Recall (Corollary 2.4.6) that we proved that the
ring of integers of a number field is noetherian, as follows.
As we saw before using norms, the ring is finitely generated as
a module over
, so it is certainly finitely generated as a ring
over
. By the Hilbert Basis Theorem, is noetherian.
If is an integral domain, the field of fractions
of is the field of all equivalence classes of formal quotients
, where with , and
if .
For example, the field of fractions of
is (canonically isomorphic
to)
and the field of fractions of
is
. The field of fractions of the ring of integers
of a number field is just the number field .
Example 3.1.1
We compute the fraction fields mentioned above.
sage: Frac(ZZ)
Rational Field
In Sage the
Frac command usually returns a field canonically
isomorphic to the fraction field (not a formal construction).
sage: K.<a> = QuadraticField(5)
sage: OK = K.ring_of_integers(); OK
Order with module basis 1/2*a + 1/2, a in Number Field
in a with defining polynomial x^2 - 5
sage: Frac(OK)
Number Field in a with defining polynomial x^2 - 5
The fraction field of an order - i.e., a subring of of
finite index - is also the number field again.
sage: O2 = K.order(2*a); O2
Order with module basis 1, 2*a in Number Field
in a with defining polynomial x^2 - 5
sage: Frac(O2)
Number Field in a with defining polynomial x^2 - 5
Definition 3.1.2 (Integrally Closed)
An integral domain
is
integrally closed in its field of
fractions if whenever
is in the field of fractions of
and
satisfies a monic polynomial
, then
.
Proposition 3.1.3
If is any number field, then is integrally closed. Also,
the ring
of all algebraic integers (in a fixed choice of
)
is integrally closed.
Proof.
We first prove that
is integrally closed. Suppose
is integral over
, so there is a monic polynomial
with
and
. The
all lie in the ring of integers
of the
number field
, and
is finitely
generated as a
-module, so
is finitely
generated as a
-module. Since
, we can write
as a
-linear combination of
for
, so
the ring
is also finitely generated as a
-module. Thus
is finitely generated as
-module
because it is a submodule of a finitely generated
-module, which
implies that
is integral over
.
Without loss we may assume that
, so that
.
Suppose
is integral over . Then since
is
integrally closed, is an element of
, so
, as required.
Definition 3.1.4 (Dedekind Domain)
An integral domain
is a
Dedekind domain if it is noetherian,
integrally closed in its field of fractions, and every nonzero prime
ideal of
is maximal.
The ring
is not a Dedekind domain because it is not an
integral domain. The ring
is not a Dedekind domain
because it is not integrally closed in its field of fractions, as
is integrally over
and lies in
,
but not in
. The ring
is a Dedekind domain, as is
any ring of integers of a number field, as we will see below.
Also, any field is a Dedekind domain, since it is an integral
domain, it is trivially integrally closed in itself, and there are no
nonzero prime ideals so the condition that they be maximal is empty.
The ring
is not noetherian, but it is integrally closed in its
field of fraction, and every nonzero prime ideal is maximal.
Proposition 3.1.5
The ring of integers of a number field is a Dedekind domain.
Proof.
By Proposition
3.1.3, the ring
is
integrally closed, and by Proposition
2.4.6 it is
noetherian. Suppose that
is a nonzero prime ideal of
.
Let
be a nonzero element, and let
be the
minimal polynomial of
. Then
so
. Since
is irreducible,
is a nonzero element of
that lies
in
. Every element of the finitely generated abelian group
is killed by
, so
is a finite set.
Since
is prime,
is an integral domain. Every finite
integral domain is a field (see Exercise
10), so
is maximal, which completes the proof.
If and are ideals in a ring , the product is the ideal
generated by all products of elements in with elements in :
Note that the set of all products , with and ,
need not be an ideal, so it is important to take the ideal generated
by that set (see Exercise 11).
Definition 3.1.6 (Fractional Ideal)
A
fractional ideal is a nonzero
-submodule
of
that
is finitely generated as an
-module.
We will sometimes call a genuine ideal
an
integral ideal. The notion of fractional ideal makes
sense for an arbitrary Dedekind domain - it is an
-module
that is finitely
generated as an -module.
Example 3.1.7
We multiply two fractional ideals in
SAGE:
sage: K.<a> = NumberField(x^2 + 23)
sage: I = K.fractional_ideal(2, 1/2*a - 1/2)
sage: J = I^2
sage: I
Fractional ideal (2, 1/2*a - 1/2) of Number Field ...
sage: J
Fractional ideal (4, 1/2*a + 3/2) of Number Field ...
sage: I*J
Fractional ideal (-1/2*a - 3/2) of Number Field ...
Since fractional ideals are finitely generated, we can clear
denominators of a generating set to see that there is some nonzero
such that
with an integral ideal. Thus dividing by , we see
that every fractional ideal is
of the form
for some and integral ideal
.
For example, the set
of rational numbers with
denominator or is a fractional ideal of
.
Theorem 3.1.8
The set of fractional ideals of a Dedekind domain is an
abelian group under ideal multiplication with identity element .
Note that fractional ideals are nonzero by definition, so it is not
necessary to write ``nonzero fractional ideals'' in the statement of
the theorem. We will only prove Theorem 3.1.8 in the
case when is the ring of integers of a number field .
Before proving Theorem 3.1.8 we prove a lemma. For the
rest of this section is the ring of integers of a number
field .
Definition 3.1.9 (Divides for Ideals)
Suppose that
are ideals of
.
Then we say that
divides if
.
To see that this notion of divides is sensible, suppose
, so
. Then and for some integer and , and
divides means that
, i.e., that there exists
an integer such that , which exactly means that divides
, as expected.
Lemma 3.1.10
Suppose is a nonzero ideal of . Then there exist prime ideals
such that
,
i.e., divides a product of prime ideals.
Proof.
Let
be the set of nonzero ideals of
that do not
satisfy the conclusion
of the lemma. The key idea is to use that
is noetherian to show that
is the empty set. If
is
nonempty, then since
is noetherian, there is an ideal
that is maximal as an element of
. If
were prime, then
would trivially contain a product of primes, so we may assume that
is not prime. Thus there exists
such that
but
and
. Let
and
. Then neither
nor
is in
, since
is
maximal, so both
and
contain a product of prime ideals,
say
and
.
Then
so
contains a product of primes. This is a contradiction,
since we assumed
. Thus
is empty, which completes
the proof.
We are now ready to prove the theorem.
Proof.
[Proof of Theorem
3.1.8]
Note that we will
only prove Theorem
3.1.8 in the
case when
is the ring of integers of a number field
.
The product of two fractional ideals is again finitely generated, so
it is a fractional ideal, and for any ideal ,
so to prove that the set of fractional ideals under multiplication is
a group it suffices to show the existence of inverses. We will first
prove that if
is a prime ideal, then
has an inverse, then we
will prove that all nonzero integral ideals have inverses, and finally
observe that every fractional ideal has an inverse. (Note: Once we
know that the set of fractional ideals is a group, it will follows
that inverses are unique; until then we will be careful to write
``an'' instead of ``the''.)
Suppose
is a nonzero prime ideal of . We will show that
the -module
is a fractional ideal of
such that
, so that
is an inverse of
.
For the rest of the proof, fix a nonzero element
. Since
is an -module,
is an ideal, hence is
a fractional ideal. Since
we have
, hence since
is maximal, either
or
. If
, we are done since then is an inverse
of
. Thus suppose that
. Our strategy is to show that
there is some , with
. Since
, such
a would leave
invariant, i.e.,
. Since
is an -module we will see that it will follow that ,
a contradiction.
By Lemma 3.1.10, we can choose a product
,
with minimal, with
If no
is contained in
, then we can choose for each
an
with
; but then
,
which contradicts that
is a prime ideal. Thus some
, say
, is contained in
, which implies that
since
every nonzero prime ideal is maximal. Because
is minimal,
is not a subset of
, so there exists
that does not lie in
. Then
,
so by definition of
we have
. However,
, since if it were then
would be in
. We have thus
found our element
that does not lie in
.
To finish the proof that
has an inverse, we observe that
preserves the -module
, and is hence in , a
contradiction. More precisely, if
is a basis for
as a
-module, then the action of on
is given by a
matrix with entries in
, so the minimal polynomial of has
coefficients in
(because satisfies the minimal polynomial of
, by the Cayley-Hamilton theorem - here we also use that
, since
is a finite set). This implies
that is integral over
, so , since is
integrally closed by Proposition 3.1.3. (Note
how this argument depends strongly on the fact that is
integrally closed!)
So far we have proved that if
is a prime ideal of , then
is the inverse of
in
the monoid of nonzero fractional ideals of
. As mentioned after
Definition
3.1.6, every nonzero fractional
ideal is of the form
for
and
an integral ideal, so
since
has inverse
, it suffices to show that every
integral ideal
has an inverse. If not, then there is a nonzero
integral ideal
that is maximal among all nonzero integral ideals
that do not have an inverse. Every ideal is contained in a maximal
ideal, so there is a nonzero prime ideal
such that
.
Multiplying both sides of this inclusion by
and using that
, we see that
If
, then arguing as in the proof that
is an
inverse of
, we see that each element of
preserves the
finitely generated
-module
and is hence integral. But then
, which, upon multiplying both sides by
,
implies that
, a contradiction.
Thus
. Because
is
maximal among ideals that do not have an inverse, the ideal
does have an inverse
.
Then
is an inverse of
, since
We can finally deduce the crucial Theorem 3.1.11, which
will allow us to show that any nonzero ideal of a Dedekind domain can
be expressed uniquely as a product of primes (up to order). Thus
unique factorization holds for ideals in a Dedekind domain, and it is
this unique factorization that initially motivated the introduction of
ideals to mathematics over a century ago.
Proof.
Suppose
is an ideal that is maximal among the set of all ideals in
that can not be written as a product of primes. Every ideal is
contained in a maximal ideal, so
is contained in a nonzero prime
ideal
. If
, then by Theorem
3.1.8 we
can cancel
from both sides of this equation to see that
, a contradiction. Since
,
we have
, and by the above observation
is strictly contained in
. By our maximality assumption on
, there are maximal
ideals
such that
.
Then
, a contradiction. Thus every ideal
can be written as a product of primes.
Suppose
. If no
is contained in
, then for each there is an
such that
. But the product of the is in
, which is a subset of
, which contradicts that
is a prime ideal. Thus
for some . We can thus
cancel
and
from both sides of the equation by multiplying
both sides by the inverse. Repeating
this argument finishes the proof of uniqueness.
Proof.
We have
for some
and integral ideal
.
Applying Theorem
3.1.11 to
,
, and
gives
an expression as claimed. For uniqueness, if one has two such product
expressions, multiply through by the denominators and use the
uniqueness part of Theorem
3.1.11
Example 3.1.13
The ring of integers of
is
.
We have
If
, with
and neither a unit, then
, so without loss
and
. If
, then
;
since the equation
has no solution with
,
there is no element in
with norm
, so
is
irreducible. Also,
is not a unit times
or times
,
since again the norms would not match up. Thus
can not be written
uniquely as a product of irreducibles in
.
Theorem
3.1.12, however, implies that the principal
ideal
can, however, be
written uniquely as a product of prime ideals.
An explicit decomposition is
|
(3.1) |
where each of the ideals
and
is
prime. We will discuss algorithms for computing such a decomposition
in detail in Chapter
4. The first idea is to
write
, and hence reduce to the case of writing the
,
for
prime, as a product of primes. Next one decomposes the
finite (as a set) ring
.
The factorization (3.1.1) can be compute using
SAGE as follows:
sage: K.<a> = NumberField(x^2 + 6); K
Number Field in a with defining polynomial x^2 + 6
sage: K.factor_integer(6)
(Fractional ideal (2, a) of Number Field ...)^2 *
(Fractional ideal (3, a) of Number Field ...)^2
William Stein
2012-09-24