partial fractions in Euclidean domains

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.

See

partial fractions of fractional numbers

(

http://planetmath.org/PartialFractions

) for the case when

E

consists of the

integers

and

ν

⁢

(

k

)

=

|

k

|

for

k

≠

0

.

2.

See

partial fractions of expressions

for the case when

E

consists of

polynomials

over the

complex

field,
with

ν

⁢

(

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

ν

is the degree
of polynomials.

Theorem 1

.

Let

p

,

q

1

≠

0

and

q

2

≠

0

be elements of an Euclidean domain

E

,
with

q

1

and

q

2

be relatively prime.
Then there exist

α

1

and

α

2

in

E

such
that

p

q

1

⁢

q

2

=

α

1

q

1

+

α

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

p

q

1

⁢

q

2

=

p

⁢

s

2

q

1

+

p

⁢

s

1

q

2

,

so we can take

α

1

=

p

⁢

s

2

and

α

2

=

p

⁢

s

1

.
∎

Theorem 2

.

Let

p

and

q

≠

0

be elements of an Euclidean domain

E

,
and

n

be any

positive integer

.
Then there exist elements

α

1

,

…

,

α

n

,

β

in

E

such that

p

q

n

=

β

+

α

1

q

+

α

2

q

2

+

⋯

+

α

n

q

n

,

ν

⁢

(

α

j

)

<

ν

⁢

(

q

)

.

Proof.

Let

r

0

=

p

.
Iterating through

k

=

1

,

…

,

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

,

ν

⁢

(

s

k

)

<

ν

⁢

(

q

)

.

Then

p

=

r

0

=

r

1

⁢

q

+

s

1

=

(

r

2

⁢

q

+

s

2

)

⁢

q

+

s

1

=

…

=

r

n

⁢

q

n

+

s

n

⁢

q

n

-

1

+

s

n

-

1

⁢

q

n

-

2

+

⋯

+

s

2

⁢

q

+

s

1

p

q

n

=

r

n

+

s

n

q

+

s

n

-

1

q

2

+

⋯

+

s

2

q

n

-

1

+

s

1

q

n

.

So set

β

=

r

n

and

α

j

=

s

n

-

j

+

1

.
∎

Theorem 3

.

Let

p

and

q

≠

0

be elements of an Euclidean domain

E

.
Let

q

=

ϕ

1

n

1

⁢

ϕ

2

n

2

⁢

⋯

⁢

ϕ

k

n

k

be a

factorization

of

q

to

prime factors

ϕ

i

.
Then there exist elements

α

i

⁢

j

,

β

in

E

such
that

p

q

=

β

+

∑

i

=

1

k

∑

j

=

1

n

i

α

i

⁢

j

ϕ

i

j

,

ν

⁢

(

α

i

⁢

j

)

<

ν

⁢

(

ϕ

i

)

.

Proof.

Apply

Theorem

1

inductively to obtain
elements

s

i

in

E

such that

p

q

=

∑

i

=

1

k

s

i

ϕ

i

n

i

(the factors

ϕ

i

are relatively prime).
Then apply Theorem

2

to obtain elements

α

i

⁢

j

and

β

i

in

E

such that

s

i

ϕ

i

n

i

=

β

i

+

∑

j

=

1

n

i

α

i

⁢

j

ϕ

i

j

with

ν

⁢

(

α

i

⁢

j

)

<

ν

⁢

(

ϕ

i

)

.
Take

β

=

β

1

+

⋯

+

β

k

.
∎

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

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