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. See partial fractions of fractional numbers for the case when $E$ consists of the integers and $\nu(k) = \abs{k}$ for $k \neq 0$ .
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 $E$ 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 in $E$ such that

Proof. Let $r_0 = p$ . Iterating through 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

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 be a factorization of $q$ to prime factors $\phi_i$ . Then there exist elements $\alpha_{ij}, \beta$ in $E$ such that \begin{equation*} \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)\,. \end{equation*}

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 .