aarjaneiro/planetmath huggingface dataset:
cyclic ring

cyclic ring

A ring is a

cyclic ring

if its

additive group

is cyclic.

Every cyclic ring is

commutative

under

multiplication

. For if

R

is a cyclic ring,

r

is a

generator

(

http://

planetmath

.org/Generator

) of the additive group of

R

, and

s

,

t

∈

R

, then there exist

a

,

b

∈

ℤ

such that

s

=

a

⁢

r

and

t

=

b

⁢

r

. As a result,

s

⁢

t

=

(

a

⁢

r

)

⁢

(

b

⁢

r

)

=

(

a

⁢

b

)

⁢

r

2

=

(

b

⁢

a

)

⁢

r

2

=

(

b

⁢

r

)

⁢

(

a

⁢

r

)

=

t

⁢

s

.

(Note the disguised use of the

distributive property

(

http://planetmath.org/Distributive

).)

A result of the fundamental theorem of finite abelian groups (

http://planetmath.org/FundamentalTheoremOfFinitelyGeneratedAbelianGroups

) is that every ring with squarefree order is a cyclic ring.

If

n

is a positive integer, then, up to

isomorphism

, there are exactly

τ

⁢

(

n

)

cyclic rings of order

n

, where

τ

refers to the tau function. Also, if a cyclic ring has order

n

, then it has exactly

τ

⁢

(

n

)

subrings

. This result mainly follows from Lagrange’s

theorem

and its

converse

. Note that the converse of Lagrange’s theorem does not hold in general, but it does hold for finite

cyclic groups

.

Every subring of a cyclic ring is a cyclic ring. Moreover, every subring of a cyclic ring is an ideal.

R

is a finite cyclic ring of order

n

if and only if there exists a positive

divisor

k

of

n

such that

R

is isomorphic to

k

⁢

ℤ

k

⁢

n

.

R

is an cyclic ring that has no

zero divisors

if and only if there exists a positive integer

k

such that

R

is isomorphic to

k

⁢

ℤ

. (See

behavior

and its attachments for details.) Finally,

R

is an cyclic ring that has zero divisors if and only if it is isomorphic to the following subset of

𝐌

2

⁢

x

⁡

2

⁢

(

ℤ

)

:

{

(

c

-

c

c

-

c

)

|

c

∈

ℤ

}

Thus, any cyclic ring that has zero divisors is a

zero ring

.

References

1

Buck, Warren.

http://planetmath.org/?op=getobj&from=papers&id=336

Cyclic Rings

. Charleston, IL: Eastern Illinois University, 2004.

2

Kruse, Robert L. and Price, David T.

Nilpotent Rings

. New York: Gordon and Breach, 1969.

3

Maurer, I. Gy. and Vincze, J. “Despre Inele Ciclice.”

Studia Universitatis Babeş-Bolyai. Series Mathematica-Physica

, vol. 9 #1. Cluj, Romania: Universitatea Babeş-Bolyai, 1964, pp. 25-27.

4

Peinado, Rolando E. “On

Finite Rings

.”

Mathematics Magazine

, vol. 40 #2. Buffalo: The

Mathematical Association of America

, 1967, pp. 83-85.

Title

cyclic ring

Canonical name

CyclicRing

Date of creation

2013-03-22 13:30:13

Last modified on

2013-03-22 13:30:13

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

33

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 13A99

Classification

msc 16U99

Related topic

CyclicGroup

Related topic

ProofOfTheConverseOfLagrangesTheoremForCyclicGroups

Related topic

CriterionForCyclicRingsToBePrincipalIdealRings

Related topic

MultiplicativeIdentityOfACyclicRingMustBeAGenerator

Generated on Fri Feb 9 17:23:59 2018 by

LaTeXML
cyclic rings of behavior one

cyclic rings of behavior one

Theorem

.

A

cyclic ring

has a

multiplicative identity

if and only if it has

behavior

one.

Proof.

For a proof that a cyclic ring with a multiplicative identity has behavior one, see this

theorem

(

http://planetmath.org/MultiplicativeIdentityOfACyclicRingMustBeAGenerator

).

Let

R

be a cyclic ring with behavior one. Let

r

be a

generator

(

http://planetmath.org/Generator

) of the

additive group

of

R

such that

r

2

=

r

. Let

s

∈

R

. Then there exists

a

∈

R

with

s

=

a

⁢

r

. Since

r

⁢

s

=

r

⁢

(

a

⁢

r

)

=

a

⁢

r

2

=

a

⁢

r

=

s

and

multiplication

in cyclic rings is

commutative

, then

r

is a multiplicative identity.
∎

Title

cyclic rings of behavior one

Canonical

name

CyclicRingsOfBehaviorOne

Date of creation

2013-03-22 16:03:10

Last modified on

2013-03-22 16:03:10

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 13A99

Classification

msc 16U99

Classification

msc 13F10

Related topic

MultiplicativeIdentityOfACyclicRingMustBeAGenerator

Related topic

CriterionForCyclicRingsToBePrincipalIdealRings

Generated on Fri Feb 9 17:24:04 2018 by

LaTeXML
cyclic rings that are isomorphic to k⁢ℤ

cyclic rings that are isomorphic to

k

⁢

ℤ

Corollary

.

An

infinite cyclic

ring (

http://planetmath.org/CyclicRing3

) with positive

behavior

k

is

isomorphic

to

k

⁢

Z

.

Proof.

Note that

k

⁢

ℤ

is an and that

k

is a

generator

(

http://planetmath.org/Generator

) of its

additive group

. Since

k

2

=

k

⁢

(

k

)

, then

k

⁢

ℤ

has behavior

k

.
∎

Title

cyclic rings

that are isomorphic to

k

⁢

ℤ

Canonical name

CyclicRingsThatAreIsomorphicToKmathbbZ

Date of creation

2013-03-22 16:02:42

Last modified on

2013-03-22 16:02:42

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Corollary

Classification

msc 13A99

Classification

msc 16U99

Related topic

MathbbZ

Generated on Fri Feb 9 17:24:08 2018 by

LaTeXML
Lagrange’s identity

Lagrange’s identity

Let

R

be a commutative ring, and let

x

1

,

…

,

x

n

,

y

1

,

…

,

y

n

be arbitrary

elements

in

R

. Then

(

∑

k

=

1

n

x

k

⁢

y

k

)

2

=

(

∑

k

=

1

n

x

k

2

)

⁢

(

∑

k

=

1

n

y

k

2

)

-

∑

1

≤

k

<

i

≤

n

(

x

k

⁢

y

i

-

x

i

⁢

y

k

)

2

⁢

.

Proof.

Since

R

is

commutative

, we can apply the binomial formula.We start out with

(

∑

i

=

1

n

x

i

⁢

y

i

)

2

=

∑

i

=

1

n

(

x

i

2

⁢

y

i

2

)

+

∑

1

≤

i

<

j

≤

n

2

⁢

x

i

⁢

y

j

⁢

x

j

⁢

y

i

(1)

Using the binomial formula, we see that

(

x

i

⁢

y

j

-

x

j

⁢

y

i

)

2

=

x

i

2

⁢

y

j

2

-

2

⁢

x

i

⁢

x

j

⁢

y

i

⁢

y

j

+

x

j

2

⁢

y

i

2

.

So we get

(

∑

i

=

1

n

x

i

⁢

y

i

)

2

+

∑

1

≤

i

<

j

≤

n

n

(

x

i

⁢

y

j

-

x

j

⁢

y

i

)

2

=

∑

i

=

1

n

(

x

i

2

⁢

y

i

2

)

+

∑

1

≤

i

<

j

≤

n

n

(

x

i

2

⁢

y

j

2

+

x

j

2

⁢

y

i

2

)

(2)

=

(

∑

i

=

1

n

x

i

2

)

⁢

(

∑

i

=

1

n

y

i

2

)

(3)

Note that changing the roles of

i

and

j

in

x

i

⁢

y

j

-

x

j

⁢

y

i

, we get

x

j

⁢

y

i

-

x

i

⁢

y

j

=

-

(

x

i

⁢

y

j

-

x

j

⁢

y

i

)

,

but the

negative

sign will disappear when we square. So we can rewrite the last

equation

to

(

∑

i

=

1

n

x

i

⁢

y

i

)

2

+

∑

1

≤

i

<

j

≤

n

(

x

i

⁢

y

j

-

x

j

⁢

y

i

)

2

=

(

∑

i

=

1

n

x

i

2

)

⁢

(

∑

i

=

1

n

y

i

2

)

.

(4)

This is

equivalent

to the stated

identity

.
∎

Title

Lagrange’s identity

Canonical name

LagrangesIdentity

Date of creation

2013-03-22 13:18:01

Last modified on

2013-03-22 13:18:01

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

21

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 13A99

Generated on Fri Feb 9 17:24:14 2018 by

LaTeXML
Marot ring

Marot ring

A

Marot ring

is any

commutative ring

with

non-zero unity

, where every

regular ideal

may be generated by

regular elements

.

Note.

ℤ

m

is trivially a Marot ring (see the entry “regular ideal (

http://planetmath.org/RegularIdeal

)”); an

integral domain

is another example because it has no

zero divisors

.

Title

Marot ring

Canonical name

MarotRing

Date of creation

2013-03-22 15:43:03

Last modified on

2013-03-22 15:43:03

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

8

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 13A99

Generated on Fri Feb 9 17:24:17 2018 by

LaTeXML
polynomial function

polynomial function

Definition.

Let

R

be a

commutative ring

.  A

function

f

:

R

→

R

is called a

polynomial function

of

R

, if there are some

elements

a

0

,

a

1

,

…

,

a

m

of

R

such that

f

⁢

(

x

)

=

a

0

+

a

1

⁢

x

+

⋯

+

a

m

⁢

x

m

⁢

∀

x

∈

R

.

Remark.

The

coefficients

a

i

in a polynomial function need not be unique; e.g. if

R

=

{

0

,

1

}

is the ring (and field) of two elements, then the

polynomials

X

and

X

2

both may be used for the same polynomial function.  However, if we stipulate that

R

is an

infinite

integral domain

, the coefficients are guaranteed to be unique.

The set of all polynomial functions of

R

, being a subset of the set

R

R

of all functions from

R

to

R

, is here denoted by

R

/

R

.

Theorem

.

If

R

is a commutative ring, then the set

R

/

R

of all polynomial functions of

R

, equipped with the

operations

(

f

+

g

)

⁢

(

x

)

:=

f

⁢

(

x

)

+

g

⁢

(

x

)

,

(

f

⋅

g

)

⁢

(

x

)

:=

f

⁢

(

x

)

⁢

g

⁢

(

x

)

∀

x

∈

R

,

(1)

is a commutative ring.

Proof.

It’s straightforward to show that the function set

R

R

forms a commutative ring when equipped with the operations “

+

” and “

⋅

” defined as (1).  We show now that

R

/

R

forms a

subring

of

R

R

.  Let

f

and

g

be any two polynomial functions given by

f

⁢

(

x

)

=

a

0

+

a

1

⁢

x

+

⋯

+

a

m

⁢

x

m

,

g

⁢

(

x

)

=

b

0

+

b

1

⁢

x

+

⋯

+

b

n

⁢

x

n

.

Then we can give

f

+

g

by

(

f

+

g

)

⁢

(

x

)

=

∑

i

=

0

k

(

a

i

+

b

i

)

⁢

x

i

where

k

=

max

⁡

{

m

,

n

}

and

a

i

=

0

(resp.

b

i

=

0

) for

i

>

m

(resp.

i

>

n

).  This means that

f

+

g

∈

R

/

R

.  Secondly, the equation

(

f

⋅

g

)

⁢

(

x

)

=

a

0

⁢

b

0

+

(

a

0

⁢

b

1

+

a

1

⁢

b

0

)

⁢

x

+

(

a

0

⁢

b

2

+

a

1

⁢

b

1

+

a

2

⁢

b

0

)

⁢

x

2

+

⋯

+

a

m

⁢

b

n

⁢

x

m

+

n

signifies that

f

⋅

g

∈

R

/

R

.  Because also the function

-

f

given by

(

-

f

)

⁢

(

x

)

=

-

a

0

-

a

1

⁢

x

-

⋯

-

a

m

⁢

x

m

and satisfying

-

f

+

f

=

0

:

x

↦

0

belongs to

R

/

R

, the subset

R

/

R

is a subring of

R

R

.

Title

polynomial function

Canonical name

PolynomialFunction

Date of creation

2013-03-22 15:40:34

Last modified on

2013-03-22 15:40:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Definition

Classification

msc 13A99

Synonym

ring of polynomial functions

Related topic

NotationInSetTheory

Related topic

ProductAndQuotientOfFunctionsSum

Related topic

ZeroOfPolynomial

Related topic

PolynomialFunctionIsAProperMap

Related topic

DerivativeOfPolynomial

Defines

polynomial function

Generated on Fri Feb 9 17:24:22 2018 by

LaTeXML
proof of Euler four-square identity

proof of Euler four-square identity

Using Lagrange’s

identity

, we have

(

∑

k

=

1

4

x

k

⁢

y

k

)

2

=

(

∑

k

=

1

4

x

k

2

)

⁢

(

∑

k

=

1

4

y

k

2

)

-

∑

1

≤

k

<

i

≤

4

(

x

k

⁢

y

i

-

x

i

⁢

y

k

)

2

⁢

.

(1)

We group the six squares into 3 groups of two squares and rewrite:

(

x

1

⁢

y

2

-

x

2

⁢

y

1

)

2

+

(

x

3

⁢

y

4

-

x

4

⁢

y

3

)

2

(2)

=

(

(

x

1

⁢

y

2

-

x

2

⁢

y

1

)

+

(

x

3

⁢

y

4

-

x

4

⁢

y

3

)

)

2

-

2

⁢

(

(

x

1

⁢

y

2

-

x

2

⁢

y

1

)

⁢

(

x

3

⁢

y

4

-

x

4

⁢

y

3

)

)

(3)

(

x

1

⁢

y

3

-

x

3

⁢

y

1

)

2

+

(

x

2

⁢

y

4

-

x

4

⁢

y

2

)

2

=

(

(

x

1

⁢

y

3

-

x

3

⁢

y

1

)

-

(

x

2

⁢

y

4

-

x

4

⁢

y

2

)

)

2

+

2

⁢

(

x

1

⁢

y

3

-

x

3

⁢

y

1

)

⁢

(

x

2

⁢

y

4

-

x

4

⁢

y

2

)

(4)

(

x

1

⁢

y

4

-

x

4

⁢

y

1

)

2

+

(

x

2

⁢

y

3

-

x

3

⁢

y

2

)

2

=

(

(

x

1

⁢

y

4

-

x

4

⁢

y

1

)

+

(

x

2

⁢

y

3

-

x

3

⁢

y

2

)

)

2

-

2

⁢

(

x

1

⁢

y

4

-

x

4

⁢

y

1

)

⁢

(

x

2

⁢

y

3

-

x

3

⁢

y

2

)

⁢

.

(5)

Using

-

2

⁢

(

(

x

1

⁢

y

2

-

x

2

⁢

y

1

)

⁢

(

x

3

⁢

y

4

-

x

4

⁢

y

3

)

)

+

2

⁢

(

x

1

⁢

y

3

-

x

3

⁢

y

1

)

⁢

(

x

2

⁢

y

4

-

x

4

⁢

y

2

)

(6)

-

2

⁢

(

x

1

⁢

y

4

-

x

4

⁢

y

1

)

⁢

(

x

2

⁢

y

3

-

x

3

⁢

y

2

)

=

0

we get

∑

1

≤

k

<

i

≤

4

(

x

k

⁢

y

i

-

x

i

⁢

y

k

)

2

=

(

(

x

1

y

2

-

x

2

y

1

)

+

(

x

3

y

4

-

x

4

y

3

)

)

2

(7)

+

(

(

x

1

⁢

y

3

-

x

3

⁢

y

1

)

-

(

x

2

⁢

y

4

-

x

4

⁢

y

2

)

)

2

(8)

+

(

(

x

1

⁢

y

4

-

x

4

⁢

y

1

)

+

(

x

2

⁢

y

3

-

x

3

⁢

y

2

)

)

2

by adding equations

2

-

4

. We put the result of equation

7

into

1

and get

(

∑

k

=

1

4

x

k

⁢

y

k

)

2

(9)

=

(

∑

k

=

1

4

x

k

2

)

⁢

(

∑

k

=

1

4

y

k

2

)

-

(

(

x

1

y

2

-

x

2

y

1

+

x

3

y

4

-

x

4

y

3

)

2

-

(

x

1

⁢

y

3

-

x

3

⁢

y

1

+

x

4

⁢

y

2

-

x

2

⁢

y

4

)

2

-

(

x

1

⁢

y

4

-

x

4

⁢

y

1

+

x

2

⁢

y

3

-

x

3

⁢

y

2

)

2

which is

equivalent

to the claimed identity.

Title

proof of Euler four-square identity

Canonical name

ProofOfEulerFoursquareIdentity

Date of creation

2013-03-22 13:18:10

Last modified on

2013-03-22 13:18:10

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

7

Author

Thomas Heye (1234)

Entry type

Proof

Classification

msc 13A99

Generated on Fri Feb 9 17:24:38 2018 by

LaTeXML
proof that every subring of a cyclic ring is a cyclic ring

proof that every subring of a cyclic ring is a cyclic ring

The following is a proof that every subring of a cyclic ring is a cyclic ring.

Proof.

Let

R

be a cyclic ring and

S

be a subring of

R

. Then the

additive group

of

S

is a

subgroup

of the additive group of

R

. By definition of

cyclic group

, the additive group of

R

is cyclic (

http://planetmath.org/CyclicGroup

). Thus, the additive group of

S

is cyclic. It follows that

S

is a cyclic ring.
∎

Title

proof that every subring of a cyclic ring is a cyclic ring

Canonical name

ProofThatEverySubringOfACyclicRingIsACyclicRing

Date of creation

2013-03-22 13:30:50

Last modified on

2013-03-22 13:30:50

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

8

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 13A99

Classification

msc 16U99

Related topic

ProofThatAllSubgroupsOfACyclicGroupAreCyclic

Related topic

ProofThatEverySubringOfACyclicRingIsAnIdeal

Generated on Fri Feb 9 17:24:43 2018 by

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proof that every subring of a cyclic ring is an ideal

proof that every subring of a cyclic ring is an ideal

The following is a proof that every subring of a cyclic ring is an ideal.

Proof.

Let

R

be a cyclic ring and

S

be a subring of

R

. Then

R

and

S

are both cyclic rings. Let

r

be a

generator

(

http://planetmath.org/Generator

) of the

additive group

of

R

and

s

be a generator of the additive group of

S

. Then

s

∈

R

. Thus, there exists

z

∈

ℤ

with

s

=

z

⁢

r

.

Let

t

∈

R

and

u

∈

S

. Then

u

∈

R

. Since

multiplication

is

commutative

in a cyclic ring,

t

⁢

u

=

u

⁢

t

. Since

t

∈

R

, there exists

a

∈

ℤ

with

t

=

a

⁢

r

. Since

u

∈

S

, there exists

b

∈

ℤ

with

u

=

b

⁢

s

.

Since

R

is a ring,

r

2

∈

R

. Thus, there exists

k

∈

ℤ

with

r

2

=

k

⁢

r

. Since

t

⁢

u

=

(

a

⁢

r

)

⁢

(

b

⁢

s

)

=

(

a

⁢

r

)

⁢

[

b

⁢

(

z

⁢

r

)

]

=

(

a

⁢

b

⁢

z

)

⁢

r

2

=

(

a

⁢

b

⁢

z

)

⁢

(

k

⁢

r

)

=

(

a

⁢

b

⁢

k

⁢

z

)

⁢

r

=

(

a

⁢

b

⁢

k

)

⁢

(

z

⁢

r

)

=

(

a

⁢

b

⁢

k

)

⁢

s

∈

S

, it follows that

S

is an ideal of

R

.
∎

References

1

Buck, Warren.

http://planetmath.org/?op=getobj&from=papers&id=336

Cyclic Rings

. Charleston, IL: Eastern Illinois University, 2004.

2

Maurer, I. Gy. and Vincze, J. “Despre Inele Ciclece.”

Studia Universitatis Babeş-Bolyai. Series Mathematica-Physica

, vol. 9 #1. Cluj, Romania: Universitatea Babeş-Bolyai, 1964, pp. 25-27.

Title

proof that every subring of a cyclic ring is an ideal

Canonical

name

ProofThatEverySubringOfACyclicRingIsAnIdeal

Date of creation

2013-03-22 13:30:52

Last modified on

2013-03-22 13:30:52

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 13A99

Classification

msc 16U99

Related topic

ProofThatEverySubringOfACyclicRingIsACyclicRing

Generated on Fri Feb 9 17:24:47 2018 by

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symmetric multilinear function

symmetric multilinear function

Let

R

be a

commutative ring

with

identity

and

M

,

N

be unital

R

-modules.

Suppose that

ϕ

:

M

×

⋯

×

M

→

N

is a multilinear map, where there
are

n

copies of

M

.

Let

H

be a

subgroup

of

S

n

, the

symmetric group

on

{

1

,

…

,

n

}

, and

χ

:

H

→

R

satisfy

1.

χ

⁢

(

e

)

=

1

2.

χ

⁢

(

g

1

⁢

g

2

)

=

χ

⁢

(

g

1

)

⁢

χ

⁢

(

g

2

)

for all

g

1

,

g

2

∈

H

We say that

ϕ

is

symmetric

with respect to

H

and

χ

if

ϕ

⁢

(

m

σ

⁢

(

1

)

,

…

,

m

σ

⁢

(

n

)

)

=

χ

⁢

(

σ

)

⁢

ϕ

⁢

(

m

1

,

…

,

m

n

)

holds for all

σ

∈

H

and all

m

i

∈

M

.

Now suppose that

H

=

S

n

.

If

χ

=

1

then we say that

ϕ

is a

symmetric multilinear function

.

If

χ

=

ϵ

, the sign of the

permutation

σ

, we say that

ϕ

is a

skew-symmetric multilinear function

.

For example, the permanent is a symmetric multilinear function of its rows (columns).

The determinant is a skew-symmetric multilinear function of its rows (columns).

Title

symmetric multilinear function

Canonical name

SymmetricMultilinearFunction

Date of creation

2013-03-22 16:10:53

Last modified on

2013-03-22 16:10:53

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

11

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 13A99

Defines

skew-symmetric multilinear function

Generated on Fri Feb 9 17:24:51 2018 by

LaTeXML
algebraic

algebraic

Let

B

be a ring with a subring

A

. An element

x

∈

B

is

algebraic

over

A

if there exist elements

a

1

,

…

,

a

n

∈

A

, with

a

n

≠

0

, such that

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

⋯

+

a

1

⁢

x

+

a

0

=

0

.

An element

x

∈

B

is

transcendental

over

A

if it is not algebraic.

The ring

B

is

algebraic

over

A

if every element of

B

is algebraic over

A

.

Title

algebraic

Canonical name

Algebraic1

Date of creation

2013-03-22 12:07:50

Last modified on

2013-03-22 12:07:50

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Definition

Classification

msc 13B02

Related topic

AlgebraicExtension

Defines

transcendental

Generated on Fri Feb 9 17:24:54 2018 by

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decomposition group

decomposition group

1

Decomposition Group

Let

A

be a

Noetherian

integrally closed

integral domain

with

field
of fractions

K

. Let

L

be a

Galois extension

of

K

and denote by

B

the

integral closure

of

A

in

L

. Then, for any

prime ideal

𝔭

⊂

A

, the

Galois group

G

:=

Gal

⁡

(

L

/

K

)

acts transitively on the
set of all prime ideals

𝔓

⊂

B

containing

𝔭

. If we fix a
particular prime ideal

𝔓

⊂

B

lying over

𝔭

, then the

stabilizer

of

𝔓

under this

group action

is a

subgroup

of

G

, called the

decomposition group

at

𝔓

and denoted

D

⁢

(

𝔓

/

𝔭

)

. In other words,

D

⁢

(

𝔓

/

𝔭

)

:=

{

σ

∈

G

∣

σ

⁢

(

𝔓

)

=

(

𝔓

)

}

.

If

𝔓

′

⊂

B

is another prime ideal of

B

lying over

𝔭

, then
the decomposition groups

D

⁢

(

𝔓

/

𝔭

)

and

D

⁢

(

𝔓

′

/

𝔭

)

are

conjugate

in

G

via any Galois

automorphism

mapping

𝔓

to

𝔓

′

.

2

Inertia Group

Write

l

for the

residue field

B

/

𝔓

and

k

for the residue field

A

/

𝔭

. Assume that the

extension

l

/

k

is

separable

(if it is not,
then this development is still possible, but considerably more
complicated; see

[

serre

, p. 20]

). Any

element

σ

∈

D

⁢

(

𝔓

/

𝔭

)

, by definition, fixes

𝔓

and hence descends to a well
defined automorphism of the field

l

. Since

σ

also fixes

A

by virtue of being in

G

, it induces an automorphism of the extension

l

/

k

fixing

k

. We therefore have a

group homomorphism

D

⁢

(

𝔓

/

𝔭

)

⟶

Gal

⁡

(

l

/

k

)

,

and the kernel (

http://planetmath.org/KernelOfAGroupHomomorphism

) of this

homomorphism

is called the

inertia group

of

𝔓

, and written

T

⁢

(

𝔓

/

𝔭

)

. It turns out that this homomorphism is
actually

surjective

, so there is an

exact sequence

(1)

Generated on Fri Feb 9 17:25:04 2018 by

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module-finite

module-finite

Let

S

be a ring with

subring

R

.

We say that

S

is

module-finite

over

R

if

S

is

finitely generated

as an

R

-module.

We say that

S

is

ring-finite

over

R

if

S

=

R

⁢

[

v

1

,

…

,

v

n

]

for some

v

1

,

…

,

v

n

∈

S

.

Note that module-finite implies ring-finite, but the converse is false.

If

L

is ring-finite over

K

, with

L

,

K

fields, then

L

is a

finite extension

of

K

.

Title

module-finite

Canonical name

Modulefinite

Date of creation

2013-03-22 12:36:56

Last modified on

2013-03-22 12:36:56

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

6

Author

yark (2760)

Entry type

Definition

Classification

msc 13B02

Classification

msc 13C05

Classification

msc 16D10

Related topic

FinitelyGeneratedRModule

Defines

ring-finite

Generated on Fri Feb 9 17:25:08 2018 by

LaTeXML
algebraic

algebraic

Let

K

be an

extension field

of

F

and let

a

∈

K

.

If there is a nonzero

polynomial

f

∈

F

⁢

[

x

]

such that

f

⁢

(

a

)

=

0

(in

K

) we say that

a

is

algebraic over

F

.

For example,

2

∈

ℝ

is algebraic over

ℚ

since there is a nonzero polynomial with

rational

coefficients, namely

f

⁢

(

x

)

=

x

2

-

2

, such that

f

⁢

(

2

)

=

0

.

If all elements of

K

are algebraic over

F

, one says that
the

field extension

K

/

F

is

algebraic

.

Title

algebraic

Canonical name

Algebraic

Date of creation

2013-11-05 18:32:06

Last modified on

2013-11-05 18:32:06

Owner

drini (3)

Last modified by

pahio (2872)

Numerical id

8

Author

drini (2872)

Entry type

Definition

Classification

msc 13B05

Classification

msc 11R04

Classification

msc 11R32

Related topic

AlgebraicNumber

Related topic

FiniteExtension

Related topic

ProofOfTranscendentalRootTheorem

Generated on Fri Feb 9 17:25:11 2018 by

LaTeXML
anti-isomorphism

anti-isomorphism

Let

R

and

S

be rings and

f

:

R

⟶

S

be a function such
that

f

⁢

(

r

1

⁢

r

2

)

=

f

⁢

(

r

2

)

⁢

f

⁢

(

r

1

)

for all

r

1

,

r

2

∈

R

.

If

f

is a

homomorphism

of the

additive groups

of

R

and

S

,
then

f

is called an

anti-homomorphsim

.

If

f

is a bijection and

anti-homomorphism

,
then

f

is called an

anti-isomorphism

.

If

f

is an anti-homomorphism and

R

=

S

then

f

is called an

anti-endomorphism

.

If

f

is an anti-isomorphism and

R

=

S

then

f

is called an

anti-automorphism

.

As an example, when

m

≠

n

, the mapping that sends a matrix to its

transpose

(or to its

conjugate transpose

if the matrix is

complex

) is an anti-isomorphism
of

M

m

,

n

→

M

n

,

m

.

R

and

S

are

anti-isomorphic

if there is an anti-isomorphism

R

→

S

.

All of the things defined in this entry are also defined for groups.

Title

anti-isomorphism

Canonical name

Antiisomorphism

Date of creation

2013-03-22 16:01:08

Last modified on

2013-03-22 16:01:08

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

15

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 13B10

Classification

msc 16B99

Defines

anti-endomorphism

Defines

anti-homomorphism

Defines

anti-isomorphic

Defines

anti-automorphism

Generated on Fri Feb 9 17:25:14 2018 by

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