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 ν to denote the Euclidean valuation function of an Euclidean domain E, with the convention that ν(0)=-.

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 ν(k)=|k| for k0.

  2. 2.

    See partial fractions of expressions for the case when E consists of polynomialsMathworldPlanetmathPlanetmathPlanetmath over the complex field, with ν(p) being the degree of the polynomial p.

  3. 3.

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

Theorem 1.

Let p, q10 and q20 be elements of an Euclidean domain E, with q1 and q2 be relatively prime. Then there exist α1 and α2 in E such that

pq1q2=α1q1+α2q2.
Proof.

By the Euclidean algorithmMathworldPlanetmath, we can obtain elements s1 and s2 in E such that

1=s1q1+s2q2.

Then

pq1q2=ps2q1+ps1q2,

so we can take α1=ps2 and α2=ps1. ∎

Theorem 2.

Let p and q0 be elements of an Euclidean domain E, and n be any positive integer. Then there exist elements α1,,αn,β in E such that

pqn=β+α1q+α2q2++αnqn,ν(αj)<ν(q).
Proof.

Let r0=p. Iterating through k=1,,n in order, using the division algorithmPlanetmathPlanetmath, we can find elements rk and sk such that

rk-1=rkq+sk,ν(sk)<ν(q).

Then

p=r0 =r1q+s1
=(r2q+s2)q+s1
=
=rnqn+snqn-1+sn-1qn-2++s2q+s1
pqn =rn+snq+sn-1q2++s2qn-1+s1qn.

So set β=rn and αj=sn-j+1. ∎

Theorem 3.

Let p and q0 be elements of an Euclidean domain E. Let q=ϕ1n1ϕ2n2ϕknk be a factorization of q to prime factorsMathworldPlanetmathPlanetmath ϕi. Then there exist elements αij,β in E such that

pq=β+i=1kj=1niαijϕij,ν(αij)<ν(ϕi).
Proof.

Apply Theorem 1 inductively to obtain elements si in E such that

pq=i=1ksiϕini

(the factors ϕi are relatively prime). Then apply Theorem 2 to obtain elements αij and βi in E such that

siϕini=βi+j=1niαijϕij

with ν(αij)<ν(ϕi). Take β=β1++βk. ∎

Title partial fractions in Euclidean domains
Canonical name PartialFractionsInEuclideanDomains
Date of creation 2013-03-22 15:40:18
Last modified on 2013-03-22 15:40:18
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 4
Author stevecheng (10074)
Entry type Result
Classification msc 13F07
Synonym partial fraction decomposition in Euclidean domains