partial fractions in Euclidean domains

This entry states and proves the existence of partial fraction decompositionson an Euclidean domain.

In the following, we use $\nu$ to denote the Euclidean valuation functionof an Euclidean domain $E$, with the convention that $\nu (0)=-\mathrm{\infty }$.

For a gentle introduction:

1. See partial fractions of fractional numbers (http://planetmath.org/PartialFractions) for the case when $E$ consists of theintegers and $\nu (k)=|k|$for $k\ne 0$.

2. See partial fractions of expressions for the case when $E$consists of polynomials over the complex field,with $\nu (p)$ being the degree of the polynomial $p$.

3. See partial fractions for polynomials for the case when $E$ isthe ring of polynomials over any field, and $\nu$ is the degreeof polynomials.

Theorem 1.

Let $p$, $q_{\mathrm{1}}\mathrm{\ne }\mathrm{0}$ and $q_{\mathrm{2}}\mathrm{\ne }\mathrm{0}$ be elements of an Euclidean domain $E$,with $q_{\mathrm{1}}$ and $q_{\mathrm{2}}$ be relatively prime.Then there exist ${\alpha }_{\mathrm{1}}$ and ${\alpha }_{\mathrm{2}}$ in $E$suchthat

$$\frac{p}{q_1{q}_2}=\frac{{\alpha }_1}{q_1}+\frac{{\alpha }_2}{q_2}.$$

Proof.

By the Euclidean algorithm, we can obtain elements $s_1$ and $s_2$ in $E$such that

$$1=s_1{q}_1+s_2{q}_2.$$

Then

$$\frac{p}{q_1{q}_2}=\frac{p{s}_2}{q_1}+\frac{p{s}_1}{q_2},$$

so we can take ${\alpha }_1=p{s}_2$ and ${\alpha }_2=p{s}_1$.∎

Theorem 2.

Let $p$ and $q\mathrm{\ne }\mathrm{0}$ be elements of an Euclidean domain $E$,and $n$ be any positive integer.Then there exist elements${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_n\mathrm{,}\beta$ in $E$ such that

Proof.

Let $r_0=p$.Iterating through $k=1,\mathrm{\dots },n$ in order,using the division algorithm,we can find elements $r_k$ and $s_k$such that

Then

$p=r_0$$=r_{1}q+s_1$

$=(r_{2}q+s_2)q+s_1$

$=\mathrm{\dots }$

$=r_n{q}^n+s_n{q}^{n-1}+s_{n-1}{q}^{n-2}+\mathrm{\cdots }+s_{2}q+s_1$

${\displaystyle \frac{p}{q^n}}$$=r_n+{\displaystyle \frac{s_n}{q}}+{\displaystyle \frac{s_{n-1}}{q^2}}+\mathrm{\cdots }+{\displaystyle \frac{s_2}{q^{n-1}}}+{\displaystyle \frac{s_1}{q^n}}.$

So set $\beta =r_n$ and ${\alpha }_j=s_{n-j+1}$.∎

Theorem 3.

Let $p$ and $q\mathrm{\ne }\mathrm{0}$ be elements of an Euclidean domain $E$.Let $q\mathrm{=}{\varphi }_{\mathrm{1}}^{n_{\mathrm{1}}}\mathit{}{\varphi }_{\mathrm{2}}^{n_{\mathrm{2}}}\mathit{}\mathrm{\cdots }\mathit{}{\varphi }_k^{n_k}$be a factorization of $q$ to prime factors ${\varphi }_i$.Then there exist elements ${\alpha }_{i\mathit{}j}\mathrm{,}\beta$ in $E$suchthat

Proof.

Apply Theorem 1 inductively to obtainelements $s_i$ in $E$ such that

$$\frac{p}{q}=\sum _{i=1}^k\frac{s_i}{{\varphi }_i^{n_i}}$$

(the factors ${\varphi }_i$ are relatively prime).Then apply Theorem 2 to obtain elements${\alpha }_{ij}$ and ${\beta }_i$ in $E$ such that

$$\frac{s_i}{{\varphi }_i^{n_i}}={\beta }_i+\sum _{j=1}^{n_i}\frac{{\alpha }_{ij}}{{\varphi }_i^j}$$

with .Take $\beta ={\beta }_1+\mathrm{\cdots }+{\beta }_k$.∎