aarjaneiro/planetmath huggingface dataset:
example of PID

example of PID

Important examples of

principal ideal domains

:

•

The ring of the

integers

ℤ

.

•

The ring of polynomials in one variable over a field, i.e. a ring of
the form

𝔽

⁢

[

X

]

, where

𝔽

is a field. Note that the ring of polynomials in more than one variable over a field is never a PID.

Both of these examples are actually examples of

Euclidean rings

, which are
always PIDs. There are, however, more complicated

examples of PIDs

which are
not Euclidean rings.

Title

example of PID

Canonical name

ExampleOfPID

Date of creation

2013-03-22 13:33:56

Last modified on

2013-03-22 13:33:56

Owner

sleske (997)

Last modified by

sleske (997)

Numerical id

5

Author

sleske (997)

Entry type

Example

Classification

msc 13G05

Generated on Fri Feb 9 17:32:38 2018 by

LaTeXML
example of ring which is not a UFD

example of ring which is not a UFD

Example 1

.

We define a ring

R

=

ℤ

⁢

[

-

5

]

=

{

n

+

m

⁢

-

5

:

n

,

m

∈

ℤ

}

with

addition

and

multiplication

inherited from

ℂ

(notice that

R

is the

ring of integers

of the

quadratic number field

ℚ

⁢

(

-

5

)

). Notice that the only units (

http://planetmath.org/UnitsOfQuadraticFields

) of

R

are

R

×

=

{

±

1

}

. Then:

6

=

2

⋅

3

=

(

1

+

-

5

)

⋅

(

1

-

-

5

)

.

(1)

Moreover,

2

,

3

,

1

+

-

5

and

1

-

-

5

are

irreducible elements

of

R

and they are not

associates

(to see this, one can compare the norm of every element). Therefore,

R

is not a UFD.

However, the ideals of

R

factor (

http://planetmath.org/DivisibilityInRings

) uniquely into

prime ideals

. For example:

(

6

)

=

(

2

,

1

+

-

5

)

2

⋅

(

3

,

1

+

-

5

)

⋅

(

3

,

1

-

-

5

)

where

𝔓

=

(

2

,

1

+

-

5

)

,

𝔔

=

(

3

,

1

+

-

5

)

, and

𝔔

¯

=

(

3

,

1

-

-

5

)

are all prime ideals (see prime ideal decomposition of quadratic extensions of

ℚ

(

http://planetmath.org/PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ

)). Notice that:

𝔓

2

=

(

2

)

,

𝔔

⋅

𝔔

¯

=

(

3

)

,

𝔓

⋅

𝔔

=

(

1

+

-

5

)

,

𝔓

⋅

𝔔

¯

=

(

1

-

-

5

)

.

Thus, Eq. (

1

) above is the outcome of different rearrangements of the product of prime ideals:

(

6

)

=

𝔓

2

⋅

(

𝔔

⋅

𝔔

¯

)

=

(

𝔓

⋅

𝔔

)

⋅

(

𝔓

⋅

𝔔

¯

)

.

Notice also that if

𝔓

was a

principal ideal

then there would be an element

α

∈

R

with

(

α

)

=

𝔓

and

(

α

)

2

=

(

2

)

. Thus such a number

α

would have norm

2

, but the norm of

n

+

m

⁢

-

5

is

n

2

+

5

⁢

m

2

so it is clear that there are no

algebraic integers

of norm

2

. Therefore

𝔓

is not principal. Thus

R

is not a PID.

Title

example of ring which is not a UFD

Canonical name

ExampleOfRingWhichIsNotAUFD

Date of creation

2013-03-22 15:08:19

Last modified on

2013-03-22 15:08:19

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Example

Classification

msc 13G05

Synonym

example of a ring of integers which is not a UFD

Related topic

DeterminingTheContinuationsOfExponent

Defines

example of a

number ring

which is not a UFD

Generated on Fri Feb 9 17:32:40 2018 by

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finite ring has no proper overrings

finite ring has no proper overrings

The

regular elements

of a finite

commutative ring

R

are the units of the ring (see the parent (

http://planetmath.org/NonZeroDivisorsOfFiniteRing

) of this entry).  Generally, the largest

overring

of

R

, the

total ring of fractions

T

, is obtained by forming

S

-

1

⁢

R

, the

extension by localization

, using as the

multiplicative set

S

the set of all regular elements, which in this case is the

unit group

of

R

.  The ring

R

may be considered as a

subring

of

T

, which consists formally of the fractions

a

s

=

a

⁢

s

-

1

with

a

∈

R

and

s

∈

S

.  Since every

s

has its own group inverse

s

-

1

in

S

and so in

R

, it’s evident that

T

no other

elements

than the elements of

R

.  Consequently,

T

=

R

,  and therefore also any overring of

R

coincides with

R

.

Accordingly, one can not extend a finite commutative ring by using a

localization

.  Possible

extensions

must be made via some kind of adjunction (

http://planetmath.org/RingAdjunction

).  A more known special case is a finite

integral domain

(

http://planetmath.org/AFiniteIntegralDomainIsAField

) — it is always a field and thus

closed under

the divisions.

Title

finite ring has no proper overrings

Canonical name

FiniteRingHasNoProperOverrings

Date of creation

2013-03-22 15:11:12

Last modified on

2013-03-22 15:11:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Result

Classification

msc 13G05

Related topic

ExtensionByLocalization

Related topic

ClassicalRingOfQuotients

Related topic

AFiniteIntegralDomainIsAField

Related topic

RingAdjunction

Related topic

FormalPowerSeries

Generated on Fri Feb 9 17:32:42 2018 by

LaTeXML
half-factorial ring

half-factorial ring

An

integral domain

D

is called a

half-factorial ring

(HFD) if it satisfies the following conditions:

•

Every nonzero element of

D

that is not a unit can be factored into a product of a finite number of

irreducibles

.

•

If

p

1

⁢

p

2

⁢

⋯

⁢

p

m

and

q

1

⁢

q

2

⁢

⋯

⁢

q

n

are two factorizations of the same element

a

into irreducibles, then

m

=

n

.

If, in , the irreducibles

p

i

and

q

j

are always pairwise

associates

, then

D

is a

factorial ring

(UFD).

For example, many orders (

http://planetmath.org/OrderInAnAlgebra

) in the

maximal order

of an

algebraic number field

are half-factorial rings, e.g.

ℤ

⁢

[

3

⁢

2

]

is a HFD but not a UFD (see

http://www.math.ndsu.nodak.edu/faculty/coykenda/paper6b.pdf

this paper).

Title

half-factorial ring

Canonical

name

HalffactorialRing

Date of creation

2013-03-22 18:31:14

Last modified on

2013-03-22 18:31:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Definition

Classification

msc 13G05

Synonym

half-factorial domain

Defines

HFD

Generated on Fri Feb 9 17:32:47 2018 by

LaTeXML
integral domain

integral domain

An

integral domain

, or

domain

, is a

commutative

cancellation ring

with an

identity element

1

≠

0

.

Integral domains are sometimes allowed to be noncommutative, but we adopt the convention that an integral domain is commutative unless otherwise specified.

This notion has essentially nothing to do with the domain of a function (

http://

planetmath

.org/Domain

). It is also not very closely related to the notion of

integral

, which is applied to ring

elements

, or that of

integral closure

, which is applied to

extensions

of rings, although these concepts are normally applied to integral domains. An integral domain shares some of the properties of the

integers

(more than other kinds of rings, but by no means all those of interest). Integral domains have

fraction fields

, which play the role of the

rational numbers

, and they each have a

characteristic

(which is either a

prime number

or zero).

Title

integral domain

Canonical name

IntegralDomain

Date of creation

2013-03-22 11:50:24

Last modified on

2013-03-22 11:50:24

Owner

djao (24)

Last modified by

djao (24)

Numerical id

16

Author

djao (24)

Entry type

Definition

Classification

msc 13G05

Synonym

domain

Related topic

CancellationRing

Related topic

ZeroDivisor

Related topic

WhyEuclideanDomains

Generated on Fri Feb 9 17:32:48 2018 by

LaTeXML
irreducible

irreducible

Let

r

be a non-unit of an

integral domain

D

.  We say that

r

is

irreducible

in

D

, if any

factorization

r

=

a

⁢

b

in

D

requires that

a

or

b

is a unit.

Title

irreducible

Canonical name

Irreducible

Date of creation

2013-03-22 11:56:13

Last modified on

2013-03-22 11:56:13

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Definition

Classification

msc 13G05

Related topic

UFD

Related topic

DivisibilityInRings

Related topic

PID

Related topic

IrreduciblePolynomial2

Related topic

IrreducibleIdeal

Defines

irreducible element

Generated on Fri Feb 9 17:32:50 2018 by

LaTeXML
irreducible of a UFD is prime

irreducible of a UFD is prime

Any

irreducible element

of a

factorial ring

D

is a

prime element

of

D

.

Proof.

Let

p

be an arbitrary irreducible element of

D

.  Thus

p

is a non-unit.  If

a

⁢

b

∈

(

p

)

∖

{

0

}

,  then

a

⁢

b

=

c

⁢

p

with

c

∈

D

.  We write

a

,

b

,

c

as products of

irreducibles

:

a

=

p

1

⁢

⋯

⁢

p

l

,

b

=

q

1

⁢

⋯

⁢

q

m

,

c

=

r

1

⁢

⋯

⁢

r

n

Here, one of those first two products may me empty, i.e. it may be a unit.  We have

p

1

⁢

⋯

⁢

p

l

⁢

q

1

⁢

⋯

⁢

q

m

=

r

1

⁢

⋯

⁢

r

n

⁢

p

.

(1)

Due to the uniqueness of

prime factorization

, every factor

r

k

is an

associate

of certain of the

l

+

m

irreducibles on the

left hand side

of (1).  Accordingly,

p

has to be an associate of one of the

p

i

’s or

q

j

’s.  It means that either

a

∈

(

p

)

or

b

∈

(

p

)

.  Thus,

(

p

)

is a

prime ideal

of

D

, and its

generator

must be a prime element.

Title

irreducible of a UFD is prime

Canonical name

IrreducibleOfAUFDIsPrime

Date of creation

2013-03-22 18:04:35

Last modified on

2013-03-22 18:04:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13G05

Classification

msc 13F15

Related topic

PrimeElementIsIrreducibleInIntegralDomain

Generated on Fri Feb 9 17:32:52 2018 by

LaTeXML
orders of elements in integral domain

orders of elements in integral domain

Theorem

.

Let

(

D

,

+

,

⋅

)

be an

integral domain

, i.e. a

commutative ring

with

non-zero unity

1 and no

zero divisors

.  All non-zero elements of

D

have the same order (

http://planetmath.org/OrderGroup

) in the

additive group

(

D

,

+

)

.

Proof.

Let

a

be arbitrary non-zero element.  Any

multiple

(

http://planetmath.org/GeneralAssociativity

)

n

⁢

a

may be written as

n

⁢

a

=

n

⁢

(

1

⁢

a

)

=

1

⁢

a

+

1

⁢

a

+

⋯

+

1

⁢

a

⏟

n

=

(

1

+

1

+

⋯

+

1

⏟

n

)

⁢

a

=

(

n

⁢

1

)

⁢

a

.

Thus, because

a

≠

0

and there are no zero divisors, an

equation

n

⁢

a

=

0

is equivalent (

http://planetmath.org/Equivalent3

) with the equation

n

⁢

1

=

0

.  So

a

must have the same as the unity of

D

.

Note.

The of the unity element is the

characteristic

(

http://planetmath.org/Characteristic

) of the integral domain, which is 0 or a

positive

prime number

.

Title

orders of elements in integral domain

Canonical name

OrdersOfElementsInIntegralDomain

Date of creation

2013-03-22 15:40:28

Last modified on

2013-03-22 15:40:28

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13G05

Related topic

OrderGroup

Related topic

IdealOfElementsWithFiniteOrder

Generated on Fri Feb 9 17:32:56 2018 by

LaTeXML
prime element is irreducible in integral domain

prime element is irreducible in integral domain

Theorem

.

Every

prime element

of an

integral domain

is

irreducible

.

Proof.

Let

D

be an integral domain, and let

a

∈

D

be a prime element. Assume

a

=

b

⁢

c

for some

b

,

c

∈

D

.

Clearly

a

∣

b

c

, so since

a

is prime,

a

∣

b

or

a

∣

c

. Without loss of generality, assume

a

∣

b

, and say

a

⁢

t

=

b

for some

t

∈

D

.

If

1

is the unity of

D

, then

1

⁢

b

=

b

=

a

⁢

t

=

(

b

⁢

c

)

⁢

t

=

b

⁢

(

c

⁢

t

)

.

Since

D

is an integral domain,

b

can be cancelled, giving

1

=

c

⁢

t

, so

c

is a unit.
∎

Title

prime element is irreducible in integral domain

Canonical name

PrimeElementIsIrreducibleInIntegralDomain

Date of creation

2013-03-22 17:15:29

Last modified on

2013-03-22 17:15:29

Owner

me_and (17092)

Last modified by

me_and (17092)

Numerical id

7

Author

me_and (17092)

Entry type

Theorem

Classification

msc 13G05

Related topic

IrreducibleOfAUFDIsPrime

Generated on Fri Feb 9 17:32:58 2018 by

LaTeXML
proof that a gcd domain is integrally closed

proof that a gcd domain is integrally closed

Proposition 1

.

Every

gcd domain

is

integrally closed

.

Proof.

Let

D

be a gcd domain. For any

a

,

b

∈

D

, let

GCD

⁡

(

a

,

b

)

be the collection of all gcd’s of

a

and

b

. For this proof, we need two facts:

1.

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

=

m

⁢

GCD

⁡

(

a

,

b

)

.

2.

If

GCD

⁡

(

a

,

b

)

=

[

1

]

and

GCD

⁡

(

a

,

c

)

=

[

1

]

, then

GCD

⁡

(

a

,

b

⁢

c

)

=

[

1

]

.

The proof of the two properties above can be found here (

http://planetmath.org/PropertiesOfAGcdDomain

). For convenience, we let

gcd

⁡

(

a

,

b

)

be any one of the representatives in

GCD

⁡

(

a

,

b

)

.

Let

K

be the

field of fraction

of

D

, and

a

/

b

∈

K

(

a

,

b

∈

D

and

b

≠

0

) is a root of a

monic polynomial

p

⁢

(

x

)

∈

D

⁢

[

x

]

. We may, from property (1) above, assume that

gcd

⁡

(

a

,

b

)

=

1

.

Write

f

⁢

(

x

)

=

x

n

+

c

n

-

1

⁢

x

n

-

1

+

⋯

+

c

0

.

So we have

0

=

(

a

/

b

)

n

+

c

n

-

1

⁢

(

a

/

b

)

n

-

1

+

⋯

+

c

0

.

Multiply the equation by

b

n

then rearrange, and we get

-

a

n

=

c

n

-

1

⁢

b

⁢

a

n

-

1

+

⋯

+

c

0

⁢

b

n

=

b

⁢

(

c

n

-

1

⁢

a

n

-

1

+

⋯

+

c

0

⁢

b

n

-

1

)

.

Therefore,

b

∣

a

n

. Since

gcd

⁡

(

a

,

b

)

=

1

,

1

=

gcd

⁡

(

a

n

,

b

)

=

b

, by repeated applications of property (2), and one application of property (1) above. Therefore

b

is an

associate

of 1, hence a unit and we have

a

/

b

∈

D

.

∎

Together with the additional property (call it property 3)

if

GCD

⁡

(

a

,

b

)

=

[

1

]

and

a

∣

b

c

, then

a

∣

c

(proof found here (

http://planetmath.org/PropertiesOfAGcdDomain

)),

we have the following

Proposition 2

.

Every gcd domain is a

Schreier domain

.

Proof.

That a gcd domain is integrally closed is clear from the previous paragraph. We need to show that

D

is pre-Schreier, that is, every non-zero element is primal. Suppose

c

is non-zero in

D

, and

c

∣

a

b

with

a

,

b

∈

D

.
Let

r

=

gcd

⁡

(

a

,

c

)

and

r

⁢

t

=

a

,

r

⁢

s

=

c

. Then

1

=

gcd

⁡

(

s

,

t

)

by property (1) above. Next, since

c

∣

a

b

, write

c

⁢

d

=

a

⁢

b

so that

r

⁢

s

⁢

d

=

r

⁢

t

⁢

b

. This implies that

s

⁢

d

=

t

⁢

b

. So

s

∣

t

b

together with

gcd

⁡

(

s

,

t

)

=

1

show that

s

∣

b

by property (3). So we have just shown the existence of

r

,

s

∈

D

with

c

=

r

⁢

s

,

r

∣

a

and

s

∣

b

. Therefore,

c

is primal and

D

is a Schreier domain.

∎

Title

proof that a gcd domain is integrally closed

Canonical name

ProofThatAGcdDomainIsIntegrallyClosed

Date of creation

2013-03-22 18:19:27

Last modified on

2013-03-22 18:19:27

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Derivation

Classification

msc 13G05

Generated on Fri Feb 9 17:33:12 2018 by

LaTeXML
properties of a gcd domain

properties of a gcd domain

Let

D

be a

gcd domain

. For any

a

∈

D

, denote

[

a

]

the set of all

elements

in

D

that are

associates

of

a

,

GCD

⁡

(

a

,

b

)

the set of all gcd’s of elements

a

and

b

in

D

, and any

S

⊆

D

,

m

⁢

S

:=

{

m

⁢

s

∣

s

∈

S

}

. Then

1.

GCD

⁡

(

a

,

b

)

=

[

a

]

iff

a

∣

b

.

2.

m

⁢

GCD

⁡

(

a

,

b

)

=

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

.

3.

If

GCD

⁡

(

a

⁢

b

,

c

)

=

[

1

]

, then

GCD

⁡

(

a

,

c

)

=

[

1

]

4.

If

GCD

⁡

(

a

,

b

)

=

[

1

]

and

GCD

⁡

(

a

,

c

)

=

[

1

]

, then

GCD

⁡

(

a

,

b

⁢

c

)

=

[

1

]

.

5.

If

GCD

⁡

(

a

,

b

)

=

[

1

]

and

a

∣

b

c

, then

a

∣

c

.

Proof.

To aid in the proof of these

properties

, let us denote, for

a

∈

D

and

S

⊆

D

,

a

|

S

to mean that every element of

S

is divisible by

a

, and

S

|

a

to mean that every element in

S

divides

a

.
We take the following four steps:

1.

One direction is

obvious

from the

definition

. So now suppose

a

∣

b

. Then

a

∣

GCD

(

a

,

b

)

. But by
definition,

GCD

(

a

,

b

)

∣

a

, so

[

a

]

=

GCD

⁡

(

a

,

b

)

.

2.

Pick

d

∈

GCD

⁡

(

a

,

b

)

and

x

∈

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

. We want to show that

m

⁢

d

and

x

are associates. By

assumption

,

d

∣

a

and

d

∣

b

, so

m

d

∣

m

a

and

m

d

∣

m

b

, which implies that

m

d

∣

x

. Write

x

=

m

⁢

n

for some

n

∈

D

. Then

m

n

∣

m

a

and

m

n

∣

m

b

imply that

n

∣

a

and

n

∣

b

, and therefore

n

∣

d

since

d

is a gcd of

a

and

b

. As a result,

m

n

∣

m

d

, or

x

∣

m

d

, showing that

x

and

m

⁢

d

are associates. As a result, the map

f

:

m

⁢

GCD

⁡

(

a

,

b

)

→

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

given by

f

⁢

(

d

)

=

m

⁢

d

is a

bijection

.

3.

If

d

∣

a

and

d

∣

c

, then

d

∣

a

b

and

d

∣

c

. So

d

∣

GCD

(

a

b

,

c

)

=

[

1

]

, hence

d

is a unit and
the result follows.

4.

Suppose

d

∣

a

and

d

∣

b

c

. Then

d

∣

a

b

and

d

∣

b

c

and hence

d

∣

GCD

(

a

b

,

b

c

)

=

b

GCD

(

a

,

c

)

=

[

b

]

. But

d

∣

a

also, so

d

∣

GCD

(

a

,

b

)

=

[

1

]

and

d

is a unit.

5.

GCD

⁡

(

a

,

b

)

=

[

1

]

implies

[

c

]

=

GCD

⁡

(

a

⁢

c

,

b

⁢

c

)

. Now,

a

∣

a

c

and by assumption,

a

∣

b

c

. Therefore,

a

∣

GCD

(

a

c

,

b

c

)

=

[

c

]

.

∎

The second property above can be generalized to arbitrary

integral domain

: let

D

be an integral domain,

a

,

b

∈

D

, with

GCD

⁡

(

a

,

b

)

≠

∅

≠

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

, then

d

∈

GCD

⁡

(

a

,

b

)

iff

m

⁢

d

∈

GCD

⁡

(

m

⁢

a

,

m

⁢

b

)

.

Title

properties of a gcd domain

Canonical

name

PropertiesOfAGcdDomain

Date of creation

2013-03-22 18:18:44

Last modified on

2013-03-22 18:18:44

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Result

Classification

msc 13G05

Generated on Fri Feb 9 17:33:22 2018 by

LaTeXML
regular elements of finite ring

regular elements of finite ring

Theorem

.

If the

finite ring

R

has

regular elements

, then it has a unity.  All regular elements of

R

form a group under the

ring multiplication

and with

identity element

the unity of

R

.  Thus the regular elements are exactly the units of the ring; the rest of the elements are the zero and the

zero divisors

.

Proof.

Obviously, the set of the regular elements is non-empty and closed under the

multiplication

.  Let’s think the multiplication table of this set.  It is a finite distinct elements (any

equation

a

⁢

x

=

a

⁢

y

reduces to

x

=

y

).  Hence, for every regular element

a

, the square

a

2

determines another

a

′

such that

a

2

⁢

a

′

=

a

.  This implies

a

′

⁢

(

a

2

⁢

a

′

)

⁢

(

a

⁢

a

′

)

=

a

′

⁢

a

⁢

(

a

⁢

a

′

)

,  i.e.

(

a

′

⁢

a

)

⁢

(

a

⁢

a

′

)

2

=

(

a

′

⁢

a

)

⁢

(

a

⁢

a

′

)

,  and since

a

′

⁢

a

is

regular

(

http://planetmath.org/ZeroDivisor

),  we obtain that

(

a

⁢

a

′

)

2

=

a

⁢

a

′

.  So

a

⁢

a

′

is

idempotent

, and because it also is , it must be the unity of the ring (

http://planetmath.org/Unity

):

a

⁢

a

′

=

1

.  Thus we see that

R

has a unity which is a regular element and that

a

has a multiplicative inverse

a

′

, also regular.  Consequently the regular elements form a group.

Title

regular elements of finite ring

Canonical

name

RegularElementsOfFiniteRing

Date of creation

2013-03-22 15:11:09

Last modified on

2013-03-22 15:11:09

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

19

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13G05

Classification

msc 16U60

Related topic

GroupOfUnits

Related topic

PrimeResidueClass

Related topic

WedderburnsTheorem

Related topic

Unity

Generated on Fri Feb 9 17:33:25 2018 by

LaTeXML
ring without irreducibles

ring without irreducibles

An

integral domain

may not any

irreducible elements

.  One such example is the ring of all

algebraic integers

.  Any nonzero non-unit

ϑ

of this ring satisfies an equation

x

n

+

a

1

⁢

x

n

-

1

+

⋯

+

a

n

-

1

⁢

x

+

a

n

=

0

with

integer

coefficients

a

j

, since it is an algebraic
integer; moreover,  we can assume that

a

n

=

N

⁢

(

ϑ

)

≠

±

1

(see

norm and trace of
algebraic number

: 2).  The element

ϑ

has the

ϑ

=

ϑ

⋅

ϑ

.

Here,

ϑ

belongs to the ring because it
satisfies the equation

x

2

⁢

n

+

a

1

⁢

x

2

⁢

n

-

2

+

⋯

+

a

n

-

1

⁢

x

2

+

a

n

=

0

,

and it is no unit.  Thus the element

ϑ

is not irreducible.

Title

ring without irreducibles

Canonical name

RingWithoutIrreducibles

Date of creation

2014-05-29 11:39:19

Last modified on

2014-05-29 11:39:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Example

Classification

msc 13G05

Related topic

FieldOfAlgebraicNumbers

Generated on Fri Feb 9 17:33:26 2018 by

LaTeXML
Schreier domain

Schreier domain

An

integral domain

D

is a

pre-Schreier domain

if every non-zero element of

D

is primal. If in

addition

D

is

integrally closed

, then

D

is called a

Schreier domain

.

Remarks.

1.

Every

irreducible element

of a pre-Schreier domain is prime.

2.

A

gcd domain

is a Schreier domain (a proof of this can be found here (

http://planetmath.org/ProofThatAGcdDomainIsIntegrallyClosed

)).

Title

Schreier domain

Canonical name

SchreierDomain

Date of creation

2013-03-22 14:50:41

Last modified on

2013-03-22 14:50:41

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13G05

Synonym

pre-Schreier

Defines

pre-Schreier domain

Generated on Fri Feb 9 17:33:28 2018 by

LaTeXML