proof of prime ideal decomposition in quadratic extensions of ℚ

proof of prime ideal decomposition in quadratic extensions of

ℚ

Much of the proof of this theorem is given in Marcus’

Number Fields

(

http://planetmath.org/NumberField

); however, all of the details will be filled in here, and some aspects of the proof here will differ from those of Marcus.

Note that

gcd

⁡

(

a

,

b

)

refers to the

greatest common divisor

in

Z

of

a

and

b

(which must necessarily be

rational integers

).

Proof.

Let

d

be a

squarefree

integer

with

d

≠

1

and

K

=

ℚ

⁢

(

d

)

.

If

p

is a

rational prime

that divides

d

, then

⟨

p

,

d

⟩

2

=

⟨

p

2

,

p

⁢

d

,

d

⟩

=

⟨

gcd

⁡

(

p

2

,

d

)

,

p

⁢

d

⟩

=

⟨

p

,

p

⁢

d

⟩

=

⟨

p

⟩

=

p

⁢

𝒪

K

.

Note that

⟨

p

,

d

⟩

≠

𝒪

K

. (If they were equal, then

⟨

p

,

d

⟩

2

would equal

𝒪

K

.)

If

d

≡

3

⁢

mod

⁡

4

, then

disc

⁡

(

K

)

=

4

⁢

d

. Note that

2

divides

disc

⁡

(

K

)

. Thus,

2

ramifies in

𝒪

K

. Therefore,

2

⁢

𝒪

K

=

P

2

for some

prime ideal

P

of

𝒪

K

. Moreover,

P

is the unique ideal of

𝒪

K

of norm (

http://planetmath.org/IdealNorm

)

2

. Since

d

≡

-

1

⁢

mod

⁡

⟨

2

,

1

+

d

⟩

, then

𝒪

K

/

⟨

2

,

1

+

d

⟩

=

{

a

+

b

⁢

d

+

⟨

2

,

1

+

d

⟩

:

a

,

b

∈

ℤ

}

=

{

a

-

b

+

⟨

2

,

1

+

d

⟩

:

a

,

b

∈

ℤ

}

=

{

0

+

⟨

2

,

1

+

d

⟩

,

1

+

⟨

2

,

1

+

d

⟩

}

.

Since

⟨

2

,

1

+

d

⟩

has

2

, it follows that

P

=

⟨

2

,

1

+

d

⟩

and

2

⁢

𝒪

K

=

⟨

2

,

1

+

d

⟩

2

.

If

d

≡

1

⁢

mod

⁡

8

, then

disc

⁡

(

K

)

=

d

. Note that

2

does not divide

disc

⁡

(

K

)

. Thus,

2

does not ramify in

𝒪

K

. Since

⟨

2

,

1

+

d

2

⟩

⁢

⟨

2

,

1

-

d

2

⟩

=

⟨

4

,

1

+

d

,

1

-

d

,

1

-

d

4

⟩

=

⟨

4

,

2

,

2

⁢

(

1

-

d

2

)

,

1

-

d

4

⟩

=

⟨

2

⟩

=

2

⁢

𝒪

K

,

we have that

⟨

2

,

1

+

d

2

⟩

and

⟨

2

,

1

-

d

2

⟩

must be distinct. Proving that these ideals are indeed given below.

If

d

≡

5

⁢

mod

⁡

8

, then consider the

minimal polynomial

f

⁢

(

x

)

∈

ℤ

⁢

[

x

]

for

1

+

d

2

. Since

1

+

d

2

∉

ℚ

, it must be the case that

deg

⁡

f

≥

2

.

α

=

1

+

d

2

2

⁢

α

-

1

=

d

(

2

⁢

α

-

1

)

2

=

d

4

⁢

α

2

-

4

⁢

α

+

1

=

d

4

⁢

α

2

-

4

⁢

α

+

1

-

d

=

0

α

2

-

α

+

1

-

d

4

=

0

Thus,

f

⁢

(

x

)

=

x

2

-

x

+

1

-

d

4

.

Let

P

be a lying over

2

in

𝒪

K

. Note that

f

⁢

(

x

)

has a root (

http://planetmath.org/Root

) in

𝒪

K

and thus in

𝒪

K

/

P

. On the other hand, since

f

⁢

(

x

)

≡

x

2

+

x

+

1

⁢

mod

⁡

2

,

f

⁢

(

x

)

considered as an element of

𝔽

2

⁢

[

x

]

has no root in

𝔽

2

. Thus,

𝒪

K

/

P

and

𝔽

2

are not

isomorphic

. Therefore,

[

𝒪

K

/

P

:

𝔽

2

]

>

1

. Since

1

<

[

𝒪

K

/

P

:

𝔽

2

]

=

f

(

P

|

2

)

≤

[

K

:

ℚ

]

=

2

, we have that

f

(

P

|

2

)

=

2

. Thus,

2

is inert in

𝒪

K

. It follows that

2

⁢

𝒪

K

is in

𝒪

K

.

If

p

is an

odd prime

(

http://planetmath.org/Prime

) that does not divide

d

and

d

≡

n

2

⁢

mod

⁡

p

, then

p

does not divide

disc

⁡

(

K

)

(which equals either

d

or

4

⁢

d

). Thus,

p

does not ramify in

𝒪

K

. Also,

p

does not divide

n

. Since

⟨

p

,

n

+

d

⟩

⁢

⟨

p

,

n

-

d

⟩

=

⟨

p

2

,

p

⁢

n

+

p

⁢

d

,

p

⁢

n

-

p

⁢

d

,

n

2

-

d

⟩

=

⟨

p

2

,

2

⁢

p

⁢

n

,

p

⁢

n

-

p

⁢

d

,

n

2

-

d

⟩

=

⟨

gcd

⁡

(

p

2

,

2

⁢

p

⁢

n

)

,

p

⁢

n

-

p

⁢

d

,

n

2

-

d

⟩

=

⟨

p

,

p

⁢

n

-

p

⁢

d

,

n

2

-

d

⟩

=

⟨

p

⟩

=

p

⁢

𝒪

K

,

we have that

⟨

p

,

n

+

d

⟩

and

⟨

p

,

n

-

d

⟩

must be distinct. It will be proven that these ideals are indeed .

Let

∥

I

∥

denote the norm of the ideal

I

(

http://planetmath.org/IdealNorm

) of

𝒪

K

and

σ

∈

Gal

⁡

(

K

/

ℚ

)

with

σ

⁢

(

d

)

=

-

d

. Then

∥

⟨

p

,

n

+

d

⟩

∥

=

σ

⁢

(

∥

⟨

p

,

n

+

d

⟩

∥

)

=

∥

σ

⁢

(

⟨

p

,

n

+

d

⟩

)

∥

=

∥

⟨

σ

⁢

(

p

)

,

σ

⁢

(

n

+

d

)

⟩

∥

=

∥

⟨

p

,

n

-

d

⟩

∥

.

Note that

p

2

=

∥

p

⁢

𝒪

K

∥

=

∥

⟨

p

,

n

+

d

⟩

∥

⁢

∥

⟨

p

,

n

-

d

⟩

∥

=

∥

⟨

p

,

n

-

d

⟩

∥

2

. Therefore,

∥

⟨

p

,

n

+

d

⟩

∥

=

∥

⟨

p

,

n

-

d

⟩

∥

=

p

. It follows that the indicated ideals are .

Finally, if

p

is an odd prime that does not divide

d

and

d

is not a square

mod

⁡

p

, then consider the minimal polynomial

g

⁢

(

x

)

=

x

2

-

d

for

d

over

ℚ

. Let

P

be a lying over

p

in

𝒪

K

. Note that

g

⁢

(

x

)

has a root in

𝒪

K

and thus in

𝒪

K

/

P

. On the other hand, since

disc

⁡

g

⁢

(

x

)

=

-

4

⁢

(

-

d

)

=

4

⁢

d

, which is not a square in

𝔽

p

, then

g

⁢

(

x

)

considered as an element of

𝔽

p

⁢

[

x

]

has no root in

𝔽

p

. Thus,

𝒪

K

/

P

and

𝔽

p

are not isomorphic. Therefore,

[

𝒪

K

/

P

:

𝔽

p

]

>

1

. Note that

1

<

[

𝒪

K

/

P

:

𝔽

p

]

=

f

(

P

|

p

)

≤

[

K

:

ℚ

]

=

2

. Thus,

f

(

P

|

p

)

=

2

. Therefore,

p

is inert in

𝒪

K

. It follows that

p

⁢

𝒪

K

is in

𝒪

K

.
∎

References

1

Marcus, Daniel A.

Number Fields

. New York: Springer-Verlag, 1977.

Title

proof of prime ideal

decomposition

in

quadratic extensions

of

ℚ

Canonical name

ProofOfPrimeIdealDecompositionInQuadraticExtensionsOfmathbbQ

Date of creation

2013-03-22 15:59:06

Last modified on

2013-03-22 15:59:06

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

20

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11R11

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