squarefree factorization

squarefree factorization

Given a

polynomial

A

∈

𝔽

p

⁢

[

X

]

, where

p

is a prime and

𝔽

p

is the field with

p

elements

, we want to find a decomposition

A

=

∏

i

=

1

n

A

i

i

with

squarefree

polynomials

A

i

, which are

pairwise coprime

. Since we are in a field, we can assume that

A

is monic.
Since

A

has a

unique factorization

we can take

A

i

to be the

product

of all

irreducible

divisors

P

of

A

with

v

P

⁢

(

A

)

=

i

, where

v

P

⁢

(

A

)

is the number such that

P

v

P

⁢

(

A

)

divides

A

but

P

v

P

⁢

(

A

)

+

1

does not, so such a decomposition exists.

If

A

is in the desired form, we have for the

derivative

A

′

of

A

:

A

′

=

∑

i

=

1

n

∏

j

≠

i

i

⁢

A

j

j

⁢

A

i

′

⁢

A

i

i

-

1

.

(1)

Now let

T

:=

gcd

⁡

(

A

,

A

′

)

. For every

irreducible polynomial

P

dividing

T

we can determine

v

P

⁢

(

T

)

in the following way:

P

must divide

A

m

for some value of

m

(then it does not divide any other

A

i

). Then for all

i

≠

m

the

v

P

of the

i

th summand in equation (

1

) is at least

m

and for the

m

th summand it is

m

-

1

(

A

m

′

is not

divisible

by

P

since

A

m

is squarefree) if

p

∤

m

and 0 if

p

|

m

(because of the factor

m

in that summand). So we find

v

P

⁢

(

T

)

=

m

-

1

if

p

∤

m

and

v

P

⁢

(

T

)

=

m

if

p

|

m

(equality holds because

T

|

A

). So we obtain

T

=

gcd

⁡

(

A

,

A

′

)

=

∏

p

∤

i

A

i

i

-

1

⁢

∏

p

|

i

A

i

i

.

Now we define two

sequences

(

T

i

)

and

(

V

i

)

by

T

1

=

T

and

V

1

=

A

T

=

∏

p

∤

i

A

i

.

Then set

V

k

+

1

=

gcd

⁡

(

T

k

,

V

k

)

if

p

∤

k

and

V

k

+

1

=

V

k

if

p

|

k

and set

T

k

+

1

=

T

k

V

k

+

1

. By

induction

one finds

V

k

=

∏

i

≥

k

,

p

∤

i

A

i

;

T

k

=

∏

i

≥

k

-

1

,

p

∤

i

A

i

i

-

k

⁢

∏

p

|

i

A

i

i

.

So for

p

∤

k

it follows

A

k

=

V

k

V

k

+

1

, so for

p

∤

k

we can continue as long as

V

k

is non-constant. When

V

k

is

constant

, we have

T

k

-

1

=

∏

p

|

i

A

i

i

.

Assume

A

i

⁢

(

X

)

=

a

k

⁢

X

k

+

…

+

a

1

⁢

X

+

a

0

, then with

i

=

l

⁢

p

A

i

l

⁢

p

=

a

k

l

⁢

p

⁢

X

k

⁢

l

⁢

p

+

…

+

a

1

l

⁢

p

⁢

X

l

⁢

p

+

a

0

l

⁢

p

=

a

k

⁢

X

k

⁢

l

⁢

p

+

…

+

a

1

⁢

X

l

⁢

p

+

a

0

,

so we can set

Y

:=

X

p

and have

T

k

-

1

=

U

p

⁢

(

X

)

=

U

⁢

(

Y

)

. Now we can decompose

U

by using the

algorithm

recursively.

In practice one can easily compute

T

using Euclid’s algorithm. Then one computes the sequences

(

V

i

)

and

(

T

i

)

to get the

A

k

. The algorithm is needed as a first step when one wants to find the prime decomposition of a polynomial, because it reduces the problem to the problem of

factoring

a squarefree polynomial.

Title

squarefree factorization

Canonical name

SquarefreeFactorization

Date of creation

2013-03-22 14:54:20

Last modified on

2013-03-22 14:54:20

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Algorithm

Classification

msc 11Y99

Classification

msc 11C99

Classification

msc 13P05

Classification

msc 13M10

Related topic

CantorZassenhausSplit

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