resultant (alternative treatment)

resultant (alternative treatment)

Summary.

The

resultant

of two

polynomials

is a number, calculated from
the

coefficients

of those polynomials, that vanishes if and only if
the two polynomials share a common root.

Conversely

, the resultant is
non-zero if and only if the two polynomials are mutually prime.

Definition.

Let

𝕂

be a field and let

p

⁢

(

x

)

=

a

0

⁢

x

n

+

a

1

⁢

x

n

-

1

+

…

+

a

n

,

q

⁢

(

x

)

=

b

0

⁢

x

m

+

b

1

⁢

x

m

-

1

+

…

+

b

m

be two polynomials over

𝕂

of degree

n

and

m

, respectively.
We define

Res

⁢

[

p

,

q

]

∈

𝕂

, the resultant of

p

⁢

(

x

)

and

q

⁢

(

x

)

,
to be the

determinant

of a

n

+

m

square matrix

with columns 1 to

m

formed by shifted

sequences

consisting of the coefficients of

p

⁢

(

x

)

,
and columns

m

+

1

to

n

+

m

formed by shifted sequences consisting of
coefficients of

q

⁢

(

x

)

, i.e.

Res

⁢

[

p

,

q

]

=

|

a

0

0

…

0

b

0

0

…

0

a

1

a

0

…

0

b

1

b

0

…

0

a

2

a

1

…

0

b

2

b

1

…

0

⋮

⋮

⋱

⋮

⋮

⋮

⋱

⋮

0

0

…

a

n

-

1

0

0

…

b

m

-

1

0

0

…

a

n

0

0

…

b

m

|

Proposition 1

The resultant of two polynomials is non-zero if and only if the
polynomials are relatively prime.

Proof.

Let

p

⁢

(

x

)

,

q

⁢

(

x

)

∈

𝕂

⁢

[

x

]

be two arbitrary
polynomials of degree

n

and

m

, respectively. The polynomials are
relatively prime if and only if every polynomial — including the
unit polynomial 1 — can be formed as a

linear combination

of

p

⁢

(

x

)

and

q

⁢

(

x

)

. Let

r

⁢

(

x

)

=

c

0

⁢

x

m

-

1

+

c

1

⁢

x

m

-

2

+

…

+

c

m

-

1

,

s

⁢

(

x

)

=

d

0

⁢

x

n

-

1

+

b

1

⁢

x

n

-

2

+

…

+

d

n

-

1

be polynomials of degree

m

-

1

and

n

-

1

, respectively. The coefficients of the linear
combination

r

⁢

(

x

)

⁢

p

⁢

(

x

)

+

s

⁢

(

x

)

⁢

q

⁢

(

x

)

are given by the following matrix–

vector
multiplication

:

[

a

0

0

…

0

b

0

0

…

0

a

1

a

0

…

0

b

1

b

0

…

0

a

2

a

1

…

0

b

2

b

1

…

0

⋮

⋮

⋱

⋮

⋮

⋮

⋱

⋮

0

0

…

a

n

-

1

0

0

…

b

m

-

1

0

0

…

a

n

0

0

…

b

m

]

⁢

[

c

0

c

1

c

2

⋮

c

m

-

1

d

0

d

1

d

2

⋮

d

n

-

1

]

In

consequence

of the preceding remarks,

p

⁢

(

x

)

and

q

⁢

(

x

)

are
relatively prime if and only if the matrix above is

non-singular

,
i.e. the resultant is non-vanishing. Q.E.D.

Alternative Characterization.

The following

Proposition

describes the resultant of two polynomials
in terms of the polynomials’ roots. Indeed this property uniquely
characterizes the resultant, as can be seen by carefully studying the
appended proof.

Proposition 2

Let

p

⁢

(

x

)

,

q

⁢

(

x

)

be as above and let

x

1

,

…

,

x

n

and

y

1

,

…

,

y

m

be their respective roots in the

algebraic closure

of

K

. Then,

Res

⁢

[

p

,

q

]

=

a

0

m

⁢

b

0

n

⁢

∏

i

=

1

n

∏

j

=

1

m

(

x

i

-

y

j

)

Proof.

The

multilinearity

property of determinants implies that

Res

⁢

[

p

,

q

]

=

a

0

m

⁢

b

0

n

⁢

|

1

0

…

0

1

0

…

0

A

1

1

…

0

B

1

1

…

0

A

2

A

1

…

0

B

2

B

1

…

0

⋮

⋮

⋱

⋮

⋮

⋮

⋱

⋮

0

0

…

A

n

-

1

0

0

…

B

m

-

1

0

0

…

A

n

0

0

…

B

m

|

where

A

i

=

a

i

a

0

,

i

=

1

,

…

⁢

n

,

B

j

=

b

j

b

0

,

j

=

1

,

…

⁢

m

.

It therefore suffices to prove the proposition for

monic polynomials

.
Without loss of generality we can also assume that the roots in
question are

algebraically independent

.

Thus, let

X

1

,

…

,

X

n

,

Y

1

,

…

,

Y

m

be

indeterminates

and set

F

⁢

(

X

1

,

…

,

X

n

,

Y

1

,

…

,

Y

m

)

=

∏

i

=

1

n

∏

j

=

1

m

(

X

i

-

Y

j

)

P

⁢

(

x

)

=

(

x

-

X

1

)

⁢

…

⁢

(

x

-

X

n

)

,

Q

⁢

(

x

)

=

(

x

-

Y

1

)

⁢

…

⁢

(

x

-

Y

m

)

,

G

⁢

(

X

1

,

…

,

X

n

,

Y

1

,

…

,

Y

m

)

=

Res

⁢

[

P

,

Q

]

Now by Proposition 1,

G

vanishes if we replace any of the

Y

1

,

…

,

Y

m

by any of

X

1

,

…

,

X

n

and hence

F

divides

G

.

Next, consider the main diagonal of the matrix whose determinant gives

Res

⁢

[

P

,

Q

]

. The first

m

entries of the

diagonal

are equal to

1

,
and the next

n

entries are equal to

(

-

1

)

m

⁢

Y

1

⁢

…

⁢

Y

m

. It
follows that the expansion of

G

contains a term of the form

(

-

1

)

m

⁢

n

⁢

Y

1

n

⁢

…

⁢

Y

m

n

. However, the expansion of

F

contains
exactly the same term, and therefore

F

=

G

. Q.E.D.

Title

resultant (alternative treatment)

Canonical name

ResultantalternativeTreatment

Date of creation

2013-03-22 12:32:52

Last modified on

2013-03-22 12:32:52

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

6

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 12E05

Related topic

Discriminant

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