 y_1, \ldots, x_k y_k). \end{align*} The multiplicative identity is $(1 + m_1\mathbb{Z}, \ldots, 1 + m_k\mathbb{Z})$. For $a, b \in \mathbb{Z}$, \begin{align*} \Theta((a + b) + M\mathbb{Z}) &= ((a+b) + m_1\mathbb{Z},\, \ldots,\, (a+b) + m_k\mathbb{Z}) \\ &= (a + m_1\mathbb{Z},\, \ldots,\, a + m_k\mathbb{Z}) + (b + m_1\mathbb{Z},\, \ldots,\, b + m_k\mathbb{Z}) \\ &= \Theta(a + M\mathbb{Z}) + \Theta(b + M\mathbb{Z}), \end{align*} and identically for multiplication. Finally, $\Theta(1 + M\mathbb{Z}) = (1 + m_1\mathbb{Z}, \ldots, 1 + m_k\mathbb{Z})$ is the identity of the product ring. Hence $\Theta$ is a ring homomorphism. [/step] [step:Reduce bijectivity to the Chinese Remainder Theorem] Both source and target are finite rings. The source has cardinality $M = m_1 \cdots m_k$, and the target has cardinality $m_1 \cdot m_2 \cdots m_k = M$ as well. Because the two sets have equal finite cardinality, it suffices to show that $\Theta$ is surjective (equivalently, injective). [guided] The source $\mathbb{Z}/M\mathbb{Z}$ has $M$ elements. The target is the Cartesian product of $\mathbb{Z}/m_i\mathbb{Z}$ for $i = 1, \ldots, k$, each of which has $m_i$ elements, so the product has $m_1 m_2 \cdots m_k = M$ elements. Since the pairwise coprimality $(m_i, m_j) = 1$ for $i \neq j$ implies the product formula $M = \prod_i m_i$ holds in the natural-number sense (no cancellation), both rings are finite sets of equal size $M$. For a function between finite sets of equal cardinality, injectivity and surjectivity are equivalent. Therefore we need only prove one of them; we choose surjectivity (existence of simultaneous solutions), since that is the standard content of the [Chinese Remainder Theorem](/theorems/???). [/guided] [/step] [step:Prove surjectivity via the Chinese Remainder Theorem] Let $(b_1 + m_1\mathbb{Z}, \ldots, b_k + m_k\mathbb{Z})$ be an arbitrary element of the product ring, with $b_i \in \mathbb{Z}$. We apply the [Chinese Remainder Theorem](/theorems/???), which asserts: given integers $b_1, \ldots, b_k$ and pairwise coprime positive integers $m_1, \ldots, m_k$, there exists $a \in \mathbb{Z}$ such that \begin{align*} a \equiv b_i \pmod{m_i} \quad \text{for each } i \in \{1, \ldots, k\}. \end{align*} The hypotheses are met: the $m_i$ are pairwise coprime by assumption. Hence such an $a$ exists, and $\Theta(a + M\mathbb{Z}) = (a + m_1\mathbb{Z}, \ldots, a + m_k\mathbb{Z}) = (b_1 + m_1\mathbb{Z}, \ldots, b_k + m_k\mathbb{Z})$. Therefore $\Theta$ is surjective. Since it is a map between finite sets of equal cardinality, it is also injective, hence bijective. A bijective ring homomorphism is a ring isomorphism, completing the proof. [guided] Given any target tuple $(b_1 + m_1\mathbb{Z}, \ldots, b_k + m_k\mathbb{Z})$, we must exhibit a preimage — that is, an integer $a$ satisfying \begin{align*} a \equiv b_1 \pmod{m_1}, \quad a \equiv b_2 \pmod{m_2}, \quad \ldots, \quad a \equiv b_k \pmod{m_k}. \end{align*} The existence of such an $a$ is precisely what the [Chinese Remainder Theorem](/theorems/???) guarantees, under the hypothesis that $m_1, \ldots, m_k$ are pairwise coprime. We verify this hypothesis: by assumption, the factorisation $M = m_1 \cdots m_k$ has pairwise coprime factors. Therefore such an $a$ exists, and by construction $\Theta(a + M\mathbb{Z})$ is our target tuple. Once surjectivity is established, injectivity follows from the pigeonhole principle applied to a surjection between finite sets of equal size. Explicitly: if $\Theta(a + M\mathbb{Z}) = \Theta(a' + M\mathbb{Z})$, then $m_i \mid (a - a')$ for all $i$. Since the $m_i$ are pairwise coprime, their least common multiple is their product $M$, so $M \mid (a - a')$, i.e., $a + M\mathbb{Z} = a' + M\mathbb{Z}$. Either argument suffices; together they show $\Theta$ is bijective. A bijective ring homomorphism is a ring isomorphism, so $\Theta$ is an isomorphism of rings. [/guided] [/step]