examples of prime ideal decomposition in number fields

examples of prime ideal decomposition in number fields

Here we follow the notation of the entry on the

decomposition
group

. See also

http://planetmath.org/encyclopedia/PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ.html

this entry.

Example 1

Let

K

=

ℚ

⁢

(

-

7

)

; then

Gal

⁡

(

K

/

ℚ

)

=

{

Id

,

σ

}

≅

ℤ

/

2

⁢

ℤ

, where

σ

is the

complex conjugation

map. Let

𝒪

K

be the

ring of integers

of

K

. In this case:

𝒪

K

=

ℤ

⁢

[

1

+

-

7

2

]

The

discriminant

of this field is

D

K

/

ℚ

=

-

7

. We look at the
decomposition in

prime ideals

of some prime ideals in

ℤ

:

1.

The only prime ideal in

ℤ

that ramifies is

(

7

)

:

(

7

)

⁢

𝒪

K

=

(

-

7

)

2

and we have

e

=

2

,

f

=

g

=

1

. Next
we compute the decomposition and inertia groups from the
definitions. Notice that both

Id

,

σ

fix the
ideal

(

-

7

)

. Thus:

D

⁢

(

(

-

7

)

/

(

7

)

)

=

Gal

⁡

(

K

/

ℚ

)

For the inertia group, notice that

σ

≡

Id

⁡

mod

⁡

(

-

7

)

. Hence:

T

⁢

(

(

-

7

)

/

(

7

)

)

=

Gal

⁡

(

K

/

ℚ

)

Also note that this is trivial if we use the properties of the

fixed field

of

D

⁢

(

(

-

7

)

/

(

7

)

)

and

T

⁢

(

(

-

7

)

/

(

7

)

)

(see
the

section

on “decomposition of

extensions

” in the entry on
decomposition group), and the fact that

e

⋅

f

⋅

g

=

n

, where

n

is the degree of the extension (

n

=

2

in our case).

2.

The primes

(

5

)

,

(

13

)

are inert, i.e. they are prime ideals
in

𝒪

K

. Thus

e

=

1

=

g

,

f

=

2

. Obviously the

conjugation

map

σ

fixes the ideals

(

5

)

,

(

13

)

, so

D

⁢

(

5

⁢

𝒪

K

/

(

5

)

)

=

Gal

⁡

(

K

/

ℚ

)

=

D

⁢

(

13

⁢

𝒪

K

/

(

13

)

)

On the other hand

σ

⁢

(

-

7

)

≡

-

-

7

⁢

mod

⁡

(

5

)

,

(

13

)

, so

σ

≠

Id

⁡

mod

⁡

(

5

)

,

(

13

)

and

T

⁢

(

5

⁢

𝒪

K

/

(

5

)

)

=

{

Id

}

=

T

⁢

(

13

⁢

𝒪

K

/

(

13

)

)

3.

The primes

(

2

)

,

(

29

)

are split:

2

⁢

𝒪

K

=

(

2

,

1

+

-

7

2

)

⁢

(

2

,

1

-

-

7

2

)

=

𝒫

⋅

𝒫

′

29

⁢

𝒪

K

=

(

29

,

14

+

-

7

)

⁢

(

29

,

14

-

-

7

)

=

ℛ

⋅

ℛ

′

so

e

=

f

=

1

,

g

=

2

and

D

⁢

(

𝒫

/

(

2

)

)

=

T

⁢

(

𝒫

/

(

2

)

)

=

{

Id

}

=

D

⁢

(

ℛ

/

(

29

)

)

=

T

⁢

(

ℛ

/

(

29

)

)

Example 2

Let

ζ

7

=

e

2

⁢

π

⁢

i

7

, i.e. a

7

t

⁢

h

-

root of unity

,
and let

L

=

ℚ

⁢

(

ζ

7

)

. This is a

cyclotomic extension

of

ℚ

with

Galois group

Gal

⁡

(

L

/

ℚ

)

≅

(

ℤ

/

7

⁢

ℤ

)

×

≅

ℤ

/

6

⁢

ℤ

Moreover

Gal

(

L

/

ℚ

)

=

{

σ

a

:

L

→

L

∣

σ

a

(

ζ

7

)

=

ζ

7

a

,

a

∈

(

ℤ

/

7

ℤ

)

×

}

Galois theory

gives us the

subfields

of

L

:

\xymatrix

⁢

@

⁢

d

⁢

r

⁢

@

⁢

C

=

1

⁢

p

⁢

c

⁢

L

=

ℚ

⁢

(

ζ

7

)

⁢

\ar

⁢

@

-

[

r

]

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

ℚ

⁢

(

ζ

7

+

ζ

7

6

)

⁢

\ar

⁢

@

-

[

d

]

⁢

ℚ

⁢

(

-

7

)

⁢

\ar

⁢

@

-

[

r

]

⁢

&

⁢

ℚ

The discriminant of the extension

L

/

ℚ

is

D

L

/

ℚ

=

-

7

5

. Let

𝒪

L

denote the ring of
integers of

L

, thus

𝒪

L

=

ℤ

⁢

[

ζ

7

]

. We use the results of

http://planetmath.org/encyclopedia/PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ.html

this entry to find the decomposition of the primes

2

,

5

,

7

,

13

,

29

:

\xymatrix

⁢

L

=

ℚ

⁢

(

ζ

7

)

⁢

\ar

⁢

@

-

[

d

]

3

⁢

&

⁢

(

1

-

ζ

7

)

6

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝔓

⋅

𝔓

′

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

(

5

)

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

𝔔

1

⋅

𝔔

2

⋅

𝔔

3

⁢

\ar

⁢

@

-

[

d

]

⁢

K

=

ℚ

⁢

(

-

7

)

⁢

\ar

⁢

@

-

[

d

]

2

⁢

&

⁢

(

-

7

)

2

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

(

2

,

1

+

-

7

2

)

⁢

(

2

,

1

-

-

7

2

)

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

(

5

)

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

(

13

)

⁢

\ar

⁢

@

-

[

d

]

⁢

ℚ

⁢

&

⁢

(

7

)

⁢

&

⁢

(

2

)

⁢

&

⁢

(

5

)

⁢

&

⁢

(

13

)

1.

The prime ideal

7

⁢

ℤ

is

totally ramified

in

L

, and the
only prime ideal that ramifies:

7

⁢

𝒪

L

=

(

1

-

ζ

7

)

6

=

𝔗

6

Thus

e

⁢

(

𝔗

/

(

7

)

)

=

6

,

f

⁢

(

𝔗

/

(

7

)

)

=

g

⁢

(

𝔗

/

(

7

)

)

=

1

Note that, by the properties of the fixed fields of decomposition
and inertia groups, we must have

L

T

⁢

(

𝔗

/

(

7

)

)

=

ℚ

=

L

D

⁢

(

𝔗

/

(

7

)

)

, thus, by
Galois theory,

D

⁢

(

𝔗

/

(

7

)

)

=

T

⁢

(

𝔗

/

(

7

)

)

=

Gal

⁡

(

L

/

ℚ

)

2.

The ideal

2

⁢

ℤ

factors in

K

as above,

2

⁢

𝒪

K

=

𝒫

⋅

𝒫

′

, and each of the
prime ideals

𝒫

,

𝒫

′

remains inert from

K

to

L

, i.e.

𝒫

⁢

𝒪

L

=

𝔓

, a prime ideal
of

L

. Note also that the order of

2

⁢

mod

⁡

7

is

3

, and since

g

is at least

2

,

2

⋅

3

=

6

, so

e

must equal

1

(recall that

e

⁢

f

⁢

g

=

n

):

e

⁢

(

𝔓

/

(

2

)

)

=

1

,

f

⁢

(

𝔓

/

(

2

)

)

=

3

,

g

⁢

(

𝔓

/

(

2

)

)

=

2

Since

e

=

1

,

L

T

⁢

(

𝔓

/

(

2

)

)

=

L

, and

[

L

:

L

D

⁢

(

𝔓

/

(

2

)

)

]

=

3

, so

D

⁢

(

𝔓

/

(

2

)

)

=

<

σ

2

>

≅

ℤ

/

3

⁢

ℤ

,

T

⁢

(

𝔓

/

(

2

)

)

=

{

Id

}

3.

The ideal

(

5

)

is inert,

5

⁢

𝒪

L

=

𝔖

is
prime and the order of

5

modulo

7

is

6

. Thus:

e

⁢

(

𝔖

/

(

5

)

)

=

1

,

f

⁢

(

𝔖

/

(

5

)

)

=

6

,

g

⁢

(

𝔖

/

(

5

)

)

=

1

D

⁢

(

𝔖

/

(

5

)

)

=

Gal

⁡

(

L

/

ℚ

)

,

T

⁢

(

𝔖

/

(

5

)

)

=

{

Id

}

4.

The prime ideal

13

⁢

ℤ

is inert in

K

but it splits in

L

,

13

⁢

𝒪

L

=

𝔔

1

⋅

𝔔

2

⋅

𝔔

3

, and

13

≡

6

≡

-

1

⁢

mod

⁡

7

, so the order of

13

is

2

:

e

⁢

(

𝔔

i

/

(

13

)

)

=

1

,

f

⁢

(

𝔔

i

/

(

13

)

)

=

2

,

g

⁢

(

𝔔

i

/

(

13

)

)

=

3

D

⁢

(

𝔔

i

/

(

13

)

)

=

<

σ

6

>

≅

ℤ

/

2

⁢

ℤ

,

T

⁢

(

𝔔

i

/

(

13

)

)

=

{

Id

}

5.

The prime ideal

29

⁢

ℤ

is splits completely in

L

,

29

⁢

𝒪

L

=

ℜ

1

⋅

ℜ

2

⋅

ℜ

3

⋅

ℜ

′

1

⋅

ℜ

′

2

⋅

ℜ

′

3

Also

29

≡

1

⁢

mod

⁡

7

, so

f

=

1

,

e

(

ℜ

i

/

(

29

)

)

=

1

,

f

(

ℜ

i

/

(

29

)

=

1

,

g

(

ℜ

i

/

(

29

)

)

=

6

D

⁢

(

ℜ

i

/

(

29

)

)

=

T

⁢

(

ℜ

i

/

(

29

)

)

=

{

Id

}

Title

examples of prime ideal decomposition in number fields

Canonical name

ExamplesOfPrimeIdealDecompositionInNumberFields

Date of creation

2013-03-22 13:53:05

Last modified on

2013-03-22 13:53:05

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

12

Author

alozano (2414)

Entry type

Example

Classification

msc 11S15

Related topic

DecompositionGroup

Related topic

Discriminant

Related topic

NumberField

Related topic

PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ

Related topic

PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ

Related topic

ExamplesOfRamificationOfArchimedeanPlaces

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