aarjaneiro/planetmath huggingface dataset:
lattice of fields

lattice of fields

Let

K

be a field and

K

¯

be its

algebraic closure

. The set

Latt

⁡

(

K

)

of all intermediate fields

E

(where

K

⊆

E

⊆

K

¯

), ordered by set theoretic

inclusion

, is a poset. Furthermore, it is a

complete lattice

, where

K

is the bottom and

K

¯

is the top.

This is the direct result of the fact that any

topped intersection structure

is a complete lattice, and

Latt

⁡

(

K

)

is such a

structure

. However, it can be easily proved directly: for any

collection

of intermediate fields

{

E

i

∣

i

∈

I

}

, the

intersection

is clearly an intermediate field, and is the

infimum

of the collection. The

compositum

of these fields, which is the smallest intermediate field

E

such that

E

i

⊆

E

, is the

supremum

of the collection.

It is not hard to see that

Latt

⁡

(

K

)

is an

algebraic lattice

, since the union of any directed family of intermediate fields between

K

and

K

¯

is an intermediate field. The

compact elements

in

Latt

⁡

(

K

)

are the finite

algebraic extensions

of

K

. The set of all compact elements in

Latt

⁡

(

K

)

, denoted by

Latt

F

⁡

(

K

)

, is a

lattice ideal

, for any

subfield

of a finite algebraic extension of

K

is finite algebraic over

K

. However,

Latt

F

⁡

(

K

)

, as a

sublattice

, is usually not

complete

(take the compositum of all

simple extensions

ℚ

⁢

(

p

)

, where

p

∈

ℤ

are

rational primes

).

Title

lattice of fields

Canonical name

LatticeOfFields

Date of creation

2013-03-22 17:13:26

Last modified on

2013-03-22 17:13:26

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Definition

Classification

msc 12F99

Generated on Fri Feb 9 16:44:40 2018 by

LaTeXML
non-constant element of rational function field

non-constant element of rational function field

Let

K

be a field.   Every simple (

http://planetmath.org/SimpleFieldExtension

) transcendent

field extension

K

⁢

(

α

)

/

K

may be represented by the extension

K

⁢

(

X

)

/

K

, where

K

⁢

(

X

)

is the

field of fractions

of the

polynomial ring

K

⁢

[

X

]

in one

indeterminate

X

.  The elements of

K

⁢

(

X

)

are

rational functions

, i.e. rational expressions

ϱ

=

f

⁢

(

X

)

g

⁢

(

X

)

(1)

with

f

⁢

(

X

)

and

g

⁢

(

X

)

polynomials

in

K

⁢

[

X

]

.

Theorem

.

Let the non-constant rational function (1) be reduced to

lowest terms

and let the greater of the degrees of its numerator and denominator be

n

.  This element

ϱ

is

transcendental

with respect to the

base field

K

.  The field extension

K

⁢

(

X

)

/

K

⁢

(

ϱ

)

is

algebraic

and of degree

n

.

Proof.

The element

X

satisfies the equation

ϱ

⁢

g

⁢

(

X

)

-

f

⁢

(

X

)

=

0

,

(2)

the

coefficients

of which are in the field

K

⁢

(

ϱ

)

, actually in the ring

K

⁢

[

ϱ

]

.  If all these coefficients were zero, we could take one non-zero coefficient

b

ν

in

g

⁢

(

X

)

and the coefficient

a

ν

of the same power of

X

in

f

⁢

(

X

)

, and then we would have especially

ϱ

⁢

b

ν

-

a

ν

=

0

;  this would mean that

ϱ

=

a

ν

b

ν

= constant, contrary to the supposition.  Thus at least one coefficient in (2) differs from zero, and we conclude that

X

is algebraic with respect to

K

⁢

(

ϱ

)

.  If

K

⁢

(

ϱ

)

were algebraic with respect to

K

, then also

X

should be algebraic with respect to

K

.  This is not true, and therefore we see that

K

⁢

(

ϱ

)

is transcendental, Q.E.D.

Further,

X

is a zero of the

n

th

degree polynomial

h

⁢

(

Y

)

=

ϱ

⁢

g

⁢

(

Y

)

-

f

⁢

(

Y

)

of the ring

K

⁢

(

ϱ

)

⁢

[

Y

]

, actually of the ring

K

⁢

[

ϱ

]

⁢

[

Y

]

, i.e. of

K

[

ϱ

, Y].  The polynomial is

irreducible

in this ring, since otherwise it would have there two factors, and because

h

⁢

(

Y

)

is linear in

ϱ

, the other factor should depend only on

Y

; but there can not be such a factor, for the polynomials

f

⁢

(

Z

)

and

g

⁢

(

Z

)

are relatively prime.  The conclusion is that

X

is an

algebraic element

over

K

⁢

(

ϱ

)

of degree

n

and therefore also

(

K

(

X

)

:

K

(

ϱ

)

)

=

n

,

Q.E.D.

References

1

B. L. van der Waerden:

Algebra

.  Siebte Auflage der

Modernen Algebra

.  Erster Teil.

— Springer-Verlag. Berlin, Heidelberg (1966).

Title

non-constant element of rational function field

Canonical

name

NonconstantElementOfRationalFunctionField

Date of creation

2013-03-22 15:02:50

Last modified on

2013-03-22 15:02:50

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12F99

Synonym

field of rational functions

Synonym

rational function field

Generated on Fri Feb 9 16:44:42 2018 by

LaTeXML
simple field extension

simple field extension

Let

K

⁢

(

α

)

be obtained from the field

K

via the of the element

α

, which is called the

primitive element

of the

field extension

K

⁢

(

α

)

/

K

.  We shall settle the of the field

K

⁢

(

α

)

.

We consider the

substitution homomorphism

φ

:

K

⁢

[

X

]

→

K

⁢

[

α

]

, where

∑

a

ν

⁢

X

ν

↦

∑

a

ν

⁢

α

ν

.

According to the

ring homomorphism

theorem, the image ring

K

⁢

[

α

]

is

isomorphic

with the

residue class ring

K

⁢

[

X

]

/

𝔭

, where

𝔭

is the ideal of

polynomials

having

α

as their zero.  Because

K

⁢

[

α

]

is, as

subring

of the field

K

⁢

(

α

)

, an

integral domain

, then also

K

⁢

[

X

]

/

𝔭

has no

zero divisors

, and hence

𝔭

is a

prime ideal

.  It must be principal, for

K

⁢

[

X

]

is a

principal ideal ring

.

There are two possibilities:

1.

𝔭

=

(

p

⁢

(

X

)

)

, where

p

⁢

(

X

)

is an

irreducible polynomial

with

p

⁢

(

α

)

=

0

.  Because every non-zero prime ideal of

K

⁢

[

X

]

is maximal, the isomorphic image

K

⁢

[

X

]

/

(

p

⁢

(

X

)

)

of

K

⁢

[

α

]

is a field, and it must give the of

K

⁢

(

α

)

=

K

⁢

[

α

]

.  We say that

α

is

algebraic

with respect to

K

(or

over

K

).  In this case, we have a

finite field extension

K

⁢

(

α

)

/

K

.

2.

𝔭

=

(

0

)

.  This means that the

homomorphism

φ

is an

isomorphism

between

K

⁢

[

X

]

and

K

⁢

[

α

]

, i.e. all expressions

∑

a

ν

⁢

α

ν

behave as the polynomials

∑

a

ν

⁢

X

ν

.  Now,

K

⁢

[

α

]

is no field because

K

⁢

[

X

]

is not such, but the isomorphy of the rings implies the isomorphy of the corresponding

fields of fractions

.  Thus the

simple extension

field

K

⁢

(

α

)

is isomorphic with the field

K

⁢

(

X

)

of

rational functions

in one

indeterminate

X

.  We say that

α

is

transcendental

(

http://planetmath.org/Algebraic

) with respect to

K

(or

over

K

).  This time we have a

simple infinite field extension

K

⁢

(

α

)

/

K

.

Title

simple field extension

Canonical name

SimpleFieldExtension

Date of creation

2013-03-22 14:23:06

Last modified on

2013-03-22 14:23:06

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

25

Author

pahio (2872)

Entry type

Definition

Classification

msc 12F99

Related topic

PrimitiveElementTheorem

Related topic

CanonicalFormOfElementOfNumberField

Defines

primitive element

Generated on Fri Feb 9 16:44:49 2018 by

LaTeXML
simple transcendental field extension

simple transcendental field extension

The

extension field

K

⁢

(

α

)

of a

base field

K

, where

α

is a

transcendental element

with respect to

K

, is a

simple (

http://planetmath.org/SimpleFieldExtension

)

transcendental extension

of

K

.  All such extension fields are isomorphic to the field

K

⁢

(

X

)

of

rational functions

in one

indeterminate

X

over

K

, and thus to each other.

Example.

The

subfields

ℚ

⁢

(

π

)

and

ℚ

⁢

(

e

)

of

ℝ

, where

π

is Ludolph’s constant (

http://planetmath.org/Pi

) and

e

Napier’s constant, are isomorphic.

Title

simple transcendental field extension

Canonical

name

SimpleTranscendentalFieldExtension

Date of creation

2013-03-22 15:02:20

Last modified on

2013-03-22 15:02:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Corollary

Classification

msc 12F99

Synonym

simple infinite field extension

Related topic

FunctionField

Generated on Fri Feb 9 16:44:50 2018 by

LaTeXML
the compositum of a Galois extension and another extension is Galois

the compositum of a Galois extension and another extension is Galois

Theorem 1

.

Let

E

/

K

be a

Galois extension

of fields, let

F

/

K

be an arbitrary

extension

and assume that

E

and

F

are both

subfields

of some other larger field

T

. The

compositum

of

E

and

F

is here denoted by

E

⁢

F

. Then:

1.

E

⁢

F

is a Galois extension of

F

and

E

is Galois over

E

∩

F

;

2.

Let

H

=

Gal

⁡

(

E

⁢

F

/

F

)

. The

restriction

map:

H

=

Gal

⁡

(

E

⁢

F

/

F

)

⟶

Gal

⁡

(

E

/

E

∩

F

)

σ

⟶

σ

|

E

is an

isomorphism

, where

σ

|

E

denotes the restriction of

σ

to

E

.

Remark 1

.

Notice, however, that if

E

/

F

and

F

/

K

are both Galois extensions, the extension

E

/

K

need not be Galois. See example of

normal extension

for a

counterexample

.

Title

the compositum of a Galois extension and another extension is Galois

Canonical

name

TheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois

Date of creation

2013-03-22 15:04:13

Last modified on

2013-03-22 15:04:13

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F99

Classification

msc 11R32

Related topic

FundamentalTheoremOfGaloisTheory

Related topic

GaloisExtension

Related topic

ExampleOfNormalExtension

Related topic

ClassNumberDivisibilityInExtensions

Related topic

GaloisGroupOfTheCompositumOfTwoGaloisExtensions

Related topic

ExtensionsWithoutUnramifiedSubextensionsAndClassNumberDivisibility

Generated on Fri Feb 9 16:44:55 2018 by

LaTeXML
the ramification index and the inertial degree are multiplicative in towers

the ramification index and the inertial degree are multiplicative in towers

Theorem

.

Let

E

,

F

and

K

be

number fields

in a tower:

K

⊆

F

⊆

E

and let

O

E

,

O

F

and

O

K

be their

rings of integers

respectively. Suppose

p

is a

prime ideal

of

O

K

and let

P

be a prime ideal of

O

F

lying above

p

, and

P

is a prime ideal of

O

E

lying above

P

.

\xymatrix

⁢

E

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝒪

E

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝒫

⁢

\ar

⁢

@

-

[

d

]

⁢

F

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝒪

F

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝔓

⁢

\ar

⁢

@

-

[

d

]

⁢

K

⁢

&

⁢

𝒪

K

⁢

&

⁢

𝔭

Then the indices of the

extensions

, the

ramification

indices and

inertial degrees

satisfy:

[

E

:

K

]

=

[

E

:

F

]

⋅

[

F

:

K

]

,

(1)

e

(

𝒫

|

𝔭

)

=

e

(

𝒫

|

𝔓

)

⋅

e

(

𝔓

|

𝔭

)

,

(2)

f

(

𝒫

|

𝔭

)

=

f

(

𝒫

|

𝔓

)

⋅

f

(

𝔓

|

𝔭

)

.

(3)

Title

the ramification index and the inertial degree are

multiplicative

in towers

Canonical name

TheRamificationIndexAndTheInertialDegreeAreMultiplicativeInTowers

Date of creation

2013-03-22 15:06:34

Last modified on

2013-03-22 15:06:34

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F99

Classification

msc 13B02

Classification

msc 11S15

Related topic

Ramify

Related topic

InertialDegree

Generated on Fri Feb 9 16:44:57 2018 by

LaTeXML
Gelfand–Tornheim theorem

Gelfand–Tornheim theorem

Theorem

.

Any

normed field

is

isomorphic

either to the field

ℝ

of

real numbers

or to the field

ℂ

of

complex numbers

.

The

normed field

means a field

K

having a

subfield

R

isomorphic to

ℝ

and satisfying the following:
There is a mapping

∥

⋅

∥

from

K

to the set of non-negative reals such that

•

∥

a

∥

=

0

iff

a

=

0

•

∥

a

⁢

b

∥

≦

∥

a

∥

⋅

∥

b

∥

•

∥

a

+

b

∥

≦

∥

a

∥

+

∥

b

∥

•

∥

a

⁢

b

∥

=

|

a

|

⋅

∥

b

∥

when

a

∈

R

and

b

∈

K

Using the Gelfand–Tornheim theorem, it can be shown that the only fields with

archimedean valuation

are isomorphic to subfields of

ℂ

and that the

valuation

is the usual

absolute value

(

modulus

) or some

positive

power of the absolute value.

References

1

Emil Artin: .  Lecture notes.  Mathematisches Institut, Göttingen (1959).

Title

Gelfand–Tornheim theorem

Canonical name

GelfandTornheimTheorem

Date of creation

2013-03-22 14:11:49

Last modified on

2013-03-22 14:11:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

40

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12J05

Synonym

Gelfand-Tornheim theorem

Related topic

ExtensionOfKrullValuation

Related topic

TopicEntryOnRealNumbers

Related topic

BanachAlgebra

Related topic

NormedAlgebra

Related topic

ArchimedeanOrderedFieldsAreReal

Defines

normed field

Generated on Fri Feb 9 16:44:58 2018 by

LaTeXML
complete ultrametric field

complete ultrametric field

A field

K

equipped with a

non-archimedean valuation

|

⋅

|

is called a

non-archimedean field

or also an

ultrametric field

, since the

valuation

the

ultrametric

d

⁢

(

x

,

y

)

:=

|

x

-

y

|

of

K

.

Theorem

.

Let

(

K

,

d

)

be a

complete

(

http://planetmath.org/Complete

) ultrametric field.  A necessary and

sufficient

condition for the convergence of the

series

a

1

+

a

2

+

a

3

+

…

(1)

in

K

is that

lim

n

→

∞

⁡

a

n

=

0

.

(2)

Proof.

Let

ε

be any positive number.  When (1)

converges

, it

satisfies

the Cauchy condition and therefore exists a number

m

ε

such that surely

|

a

m

+

1

|

=

|

∑

j

=

1

m

+

1

a

j

-

∑

j

=

1

m

a

j

|

<

ε

for all

m

≧

m

ε

;  thus (2) is necessary.  On the contrary, suppose the

validity

of (2).  Now one may determine such a great number

n

ε

that

|

a

m

|

<

ε

∀

m

≧

n

ε

.

No matter how great is the

natural number

n

, the ultrametric then guarantees the

inequality

|

a

m

+

a

m

+

1

+

…

+

a

m

+

n

|

≦

max

⁡

{

|

a

m

|

,

|

a

m

+

1

|

,

…

,

|

a

m

+

n

|

}

<

ε

always when

m

≧

n

ε

.  Thus the partial sums of (1) form a

Cauchy sequence

, which converges in the complete field.  Hence the series (1) converges, and (2) is sufficient.

Title

complete ultrametric field

Canonical name

CompleteUltrametricField

Date of creation

2013-03-22 14:55:37

Last modified on

2013-03-22 14:55:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12J10

Classification

msc 54E35

Related topic

Series

Related topic

NecessaryConditionOfConvergence

Related topic

ExtensionOfValuationFromCompleteBaseField

Related topic

PropertiesOfNonArchimedeanValuations

Defines

ultrametric field

Defines

non-archimedean field

Generated on Fri Feb 9 16:45:01 2018 by

LaTeXML
complex p-adic numbers

complex p-adic numbers

First, we review a possible construction of the

complex numbers

. We start from the

rational numbers

,

ℚ

, which we consider as a

metric space

, where the distance is given by the usual

absolute value

|

⋅

|

, e.g.

|

-

3

/

2

|

=

3

/

2

. As we know, the

field of rational

numbers is not an algebraically closed field (e.g.

i

=

-

1

∉

ℚ

). Let

ℚ

¯

be a fixed

algebraic closure

of

ℚ

. The absolute value in

ℚ

extends uniquely to

ℚ

¯

. However,

ℚ

¯

is not

complete

with respect to

|

⋅

|

(e.g.

e

=

∑

n

≥

0

1

/

n

!

∉

ℚ

¯

because e is

transcendental

). The

completion

of

ℚ

¯

with respect to

|

⋅

|

is

ℂ

, the field of complex numbers.

Construction of

ℂ

p

We follow the construction of

ℂ

above to build

ℂ

p

. Let

p

be a

prime number

and let

ℚ

p

be the

p

-adic rationals (

http://planetmath.org/PAdicIntegers

) or (

p

-adic numbers). The

p

-adics,

ℚ

p

, are the completion of

ℚ

with respect to the usual

p

-adic

valuation

(

http://planetmath.org/PAdicValuation

)

|

⋅

|

p

. Thus, we regard

(

ℚ

p

,

|

⋅

|

p

)

as a

complete metric space

. However, the field

ℚ

p

is not

algebraically closed

(e.g.

i

=

-

1

∈

ℚ

p

if and only if

p

≡

1

mod

4

). Let

ℚ

¯

p

be a fixed algebraic closure of

ℚ

p

. The

p

-adic valuation

|

⋅

|

p

extends uniquely to

ℚ

¯

p

. However:

Proposition

.

The field

Q

¯

p

is not complete with respect to

|

⋅

|

p

.

Proof.

Let

β

n

be defined as:

β

n

=

{

e

2

⁢

π

⁢

i

/

n

,

if

⁢

(

n

,

p

)

=

1

;

1

,

otherwise.

One can prove that if we define:

α

=

∑

n

=

1

∞

β

n

⁢

p

n

then

α

∉

ℚ

¯

p

, although

∑

n

=

m

∞

β

n

⁢

p

n

→

0

as

m

→

∞

(see

[

1

]

, p. 48, for details). Thus,

ℚ

¯

p

is not complete with respect to

|

⋅

|

p

.
∎

Definition

.

The field of

complex

p

-adic numbers is defined to be the completion of

Q

¯

p

with respect to the

p

-adic absolute value

|

⋅

|

p

.

Proposition

(Properties of

C

p

).

The field

C

p

enjoys the following properties:

1.

ℂ

p

is algebraically closed.

2.

The absolute value

|

⋅

|

p

extends uniquely to

ℂ

p

, which becomes an algebraically closed, complete metric space.

3.

ℂ

p

is a

complete ultrametric field

.

4.

ℚ

¯

p

is

dense in

ℂ

p

.

5.

ℂ

p

is

isomorphic

to

ℂ

as fields, although they are not isomorphic as

topological spaces

.

References

1

L. C. Washington,

Introduction to

Cyclotomic Fields

,
Springer-Verlag, New York.

Title

complex p-adic numbers

Canonical name

ComplexPadicNumbers

Date of creation

2013-03-22 15:13:44

Last modified on

2013-03-22 15:13:44

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Definition

Classification

msc 12J12

Classification

msc 11S99

Synonym

complex

p

-adic numbers

Generated on Fri Feb 9 16:45:03 2018 by

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theorems on continuation

theorems on continuation

Theorem 1.

When

ν

0

is an

exponent valuation

of the field

k

and

K

/

k

is a

finite field extension

,

ν

0

has a

continuation

to the

extension field

K

.

Theorem 2.

If the degree (

http://planetmath.org/ExtensionField

) of the field extension

K

/

k

is

n

and

ν

0

is an arbitrary exponent (

http://planetmath.org/ExponentValuation2

) of

k

, then

ν

0

has at most

n

continuations to the extension field

K

.

Theorem 3.

Let

ν

0

be an exponent valuation of the field

k

and

𝔬

the ring of the exponent

ν

0

.  Let

K

/

k

be a

finite extension

and

𝔒

the

integral closure

of

𝔬

in

K

.  If

ν

1

,

…

,

ν

m

are all different continuations of

ν

0

to the field

K

and

𝔒

1

,

…

,

𝔒

m

their rings (

http://planetmath.org/RingOfExponent

), then

𝔒

=

⋂

i

=

1

m

𝔒

i

.

The proofs of those theorems are found in [1], which is available also in Russian (original), English and French.

Corollary.

The ring

𝔒

(of theorem 3) is a UFD.  The exponents of

K

, which are determined by the

pairwise coprime

prime elements

of

𝔒

, coincide with the continuations

ν

1

,

…

,

ν

m

of

ν

0

.  If

π

1

,

…

,

π

m

are the pairwise coprime prime elements of

𝔒

such that

ν

i

⁢

(

π

1

)

=

1

for all

i

’s and if the prime element

p

of the ring

𝔬

has the

p

=

ε

⁢

π

1

e

1

⁢

⋯

⁢

π

m

e

m

with

ε

a unit of

𝔒

, then

e

i

is the

ramification index

of the exponent

ν

i

with respect to

ν

0

(

i

=

1

,

…

,

m

).

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

theorems on continuation

Canonical name

TheoremsOnContinuation

Date of creation

2013-03-22 17:59:51

Last modified on

2013-03-22 17:59:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12J20

Classification

msc 13A18

Classification

msc 13F30

Classification

msc 11R99

Synonym

theorems on continuations of exponents

Generated on Fri Feb 9 16:45:07 2018 by

LaTeXML
trivial valuation

trivial valuation

The

trivial valuation

of a field

K

is the

Krull valuation

|

⋅

|

of

K

such that

|

0

|

=

0

and

|

x

|

=

1

for other elements

x

of

K

.

Properties

1.

Every field has the trivial valuation.

2.

The trivial valuation is

non-archimedean

.

3.

The

valuation ring

of the trivial valuation is the whole field and the corresponding

maximal ideal

is the

zero ideal

.

4.

The field is

complete

(

http://planetmath.org/Complete

) with respect to (the metric given by) its trivial valuation.

5.

A

finite field

has only the trivial valuation.  (Let

a

be the

primitive element

of the

multiplicative group

of the field, which is cyclic (

http://planetmath.org/CyclicGroup

).  If

|

⋅

|

is any

valuation

of the field, then one must have

|

a

|

=

1

since otherwise

|

1

|

≠

1

.  Consequently,

|

x

|

=

|

a

m

|

=

|

a

|

m

=

1

m

=

1

for all non-zero elements

x

.)

6.

Every

algebraic extension

of finite fields has only the trivial valuation, but every field of

characteristic

0 has non-trivial valuations.

Title

trivial valuation

Canonical name

TrivialValuation

Date of creation

2013-03-22 14:20:23

Last modified on

2013-03-22 14:20:23

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Definition

Classification

msc 12J20

Classification

msc 11R99

Related topic

IndependenceOfTheValuations

Related topic

KrullValuation

Generated on Fri Feb 9 16:45:09 2018 by

LaTeXML
examples for Hensel’s lemma

examples for Hensel’s lemma

Example 1

.

Let

p

be a

prime number

greater than

2

. Are there solutions to

x

2

+

7

=

0

in the field

ℚ

p

(the

p

-adic numbers (

http://planetmath.org/PAdicIntegers

))? If there are,

-

7

must be a

quadratic residue

modulo

p

. Thus, let

p

be a prime such that

(

-

7

p

)

=

1

where

(

⋅

p

)

is the

Legendre symbol

. Hence, there exist

α

0

∈

ℤ

such that

α

0

2

≡

-

7

mod

p

. We claim that

x

2

+

7

=

0

has a solution in

ℚ

p

if and only if

-

7

is a quadratic residue modulo

p

. Indeed, if we let

f

⁢

(

x

)

=

x

2

+

7

(so

f

′

⁢

(

x

)

=

2

⁢

x

), the element

α

0

∈

ℤ

p

satisfies the conditions of the (trivial case of) Hensel’s lemma. Therefore there exist a root

α

∈

ℚ

p

of

x

2

+

7

=

0

.

Example 2

.

Let

p

=

2

. Are there any solutions to

x

2

+

7

=

0

in

ℚ

2

? Notice that if we let

f

⁢

(

x

)

=

x

2

+

7

, then

f

′

⁢

(

x

)

=

2

⁢

x

and for any

α

0

∈

ℤ

2

, the number

f

′

⁢

(

α

0

)

=

2

⁢

α

0

is

congruent

to

0

modulo

2

. Thus, we cannot use the trivial case of Hensel’s lemma.

Let

α

0

=

1

∈

ℤ

2

. Notice that

f

⁢

(

1

)

=

8

and

f

′

⁢

(

1

)

=

2

. Thus

|

8

|

2

<

|

2

2

|

2

and the general case of Hensel’s lemma applies. Hence, there exist a

2

-adic solution to

x

2

+

7

=

0

. The following is the

2

-adic

canonical form

(

http://planetmath.org/PAdicCanonicalForm

) for one of the solutions:

α

=

1

+

1

⋅

2

3

+

1

⋅

2

4

+

…

=

…

⁢

11001

Title

examples for Hensel’s lemma

Canonical name

ExamplesForHenselsLemma

Date of creation

2013-03-22 15:08:36

Last modified on

2013-03-22 15:08:36

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Example

Classification

msc 12J99

Classification

msc 11S99

Classification

msc 13H99

Generated on Fri Feb 9 16:45:11 2018 by

LaTeXML