# partial fractions in Euclidean domains

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

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

For a gentle introduction:

1. 1.

See partial fractions of fractional numbers (http://planetmath.org/PartialFractions) for the case when $E$ consists of the integers and $\nu(k)=\lvert k\rvert$ for $k\neq 0$.

2. 2.
3. 3.

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

###### Theorem 1.

Let $p$, $q_{1}\neq 0$ and $q_{2}\neq 0$ be elements of an Euclidean domain $E$, with $q_{1}$ and $q_{2}$ be relatively prime. Then there exist $\alpha_{1}$ and $\alpha_{2}$ in $E$ such that

 $\frac{p}{q_{1}\,q_{2}}=\frac{\alpha_{1}}{q_{1}}+\frac{\alpha_{2}}{q_{2}}\,.$
###### 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 $q\neq 0$ be elements of an Euclidean domain $E$, and $n$ be any positive integer. Then there exist elements $\alpha_{1},\ldots,\alpha_{n},\beta$ in $E$ such that

 $\frac{p}{q^{n}}=\beta+\frac{\alpha_{1}}{q}+\frac{\alpha_{2}}{q^{2}}+\cdots+% \frac{\alpha_{n}}{q^{n}}\,,\quad\nu(\alpha_{j})<\nu(q)\,.$
###### Proof.

Let $r_{0}=p$. Iterating through $k=1,\ldots,n$ in order, using the division algorithm  , we can find elements $r_{k}$ and $s_{k}$ such that

 $r_{k-1}=r_{k}\,q+s_{k}\,,\quad\nu(s_{k})<\nu(q)\,.$

Then

 $\displaystyle p=r_{0}$ $\displaystyle=r_{1}\,q+s_{1}$ $\displaystyle=(r_{2}q+s_{2})\,q+s_{1}$ $\displaystyle=\ldots$ $\displaystyle=r_{n}\,q^{n}+s_{n}\,q^{n-1}+s_{n-1}\,q^{n-2}+\cdots+s_{2}\,q+s_{1}$ $\displaystyle\frac{p}{q^{n}}$ $\displaystyle=r_{n}+\frac{s_{n}}{q}+\frac{s_{n-1}}{q^{2}}+\cdots+\frac{s_{2}}{% q^{n-1}}+\frac{s_{1}}{q^{n}}\,.$

So set $\beta=r_{n}$ and $\alpha_{j}=s_{n-j+1}$. ∎

###### Theorem 3.

Let $p$ and $q\neq 0$ be elements of an Euclidean domain $E$. Let $q=\phi_{1}^{n_{1}}\,\phi_{2}^{n_{2}}\,\cdots\,\phi_{k}^{n_{k}}$ be a factorization of $q$ to prime factors   $\phi_{i}$. Then there exist elements $\alpha_{ij},\beta$ in $E$ such that

 $\frac{p}{q}=\beta+\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}\frac{\alpha_{ij}}{\phi_{i}^% {j}}\,,\quad\nu(\alpha_{ij})<\nu(\phi_{i})\,.$
###### Proof.

Apply Theorem 1 inductively to obtain elements $s_{i}$ in $E$ such that

 $\frac{p}{q}=\sum_{i=1}^{k}\frac{s_{i}}{\phi_{i}^{n_{i}}}$

(the factors $\phi_{i}$ are relatively prime). Then apply Theorem 2 to obtain elements $\alpha_{ij}$ and $\beta_{i}$ in $E$ such that

 $\frac{s_{i}}{\phi_{i}^{n_{i}}}=\beta_{i}+\sum_{j=1}^{n_{i}}\frac{\alpha_{ij}}{% \phi_{i}^{j}}$

with $\nu(\alpha_{ij})<\nu(\phi_{i})$. Take $\beta=\beta_{1}+\cdots+\beta_{k}$. ∎

Title partial fractions in Euclidean domains PartialFractionsInEuclideanDomains 2013-03-22 15:40:18 2013-03-22 15:40:18 stevecheng (10074) stevecheng (10074) 4 stevecheng (10074) Result msc 13F07 partial fraction decomposition in Euclidean domains