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# Chinese Remainder Theorem

> 礙於技術限制，各項公式未能個別標註出處。
> 參考文獻統一排列於「[Reference](#Reference)」節。
> 學術引用時務必完整標示所有文獻。

<div class="invisible-box">

$\newcommand{\ideal}[1]{\mathfrak{#1}}$
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> :::spoiler Table of Contents
> [TOC]
> :::

## Theorems

### sums and products
<div class="definition" data-info="direct product of rings">

Let $(R_j)$ be any family of rings.
Then their Cartesian product $R = \prod_j R_j$ forms a ring, which is called the *direct product* of $(R_j)$, with respect to the following componentwise operations:
- $(a_j) + (b_j) := (a_j + b_j)$,
- $(a_j) \cdot (b_j) := (a_j \cdot b_j)$.

If each ring $R_j$ has a multiplicative identity $1_j$, then $1 = (1_j)$ is the multiplicative identity of $R$.
</div>

The natural projections $\pi_j\colon R \to R_j$ and injections $\iota_j\colon R_j \to R$ are homomorphisms which preserve addition and multiplication.
Let $\ideal{a}_j = \operatorname{im}\iota_j$ be the isomorphic image of $R_j$ in $R$, then $\ideal{a}_j$ are ideals of $R$.

However, if each ring $R_j$ has $1_j$, then $\pi_j(1) = 1_j$, but $\iota_j(1_j) \ne 1$.
Some authors may say that $\iota_j$ are not homomorphisms, and say that $\ideal{a}_j$ are not subrings of $R$.

<div class="definition" data-info="sum of ideals">

Let $\ideal{a}_1, \ideal{a}_2, \dotsc, \ideal{a}_t$ be any ideals of a rings $R$.
The set
\\[
\sum_{j=1}^t \ideal{a}_j := \{ r_1 + r_2 + \dotsb + r_t \mid r_j \in \ideal{a}_j \}
\\]
forms an ideal of $R$ which is called the *sum* of $(\ideal{a}_j)$.

We write the sum as $\bigoplus_{j=1}^t \ideal{a}_j$ if each element is uniquely expressible in the form $\sum_{j=1}^t r_j$ with $r_j \in \ideal{a}_j$.
</div>

If $R = \bigoplus_{j=1}^t \ideal{a}_j$, then
\\[
\ideal{a}_i \ideal{a}_j \subseteq \ideal{a}_i \cap \ideal{a}_j = 0
\quad \text{if $i \ne j$},
\qquad
\ideal{a}_i \ideal{a}_i \subseteq \ideal{a}_i.
\\]

If $R = \sum_{j=1}^t \ideal{a}_j$, then $R = \bigoplus_{j=1}^t \ideal{a}_j$ iff the underlying additive group of $R$ is the direct sum of the underlying additive group of $(\ideal{a}_j)$.
Thus, we sometimes call $\bigoplus_{j=1}^t \ideal{a}_j$ the *direct sum* of $(\ideal{a}_j)$, but this is rather misleading from the point of view of category theory since it is usually not a coproduct in the category of rings with $1$.

<div class="theorem">

Let $R = \prod_{j=1}^t R_j$ be the direct product of rings $R_1, R_2, \dotsc, R_t$ with injections $\iota_j$ and $\mathfrak a_j = \operatorname{im}\iota_j$.
Then $R = \bigoplus_{j=1}^t \mathfrak a_j$.

Conversely, any ring $R = \bigoplus_{j=1}^t \mathfrak a_j$, where each $\mathfrak a_j$ is an ideal in $R$, may be expressed as $\prod_{j=1}^t R_j$, where $R_j$ is isomorphic to $\mathfrak a_j$.
</div>
<div class="proof">

Since finite direct product of abelian groups is isomorphic to their direct sum, $R = \bigoplus_{j=1}^t \mathfrak a_j$ follows from $R = \prod_{j=1}^t R_j$.

Conversely, any ring of the form $R = \bigoplus_{j=1}^t \mathfrak a_j$ is a direct product of $\ideal{a}_j$ as abelian groups.
Since $\ideal{a}_i \ideal{a}_j = 0$ if $i \ne j$ and $\ideal{a}_i \ideal{a}_i \subseteq \ideal{a}_i$, the multiplication is performed componentwise.
Moreover, the natural homomorphism from $R$ with kernel $\sum_{i\ne j}\ideal{a}_i$ has image $\ideal{a}_j$ which is therefore a ring, say $R_j$, and so $R \cong \prod_{j=1}^t R_j$.
</div>

[Cohn, pp.102]
[Lang, pp.94]

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## Examples

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> *Work In Progress*...
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## Reference
> 礙於技術限制，各項公式未能個別標註出處。
> 參考文獻按數學論文慣例，以字母順序統一排列於下。
> 學術引用時務必完整標示所有文獻。

- P. M. Cohn, *Basic Algebra*, 2003.
- S. Lang, *Algebra*, Springer GTM 211, 2002.