aarjaneiro/planetmath huggingface dataset:
UFD

UFD

An

integral domain

D

satisfying

•

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

r

and

q

1

⁢

q

2

⁢

⋯

⁢

q

s

are two factorizations of the same element

a

into irreducibles, then

r

=

s

and we can reorder the

q

j

’s in a way that

q

j

is an

associate

element of

p

j

for all

j

is called a

unique factorization domain

(UFD), also a

factorial ring

.

The factors

p

1

,

p

2

,

…

,

p

r

are called the

prime factors

of

a

.

Some of the classic results about UFDs:

•

On a UFD, the

concept

of

prime element

and

irreducible element

coincide.

•

If

F

is a field, then

F

⁢

[

x

]

is a UFD.

•

If

D

is a UFD, then

D

⁢

[

x

]

(the ring of polynomials on the variable

x

over

D

) is also a UFD.

Since

R

⁢

[

x

,

y

]

≅

R

⁢

[

x

]

⁢

[

y

]

,  these results can be extended to rings of polynomials with a finite number of variables.

•

If

D

is a

principal ideal domain

, then it is also a UFD.

The converse is, however, not true.  Let

F

a field and consider the UFD

F

⁢

[

x

,

y

]

.
Let

I

the ideal consisting of all the elements of

F

⁢

[

x

,

y

]

whose constant term is

0

.  Then it can be proved that

I

is not a

principal ideal

.  Therefore not every UFD is a PID.

Title

UFD

Canonical name

UFD

Date of creation

2013-03-22 11:56:22

Last modified on

2013-03-22 11:56:22

Owner

drini (3)

Last modified by

drini (3)

Numerical id

19

Author

drini (3)

Entry type

Definition

Classification

msc 13G05

Synonym

unique factorization domain

Related topic

IntegralDomain

Related topic

Irreducible

Related topic

EuclideanRing

Related topic

EuclideanValuation

Related topic

ProofThatAnEuclideanDomainIsAPID

Related topic

WhyEuclideanDomains

Related topic

Y2X32

Related topic

PID

Related topic

PIDsAreUFDs

Related topic

FundamentalTheoremOfArithmetic

Defines

factorial ring

Defines

prime factor

Defines

UFD

Defines

unique factorization

Generated on Fri Feb 9 17:33:30 2018 by

LaTeXML
UFD’s are integrally closed

UFD’s are integrally closed

Theorem:

Every UFD is

integrally closed

.

Proof:

Let

R

be a UFD,

K

its

field of fractions

,

u

∈

K

,

u

integral over

R

. Then for some

c

0

,

…

,

c

n

-

1

∈

R

,

u

n

+

c

n

-

1

⁢

u

n

-

1

+

…

+

c

0

=

0

Write

u

=

a

b

,

a

,

b

∈

R

, where

a

,

b

have no non-unit common

divisor

(which we can assume since

R

is a UFD). Multiply the above

equation

by

b

n

to get

a

n

+

c

n

-

1

⁢

b

⁢

a

n

-

1

+

…

+

c

0

⁢

b

n

=

0

Let

d

be an

irreducible

divisor of

b

. Then

d

is prime since

R

is a UFD. Now,

d

|

a

n

since it divides all the other terms and thus (since

d

is prime)

d

|

a

. But

a

,

b

have no non-unit common divisors, so

d

is a unit. Thus

b

is a unit and hence

u

∈

R

.

Title

UFD’s are integrally closed

Canonical name

UFDsAreIntegrallyClosed

Date of creation

2013-03-22 15:49:25

Last modified on

2013-03-22 15:49:25

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13G05

Generated on Fri Feb 9 17:33:32 2018 by

LaTeXML
zero divisor

zero divisor

Let

a

be a nonzero element of a ring

R

.

The element

a

is a

left zero divisor

if there exists a nonzero element

b

∈

R

such that

a

⋅

b

=

0

. Similarly,

a

is a

right zero divisor

if there exists a nonzero element

c

∈

R

such that

c

⋅

a

=

0

.

The element

a

is said to be a

zero divisor

if it is both a left and right zero divisor. A nonzero element

a

∈

R

is said to be a

regular element

if it is neither a left nor a right zero divisor.

Example:

Let

R

=

ℤ

6

. Then the elements

2

and

3

are zero divisors, since

2

⋅

3

≡

6

≡

0

(

mod

6

)

.

Title

zero divisor

Canonical name

ZeroDivisor

Date of creation

2013-03-22 12:49:59

Last modified on

2013-03-22 12:49:59

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

9

Author

cvalente (11260)

Entry type

Definition

Classification

msc 13G05

Related topic

CancellationRing

Related topic

IntegralDomain

Related topic

Unity

Defines

left zero divisor

Defines

right zero divisor

Defines

regular element

Generated on Fri Feb 9 17:33:34 2018 by

LaTeXML
zero rule of product

zero rule of product

For real and

complex numbers

, and more generally for elements of an

integral domain

, a

product

equals to zero if and only if at least one of the equals to zero.  For two elements

a

and

b

, we have

a

⁢

b

=

0

⟺

a

=

0

∨

b

=

0

.

For example, this rule can be used in solving polynomial equations:

x

3

-

x

2

-

2

⁢

x

+

2

=

0

(

x

3

-

x

2

)

+

(

-

2

⁢

x

+

2

)

=

0

x

2

⁢

(

x

-

1

)

-

2

⁢

(

x

-

1

)

=

0

(

x

-

1

)

⁢

(

x

2

-

2

)

=

0

x

-

1

=

0

∨

x

2

-

2

=

0

x

=

1

∨

x

=

±

2

The used sign “

∨

” is the logical or.

Title

zero rule of product

Canonical name

ZeroRuleOfProduct

Date of creation

2013-03-22 15:06:46

Last modified on

2013-03-22 15:06:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Result

Classification

msc 13G05

Synonym

product to zero rule

Related topic

CancellationRing

Related topic

EulersDerivationOfTheQuarticFormula

Related topic

GroupingMethodForFactorizingPolynomials

Related topic

HyperbolasOrthogonalToEllipses

Generated on Fri Feb 9 17:33:36 2018 by

LaTeXML
formal power series converges if and only if it converges along every line

formal power series converges if and only if it converges along every line

Suppose

T

⁢

(

x

)

denotes the

formal power series

∑

α

a

α

⁢

x

α

,

using the

multi-index notation

,
where

x

=

(

x

1

,

…

,

x

N

)

and

a

α

∈

ℂ

.

Fixing

v

∈

ℝ

N

and we can also talk of the formal power series in

t

∈

ℝ

T

⁢

(

t

⁢

v

)

=

∑

α

a

α

⁢

(

t

⁢

v

)

α

=

∑

α

a

α

⁢

v

α

⁢

t

|

α

|

=

∑

k

=

0

∞

(

∑

|

α

|

=

k

a

α

⁢

v

α

)

⁢

t

k

.

Theorem

.

Suppose

T

⁢

(

x

)

is a formal power series in

x

∈

R

N

. Suppose

T

⁢

(

t

⁢

v

)

is a

convergent

power series

in

t

∈

R

for
all

v

∈

R

N

. Then

T

is convergent.

The other direction, if

T

⁢

(

x

)

converges then

t

↦

T

⁢

(

t

⁢

v

)

converges, is obvious.

References

1

M. Salah Baouendi,
Peter Ebenfelt,
Linda Preiss Rothschild.
,
Princeton University Press,
Princeton, New Jersey, 1999.

Title

formal power series converges if and only if it converges along every line

Canonical name

FormalPowerSeriesConvergesIfAndOnlyIfItConvergesAlongEveryLine

Date of creation

2013-03-22 17:42:11

Last modified on

2013-03-22 17:42:11

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

5

Author

jirka (4157)

Entry type

Theorem

Classification

msc 13H05

Classification

msc 13B35

Classification

msc 13J05

Classification

msc 13F25

Generated on Fri Feb 9 17:33:44 2018 by

LaTeXML
formal power series over field

formal power series over field

Theorem.

If

K

is a field, then the ring

K

⁢

[

[

X

]

]

of

formal power series

is a

discrete valuation ring

with

(

X

)

its unique

maximal ideal

.

Proof.

We show first that an arbitrary ideal

I

of

K

⁢

[

[

X

]

]

is a

principal ideal

.  If

I

=

(

0

)

,  the thing is ready.  Therefore, let

I

≠

(

0

)

.  Take an element

f

⁢

(

X

)

:=

∑

i

=

0

∞

a

i

⁢

X

i

of

I

such that it has the least possible amount of successive zero coefficients in its beginning; let its first non-zero coefficient be

a

k

.  Then

f

⁢

(

X

)

=

X

k

⁢

(

a

k

+

a

k

+

1

⁢

X

+

…

)

.

Here we have in the parentheses an

invertible formal power series

g

⁢

(

X

)

, whence get the

equation

X

k

=

f

⁢

(

X

)

⁢

[

g

⁢

(

X

)

]

-

1

implying

X

k

∈

I

and consequently

(

X

k

)

⊆

I

.

For obtaining the reverse inclusion, suppose that

h

⁢

(

X

)

:=

b

n

⁢

X

n

+

b

n

+

1

⁢

X

n

+

1

+

…

is an arbitrary nonzero element of

I

where

b

n

≠

0

.  Because

n

≥

k

,  we may write

h

⁢

(

X

)

=

X

k

⁢

(

b

n

⁢

X

n

-

k

+

b

n

+

1

⁢

X

n

-

k

+

1

+

…

)

.

This equation says that

h

⁢

(

X

)

∈

(

X

k

)

,  whence

I

⊆

(

X

k

)

.

Thus we have seen that

I

is the principal ideal

(

X

k

)

, so that

K

⁢

[

[

X

]

]

is a

principal ideal domain

.

Now, all ideals of the ring

K

⁢

[

[

X

]

]

form apparently the strictly descending chain

(

X

)

⊃

(

X

2

)

⊃

(

X

3

)

⊃

…

⊃

(

0

)

,

whence the ring has the unique maximal ideal

(

X

)

.  A principal ideal domain with only one maximal ideal is a discrete valuation ring.

Title

formal power series over field

Canonical name

FormalPowerSeriesOverField

Date of creation

2015-10-19 9:13:35

Last modified on

2015-10-19 9:13:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13H05

Classification

msc 13J05

Classification

msc 13F25

Generated on Fri Feb 9 17:33:46 2018 by

LaTeXML
invertible formal power series

invertible formal power series

Theorem

.

Let

R

be a

commutative ring

with

non-zero unity

.  A

formal power series

f

⁢

(

X

)

:=

∑

i

=

0

∞

a

i

⁢

X

i

(1)

is

invertible

in the ring

R

⁢

[

[

X

]

]

iff

a

0

is invertible in the ring

R

.

Proof.

1

∘

.  Let

f

⁢

(

X

)

have the

multiplicative inverse

g

⁢

(

X

)

:=

∑

i

=

0

∞

b

i

⁢

X

i

.  Since

f

⁢

(

X

)

⁢

g

⁢

(

X

)

=

∑

i

=

0

∞

∑

j

=

0

i

a

j

⁢

b

i

-

j

⁢

X

i

=

1

,

we see that

a

0

⁢

b

0

=

1

, i.e.

a

0

is an

invertible element

(unit) of

R

.

2

∘

.  Assume

conversely

that

a

0

is invertible in

R

.  For making from a formal power series

g

⁢

(

X

)

:=

∑

i

=

0

∞

b

i

⁢

X

i

(2)

the

inverse

of

f

⁢

(

X

)

=

∑

i

=

0

∞

a

i

⁢

X

i

, we first choose

b

0

:=

a

0

-

1

.  For all already defined

coefficients

b

0

,

b

1

,

…

,

b

i

-

1

let the next coefficient be defined as

b

i

:=

-

a

0

-

1

⁢

(

a

1

⁢

b

i

-

1

+

a

2

⁢

b

i

-

2

+

…

+

a

i

⁢

b

0

)

.

This

equation

means that

∑

j

=

0

i

a

j

⁢

b

i

-

j

=

a

0

⁢

b

i

+

a

1

⁢

b

i

-

1

+

a

2

⁢

b

i

-

2

+

…

+

a

i

⁢

b

0

vanishes for all

i

=

1

,

2

,

…

;  since

a

0

⁢

b

0

=

1

,  the

product

of the formal power series (1) and (2) becomes simply equal to 1.  Accordingly,

f

⁢

(

x

)

is invertible.

Title

invertible formal power series

Canonical name

InvertibleFormalPowerSeries

Date of creation

2016-04-27 10:47:14

Last modified on

2016-04-27 10:47:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13H05

Classification

msc 13F25

Classification

msc 13J05

Related topic

RulesOfCalculusForDerivativeOfFormalPowerSeries

Generated on Fri Feb 9 17:33:49 2018 by

LaTeXML
maximal ideals of ring of formal power series

maximal ideals of ring of formal power series

Suppose that

R

is a

commutative ring

with

non-zero unity

.

If

𝔪

is a

maximal ideal

of

R

, then

𝔐

:=

𝔪

+

(

X

)

is a maximal ideal of the ring

R

⁢

[

[

X

]

]

of

formal power series

.

Also the converse is true, i.e. if

𝔐

is a maximal ideal of

R

⁢

[

[

X

]

]

, then there is a maximal ideal

𝔪

of

R

such that

𝔐

=

𝔪

+

(

X

)

.

Note.

In the special case that

R

is a field, the only maximal ideal of which is the

zero ideal

(

0

)

, this corresponds to the only maximal ideal

(

X

)

of

R

⁢

[

[

X

]

]

(see

http://planetmath.org/node/12087

formal power series over field

).

We here prove the first assertion.  So,

𝔪

is assumed to be

maximal

.  Let

f

⁢

(

x

)

:=

a

0

+

a

1

⁢

X

+

a

2

⁢

X

2

+

…

be any formal power series in

R

⁢

[

[

X

]

]

∖

𝔐

.  Hence, the constant term

a

0

cannot lie in

𝔪

.  According to the

criterion for maximal ideal

, there is an element

r

of

R

such that

1

+

r

⁢

a

0

∈

𝔪

.  Therefore

1

+

r

⁢

f

⁢

(

X

)

=

(

1

+

r

⁢

a

0

)

+

r

⁢

(

a

1

+

a

2

⁢

X

+

a

3

⁢

X

2

+

…

)

⁢

X

∈

𝔪

+

(

X

)

=

𝔐

,

whence the same criterion says that

𝔐

is a maximal ideal of

R

⁢

[

[

X

]

]

.

Title

maximal ideals of ring of formal power series

Canonical name

MaximalIdealsOfRingOfFormalPowerSeries

Date of creation

2013-03-22 19:10:49

Last modified on

2013-03-22 19:10:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Result

Classification

msc 13H05

Classification

msc 13J05

Classification

msc 13C13

Classification

msc 13F25

Generated on Fri Feb 9 17:33:51 2018 by

LaTeXML
regular local ring

regular local ring

A

local ring

R

of

dimension

n

is

regular

if and only if its

maximal ideal

𝔪

is generated by

n

elements.

Equivalently,

R

is regular if

dim

R

/

𝔪

⁡

𝔪

/

𝔪

2

=

dim

⁡

R

, where the first dimension is that of a

vector space

, and the latter is the

Krull dimension

, since by Nakayama’s lemma, elements

generate

𝔪

if and only if their images under the

projection

generate

𝔪

/

𝔪

2

.

By Krull’s principal ideal theorem,

𝔪

cannot be generated by fewer than

n

elements, so the maximal ideals of

regular local rings

have a minimal number of

generators

.

Title

regular local ring

Canonical name

RegularLocalRing

Date of creation

2013-03-22 13:20:14

Last modified on

2013-03-22 13:20:14

Owner

mps (409)

Last modified by

mps (409)

Numerical id

6

Author

mps (409)

Entry type

Definition

Classification

msc 13H05

Generated on Fri Feb 9 17:33:57 2018 by

LaTeXML
ideals of a discrete valuation ring are powers of its maximal ideal

ideals of a discrete valuation ring are powers of its maximal ideal

Theorem 1

.

Let

R

be a

discrete valuation ring

. Then all nonzero ideals of

R

are powers of its

maximal ideal

m

.

Proof.

Let

𝔪

=

(

π

)

(that is,

π

is a uniformizer for

R

). Assume that

R

is not a field (in which case the result is trivial), so that

π

≠

0

.
Let

I

=

(

α

)

⊂

R

be any ideal; claim

(

α

)

=

𝔪

k

for some

k

. By the

Krull intersection theorem

, we have

⋂

n

≥

0

𝔪

n

=

(

0

)

so that we may choose

k

≥

0

with

α

∈

𝔪

k

-

𝔪

k

+

1

. Since

α

∈

𝔪

k

, we have

α

=

u

⁢

π

k

for

u

∈

R

.

u

∉

𝔪

, since otherwise

α

∈

𝔪

k

+

1

, so that

α

is a unit (in a DVR, the maximal ideal consists precisely of the nonunits). Thus

(

α

)

=

(

π

)

k

.

Corollary 1

.

Let

R

be a

Noetherian

local ring

with a principal maximal ideal. Then all nonzero ideals are powers of the maximal ideal

m

.

Proof.

Let

I

=

(

α

1

,

…

,

α

n

)

be an ideal of

R

. Then by the above argument, for each

i

,

α

i

=

u

i

⁢

π

k

i

for

u

i

a unit, and thus

I

=

(

π

k

1

,

…

,

π

k

n

)

=

(

π

k

)

for

k

=

min

⁡

(

k

1

,

…

,

k

n

)

.

Title

ideals of a discrete valuation ring are powers of its maximal ideal

Canonical name

IdealsOfADiscreteValuationRingArePowersOfItsMaximalIdeal

Date of creation

2013-03-22 18:00:47

Last modified on

2013-03-22 18:00:47

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13H10

Classification

msc 13F30

Related topic

PAdicCanonicalForm

Related topic

IdealDecompositionInDedekindDomain

Generated on Fri Feb 9 17:33:59 2018 by

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j-multiplicity

j-multiplicity

Let

(

R

,

𝔪

)

be a

Noetherian

local ring

with

proper ideal

I

. Define

j

⁢

(

I

)

=

lim

n

→

∞

⁡

(

d

-

1

)

!

n

d

-

1

⁢

length

R

⁡

(

H

𝔪

0

⁢

(

I

n

/

I

n

+

1

)

)

and call it the

j-multiplicity

of

I

. Here

H

𝔪

0

⁢

(

∙

)

is the 0-th local cohomology functor. When

I

is

𝔪

-

primary

, it is same as the Hilbert-Samuel multiplicity

e

I

⁢

(

R

)

.

Title

j-multiplicity

Canonical name

Jmultiplicity

Date of creation

2013-03-22 18:15:26

Last modified on

2013-03-22 18:15:26

Owner

yshen (21076)

Last modified by

yshen (21076)

Numerical id

6

Author

yshen (21076)

Entry type

Definition

Classification

msc 13H15

Generated on Fri Feb 9 17:34:00 2018 by

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dominates (local ring)

dominates (local ring)

A

local ring

(

S

,

𝔫

)

is said to

dominate

a local ring (

http://planetmath.org/LocalRing

)

(

R

,

𝔪

)

if

R

is a

subring

of

S

, and

𝔫

∩

R

=

𝔪

.

Title

dominates (local ring)

Canonical name

DominateslocalRing

Date of creation

2013-03-22 18:05:25

Last modified on

2013-03-22 18:05:25

Owner

sjm1979 (13837)

Last modified by

sjm1979 (13837)

Numerical id

4

Author

sjm1979 (13837)

Entry type

Definition

Classification

msc 13H99

Defines

dominates

Generated on Fri Feb 9 17:34:02 2018 by

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Hensel’s lemma

Hensel’s lemma

The following results are used to show the

existence

of a solution to

polynomial equations

over

local fields

. Notice the similarities with Newton’s method.

Theorem

(Hensel’s Lemma).

Let

K

be a local field (

http://planetmath.org/LocalField

),

complete

with respect to a

valuation

|

⋅

|

. Let

O

K

be the

ring of integers

in

K

(i.e. the set of elements of

K

with

|

k

|

≤

1

). Let

f

⁢

(

x

)

be a

polynomial

with

coefficients

in

O

K

and suppose there exist

α

0

∈

O

K

such that

|

f

⁢

(

α

0

)

|

<

|

f

′

⁢

(

α

0

)

2

|

.

Then there exist a root

α

∈

K

of

f

⁢

(

x

)

. Moreover, the

sequence

:

α

i

+

1

=

α

i

-

f

⁢

(

α

i

)

f

′

⁢

(

α

i

)

converges

to

α

. Furthermore:

|

α

-

α

0

|

≤

|

f

⁢

(

α

i

)

f

′

⁢

(

α

i

)

|

<

1

.

Corollary

(Trivial case of Hensel’s lemma).

Let

K

be a

number field

and let

p

be a

prime ideal

in the ring of integers

O

K

. Let

K

p

be the

completion

of

K

at the

finite place

p

and let

O

p

be the ring of integers in

K

p

. Let

f

⁢

(

x

)

be a polynomial with coefficients in

O

p

and suppose there exist

α

0

∈

O

p

such that

f

⁢

(

α

0

)

≡

0

mod

𝔭

,

f

′

⁢

(

α

0

)

≠

0

mod

𝔭

.

Then there exist a root

α

∈

K

p

of

f

⁢

(

x

)

, i.e.

f

⁢

(

α

)

=

0

.

Title

Hensel’s lemma

Canonical name

HenselsLemma

Date of creation

2013-03-22 15:08:30

Last modified on

2013-03-22 15:08:30

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Theorem

Classification

msc 13H99

Classification

msc 12J99

Classification

msc 11S99

Generated on Fri Feb 9 17:34:04 2018 by

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local field

local field

A

local field

is a

topological field

which is

Hausdorff

and

locally compact

as a

topological space

.

Examples of local fields include:

•

Any field together with the

discrete topology

.

•

The field

ℝ

of

real numbers

.

•

The field

ℂ

of

complex numbers

.

•

The field

ℚ

p

of

p

–adic

rationals

(

http://planetmath.org/PAdicIntegers

), or any

finite extension

thereof.

•

The field

𝔽

q

⁢

(

(

t

)

)

of

formal Laurent series

in one variable

t

with

coefficients

in the

finite field

𝔽

q

of

q

elements.

In fact, this list is

complete

—every local field is

isomorphic

as a topological field to one of the above fields.

1

Acknowledgements

This document is dedicated to those who made it all the way through Serre’s book

[

1

]

before realizing that nowhere within the book is there a definition of the term “local field.”

References

1

Jean–Pierre Serre,

Local Fields

, Springer–Verlag, 1979 (GTM

67

).

Title

local field

Canonical name

LocalField

Date of creation

2013-03-22 12:48:07

Last modified on

2013-03-22 12:48:07

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Definition

Classification

msc 13H99

Classification

msc 12J99

Classification

msc 11S99

Generated on Fri Feb 9 17:34:06 2018 by

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proof of Hensel’s lemma

proof of Hensel’s lemma

Lemma:

Using the setup and terminology of the statement of Hensel’s Lemma, for

i

≥

0

,

i)

|

f

′

⁢

(

α

i

)

|

=

|

f

′

⁢

(

α

0

)

|

ii)

|

f

⁢

(

α

i

)

f

′

⁢

(

α

i

)

2

|

≤

D

2

i

iii)

|

α

i

-

α

0

|

≤

D

iv)

α

i

∈

𝒪

K

where

D

=

|

f

⁢

(

α

0

)

f

′

⁢

(

α

0

)

2

|

.

Proof:

All four statements clearly hold when

i

=

0

. Suppose they are true for

i

. The proof for

i

+

1

essentially uses Taylor’s

formula

. Let

δ

=

|

-

f

⁢

(

α

i

)

f

′

⁢

(

α

i

)

|

. Then

f

′

⁢

(

α

i

+

1

)

=

f

′

⁢

(

α

i

+

δ

)

=

f

′

⁢

(

α

i

)

+

δ

⁢

u

f

⁢

(

α

i

+

1

)

=

f

⁢

(

α

i

+

δ

)

=

f

⁢

(

α

i

)

+

f

′

⁢

(

α

i

)

⁢

δ

+

δ

2

⁢

v

for

u

,

v

∈

𝒪

K

.

|

δ

|

≤

D

2

i

⁢

|

f

′

⁢

(

α

i

)

|

by

induction

, and since

D

<

1

, it follows that

|

δ

|

<

|

f

′

⁢

(

α

i

)

|

. Since the norm is

non-Archimedean

, we see that

f

′

⁢

(

α

i

+

1

)

=

f

′

⁢

(

α

i

)

proving i).

f

⁢

(

α

i

)

+

f

′

⁢

(

α

i

)

⁢

δ

=

0

by definition of

δ

, so

f

⁢

(

α

i

+

1

)

=

δ

2

⁢

v

and hence

|

f

⁢

(

α

i

+

1

)

|

≤

|

δ

2

|

. Hence

|

f

⁢

(

α

i

+

1

)

f

′

⁢

(

α

i

+

1

)

2

|

≤

|

δ

|

2

|

f

′

⁢

(

α

i

+

1

)

|

2

=

|

δ

|

2

|

f

′

⁢

(

α

i

)

|

2

=

(

|

δ

|

|

f

′

⁢

(

α

i

)

|

)

2

=

(

|

f

⁢

(

α

i

)

|

|

f

′

⁢

(

α

i

)

|

2

)

2

≤

D

2

i

+

1

where the last equality follows by induction. This proves ii).

To prove iii), note that

|

α

i

+

1

-

α

i

|

=

|

δ

|

by the definitions of

δ

and

α

i

+

1

, so

|

α

i

+

1

-

α

i

|

≤

D

2

i

|

f

′

(

α

i

)

|

=

D

2

i

|

f

′

(

α

0

)

<

D

when

i

>

0

since

D

2

<

D

=

|

f

⁢

(

α

0

)

f

′

⁢

(

α

0

)

2

|

. So by induction,

|

α

i

+

1

-

α

0

|

≤

D

.

Finally, to prove iv) and the proof of the lemma,

δ

∈

𝒪

K

since

|

δ

|

<

|

f

⁢

(

α

0

)

f

′

⁢

(

α

0

)

|

≤

1

and hence is in the

valuation ring

of

K

. So by induction,

α

i

+

1

=

α

i

+

δ

∈

𝒪

K

.

Proof of Hensel’s Lemma:

To prove Hensel’s lemma from the above lemma, note that

δ

=

δ

i

→

0

since

|

δ

|

≤

D

2

i

⁢

|

f

′

⁢

(

α

0

)

|

, so

{

α

i

}

converges

to

α

∈

𝒪

K

since

K

is

complete

. Thus

f

⁢

(

α

i

)

→

f

⁢

(

α

)

by continuity. But

|

f

⁢

(

α

i

)

|

≤

|

δ

2

|

=

D

2

i

+

1

⁢

|

f

′

⁢

(

α

0

)

|

, so

|

f

⁢

(

α

i

)

|

→

0

, so

f

⁢

(

α

)

=

0

and the proof is complete.

Title

proof of Hensel’s lemma

Canonical name

ProofOfHenselsLemma

Date of creation

2013-03-22 15:32:16

Last modified on

2013-03-22 15:32:16

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Proof

Classification

msc 13H99

Classification

msc 12J99

Classification

msc 11S99

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