This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Products of Ideals
Definition 1 For two two-sided ideals \(\mathfrak{a},\mathfrak{b}\) of a ring \(A\), their product \(\mathfrak{a}\mathfrak{b}\) is defined as the set
\[\mathfrak{a}\mathfrak{b}=\{x_1y_1+x_2y_2+\cdots+x_ny_n: x_i\in \mathfrak{a}, y_i\in \mathfrak{b}, n\geq 1\}.\]That \(\mathfrak{a}\mathfrak{b}\) is a subgroup of \(A\) under addition is clear. Moreover, for any element \(x_1y_1+\cdots+x_ny_n\) of \(\mathfrak{a}\mathfrak{b}\) and any element \(x\) of \(A\),
\[x(x_1y_1+\cdots+x_ny_n)=xx_1y_1+\cdots xx_ny_n\]and since \(xx_i\in \mathfrak{a}\), we have \(x(x_1y_1+\cdots+x_ny_n)\in \mathfrak{a}\mathfrak{b}\). A similar argument holds for multiplication on the right, so we see that \(\mathfrak{a}\mathfrak{b}\) is a two-sided ideal of \(A\).
Proposition 2 With respect to the multiplication defined above, the collection of two-sided ideals of \(A\) has a monoid structure with identity element \(A\). Moreover, the distributive laws
\[\mathfrak{a}(\mathfrak{b}+\mathfrak{c})=\mathfrak{a}\mathfrak{b}+\mathfrak{a}\mathfrak{c},\quad (\mathfrak{a}+\mathfrak{b})\mathfrak{c}=\mathfrak{a}\mathfrak{c}+\mathfrak{b}\mathfrak{c}\]also hold.
Proof
Let three two-sided ideals \(\mathfrak{a},\mathfrak{b},\mathfrak{c}\) be given. Then any element of \((\mathfrak{a}\mathfrak{b})\mathfrak{c}\) can be written in the form
\[\left(\sum_{i=1}^{n_1} x_i^{(1)}y_i^{(1)}\right)z_1+\cdots+\left(\sum_{i=1}^{n_k}x_i^{(k)}y_i^{(k)}\right)z_k\]and expanding everything using the distributive law and then grouping the rightmost two factors, we see that this element belongs to \(\mathfrak{a}(\mathfrak{b}\mathfrak{c})\). The reverse inclusion can be proved in the same way, so multiplication is associative. Also, it is clear that \(A \mathfrak{a}=\mathfrak{a}A=\mathfrak{a}\) for any two-sided ideal \(\mathfrak{a}\).
Finally, for any \(b_1+c_1,\ldots, b_n+c_n\in \mathfrak{b}+\mathfrak{c}\), expanding
\[a_1(b_1+c_1)+\cdots a_n(b_n+c_n)\]using the distributive law, one can easily show that \(\mathfrak{a}(\mathfrak{b}+\mathfrak{c})\subset \mathfrak{a}\mathfrak{b}+\mathfrak{a}\mathfrak{c}\). Conversely, for any
\[a_1b_1+\cdots+a_nb_n + a_1'c_1+\cdots +a_m'c_m\in \mathfrak{a}\mathfrak{b}+\mathfrak{a}\mathfrak{c}\]since the \(b_i\) and \(c_i\) all belong to \(\mathfrak{b}+\mathfrak{c}\), the above element lies in \(\mathfrak{a}(\mathfrak{b}+\mathfrak{c})\). Similarly, one can prove the right distributive law.
Thus the collection of two-sided ideals of \(A\) has a structure identical to that of a ring except for additive inverses. Such a structure is called a semiring, though we will not have much use for it.
For any two two-sided ideals \(\mathfrak{a},\mathfrak{b}\), the two inclusions
\[\mathfrak{a}\mathfrak{b}\subset \mathfrak{a}A\subset \mathfrak{a},\quad \mathfrak{a}\mathfrak{b}\subset A \mathfrak{b}\subset \mathfrak{b}\]both hold, so \(\mathfrak{a}\mathfrak{b}\subset \mathfrak{a}\cap \mathfrak{b}\) holds. In general equality need not hold, but it may hold in special cases.
Proposition 3 Let \(\mathfrak{a},\mathfrak{b}_1,\ldots, \mathfrak{b}_n\) be two-sided ideals of \(A\), and assume that \(A=\mathfrak{a}+\mathfrak{b}_i\) holds for all \(i\). Then
\[A=\mathfrak{a}+\mathfrak{b}_1\cdots \mathfrak{b}_n=\mathfrak{a}+(\mathfrak{b}_1\cap\cdots\cap \mathfrak{b}_n)\]holds.
Proof
Since \(\mathfrak{b}_1\cdots \mathfrak{b}_n\subset \mathfrak{b}_1\cap \cdots\cap \mathfrak{b}_n\) in any case, it suffices to show the equality \(A=\mathfrak{a}+\mathfrak{b}_1\cdots \mathfrak{b}_n\). Moreover, since the proof proceeds by induction, it is enough to consider the case \(n=2\). That is, suppose \(A=\mathfrak{a}+\mathfrak{b}_1=\mathfrak{a}+\mathfrak{b}_2\), and let us show that \(A=\mathfrak{a}+\mathfrak{b}_1 \mathfrak{b}_2\).
First, from \(A=\mathfrak{a}+\mathfrak{b}_1=\mathfrak{a}+\mathfrak{b}_2\), we can choose \(a,a'\in \mathfrak{a}, b_i\in \mathfrak{b}_i\) satisfying \(1=a+b_1=a'+b_2\). Then
\[1=a'+b_2=a'+1b_2=a'+(a+b_1)b_2=(a+a'b_2)+b_1b_2\in \mathfrak{a}+\mathfrak{b}_1 \mathfrak{b}_2\]holds.
Using this we can prove the following proposition.
Proposition 4 Let \(\mathfrak{b}_1,\ldots, \mathfrak{b}_n\) be two-sided ideals of \(A\) such that \(\mathfrak{b}_i+\mathfrak{b}_j=A\) (\(i\neq j\)) always holds. Then the equality
\[\mathfrak{b}_1\cap \cdots\cap \mathfrak{b}_n=\sum_{\sigma\in S_n} \mathfrak{b}_{\sigma(1)}\cdots \mathfrak{b}_{\sigma(n)}\]holds, and therefore in particular if \(A\) is commutative then
\[\mathfrak{b}_1\cap \cdots\cap \mathfrak{b}_n=\mathfrak{b}_1\cdots \mathfrak{b}_n\]holds.
Proof
We prove this by induction as well. First, in the case \(n=2\), we can find \(b_i\in \mathfrak{b}_i\) satisfying \(b_1+b_2=1\). Now for any \(x\in \mathfrak{b}_1\cap \mathfrak{b}_2\),
\[x=x\cdot 1=x(b_1+b_2)=xb_1+xb_2\in \mathfrak{b}_2 \mathfrak{b}_1+\mathfrak{b}_1 \mathfrak{b}_2\]holds. Now assume the desired equality holds for all integers less than \(n\). First, applying Proposition 3 to \(\mathfrak{b}_n=\mathfrak{a}\) and \(\mathfrak{b}_1,\ldots, \mathfrak{b}_{n-1}\), we obtain
\[A=\mathfrak{b}_n+(\mathfrak{b}_1\cap\cdots\cap \mathfrak{b}_{n-1})=\mathfrak{b}_n+\mathfrak{b}_1\cdots \mathfrak{b}_{n-1}\]so for the two ideals \(\mathfrak{b}_n\) and \(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1}\),
\[\mathfrak{b}_n\cap(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1})=(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1})\mathfrak{b}_n+\mathfrak{b}_n(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1})\]holds. By the induction hypothesis \(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1}=\sum_{\sigma\in S_{n-1}}\mathfrak{b}_{\sigma(1)}\cdots\mathfrak{b}_{\sigma(n-1)}\) holds, so from this
\[\mathfrak{b}_n\cap(\mathfrak{b}_1\cap\cdots\cap\mathfrak{b}_{n-1})=\left(\sum_{\sigma\in S_{n-1}}\mathfrak{b}_{\sigma(1)}\cdots\mathfrak{b}_{\sigma(n-1)}\right)\mathfrak{b}_n+\mathfrak{b}_n\left(\sum_{\sigma\in S_{n-1}}\mathfrak{b}_{\sigma(1)}\cdots \mathfrak{b}_{\sigma(n-1)}\right)\]and here the right-hand side is a partial sum of \(\sum_{\sigma\in S_n}\mathfrak{b}_{\sigma(1)}\cdots \mathfrak{b}_{\sigma(n)}\), so we obtain the desired conclusion.
Chinese Remainder Theorem
Let \(A\) be a ring, and let \(\mathfrak{a}_i\) be two-sided ideals of \(A\). Then there exist projections \(\pi_i:A \rightarrow A/\mathfrak{a}_i\), and from these a ring homomorphism \(\pi:A \rightarrow\prod A/\mathfrak{a}_i\) is defined.
Proposition 5 Let \(A\) be a ring, and let \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\) be two-sided ideals of \(A\). If \(\mathfrak{a}_i+\mathfrak{a}_j=A\) always holds for \(i\neq j\), then the \(\pi:A \rightarrow \prod_1^n A/\mathfrak{a}_i\) defined above is surjective, and the kernel of this map equals \(\bigcap \mathfrak{a}_i\).
Proof
First, \(\ker\pi=\bigcap \mathfrak{a}_i\) is clear, so it suffices to show that \(\pi\) is surjective. This can be shown by induction as follows.
First, the case \(n=1\) is clear from the properties of the quotient ring. Now suppose there exists some \(y\in A\) such that \(\pi_i(y)=x_i+\mathfrak{a}_i\) holds for all \(i=1,\ldots, n-1\). If there exists \(x\in A\) satisfying \(\pi_i(x)=x_i+\mathfrak{a}_i\) for all \(i=1,\ldots, n\), then we can write \(x=y+z\) for some \(z\in A\), and then the conditions on \(x\) and \(y\) require that \(z\in\bigcap_{i=1}^{n-1} \mathfrak{a}_i\). Also, \(z+\mathfrak{a}_n=x_n-y \mathfrak{a}_n\) must hold, and conversely, if such \(z\) exists then \(x=y+z\) is the desired \(x\). But by Proposition 3, \(\mathfrak{a}_n+\bigcap_1^{n-1} \mathfrak{a}_i=A\) holds, so we can always find such \(z\).
Therefore, by the first isomorphism theorem there exists the canonical isomorphism
\[\frac{A}{\bigcap_{i=1}^n \mathfrak{a}_i}\cong \prod_{i=1}^n A/\mathfrak{a}_i\]and if \(A\) is commutative then by Proposition 4 it can be written as
\[A/\mathfrak{a}_1\cdots \mathfrak{a}_n\cong\prod_{i=1}^n A/\mathfrak{a}_i.\]In particular, if \(\bigcap \mathfrak{a}_i=0\) then we obtain an isomorphism \(A\cong\prod A/\mathfrak{a}_i\).
In particular, consider the case \(A=\mathbb{Z}\), and for pairwise coprime \(n_1,\ldots, n_r\) let \(\mathfrak{a}_i=n_i \mathbb{Z}\). Letting \(n=n_1\cdots n_r\), by the above proposition and the first isomorphism theorem we obtain an isomorphism \(\mathbb{Z}/n \mathbb{Z}\cong\prod \mathbb{Z}/n_i \mathbb{Z}\). That is, from the above proposition we recover the classical Chinese remainder theorem.
More generally, the following are all equivalent.
Proposition 6 Let \(A\) be a ring, let \(C(A)\) be its center, and let \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\) be two-sided ideals of \(A\). The following are all equivalent.
- The \(A \rightarrow \prod A/\mathfrak{a}_i\) defined above is an isomorphism.
- For all \(i\neq j\), \(\mathfrak{a}_i+\mathfrak{a}_j=A\) and \(\bigcap \mathfrak{a}_i=0\).
- For all \(i\neq j\), \(\mathfrak{a}_i+\mathfrak{a}_j=A\) and \(\prod \mathfrak{a}_i=0\).
- There exist elements \(e_1,\ldots, e_n\) of \(C(A)\) such that \(\sum e_i=1\), \(e_i^2=e_i\) for all \(i\), \(e_ie_j=0\) for all \(i\neq j\), and \(\mathfrak{a}_i=A(1-e_i)\) for all \(i\).
Proof
The \(e_i\) in the last condition denote the elements of \(\prod A/\mathfrak{a}_i\) whose \(i\)th component is \(1\) and whose remaining components are all \(0\). With this in mind, one can easily show that the four conditions are all equivalent.
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