Chinese Remainder Theorem and Partial Fraction Decomposition

Introduction and background knowledge

In many textbooks, the partial fraction decomposition makes me confused. You know, they never show you a good reason. Thus I am writing this article, I have a good reason for that.

I will apply CRT to $\mathbb{C}[x]$$\mathbb C[x]$, and then use some linear algebra.

Theorem 1.1 Chinese Remainder Theorem (CRT)

Let $R$$R$ be a ring, $I_1,I_2,...I_n$$I_1,I_2,...I_n$ is a family of ideal satisfied $I_i+I_j=R$$I_i+I_j=R$, for $i\ne j$$i\ne j$,

then $\begin{array}{c}\frac{R}{\bigcap _{k=1}^n{I}_n}\cong \prod _{k=1}^n\frac{R}{I_k}\end{array}$$\frac{R}{\bigcap_{k=1}^n I_n}\cong\prod_{k=1}^n\frac{R}{I_k}\\$.

$\mathrm{Proof}.$$\mathrm{Proof.}$

Define $\begin{array}{c}\phi :R⟶\prod _{k=1}^{n}R/I_k\end{array}$$\varphi:R\longrightarrow\prod_{k=1}^n R/I_k\\$ by $\begin{array}{c}r⟼(r\mathrm{mod}{I}_k{)}_{i=1}^n\end{array}$$r\longmapsto(r\,\,\mathrm{mod} I_k)_{i=1}^n\\$.

Easy to see that $\begin{array}{c}\mathrm{Ker}\phi =\bigcap _{k=1}^n{I}_k\end{array}$$\mathrm{Ker}{\varphi}=\bigcap_{k=1}^n I_k\\$.

Then we only need to prove that $\phi$$\varphi$ is surjective and then we conclude the proof by the first isomorphism theorem.

Since $I_i+I_j=R,\mathrm{\exists }{r}_j\in I_i,s\in I_j,r_j+s_j=1$$I_i+I_j=R,\exists r_j\in I_i,s\in I_j,r_j+s_j=1$,

Then consider $\prod _{j\ne i}(r_j+s_j)=1$$\prod_{j\ne i}(r_j+s_j)=1$.

Define $s^i:=\prod _{j\ne i}{s}_j$$s^i:=\prod_{j\ne i} s_j$, then $\prod _{j\ne i}(r_j+s_j)=r+s^i\in I_i+\prod _{j\ne i}{I}_j$$\prod_{j\ne i}(r_j+s_j)=r+s^i\in I_i+\prod_{j\ne i}I_j$.

Thus $\begin{array}{c}{s}^i\equiv \{\begin{array}{l}1\mathrm{mod}{I}_i\\ 0\mathrm{mod}{I}_j\end{array}\end{array}$$s^i\equiv\begin{cases}1\mod I_i\\0\mod I_j\end{cases}$, $\phi (x_1{s}^1+x_2{s}^2+...+x_n{s}^n)=(x_i\mathrm{mod}{I}_i{)}_{i=1}^n$$\varphi(x_1s^1+x_2s^2+...+x_ns^n)=(x_i\mod I_i)_{i=1}^n$.

Therefore $\phi$$\varphi$ is surjective. $◼$$\blacksquare$

Theorem 1.2 Chinese Remainder Theorem (CRT) for PID

Suppose $R$$R$ is a PID, $m_1,m_2,m_3,...,m_n$$m_1,m_2,m_3,...,m_n$ is coprime, Let $M=\prod _{i=1}^n{m}_i$$M=\prod_{i=1}^n m_i$

Then $\begin{array}{c}R/MR\cong \prod R/m_{i}R\end{array}$$R/MR\cong \prod R/m_iR\\$

Define $M_i=M/m_i$$M_i=M/m_i$

$t_i{m}_i+s_i{M}_i=1$$t_im_i+s_iM_i=1$

${\phi }^{-1}:=(x_i+m_{i}R{)}_{i=1}^n=\sum x_i{s}_i{M}_i+MR$$\varphi^{-1}:=(x_i+m_iR)_{i=1}^n=\sum x_is_iM_i+MR$

Partial Fraction Decomposition and Chinese Remainder Theorem in $\mathbb{C}[x]$$\mathbb C[x]$

Here is a Proof for $K[x]$$K[x]$ is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$, $K$$K$ is any field.

https://wuyulanliulongblog.blogspot.com/2023/07/using-linear-algebra-to-prove-that-cx_4.html

Consider a polynomial $Q(x)$$Q(x)$ in $\mathbb{C}[x]$$\mathbb C[x]$, it always can be factored to $Q(x)=a(x-{\lambda }_1{)}^{i_1}...(x-{\lambda }_n{)}^{i_n}$$Q(x)=a(x-\lambda_1)^{i_1}...(x-\lambda_n)^{i_n}$

$I_1,...,I_n=((x-{\lambda }_1{)}^{i_1}),...,((x-{\lambda }_n{)}^{i_n})$$I_1,...,I_{n}=((x-\lambda_1)^{i_1}),...,((x-\lambda_n)^{i_n})$

Since $\mathbb{C}[x]$$\mathbb C[x]$ is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$, and $gcd(I_i,I_j)=1,(i\ne j)$$\gcd(I_i,I_j)=1,(i\ne j)$, according to Bezout's theorem

$\mathrm{\exists }s,t\in \mathbb{C}[x]a\in I_i,b\in I_j,sa+tb=1$$\exists s,t\in\mathbb C[x]a\in I_i,b\in I_j,sa+tb=1$, then $I_i+I_j=\mathbb{C}[x]$$I_i+I_j=\mathbb C[x]$

And then we have $s^i\in \prod _{j\ne i}{I}_j$$s^i\in\prod_{j\ne i}I_j$, $\begin{array}{c}{s}^i\equiv \{\begin{array}{l}1\mathrm{mod}{I}_i\\ 0\mathrm{mod}{I}_j\end{array}\end{array}$$s^i\equiv\begin{cases}1\mod I_i\\0\mod I_j\end{cases}$

For all $P(x)\in \mathbb{C}[x]$$P(x)\in \mathbb C[x]$, $\begin{array}{c}P(x)=\sum _{i=1}^n{a}_i{s}^i+Q(x)F(x)\end{array}$$P(x)=\sum_{i=1}^{n} a_is^i+Q(x)F(x)\\$

$s^i=s_i\prod _{j\ne i}(x-{\lambda }_j)$$s^i=s_i\prod_{j\ne i}(x-\lambda_j)$

Thus $\begin{array}{c}\frac{P(x)}{Q(x)}=F(x)+\frac{1}{a}\sum _{j=1}^n\frac{a_j{s}_j}{(x-{\lambda }_j{)}^{i_j}}\end{array}$$\frac{P(x)}{Q(x)}=F(x)+\frac{1}{a}\sum_{j=1}^n\frac{a_js_j}{(x-\lambda_j)^{i_j}}\\$

Recall the Euclidean division, $P(x)=F(x)Q(x)+R(x),\mathrm{deg}R(x)<\mathrm{deg}Q(x)$$P(x)=F(x)Q(x)+R(x),\deg R(x)<\deg Q(x)$

We will omit the $F(x)$$F(x)$ below for convenient. That is, we deal with$R(x),\mathrm{deg}R(x)<\mathrm{deg}Q(x)$$R(x),\deg R(x)<\deg Q(x)$

Remark.$a_j{s}_j$$a_js_j$ is polynomial, and $\mathrm{deg}{a}_j{s}_j<i_j$$\deg a_js_j< i_j$ , and for convenient, I will denote $\begin{array}{c}\frac{P(x)}{Q(x)}=\sum _{j=1}^n\frac{S_j}{(x-{\lambda }_j{)}^{i_j}}\end{array}$$\frac{P(x)}{Q(x)}=\sum_{j=1}^n\frac{S_j}{(x-\lambda_j)^{i_j}}\\$

For example, Let $R(x)=x^2+1,Q(x)=x(x-1{)}^2$$R(x)=x^2+1,Q(x)=x(x-1)^2$

$\begin{array}{c}\frac{R(x)}{Q(x)}=\frac{A}{x}+\frac{Bx+C}{(x-1{)}^2}\end{array}$$\frac{R(x)}{Q(x)}=\frac{A}{x}+\frac{Bx+C}{(x-1)^2}\\$

And then, Linear Algebra

Let us still consider the example, $Bx+C\in P_1$$Bx+C\in P_1$, thus we need to decompose $\begin{array}{c}\frac{Bx+C}{(x-1{)}^2}\end{array}$$\frac{Bx+C}{(x-1)^2}\\$ to $\begin{array}{c}\frac{b}{(x-1)}+\frac{c}{(x-1{)}^2}\end{array}$$\frac{b}{(x-1)}+\frac{c}{(x-1)^2}\\$

Since $\begin{array}{c}\frac{b}{(x-1)}+\frac{c}{(x-1{)}^2}=\frac{b(x-1)+c}{(x-1{)}^2}\\ \end{array}$$\frac{b}{(x-1)}+\frac{c}{(x-1)^2}=\frac{b(x-1)+c}{(x-1)^2} \\\\$, $\{(x-1),1\}$$\{(x-1),1\}$ is basis of $P_1(\mathbb{C})$$P_1(\mathbb C)$

Remark. $P_n(\mathbb{C})$$P_n(\C)$ refer to the polynomial space over complex number, $\mathrm{deg}p\le n$$\deg p\le n$

In general, $\begin{array}{c}\frac{R(x)}{Q(x)}=\sum _{j=1}^n\frac{S_j}{(x-{\lambda }_j{)}^{i_j}}\end{array}$$\frac{R(x)}{Q(x)}=\sum_{j=1}^n\frac{S_j}{(x-\lambda_j)^{i_j}}\\$

$\begin{array}{c}=[\frac{A_{1,1}}{(x-{\lambda }_1)}+...+\frac{A_1{,}_{i_1}}{(x-{\lambda }_1{)}^{i_1}}]+...+[\frac{A_{i,1}}{(x-{\lambda }_n)}+...+\frac{A_n{,}_{i_n}}{(x-{\lambda }_n{)}^{i_n}}]\end{array}$$=\left[\frac{A_{1,1}}{(x-\lambda_1)}+...+\frac{A_1,_{i_1}}{(x-\lambda_{1})^{i_{1}}}\right]+...+\left[ \frac{A_{i,1}}{(x-\lambda_n)}+...+\frac{A_n,_{i_n}}{(x-\lambda_{n})^{i_{n}}}\right]\\$

Because $\{1,(x-{\lambda }_k),(x-{\lambda }_k{)}^2,...,(x-\lambda {)}^{i_{k-1}}\}$$\{1,(x-\lambda_k),(x-\lambda_k)^2,...,(x-\lambda)^{i_{k-1}}\}$ is basis of $P_n(\mathbb{C})$$P_n(\C)$

Then we get Partial Fraction Decomposition over $\mathbb{C}[x]$$\mathbb C[x]$

Summary

And as you see, it look like the structure theorem for finitely generated modules over a principle ideal domain

$\begin{array}{c}M\cong R^f\underset{i=1}{\overset{n}{⨁}}R/(d_i),d_i|d_{i+1}\end{array}$$M\cong R^f\bigoplus_{i=1}^nR/(d_i),d_i|d_{i+1}\\$

When the first time write this essay, I think $\begin{array}{c}[\frac{A_{1,1}}{(x-{\lambda }_1)}+...+\frac{A_1{,}_{i_1}}{(x-{\lambda }_1{)}^{i_1}}]+...+[\frac{A_{i,1}}{(x-{\lambda }_n)}+...+\frac{A_n{,}_{i_n}}{(x-{\lambda }_n{)}^{i_n}}]\end{array}$$\left[\frac{A_{1,1}}{(x-\lambda_1)}+...+\frac{A_1,_{i_1}}{(x-\lambda_{1})^{i_{1}}}\right]+...+\left[ \frac{A_{i,1}}{(x-\lambda_n)}+...+\frac{A_n,_{i_n}}{(x-\lambda_{n})^{i_{n}}}\right]\\$

Is really like Jordan canonical form

But the issue is I have not learned them. After I learn them, I will write a more general and elegant proof