class number divisibility in extensions

class number divisibility in extensions

Throughout this entry we will use the following

corollary

to the

existence

of the

Hilbert class field

(see the parent entry,

unramified extensions and class number divisibility

for details of the proof).

Corollary 1

.

Let

K

be a

number field

,

h

K

is its

class number

and let

p

be a prime. Then

K

has an everywhere

unramified

Galois extension

of degree

p

if and only if

h

K

is

divisible

by

p

.

In this entry we are concerned about the

divisibility

properties

of class numbers of number fields in

extensions

.

Theorem 1

.

Let

F

/

K

be a Galois extension of number fields, let

h

F

and

h

K

be their respective class numbers and let

p

be a

prime number

such that

p

does not divide

[

F

:

K

]

, the degree of the extension. Then, if

p

divides

h

K

, the class number of

F

,

h

F

, is also divisible by

p

.

Proof.

Let

F

,

K

and

p

be as in the statement of the

theorem

. Assume that

p

|

h

K

. Thus, by the corollary above, there exists an unramified

Galois extension field

E

of

K

of degree

p

. Notice that the fact that

p

does not divide the degree of the extension

F

/

K

implies that

F

∩

E

=

K

. In particular, the

compositum

F

⁢

E

is a Galois extension of

F

and

Gal

⁡

(

F

⁢

E

/

F

)

≅

Gal

⁡

(

E

/

E

∩

F

)

≅

Gal

⁡

(

E

/

K

)

.

Thus, the extension

F

⁢

E

/

F

is of degree

p

, Galois, and therefore abelian. By the corollary above, in order to prove the theorem it suffices to show that the extension

F

⁢

E

/

F

is unramified. Suppose for a

contradiction

that

𝒬

F

is a

prime ideal

which ramifies in the extension

F

⁢

E

/

F

. Let

𝒬

E

⁢

F

be a prime lying above

𝒬

F

and let

𝒬

K

be a prime of

K

such that

𝒬

F

lies above it. Similarly, let

𝒬

E

be a prime of

E

lying above

𝒬

K

and such that the prime

𝒬

E

⁢

F

lies above

𝒬

E

. For an arbitrary extension

A

/

B

, the ramification index of a prime

𝒬

A

is denoted by

e

(

𝒬

B

|

𝒬

A

)

. Then, by the multiplicativity of the ramification index in towers, we have:

e

(

𝒬

E

⁢

F

|

𝒬

K

)

=

e

(

𝒬

E

⁢

F

|

𝒬

F

)

⋅

e

(

𝒬

F

|

𝒬

K

)

=

e

(

𝒬

E

⁢

F

|

𝒬

E

)

⋅

e

(

𝒬

E

|

𝒬

K

)

Since we assumed that

𝒬

F

is ramified in

E

⁢

F

/

F

, and the degree of the extension is

p

, we must have

e

(

𝒬

E

⁢

F

|

𝒬

F

)

=

p

. Therefore, by the equality above,

p

divides

e

(

𝒬

E

⁢

F

|

𝒬

E

)

⋅

e

(

𝒬

E

|

𝒬

K

)

. Notice that the extension

E

/

K

is everywhere unramified, therefore

e

(

𝒬

E

|

𝒬

K

)

=

1

. Also,

[

E

F

:

E

]

=

[

F

:

K

]

which, by

hypothesis

, is relatively prime to

p

. Thus

e

(

𝒬

E

⁢

F

|

𝒬

E

)

is also relatively prime to

p

, and so,

p

is not a

divisor

of

e

(

𝒬

E

⁢

F

|

𝒬

E

)

⋅

e

(

𝒬

E

|

𝒬

K

)

, which leads to the desired contradiction, finishing the proof of the theorem.
∎

Also, read the entry

extensions without unramified subextensions and class number divisibility

for a

similar

and more general result.

Title

class number divisibility in extensions

Canonical

name

ClassNumberDivisibilityInExtensions

Date of creation

2013-03-22 15:04:17

Last modified on

2013-03-22 15:04:17

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

9

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11R37

Classification

msc 11R32

Classification

msc 11R29

Related topic

IdealClass

Related topic

ExistenceOfHilbertClassField

Related topic

CompositumOfAGaloisExtensionAndAnotherExtensionIsGalois

Related topic

DecompositionGroup

Related topic

ExtensionsWithoutUnramifiedSubextensionsAndClassNumberDivisibility

Related topic

ClassNumbersAndDiscriminantsTopicsOnClassGroups

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