aarjaneiro/planetmath huggingface dataset:
content of polynomial

content of polynomial

The

content

of a

polynomial

f

may be defined in any

polynomial ring

R

⁢

[

x

]

over a

commutative ring

R

as the ideal of

R

generated by the

coefficients

of the polynomial.  It is denoted by

cont

⁡

(

f

)

or

c

⁢

(

f

)

.

Coefficient module

is a little more general

concept

.

If

R

is a unique factorisation domain (

http://planetmath.org/UFD

) and

f

,

g

∈

R

⁢

[

x

]

,  the

Gauss lemma I

implies

1

1

In a UFD, one can use as contents of

f

and

g

the

http://planetmath.org/node/5800

greatest common divisors

a

and

b

of the coefficients of these polynomials, when one has

f

⁢

(

x

)

=

a

⁢

f

1

⁢

(

x

)

,

g

⁢

(

x

)

=

b

⁢

g

1

⁢

(

x

)

with

f

1

⁢

(

x

)

and

g

1

⁢

(

x

)

primitive polynomials

.  Then

f

⁢

(

x

)

⁢

g

⁢

(

x

)

=

a

⁢

b

⁢

f

1

⁢

(

x

)

⁢

g

1

⁢

(

x

)

,  and since also

f

1

⁢

g

1

is a primitive polynomial, we see that

c

⁢

(

f

⁢

g

)

=

a

⁢

b

=

c

⁢

(

f

)

⁢

c

⁢

(

g

)

.

that

c

⁢

(

f

⁢

g

)

=

c

⁢

(

f

)

⁢

c

⁢

(

g

)

.

(1)

For an arbitrary commutative ring

R

, there is only the containment

c

⁢

(

f

⁢

g

)

⊆

c

⁢

(

f

)

⁢

c

⁢

(

g

)

(2)

(cf.

product

of

finitely generated

ideals).  The ideal

c

⁢

(

f

⁢

g

)

is called the

Gaussian ideal of

the polynomials

f

and

g

.  The polynomial

f

in

R

⁢

[

x

]

is a , if (2) becomes the equality (1) for all polynomials

g

in the ring

R

⁢

[

x

]

.  The ring

R

is a

Gaussian ring

, if all polynomials in

R

⁢

[

x

]

are .

It’s quite interessant, that the

equation

(1) multiplied by the power

[

c

⁢

(

f

)

]

n

, where

n

is the degree of the other polynomial

g

, however is true in any commutative ring

R

, thus replacing the containment (2):

[

c

⁢

(

f

)

]

n

⁢

c

⁢

(

f

⁢

g

)

=

[

c

⁢

(

f

)

]

n

+

1

⁢

c

⁢

(

g

)

.

(3)

This result is called the

Hilfssatz von Dedekind–Mertens

, i.e. the
Dedekind–Mertens lemma.  A generalised form of it is in the
entry
product of finitely generated ideals (

http://planetmath.org/ProductOfFinitelyGeneratedIdeals

).

References

1

Alberto Corso & Sarah Glaz

: “Gaussian ideals and the Dedekind–Mertens lemma” in Jürgen Herzog & Gaetana Restuccia (eds.):

Geometric and combinatorial aspects of

commutative algebra

.  Marcel Dekker Inc., New York (2001).

Title

content of polynomial

Canonical

name

ContentOfPolynomial

Date of creation

2013-11-19 18:51:57

Last modified on

2013-11-19 18:51:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Definition

Classification

msc 11C08

Related topic

CoefficientModule

Related topic

PruferRing

Related topic

GaussianPolynomials

Defines

content of polynomial

Defines

Gaussian ideal

Defines

Gaussian polynomial

Defines

Gaussian ring

Generated on Fri Feb 9 15:30:41 2018 by

LaTeXML
explicit generators of a quotient polynomial ring associated to a given polynomial

explicit generators of a quotient polynomial ring associated to a given polynomial

Let

k

be a field and consider ring of polynomials

k

⁢

[

X

]

. If

F

⁢

(

X

)

∈

k

⁢

[

X

]

and

W

⁢

(

X

)

∈

k

⁢

[

X

]

, then we will write

W

⁢

(

X

)

¯

to denote

element

in

k

⁢

[

X

]

/

(

F

⁢

(

X

)

)

represented by

W

⁢

(

X

)

.

Lemma.

Assume, that

a

1

,

…

,

a

n

∈

k

are different elements and

F

⁢

(

X

)

=

(

X

-

a

1

)

⁢

⋯

⁢

(

X

-

a

n

)

. Let

W

i

⁢

(

X

)

∈

k

⁢

[

X

]

be given by

W

i

⁢

(

X

)

=

(

X

-

a

1

)

⁢

⋯

⁢

(

X

-

a

i

-

1

)

⋅

(

X

-

a

i

+

i

)

⁢

⋯

⁢

(

X

-

a

n

)

. Then there exist

λ

1

,

…

,

λ

n

∈

k

such that

F

⁢

(

X

)

divides

polynomial

U

⁢

(

X

)

=

(

∑

i

=

1

n

λ

i

⋅

W

i

⁢

(

X

)

)

-

1

.

Proof.

Note, that

W

i

⁢

(

a

i

)

≠

0

for any

i

. Thus we may define

λ

i

=

(

W

i

⁢

(

a

i

)

)

-

1

. Then for any

i

we have

λ

i

⋅

W

i

⁢

(

a

i

)

=

1

, therefore

V

⁢

(

X

)

=

∑

i

=

1

n

λ

i

⋅

W

i

⁢

(

X

)

is such that

V

⁢

(

a

i

)

=

1

for any

i

. In particular

U

⁢

(

a

i

)

=

V

⁢

(

a

i

)

-

1

=

0

and thus

(

X

-

a

i

)

divides

U

⁢

(

X

)

for any

i

. This

completes

the proof.

□

Corollary.

Under the same

assumptions

as in lemma, we have that ideal

(

W

1

⁢

(

X

)

¯

,

…

,

W

n

⁢

(

X

)

¯

)

in

k

⁢

[

X

]

/

(

F

⁢

(

X

)

)

is equal to

k

⁢

[

X

]

/

(

F

⁢

(

X

)

)

.

Proof.

Indeed, all we need to show is that we can generate

1

¯

. Lemma implies, that there is

V

⁢

(

X

)

∈

k

⁢

[

X

]

such that

(

∑

i

=

1

n

λ

i

⋅

W

i

⁢

(

X

)

)

-

1

=

F

⁢

(

X

)

⋅

V

⁢

(

X

)

.

Now, after aplying

quotient

homomorphism

to both sides we have

∑

i

=

1

n

λ

i

⋅

W

i

⁢

(

X

)

¯

=

1

¯

.

This completes the proof.

□

Remark.

This gives us an explicit formula for

generators

of

k

⁢

[

X

]

/

(

F

⁢

(

X

)

)

. In particular the dimension over

k

of this ring is at most

deg

⁢

F

⁢

(

X

)

. It can be shown that actualy it is equal, even if

F

⁢

(

X

)

is arbitrary.

Title

explicit generators of a quotient

polynomial ring

associated to a given polynomial

Canonical name

ExplicitGeneratorsOfAQuotientPolynomialRingAssociatedToAGivenPolynomial

Date of creation

2013-03-22 19:10:01

Last modified on

2013-03-22 19:10:01

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Derivation

Classification

msc 11C08

Classification

msc 12E05

Classification

msc 13P05

Generated on Fri Feb 9 15:30:45 2018 by

LaTeXML
height of a polynomial

height of a polynomial

Let

P

=

a

0

+

a

1

⁢

x

+

…

+

a

n

⁢

x

n

∈

ℂ

⁢

[

x

]

be a

polynomial

with

complex

coefficients. The height of

P

is

H

⁢

(

P

)

=

max

⁡

{

∣

a

0

∣

,

∣

a

1

∣

,

…

,

∣

a

n

∣

}

.

Title

height of a polynomial

Canonical name

HeightOfAPolynomial

Date of creation

2013-03-22 13:33:45

Last modified on

2013-03-22 13:33:45

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

4

Author

Daume (40)

Entry type

Definition

Classification

msc 11C08

Generated on Fri Feb 9 15:30:47 2018 by

LaTeXML
length of a polynomial

length of a polynomial

Let

P

=

a

0

+

a

1

⁢

x

+

…

+

a

n

⁢

x

n

∈

ℂ

⁢

[

x

]

be a

polynomial

with

complex

coefficients. The length of

P

is

L

⁢

(

P

)

=

∣

a

0

∣

+

∣

a

1

∣

+

…

+

∣

a

n

∣

=

∑

i

=

0

n

∣

a

i

∣

Title

length of a polynomial

Canonical name

LengthOfAPolynomial

Date of creation

2013-03-22 13:33:39

Last modified on

2013-03-22 13:33:39

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

5

Author

Daume (40)

Entry type

Definition

Classification

msc 11C08

Generated on Fri Feb 9 15:30:49 2018 by

LaTeXML
minimal polynomial

minimal polynomial

Let

K

/

F

be a

field extension

and

κ

∈

K

be algebraic over

F

. The

minimal polynomial

for

κ

over

F

is a

monic polynomial

m

⁢

(

x

)

∈

F

⁢

[

x

]

such that

m

⁢

(

κ

)

=

0

and, for any other

polynomial

f

⁢

(

x

)

∈

F

⁢

[

x

]

with

f

⁢

(

κ

)

=

0

,

m

divides

f

. Note that, for any element

κ

that is algebraic over

F

, a minimal polynomial exists (

http://planetmath.org/ExistenceOfTheMinimalPolynomial

); moreover, because of the monic condition, it exists uniquely.

Given

κ

∈

K

, a polynomial

m

is the minimal polynomial of

κ

if and only if

m

⁢

(

κ

)

=

0

and

m

is both monic and

irreducible

(

http://planetmath.org/IrreduciblePolynomial

).

Title

minimal polynomial

Canonical name

MinimalPolynomial

Date of creation

2013-03-22 13:20:11

Last modified on

2013-03-22 13:20:11

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

13

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 11C08

Classification

msc 11R04

Classification

msc 12F05

Classification

msc 12E05

Related topic

DegreeOfAnAlgebraicNumber

Generated on Fri Feb 9 15:30:51 2018 by

LaTeXML
Newton-Girard formula for symmetric polynomials

Newton-Girard formula for symmetric polynomials

Let

E

k

be the

elementary symmetric polynomials

in

n

variables

and

S

k

be defined by

S

k

⁢

(

x

1

,

…

,

x

n

)

=

∑

i

=

1

n

x

i

k

.

Then the

S

k

and

E

k

are related as follows:

S

1

=

E

1

S

2

=

S

1

⁢

E

1

-

2

⁢

E

2

S

3

=

S

2

⁢

E

1

-

S

1

⁢

E

2

+

3

⁢

E

3

⋮

S

k

=

-

(

∑

j

=

1

k

-

1

(

-

1

)

j

⁢

S

k

-

j

⁢

E

j

)

-

(

-

1

)

k

⁢

k

⁢

E

k

By applying these formulas recursively,

S

k

can be expressed solely in terms of the

E

k

, which is often desirable. For example, since

S

1

=

E

1

,

S

2

=

E

1

2

-

2

⁢

E

2

, and then

S

3

=

(

E

1

2

-

2

⁢

E

2

)

⁢

E

1

-

E

1

⁢

E

2

+

3

⁢

E

3

=

E

1

3

-

3

⁢

E

1

⁢

E

2

+

3

⁢

E

3

, and so on.

Note that

E

0

=

1

and

E

k

=

0

for

k

>

n

.

Title

Newton-Girard formula

for

symmetric polynomials

Canonical name

NewtonGirardFormulaForSymmetricPolynomials

Date of creation

2013-03-22 15:32:40

Last modified on

2013-03-22 15:32:40

Owner

kschalm (9486)

Last modified by

kschalm (9486)

Numerical id

5

Author

kschalm (9486)

Entry type

Theorem

Classification

msc 11C08

Related topic

WaringsFormula

Related topic

ElementarySymmetricPolynomialInTermsOfPowerSums

Generated on Fri Feb 9 15:30:53 2018 by

LaTeXML
opposite polynomial

opposite polynomial

The

opposite

polynomial

of a polynomial

P

in a

polynomial ring

R

⁢

[

X

]

is a polynomial

-

P

such that

P

+

(

-

P

)

=

𝟎

,

where

0

denotes the

zero polynomial

.  It is clear that

-

P

is obtained by changing the signs of all of the

coefficients

of

P

, i.e. (

http://planetmath.org/Ie

)

-

∑

ν

=

0

n

a

ν

⁢

X

ν

=

∑

ν

=

0

n

(

-

a

ν

)

⁢

X

ν

.

The opposite polynomial may be used to define

subtraction

of polynomials:

P

-

Q

=

:

P

+

(

-

Q

)

Forming the opposite polynomial is a

linear mapping

R

⁢

[

X

]

→

R

⁢

[

X

]

.

Title

opposite polynomial

Canonical

name

OppositePolynomial

Date of creation

2013-03-22 14:47:41

Last modified on

2013-03-22 14:47:41

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Definition

Classification

msc 11C08

Classification

msc 12E05

Classification

msc 13P05

Related topic

OppositeNumber

Related topic

Unity

Related topic

BasicPolynomial

Related topic

MinimalPolynomialEndomorphism

Generated on Fri Feb 9 15:30:56 2018 by

LaTeXML
proof of Eisenstein criterion

proof of Eisenstein criterion

Let

f

⁢

(

x

)

∈

R

⁢

[

x

]

be a

polynomial

satisfying Eisenstein’s Criterion with prime

p

.

Suppose that

f

⁢

(

x

)

=

g

⁢

(

x

)

⁢

h

⁢

(

x

)

with

g

⁢

(

x

)

,

h

⁢

(

x

)

∈

F

⁢

[

x

]

, where

F

is the

field of fractions

of

R

. Gauss’ Lemma II there exist

g

′

⁢

(

x

)

,

h

′

⁢

(

x

)

∈

R

⁢

[

x

]

such that

f

⁢

(

x

)

=

g

′

⁢

(

x

)

⁢

h

′

⁢

(

x

)

, i.e. any

factorization

can be converted to a factorization in

R

⁢

[

x

]

.

Let

f

⁢

(

x

)

=

∑

i

=

0

n

a

i

⁢

x

i

,

g

′

⁢

(

x

)

=

∑

j

=

0

ℓ

b

j

⁢

x

j

,

h

′

⁢

(

x

)

=

∑

k

=

0

m

c

k

⁢

x

k

be the expansions of

f

⁢

(

x

)

,

g

′

⁢

(

x

)

, and

h

′

⁢

(

x

)

respectively.

Let

φ

:

R

⁢

[

x

]

→

R

/

p

⁢

R

⁢

[

x

]

be the

natural homomorphism

from

R

⁢

[

x

]

to

R

/

p

⁢

R

⁢

[

x

]

. Note that since

p

∣

a

i

for

i

<

n

and

p

∤

a

n

, we have

φ

⁢

(

a

i

)

=

0

for

i

<

n

and

φ

⁢

(

a

i

)

=

α

≠

0

φ

⁢

(

f

⁢

(

x

)

)

=

φ

⁢

(

∑

i

=

0

n

a

i

⁢

x

i

)

=

∑

i

=

0

n

φ

⁢

(

a

i

)

⁢

x

i

=

φ

⁢

(

a

n

)

⁢

x

n

=

α

⁢

x

n

Therefore we have

α

⁢

x

n

=

φ

⁢

(

f

⁢

(

x

)

)

=

φ

⁢

(

g

′

⁢

(

x

)

⁢

h

′

⁢

(

x

)

)

=

φ

⁢

(

g

′

⁢

(

x

)

)

⁢

φ

⁢

(

h

′

⁢

(

x

)

)

so we must have

φ

⁢

(

g

′

⁢

(

x

)

)

=

β

⁢

x

ℓ

′

and

φ

(

h

′

(

x

)

=

γ

x

m

′

for some

β

,

γ

∈

R

/

p

⁢

R

and some

integers

ℓ

′

,

m

′

.

Clearly

ℓ

′

≤

deg

⁡

(

g

′

⁢

(

x

)

)

=

ℓ

and

m

′

≤

deg

⁡

(

h

′

⁢

(

x

)

)

=

m

, and therefore since

ℓ

′

⁢

m

′

=

n

=

ℓ

⁢

m

, we must have

ℓ

′

=

ℓ

and

m

′

=

m

.
Thus

φ

⁢

(

g

′

⁢

(

x

)

)

=

β

⁢

x

ℓ

and

φ

⁢

(

h

′

⁢

(

x

)

)

=

γ

⁢

x

m

.

If

ℓ

>

0

, then

φ

⁢

(

b

i

)

=

0

for

i

<

ℓ

. In particular,

φ

⁢

(

b

0

)

=

0

, hence

p

∣

b

0

. Similarly if

m

>

0

, then

p

∣

c

0

.

Since

f

⁢

(

x

)

=

g

′

⁢

(

x

)

⁢

h

′

⁢

(

x

)

, by equating coefficients we see that

a

0

=

b

0

⁢

c

0

.

If

ℓ

>

0

and

m

>

0

, then

p

∣

b

0

and

p

∣

c

0

, which implies that

p

2

∣

a

0

. But this contradicts our assumptions on

f

⁢

(

x

)

, and therefore we must have

ℓ

=

0

or

m

=

0

, that is, we must have a trivial factorization. Therefore

f

⁢

(

x

)

is

irreducible

.

Title

proof of Eisenstein criterion

Canonical name

ProofOfEisensteinCriterion

Date of creation

2013-03-22 12:42:11

Last modified on

2013-03-22 12:42:11

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 11C08

Classification

msc 13F15

Related topic

GausssLemmaII

Generated on Fri Feb 9 15:31:03 2018 by

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proof of Newton-Girard formula for symmetric polynomials

proof of Newton-Girard formula for symmetric polynomials

The following is a proof of

Newton-Girard formula

using

formal
power series

. Let

z

be an

indeterminate

and

f

⁢

(

z

)

be the

polynomial

1

-

E

1

⁢

z

+

…

+

(

-

1

)

n

⁢

E

n

⁢

z

n

.

Take log and differentiate both sides of the equation

f

⁢

(

z

)

=

∏

i

=

1

n

(

1

-

x

i

⁢

z

)

.

We obtain

f

′

⁢

(

z

)

/

f

⁢

(

z

)

=

∑

i

=

1

n

-

x

i

1

-

x

i

⁢

z

,

(1)

where

f

′

⁢

(

z

)

is the derivative of

f

⁢

(

z

)

f

′

⁢

(

z

)

=

-

E

1

+

2

⁢

E

2

⁢

z

-

…

+

(

-

1

)

n

⁢

n

⁢

E

n

⁢

z

n

-

1

.

The right hand side of (

1

) is equal to

-

∑

i

=

1

n

∑

k

=

0

∞

x

i

k

+

1

⁢

z

k

=

-

∑

k

=

0

∞

S

k

+

1

⁢

z

k

.

By equating coefficients of

f

′

⁢

(

z

)

=

-

f

⁢

(

z

)

⁢

(

S

1

+

S

2

⁢

z

+

S

3

⁢

z

2

+

…

)

we get the Newton-Girard formula.

Title

proof of Newton-Girard formula for symmetric polynomials

Canonical

name

ProofOfNewtonGirardFormulaForSymmetricPolynomials

Date of creation

2013-03-22 15:34:37

Last modified on

2013-03-22 15:34:37

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

4

Author

kshum (5987)

Entry type

Proof

Classification

msc 11C08

Generated on Fri Feb 9 15:31:05 2018 by

LaTeXML
proof of Waring’s formula

proof of Waring’s formula

The following is a proof of the Waring’s formula using

formal
power series

. We will work with formal power series in

indeterminate

z

with

coefficients

in the ring

ℚ

⁢

[

x

1

,

…

,

x

n

]

. We also need the following equality

-

log

⁡

(

1

-

z

)

=

∑

j

=

1

∞

z

j

j

.

Taking log on both sides of

1

-

σ

1

⁢

z

+

…

+

(

-

1

)

n

⁢

σ

n

⁢

z

n

=

∏

m

=

1

n

(

1

-

x

m

⁢

z

)

,

we get

log

⁡

(

1

-

σ

1

⁢

z

+

…

+

(

-

1

)

n

⁢

σ

n

⁢

z

n

)

=

∑

m

=

1

n

log

⁡

(

1

-

x

m

⁢

z

)

,

(1)

Waring’s formula will follow by comparing the coefficients on both
sides.

The

right hand side

of the above equation equals

∑

m

=

1

n

∑

j

=

1

∞

x

m

j

j

⁢

z

j

or

∑

j

=

1

∞

(

∑

m

=

1

n

x

m

j

)

⁢

z

j

j

The coefficient of

z

k

is equal to

S

k

/

k

.

On the other hand, the left hand side of (

1

) can be written
as

∑

j

=

1

∞

1

j

⁢

(

σ

1

⁢

z

-

σ

2

⁢

z

2

+

…

+

(

-

1

)

n

-

1

⁢

σ

n

⁢

z

n

)

j

.

For each

j

, the coefficient of

z

k

in

(

σ

1

⁢

z

-

σ

2

⁢

z

2

+

…

+

(

-

1

)

n

-

1

⁢

σ

n

⁢

z

n

)

j

is

∑

i

1

,

…

,

i

n

(

-

1

)

i

2

+

i

4

+

i

6

+

…

⁢

j

!

i

1

!

⁢

⋯

⁢

i

n

!

⁢

σ

1

i

1

⁢

⋯

⁢

σ

n

i

n

,

where the summation is extended over all

n

-tuple

(

i

1

,

…

,

i

n

)

whose entries are non-negative integers, such
that

i

1

+

i

2

+

…

+

i

n

=

j

i

1

+

2

⁢

i

2

+

…

+

n

⁢

i

n

=

k

.

So the coefficient of

z

k

in the left hand side of (

1

) is

∑

j

=

1

∞

∑

i

1

,

…

,

i

n

(

-

1

)

i

2

+

i

4

+

i

6

+

…

⁢

(

j

-

1

)

!

i

1

!

⁢

⋯

⁢

i

n

!

⁢

σ

1

i

1

⁢

⋯

⁢

σ

n

i

n

,

or

∑

(

-

1

)

i

2

+

i

4

+

i

6

+

…

⁢

(

i

1

+

…

+

i

n

-

1

)

!

i

1

!

⁢

⋯

⁢

i

n

!

⁢

σ

1

i

1

⁢

⋯

⁢

σ

n

i

n

.

The last summation is over all

(

i

1

,

…

,

i

n

)

∈

ℤ

n

with non-negative entries such that

i

1

+

2

⁢

i

2

+

…

+

n

⁢

i

n

=

k

.

Title

proof of Waring’s formula

Canonical name

ProofOfWaringsFormula

Date of creation

2013-03-22 15:34:29

Last modified on

2013-03-22 15:34:29

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

7

Author

kshum (5987)

Entry type

Proof

Classification

msc 11C08

Generated on Fri Feb 9 15:31:08 2018 by

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sufficient condition of identical congruence

sufficient condition of identical congruence

Theorem

.

Let

f

⁢

(

X

)

:=

a

n

⁢

X

n

+

…

+

a

1

⁢

X

+

a

0

be a

polynomial

in

X

with

integer

coefficients

a

i

and

m

a

positive integer

.  If the

congruence

f

⁢

(

x

)

≡

0

(

mod

m

)

(1)

is satisfied by

m

successive integers

x

, then it is satisfied by all integers

x

, in other words it is an

identical congruence

.

Proof.

There is an integer

x

0

such that (1) is satisfied by

x

:=

x

0

+

1

,

x

0

+

2

,

…

,

x

0

+

m

.

But these values form a

complete residue system

modulo

m

.  Thus, if

x

is an arbitrary integer, one has

x

≡

x

0

+

r

(

mod

m

)

where

⁢

1

≦

r

≦

m

.

This implies

a

i

⁢

x

i

≡

a

i

⁢

(

x

0

+

r

)

i

(

mod

m

)

for

⁢

i

=

0

,

1

,

…

,

n

and consequently

∑

i

=

0

n

a

i

⁢

x

i

⏟

f

⁢

(

x

)

≡

∑

i

=

0

n

a

i

⁢

(

x

0

+

r

)

i

=

f

⁢

(

x

0

+

r

)

≡

0

(

mod

m

)

.

Accordingly, (1) is true for any integer

x

, Q.E.D.

Note.

Though the congruence (1) is identical, it need not be a question of a formal congruence

f

⁢

(

X

)

⁢

≡

¯

⁢

0

(

mod

m

)

,

(2)

i.e. all coefficients

a

i

need not be

congruent

to 0 modulo

m

.

Title

sufficient condition of identical congruence

Canonical name

SufficientConditionOfIdenticalCongruence

Date of creation

2013-03-22 18:56:03

Last modified on

2013-03-22 18:56:03

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11C08

Classification

msc 11A07

Related topic

Sufficient

Related topic

CongruenceOfArbitraryDegree

Related topic

PolynomialCongruence

Generated on Fri Feb 9 15:31:10 2018 by

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Waring’s formula

Waring’s formula

Let

x

1

,

…

,

x

n

be

n

indeterminates

. For

k

≥

1

, let

σ

k

be the

k

th

elementary symmetric polynomials

in

x

1

,

…

,

x

n

, and

S

k

be the

k

th power sum defined as

S

k

=

∑

i

=

1

n

x

i

k

.

Like the Newton’s formula, the

Waring formula

is a relation
between

σ

k

and

S

k

:

S

k

=

∑

(

-

1

)

(

i

2

+

i

4

+

i

6

+

…

)

⁢

(

i

1

+

i

2

+

…

+

i

n

-

1

)

!

⁢

k

i

1

!

⁢

i

2

!

⁢

⋯

⁢

i

n

!

⁢

σ

1

i

1

⁢

σ

2

i

2

⁢

⋯

⁢

σ

n

i

n

,

where the

summation

is over all

n

-tuples

(

i

1

,

…

,

i

n

)

∈

ℤ

n

with non-negative

components

, such that

i

1

+

2

⁢

i

2

+

…

+

n

⁢

i

n

=

k

.

In particular, when there are two indeterminates, i.e.

n

=

2

, the
Waring formula reads

x

1

k

+

x

2

k

=

∑

i

=

0

⌊

k

/

2

⌋

(

-

1

)

i

⁢

k

k

-

i

⁢

(

k

-

i

i

)

⁢

(

x

1

+

x

2

)

k

-

2

⁢

i

⁢

(

x

1

⁢

x

2

)

i

.

Title

Waring’s formula

Canonical

name

WaringsFormula

Date of creation

2013-03-22 15:34:26

Last modified on

2013-03-22 15:34:26

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

9

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11C08

Synonym

Waring formula

Related topic

NewtonGirardFormulaSymmetricPolynomials

Generated on Fri Feb 9 15:31:12 2018 by

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solving linear Diophantine equation

solving linear Diophantine equation

Here we an elementary but very comprehensible method for solving any linear Diophantine equation with two unknowns, i.e. for finding the general

integer

solution of an equation of the form

a

⁢

x

-

m

⁢

y

=

b

where

a

,

b

,

m

are known integers and

x

,

y

the unknowns.

The method is illustrated via a numerical example:

37

⁢

x

-

107

⁢

y

=

25

(1)

We solve first (1) for

x

(which has absolutely smaller

coefficient

than

y

):

x

=

25

+

107

⁢

y

37

(2)

The in the

numerator

may be split so that

division

yields a

polynomial

with integer coefficients and that the

remainder

has absolutely smaller coefficients (now

-

12

and

-

4

) than the dividend in (2) had:

x

=

1

+

3

⁢

y

-

12

+

4

⁢

y

37

Since

x

and

1

+

3

⁢

y

mean integers, also

z

:=

12

+

4

⁢

y

37

must be an integer.  Now solve this last equation for

y

and split the new numerator similarly as above:

y

=

-

12

+

37

⁢

z

4

=

-

3

+

9

⁢

z

+

z

4

Since

y

and

-

3

+

9

⁢

z

mean integers, also

t

:=

z

4

must be an integer.  It is apparent that we can give any integer value for

t

, which thus may be thought as a

parameter

determining the values of the other letters.  We obtain successively

z

=

4

⁢

t

,

y

=

-

3

+

9

⋅

4

⁢

t

+

t

=

-

3

+

37

⁢

t

,

x

=

1

+

3

⁢

(

-

3

+

37

⁢

t

)

-

4

⁢

t

=

-

8

+

107

⁢

t

.

Accordingly, we may write the of (1) as

{

x

=

-

8

+

107

⁢

t

y

=

-

3

+

37

⁢

t

where

t

=

0

,

±

1

,

±

2

,

…

This method and the use of a parameter for expressing the solution were well known in the ancient world, especially in solving astronomical cycles as noted by Brahmagupta (598–668).

Title

solving linear Diophantine equation

Canonical

name

SolvingLinearDiophantineEquation

Date of creation

2013-03-22 17:45:55

Last modified on

2013-03-22 17:45:55

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Topic

Classification

msc 11D04

Related topic

LinearCongruence

Generated on Fri Feb 9 15:31:18 2018 by

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