aarjaneiro/planetmath huggingface dataset:
Teichmüller character

Teichmüller character

Before we define the Teichmüller

character

, we begin with a

corollary

of Hensel’s lemma.

Corollary

.

Let

p

be a

prime number

. The ring of

p

-adic

integers

(

http://planetmath.org/PAdicIntegers

)

Z

p

contains exactly

p

-

1

distinct

(

p

-

1

)

th

roots of unity

. Furthermore, every

(

p

-

1

)

th root of unity is distinct modulo

p

.

Proof.

Notice that

ℚ

p

, the

p

-adic

rationals

, is a field. Therefore

f

⁢

(

x

)

=

x

p

-

1

-

1

has at most

p

-

1

roots in

ℚ

p

(see this entry (

http://planetmath.org/APolynomialOfDegreeNOverAFieldHasAtMostNRoots

)). Moreover, if we let

a

∈

ℤ

with

1

≤

a

≤

p

-

1

then

f

⁢

(

a

)

=

a

p

-

1

-

1

≡

0

mod

p

by Fermat’s little theorem. Since

f

′

⁢

(

a

)

=

(

p

-

1

)

⋅

a

p

-

2

is non-zero modulo

p

, the trivial case of Hensel’s lemma implies that there exist a root of

x

p

-

1

-

1

in

ℤ

p

which is

congruent

to

a

modulo

p

. Hence, there are at least

p

-

1

roots in

ℤ

p

, and we can conclude that there are exactly

p

-

1

roots.
∎

Definition

.

The Teichmüller character is a

homomorphism

of

multiplicative groups

:

ω

:

𝔽

p

×

→

ℤ

p

×

such that

ω

⁢

(

a

)

is the unique

(

p

-

1

)

th root of unity in

Z

p

which is congruent to

a

modulo

p

(which exists by the corollary above). The map

ω

is sometimes called the Teichmüller lift of

F

p

to

Z

p

(

0

mod

p

would lift to

0

∈

Z

p

).

Remark

.

Some authors define the Teichmüller character to be the homomorphism:

ω

^

:

ℤ

p

×

→

ℤ

p

×

defined by

ω

^

⁢

(

z

)

=

lim

n

→

∞

⁡

z

p

n

.

Notice that for any

z

∈

ℤ

p

×

,

ω

^

⁢

(

z

)

is a

(

p

-

1

)

th root of unity:

(

ω

^

⁢

(

z

)

)

p

=

(

lim

n

→

∞

⁡

z

p

n

)

p

=

lim

n

→

∞

⁡

z

p

n

+

1

=

ω

^

⁢

(

z

)

.

Thus, the value

ω

^

⁢

(

z

)

is the same than

ω

⁢

(

z

mod

p

)

.

Title

Teichmüller character

Canonical

name

TeichmullerCharacter

Date of creation

2013-03-22 15:09:04

Last modified on

2013-03-22 15:09:04

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Definition

Classification

msc 13H99

Classification

msc 11S99

Classification

msc 12J99

Synonym

Teichmuler character

Synonym

Teichmuller lift

Synonym

Teichmüller lift

Related topic

PAdicIntegers

Generated on Fri Feb 9 17:34:12 2018 by

LaTeXML
non-isomorphic completions of ℚ

non-isomorphic completions of

ℚ

No field

ℚ

p

of the

p

-adic numbers

(

p

-adic

rationals

(

http://planetmath.org/PAdicIntegers

)) is

isomorphic

with the field

ℝ

of the

real numbers

.

Proof.

Let’s assume the existence of a

field isomorphism

f

:

ℝ

→

ℚ

p

for some

positive

prime number

p

.  If we denote

f

⁢

(

p

)

=

a

,  then we obtain

a

2

=

(

f

⁢

(

p

)

)

2

=

f

⁢

(

(

p

)

2

)

=

f

⁢

(

p

)

=

p

,

because the

isomorphism

maps the elements of the

prime subfield

on themselves.  Thus, if

|

⋅

|

p

is the normed

p

-adic

valuation

(

http://planetmath.org/PAdicValuation

) of

ℚ

and of

ℚ

p

, we get

|

a

|

p

=

|

a

2

|

p

=

|

p

|

p

=

1

p

,

which value is an

irrational number

as a

square root

of a non-square (

http://planetmath.org/SquareRootOf2IsIrrationalProof

) rational.  But this is impossible, since the

value group

of the

completion

ℚ

p

must be the same as the value group

|

ℚ

∖

{

0

}

|

p

which consists of all

integer

powers of

p

.  So we conclude that there can not exist such an isomorphism.

Title

non-isomorphic completions of

ℚ

Canonical name

NonisomorphicCompletionsOfmathbbQ

Date of creation

2013-03-22 14:58:17

Last modified on

2013-03-22 14:58:17

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13J10

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 13F30

Related topic

PAdicCanonicalForm

Defines

p

-adic numbers

Generated on Fri Feb 9 17:34:14 2018 by

LaTeXML
p-ring

p-ring

Definition 1

.

Let

R

be a commutative ring with

identity element

equipped with a topology defined by a

decreasing sequence

:

…

⊂

𝔄

3

⊂

𝔄

2

⊂

𝔄

1

of ideals such that

A

n

⋅

A

m

⊂

A

n

+

m

. We say that

R

is a

p

-ring if the following conditions are satisfied:

1.

The

residue

ring

k

¯

=

R

/

𝔄

1

is a

perfect ring

of

characteristic

p

.

2.

The ring

R

is Hausdorff and

complete

for its topology.

Definition 2

.

A

p

-ring

R

is said to be strict (or a

p

-adic ring) if the topology is defined by the

p

-adic

filtration

A

n

=

p

n

⁢

R

, and

p

is not a zero-divisor of

R

.

Example 1

.

The prototype of strict

p

-ring is the ring of

p

-adic

integers

(

http://planetmath.org/PAdicIntegers

)

ℤ

p

with the usual

profinite topology

.

References

1

J. P. Serre,

Local Fields

,
Springer-Verlag, New York.

Title

p-ring

Canonical name

Pring

Date of creation

2013-03-22 15:14:28

Last modified on

2013-03-22 15:14:28

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Definition

Classification

msc 13J10

Classification

msc 13K05

Synonym

p

-ring

Synonym

p-adic ring

Synonym

p

-adic ring

Synonym

strict

p

-ring

Defines

strict p-ring

Generated on Fri Feb 9 17:34:16 2018 by

LaTeXML
partially ordered ring

partially ordered ring

A ring

R

that is a poset at the same time is called a

partially ordered ring

, or a

po-ring

, if, for

a

,

b

,

c

∈

R

,

•

a

≤

b

implies

a

+

c

≤

b

+

c

, and

•

0

≤

a

and

0

≤

b

implies

0

≤

a

⁢

b

.

Note that

R

does not have to be

associative

.

If the underlying poset of a po-ring

R

is in fact a

lattice

, then

R

is called a

lattice-ordered ring

, or an

l-ring

for short.

Remark

. The underlying

abelian group

of a po-ring (with

addition

being the

binary operation

) is a

po-group

. The same is true for l-rings.

Below are some examples of po-rings:

•

Clearly, any (totally)

ordered ring

is a po-ring.

•

The

ring of continuous functions

over a

topological space

is an l-ring.

•

Any matrix ring over an ordered field is an l-ring if we define

(

a

i

⁢

j

)

≤

(

b

i

⁢

j

)

whenever

a

i

⁢

j

≤

b

i

⁢

j

for all

i

,

j

.

Remark

. Let

R

be a po-ring. The set

R

+

:=

{

r

∈

R

∣

0

≤

r

}

is called the

positive cone

of

R

.

References

1

G. Birkhoff

Lattice Theory

, 3rd Edition, AMS Volume XXV, (1967).

Title

partially ordered ring

Canonical name

PartiallyOrderedRing

Date of creation

2013-03-22 16:55:04

Last modified on

2013-03-22 16:55:04

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13J25

Classification

msc 16W80

Classification

msc 06F25

Synonym

po-ring

Synonym

l-ring

Synonym

lattice-ordered ring

Defines

lattice ordered ring

Defines

positive cone

Generated on Fri Feb 9 17:34:25 2018 by

LaTeXML
real ring

real ring

A ring

A

is called real iff the following

identity

holds for all

n

∈

ℕ

:

a

1

2

+

…

+

a

n

2

=

0

⇔

a

1

,

…

,

a

n

=

0

(

∀

a

1

,

…

,

a

n

∈

A

)

Remark

.

If

A

is a ring then being real implies the following

•

A

can have a

partial ordering

•

A

is

reduced

Conversely

, we note that if

A

is reduced and can have a partial ordering then

A

is a real ring.
If

A

is a field then we call it a

real field

. Similarly we define

real domains

,

real (von Neumann)

regular rings

,

…

Title

real ring

Canonical name

RealRing

Date of creation

2013-03-22 18:51:35

Last modified on

2013-03-22 18:51:35

Owner

jocaps (12118)

Last modified by

jocaps (12118)

Numerical id

6

Author

jocaps (12118)

Entry type

Definition

Classification

msc 13J30

Classification

msc 13J25

Related topic

FormallyRealField

Generated on Fri Feb 9 17:34:31 2018 by

LaTeXML
weakest extension of a partial ordering

weakest extension of a partial ordering

Let

A

be a

commutative ring

with

partial ordering

A

+

and suppose that

B

is a ring that admits a partial ordering.
If

f

:

A

→

B

is a

ring monomorphism

(thus we regard

B

as an over-ring of

A

),
then any partial ordering of

B

that can contain

f

⁢

(

A

+

)

will also contain the set

B

+

⊂

B

defined by

B

+

:=

{

∑

i

=

1

n

f

⁢

(

a

i

)

⁢

b

i

2

:

n

∈

ℕ

,

a

1

,

…

,

a

n

∈

A

+

}

B

+

is itself a partial ordering and it is called the

weakest partial ordering of

B

that extends

A

+

(through

f

)

. It is called ”weakest” because this is the smallest partial ordering

B

+

of

B

that will transform

f

into a poring

monomorphism

(i.e. a monomorphism in the

category

of

partially ordered rings

)

f

:

(

A

,

A

+

)

→

(

B

,

B

+

)

(for simplicity, we abuse the symbol

f

here).

Title

weakest

extension

of a partial ordering

Canonical name

WeakestExtensionOfAPartialOrdering

Date of creation

2013-03-22 18:51:40

Last modified on

2013-03-22 18:51:40

Owner

jocaps (12118)

Last modified by

jocaps (12118)

Numerical id

9

Author

jocaps (12118)

Entry type

Definition

Classification

msc 13J30

Classification

msc 13J25

Defines

weakest extension of a partial ordering

Generated on Fri Feb 9 17:34:34 2018 by

LaTeXML
cyclic rings and zero rings

cyclic rings and zero rings

Lemma 1

.

Let

n

be a positive integer and

R

be a

cyclic ring

of order (

http://planetmath.org/OrderRing

)

R

. Then the following are

equivalent

:

1.

R

is a

zero ring

;

2.

R

has

behavior

n

;

3.

R

≅

n

⁢

ℤ

n

2

.

Proof.

To show that 1 implies 2, let

R

have behavior

k

. Then there exists a

generator

(

http://planetmath.org/Generator

)

r

of the

additive group

of

R

such that

r

2

=

k

⁢

r

. Since

R

is a zero ring,

r

2

=

0

R

. Since

k

⁢

r

=

r

2

=

0

R

=

n

⁢

r

, it must be the case that

k

≡

n

mod

n

. By definition of behavior,

k

divides (

http://planetmath.org/Divides

)

n

. Hence,

k

=

n

.

The fact that 2 implies 3 follows immediately from the

theorem

that is stated and proven at cyclic rings that are

isomorphic

to

k

⁢

ℤ

k

⁢

n

(

http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ_kn

).

The fact that 3 implies 1 follows immediately since

n

⁢

ℤ

n

2

is a zero ring.
∎

Lemma 2

.

Let

R

be an

infinite

. Then the following are equivalent:

1.

R

is a zero ring;

2.

R

has behavior

0

;

3.

R

is isomorphic (

http://planetmath.org/Isomorphism7

) to the

subring

𝐁

of

𝐌

2

⁢

x

⁡

2

⁢

(

ℤ

)

:

𝐁

=

{

(

n

-

n

n

-

n

)

|

n

∈

ℤ

}

.

Proof.

To show that 1 implies 2, the

contrapositive

of the theorem that is stated and proven at cyclic rings that are isomorphic to

k

⁢

ℤ

(

http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ

) can be used. If

R

does not have behavior

0

, then its behavior

k

must be positive by definition, in which case

R

≅

k

⁢

ℤ

. It is clear that

k

⁢

ℤ

is not a zero ring.

To show that 2 implies 3, let

r

be a generator of the additive group of

R

. It can be easily verified that

φ

:

R

→

𝐁

defined by

φ

⁢

(

n

⁢

r

)

=

(

n

-

n

n

-

n

)

is a

ring isomorphism

.

The fact that 3 implies 1 follows immediately since

𝐁

is a zero ring.
∎

Title

cyclic rings and zero rings

Canonical name

CyclicRingsAndZeroRings

Date of creation

2013-03-22 17:14:43

Last modified on

2013-03-22 17:14:43

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Result

Classification

msc 13M05

Classification

msc 13A99

Classification

msc 16U99

Generated on Fri Feb 9 17:34:41 2018 by

LaTeXML
universal derivation

universal derivation

Let

R

be a commutative ring, and let

A

be a

commutative

R

-

algebra

. A

universal derivation

of

A

over

R

is defined to be an

A

-module

Ω

A

/

R

together with an

R

-linear

derivation

d

:

A

→

Ω

A

/

R

, such that the following

universal property

holds:
for every

A

-module

M

and every

R

-linear derivation

δ

:

A

→

M

there exists a unique

A

-linear map

f

:

Ω

A

/

R

→

M

such that

δ

=

f

∘

d

.

The universal property can be illustrated by a

commutative diagram

:

\xymatrix

⁢

A

⁢

\ar

⁢

[

r

]

d

⁢

\ar

⁢

[

d

⁢

r

]

δ

⁢

&

⁢

Ω

A

/

R

⁢

\ar

⁢

@

!

⁢

[

d

]

f

⁢

⁢

&

⁢

M

An

A

-module with this property can be constructed explicitly, so

Ω

A

/

R

always exists. It is generated as an

A

-module by
the set

{

d

⁢

x

:

x

∈

A

}

, with the

relations

d

⁢

(

a

⁢

x

+

b

⁢

y

)

=

a

⁢

d

⁢

x

+

b

⁢

d

⁢

y

d

⁢

(

x

⁢

y

)

=

x

⋅

d

⁢

y

+

y

⁢

d

⁢

x

for all

a

,

b

∈

R

and

x

,

y

∈

A

.

The universal property implies that

Ω

A

/

R

is unique up to
a unique

isomorphism

. The

A

-module

Ω

A

/

R

is often called
the

module of Kähler differentials

.

Title

universal derivation

Canonical name

UniversalDerivation

Date of creation

2013-03-22 15:27:57

Last modified on

2013-03-22 15:27:57

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

9

Author

pbruin (1001)

Entry type

Definition

Classification

msc 13N15

Classification

msc 13N05

Synonym

Kähler differentials

Related topic

Derivation

Generated on Fri Feb 9 17:34:50 2018 by

LaTeXML
a polynomial of degree n over a field has at most n roots

a polynomial of degree

n

over a field has at most

n

roots

Lemma

(cf. factor theorem).

Let

R

be a commutative ring with

identity

and let

p

⁢

(

x

)

∈

R

⁢

[

x

]

be a

polynomial

with

coefficients

in

R

. The

element

a

∈

R

is a root of

p

⁢

(

x

)

if and only if

(

x

-

a

)

divides

p

⁢

(

x

)

.

Proof.

See

proof of factor theorem using division

.
∎

Theorem

.

Let

F

be a field and let

p

⁢

(

x

)

be a non-zero polynomial in

F

⁢

[

x

]

of degree

n

≥

0

. Then

p

⁢

(

x

)

has at most

n

roots in

F

(counted with

multiplicity

).

Proof.

We proceed by

induction

. The case

n

=

0

is

trivial

since

p

⁢

(

x

)

is a non-zero

constant

, thus

p

⁢

(

x

)

cannot have any roots.

Suppose that any polynomial in

F

⁢

[

x

]

of degree

n

has at most

n

roots and let

p

⁢

(

x

)

∈

F

⁢

[

x

]

be a polynomial of degree

n

+

1

. If

p

⁢

(

x

)

has no roots then the result is trivial, so let us assume that

p

⁢

(

x

)

has at least one root

a

∈

F

. Then, by the lemma above, there exist a polynomial

q

⁢

(

x

)

such that:

p

⁢

(

x

)

=

(

x

-

a

)

⋅

q

⁢

(

x

)

.

Hence,

q

⁢

(

x

)

∈

F

⁢

[

x

]

is a polynomial of degree

n

. By the

induction hypothesis

, the polynomial

q

⁢

(

x

)

has at most

n

roots. It is clear that any root of

q

⁢

(

x

)

is a root of

p

⁢

(

x

)

and if

b

≠

a

is a root of

p

⁢

(

x

)

then

b

is also a root of

q

⁢

(

x

)

. Thus,

p

⁢

(

x

)

has at most

n

+

1

roots, which concludes the proof of the

theorem

.
∎

Note: The

fundamental theorem of algebra

states that if

F

is

algebraically closed

then any polynomial of degree

n

has exactly

n

roots (counted with multiplicity).

Title

a polynomial of degree

n

over a field has at most

n

roots

Canonical name

APolynomialOfDegreeNOverAFieldHasAtMostNRoots

Date of creation

2013-03-22 15:09:01

Last modified on

2013-03-22 15:09:01

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Theorem

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

Related topic

Root

Related topic

FactorTheorem

Related topic

PolynomialCongruence

Related topic

EveryPrimeHasAPrimitiveRoot

Related topic

CongruenceOfArbitraryDegree

Generated on Fri Feb 9 17:34:53 2018 by

LaTeXML
algebraic equation

algebraic equation

The equation

f

⁢

(

x

1

,

x

2

,

…

,

x

m

)

=

0

,

where the left hand is a

polynomial

in

x

1

,

x

2

, …,

x

m

with coefficients in a certain field, is called an

algebraic equation

over that field.  Often the field in question is

ℚ

; then the coefficients may be assumed to be

integers

.

By the

degree

of an algebraic equation is meant the degree of the polynomial.

E.g.

3

⁢

x

2

-

1

=

0

and

x

3

+

x

2

⁢

y

+

x

⁢

y

2

+

y

3

=

0

are algebraic equations over the field

ℚ

, the degrees of which are 2 and 3.

Title

algebraic equation

Canonical

name

AlgebraicEquation

Date of creation

2013-03-22 15:14:07

Last modified on

2013-03-22 15:14:07

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

7

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

Related topic

Equation

Related topic

PolynomialEquationOfOddDegree

Related topic

SymmetricQuarticEquation

Defines

degree of equation

Generated on Fri Feb 9 17:34:55 2018 by

LaTeXML
derivative of polynomial

derivative of polynomial

Let

R

be an arbitrary

commutative ring

.  If

f

⁢

(

X

)

:=

∑

i

=

1

n

a

i

⁢

X

i

is a

polynomial

in the ring

R

⁢

[

X

]

, one can form in a

polynomial ring

R

⁢

[

X

,

Y

]

the polynomial

f

⁢

(

X

+

Y

)

=

∑

i

=

1

n

a

i

⁢

(

X

+

Y

)

i

.

Expanding this by the powers (

http://planetmath.org/GeneralAssociativity

) of

Y

yields uniquely the form

f

⁢

(

X

+

Y

)

:=

f

⁢

(

X

)

+

f

1

⁢

(

X

)

⁢

Y

+

f

2

⁢

(

X

,

Y

)

⁢

Y

2

,

(1)

where

f

1

⁢

(

X

)

∈

R

⁢

[

X

]

and

f

2

⁢

(

X

,

Y

)

∈

R

⁢

[

X

,

Y

]

.

We define the polynomial

f

1

⁢

(

X

)

in (1) the

derivative

of the polynomial

f

⁢

(

X

)

and denote it by

f

′

⁢

(

X

)

or

d

⁢

f

d

⁢

X

.

It is apparent that this algebraic definition of

derivative of polynomial

is in harmony with the definition of derivative (

http://planetmath.org/Derivative2

) of

analysis

when

R

is

ℝ

or

ℂ

; then we identify

substitution homomorphism

and

polynomial function

.

It is easily shown the linearity of the derivative of polynomial and the

product rule

(

f

⁢

g

)

′

=

f

′

⁢

g

+

g

′

⁢

f

with its generalisations.  Especially:

(

X

n

)

′

=

n

⁢

X

n

-

1

for

⁢

n

=

1

,

2

,

3

,

…

Remark.

The polynomial ring

R

⁢

[

X

]

may be thought to be a

subring

of

R

⁢

[

[

X

]

]

, the ring of

formal power series

in

X

.  The derivatives defined in (

http://planetmath.org/FormalPowerSeries

)

R

⁢

[

[

X

]

]

extend the

concept

of derivative of polynomial and obey laws.

If we have a polynomial

f

∈

R

⁢

[

X

1

,

X

2

,

…

,

X

m

]

,  we can analogically define the

partial derivatives

of

f

, denoting them by

∂

⁡

f

∂

⁡

X

i

.  Then, e.g. the “Euler’s

theorem

on homogeneous functions (

http://planetmath.org/EulersTheoremOnHomogeneousFunctions

)”

X

1

⁢

∂

⁡

f

∂

⁡

X

1

+

X

2

⁢

∂

⁡

f

∂

⁡

X

2

+

…

+

X

m

⁢

∂

⁡

f

∂

⁡

X

m

=

n

⁢

f

is true for a

homogeneous polynomial

f

of degree

n

.

Title

derivative of polynomial

Canonical name

DerivativeOfPolynomial

Date of creation

2013-03-22 18:20:02

Last modified on

2013-03-22 18:20:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Definition

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

Related topic

DerivativesByPureAlgebra

Related topic

PolynomialFunction

Related topic

Multiplicity

Related topic

DiscriminantOfAlgebraicNumber

Defines

derivative of the polynomial

Generated on Fri Feb 9 17:34:57 2018 by

LaTeXML
evaluation homomorphism

evaluation homomorphism

Let

R

be a

commutative ring

and let

R

⁢

[

X

]

be the ring of polynomials with

coefficients

in

R

.

Theorem 1

.

Let

S

be a commutative ring, and let

ψ

:

R

→

S

be a

homomorphism

. Further, let

s

∈

S

. Then there is a unique homomorphism

ϕ

:

R

⁢

[

X

]

→

S

taking

X

to

s

and taking every

r

∈

R

to

ψ

⁢

(

r

)

.

This amounts to saying that

polynomial rings

are free objects in the category of

R

-

algebras

; the

theorem

then states that they are projective. This is true in much greater generality; in fact, the property of being projective is intended to extract the essential property of being free.

Proof.

We first prove

existence

. Let

f

∈

R

⁢

[

X

]

. Then by definition there is some finite list of

a

i

such that

f

=

∑

i

a

i

⁢

X

i

. Then define

ϕ

⁢

(

f

)

to be

∑

i

ψ

⁢

(

a

i

)

⁢

s

i

. It is clear from the definition of

addition

and

multiplication

on

polynomials

that

ϕ

is a homomorphism; the definition makes it clear that

ϕ

⁢

(

X

)

=

s

and

ϕ

⁢

(

r

)

=

ψ

⁢

(

r

)

.

Now, to show uniqueness, suppose

γ

is any homomorphism satisfying the conditions of the theorem, and let

f

∈

R

⁢

[

X

]

. Write

f

=

∑

i

a

i

⁢

X

i

as before. Then

γ

⁢

(

a

i

)

=

ψ

⁢

(

a

i

)

and

γ

⁢

(

s

)

by

assumption

. But then since

γ

is a homomorphism,

γ

⁢

(

a

i

⁢

X

i

)

=

ψ

⁢

(

a

i

)

⁢

s

i

and

γ

⁢

(

f

)

=

∑

i

ψ

⁢

(

a

i

)

⁢

s

i

=

ϕ

⁢

(

f

)

.
∎

Title

evaluation homomorphism

Canonical name

EvaluationHomomorphism

Date of creation

2013-03-22 14:13:51

Last modified on

2013-03-22 14:13:51

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

Synonym

substitution homomorphism

Related topic

LectureNotesOnPolynomialInterpolation

Defines

evaluation homomorphism

Generated on Fri Feb 9 17:35:00 2018 by

LaTeXML