\(\textbf{Lemma 1. }\)Suppose \(I, J \)are two-sided ideals in the ring R. Then \(I\times J\) and \(I\cap J\) are two-sided ideals.
\(\textbf{Proof. }\)Since both \(I\) and \(J\) are two-sided ideals, we have Suppose that \(a\in I\cap J\), then for any \(r\in R\), we have \(ra \in I\) and \(ra \in J\), so \(ra \in I\cap J\). Similarly, suppose that \(b\in I\cap J\), for any \(r\in R\), we have \(ar \in I\) and \(ar \in J\), so \(ar \in I\cap J\). Thus, we show that \(I\cap J\) is a two-sided ideal. For \(I + J\), for any \(r\in R\), we have \[ r(I + J)r = rIr + rJr = I + J. \] Thus, we show that \(I + J\) is a two-sided ideal. \[ \tag*{$\square$} \]
\(\textbf{Lemma 2. }\)Suppose \( I, J \) are two-sided ideals in the ring \( R \). There is an injective ring homomorphism \( \phi: R/(I \cap J) \rightarrow R/I \times R/J \).
\(\textbf{Proof. }\)we define the map \(\phi: R/(I \cap J) \rightarrow R/I \times R/J\) by \[ \phi((r + I \cap J)) = (r + I, r + J) \] We show that \(\phi\) is a ring homomorphism. For any \(r, s \in R\), we have \[ \phi(r + I\cap J) + \phi(s + I\cap J) = (r + I, r + J) + (s + I, s + J) = (r + s + I, r + s + J) = \phi((r + s) + I\cap J). \] \[ \phi((r + (I \cap J))(s + (I \cap J))) = \phi(rs + (I \cap J)) = (rs + I, rs + J) = (r + I, r + J)(s + I, s + J)= \phi(r + (I \cap J))\phi(s + (I \cap J)). \] Now we define this map is well defined. Suppose that \(r_1 + I\cap J = r_2 + I\cap J\). We have \(r_1 - r_2 + I\cap J = I\cap J\), so \(r_1 - r_2 \in I\cap J\). Thus, we have \(r_1 - r_2 \in I\) and \(r_1 - r_2 \in J\). Thus, we have \(r_1 = r_2 + I\) and \(r_1 = r_2 + J\), which implies that \(r_1 + I = r_2 + I\) and \(r_1 + J = r_2 + J\). Thus, we have \[ \phi(r_1 + I\cap J) = (r_1 + I, r_1 + J) = (r_2 + I, r_2 + J) = \phi(r_2 + I\cap J). \] It show that \(\phi\) is well defined. Now we have \(\phi\) is a homomorphism, and we need to show that \(\phi\) is injective. Now suppose that \(\phi(r_1 + I\cap J) = \phi(r_2 + I\cap J)\). In other words, we have \((r_1 + I, r_1 + J) = (r_2 + I, r_2 + J)\). Thus, we have \(r_1 + I = r_2 + I\) and \(r_1 + J = r_2 + J\). Thus, we have \(r_1 - r_2 \in I\) and \(r_1 - r_2 \in J\). Thus, we have \(r_1 - r_2 \in I\cap J\) which implies that \(r_1 + I\cap J = r_2 + I\cap J\). Thus, we have \(\phi\) is injective. \[ \tag*{$\square$} \]
\(\textbf{Lemma 3. }\)Suppose \( I, J \) are two-sided ideals in the ring \( R \). If \( I + J = R \). Then there is an surjective ring homomorphism \( \phi: R/(I \cap J) \rightarrow R/I \times R/J \)
\(\textbf{Proof. }\) We already know that \(\phi\) is injective, so we need to show that \(\phi\) is surjective. Since we have \(I + J = R\), and \(1\in R\), which means there exists \(i + j = 1\) for some \(i\in I\) and \(j\in J\). Then we have \(i = i - j\) and \begin{align*} \phi(i + I\cap J) = (i + I, i + J) = (i + I , 1 + (- j) + J) = (I, 1 + J),\\ \phi(j + I\cap J) = (j + I, j + J) = (1 + (- i) + I, j + J) = (1 + I, J). \end{align*} Now for any \((s + I, r + J)\) in \(R/I \times R/J\), we have \[ \begin{align} \phi(ri + sj + I\cap J) &= \phi(ri + I\cap J) + \phi(sj + I \cap J)\\ &= \phi(r + I\cap J)\phi(i + I\cap J) + \phi(s + I\cap J)\phi(j + I\cap J)\\ &= (r + I, r + J)(I, 1 + J) + (s + I, s + J)(1 + I, J)\\ &= (I, r + J) + (s + I, J)\\ &= (r + I, s + J). \end{align} \] Thus, we have \(\phi\) is surjective. \[ \tag*{$\square$} \]
\(\textbf{Theorem. (Chinese Remainder Theorem) }\)Let \(A_1, A_2, \ldots, A_x\) be ideals in \(R\). The map \[ R \to R/A_1 \times R/A_2 \times \ldots \times R/A_x \] defined by \(r \mapsto (r + A_1, r + A_2, \ldots, r + A_x)\) is a ring homomorphism with kernel \(A_1 \cap A_2 \cap \ldots \cap A_x\). If for each \(i, j \in \{1, 2, \ldots, x\}\) with \(i \neq j\) the ideals \(A_i\) and \(A_j\) are comaximal, then this map is surjective and \[ A_1 \cap A_2 \cap \ldots \cap A_x = A_1 A_2 \ldots A_x, \] so \[ R/(A_1 \cap A_2 \cap \ldots \cap A_x) = R/(A_1 A_2 \ldots A_x) = R/A_1 \times R/A_2 \times \ldots \times R/A_x. \]
\(\textbf{Proof. }\)We first prove this for \(k = 2\); the general case will follow by induction.
Let \(A = A_1\) and \(B = A_2\). Consider the map \(\phi: R \to R/A \times R/B\) defined by \(\phi(r) = (r \mod A, r \mod B)\), where \(\mod A\) means the class in \(R/A\) containing \(r\) (that is, \(r + A\)). This map is a ring homomorphism because \(\phi\) is just the natural projection of \(R\) into \(R/A\) and \(R/B\) for the two components. The kernel of \(\phi\) consists of all the elements \(r \in R\) that are in \(A\) and in \(B\), i.e., \(A \cap B\). To complete the proof in this case it remains to show that when \(A\) and \(B\) are comaximal, \(\phi\) is surjective and \(A \cap B = AB\).
Since \(A + B = R\), there are elements \(x \in A\) and \(y \in B\) such that \(x + y = 1\). This equation shows that \(\phi(x) = (0, 1)\) and \(\phi(y) = (1, 0)\) since, for example, \(x\) is an element of \(A\) and \(x = 1 - y \in 1 + B\). If now \((r_1 \mod A, r_2 \mod B)\) is an arbitrary element in \(R/A \times R/B\), then the element \(r_2x + r_1y\) maps to this element since \[ \phi(r_2x + r_1y) = \phi(r_2)\phi(x) + \phi(r_1)\phi(y) = (r_2 \mod A, r_2 \mod B) (0, 1) + (r_1 \mod A, r_1 \mod B) (1, 0) = (0, r_2 \mod B) + (r_1 \mod A, 0) = (r_1 \mod A, r_2 \mod B). \] This shows that \(\phi\) is indeed surjective. Finally, the ideal \(AB\) is always contained in \(A \cap B\). If \(A\) and \(B\) are comaximal and \(x\) and \(y\) are as above, then for any \(c \in A \cap B\), \(c = c1 = cx + cy \in AB\). This establishes the reverse inclusion \(A \cap B \subseteq AB\) and completes the proof when \(k = 2\).
The general case follows easily by induction from the case of two ideals using \(A = A_1\) and \(B = A_2 \cdots A_x\) once we show that \(A_1\) and \(A_2 \cdots A_x\) are comaximal. By hypothesis, for each \(i \in \{2, 3, \ldots, k\}\) there are elements \(x_i \in A_1\) and \(y_i \in A_i\) such that \(x_i + y_i = 1\). Since \(x_i + y_i = y_i \mod A_s\), it follows that \(1 = (x_2 + y_2) \cdots (x_k + y_k)\) is an element in \(A_s + (A_2 \cdots A_x)\). \[ \tag*{$\square$} \]