- 12.1 Dedekind domains
- 12.2 Integral extensions
- 12.3 Primary decomposition
- 12.4 Noetherian rings

** 12.1.1. Definition.**
An integral domain D is called a
** Dedekind domain**
if each proper ideal of D can be written
as a product of a finite number of prime ideals of D.

We will show in
Theorem 12.2.4
that a Dedekind domain has some of the properties of a principal ideal domain.
Specifically, a Dedekind domain must be Noetherian,
and any nonzero prime ideal of a Dedekind domain must be maximal.
This shows that a unique factorization domain can fail to be a Dedekind domain,
since, for example,
the unique factorization domain * Q*[x,y]
contains the nonzero prime ideals
<x> <x,y>.

We will prove a number of facts about Dedekind domains by using the notion of an ``inverse'' of an ideal. This necessitates the introduction of the following concept.

** 12.1.2. Definition.**
Let D be an integral domain with quotient field F. A
** fractional ideal**
of D is a nonzero D-submodule I of F such that there exists
0d D
with dI D.

If I is a fractional ideal of D, we define

I^{-1} =
{ q F |
qI D },

and say that I is
** invertible**
if I^{-1}I = D.

** 12.1.3. Lemma.**
Let D be an integral domain with quotient field F,
and let I be an ideal of D
that is invertible when considered as a fractional ideal.
The following conditions hold.

**(a)**
The ideal I is finitely generated.

**(b)**
If I is a product of prime ideals,
then this product is unique (up to order).

** 12.1.4. Theorem.**
The following conditions hold for any Dedekind domain D.

**(a)**
Every nonzero ideal of D is invertible.

**(b)**
Every proper ideal of D can be written uniquely (up to order)
as a product of a finite number of prime ideals of D;

**(c)**
D is a Noetherian domain.

**(d)**
Every nonzero prime ideal of D is maximal.

** 12.1.5. Proposition.**
Let D be an integral domain.
A fractional ideal I of D is invertible if and only if
it is projective as a D-module.

** 12.1.6. Theorem.**
The following conditions are equivalent for an integral domain D.

**(1)**
D is a Dedekind domain;

**(2)**
every nonzero ideal of D is invertible;

**(3)**
every fractional ideal of D is invertible;

**(4)**
every nonzero ideal of D is projective as a D-module.

** 12.1.7. Proposition.**
Let D be an integral domain with quotient field F.
Assume that D is Noetherian
and that every nonzero prime ideal of D is maximal.
Then for any proper nonzero ideal I of D there exists
q F-D with
qI D.

** 12.2.1. Definition.**
Let R be a subring of the commutative ring T. An element
u T is said to be
** integral over R**
if there exists a monic polynomial
f(x) R[x]
such that f(u)=0.

The ring T is said to be an
** integral extension**
of R if each element of T is integral over R.

** 12.2.2. Proposition.**
Let R be a subring of the commutative ring T.
The following conditions are equivalent for an element
u T.

**(1)**
u is integral over R;

**(2)**
the subring R[u] generated by R and u
is finitely generated as an R-module;

**(3)**
there exists a subring R' with
R R' T
such that R[u] R' and
R' is finitely generated as an R-module;

**(4)**
there exists a faithful R[u]-submodule of T
that is finitely generated as an R-module.

** 12.2.3. Corollary.**
Let R be a subring of the commutative ring T.
If T is finitely generated as a module over R,
then T is an integral extension of R.

** 12.2.4. Corollary.**
Let R be a subring of the commutative ring T. Then

R* = { u T | u is integral over R }

is a subring of T.

** 12.2.5. Definition.**
Let R be a subring of the commutative ring T.
The subring R* of all elements integral over R is called the
** integral closure**
of R in T.

If R* = R, then we say that R is
** integrally closed**
in T.
If D is an integral domain that is integrally closed
in its quotient field,
then we simply say that D is
** integrally closed**.

** 12.2.6. Proposition.**
Let R be a subring of the integral domain D,
and assume that D is an integral extension of R.
Then D is a field if and only if R is a field.

** 12.2.7. Corollary.**
Let R be a subring of the commutative ring T,
and assume that T is an integral extension of R.
If Q is a prime ideal of T,
then Q is a maximal ideal of T if and only if
QR is a maximal ideal of R.

** 12.2.8. Proposition.**
Let R be a subring of the integral domain D,
and assume that D is an integral extension of R.
If S is any multiplicative set of R, then
D_{S}
is an integral extension of
R_{S}.

** 12.2.9. Theorem.**
Let R be a subring of the integral domain D,
and assume that D is an integral extension of R.

**(a) (Incomparability)**
Let Q, Q' be prime ideals of D such that
Q Q '.
Then QQ' implies
Q R
Q' R.

**(b) (Lying-over)**
For each prime ideal P of R
there exists a prime ideal Q of D such that
Q R = P.

**(c) (Going up)**
If P P'
are prime ideals of R, and Q is a prime ideal of D with
Q R = P,
then there exists a prime ideal Q' of D such that
Q Q' and
Q' R = P'.

In the setting of the previous theorem, if we assume in addition that R is an integrally closed domain, then a further condition holds, known as ``Going down'': Let R be a subring of the integral domain D, assume that D is an integral extension of R, and that R is an integrally closed domain. If P' P are prime ideals of R, and Q is a prime ideal of D with QR = P, then there exists a prime ideal Q' of D such that Q' Q and Q'R = P'.

** 12.2.10. Theorem.**
The following conditions are equivalent for an integral domain D.

**(1)**
D is a Dedekind domain;

**(2)**
D is Noetherian,
integrally closed in its quotient field,
and each nonzero prime ideal of D is maximal.

** 12.2.11. Theorem.**
Let D be an integral domain with quotient field Q,
and let F be a finite extension field of Q.
If D* is the set of all elements of F that are integral over D,
then D* is a Dedekind domain.

Let p be a prime,
and let
be a primitive pth root of unity.
If * Z*[]
denotes the subring of

To obtain an appropriate generalization of unique factorization of ideals in polynomial rings, it is necessary to replace products of ideals with intersections of ideals, and to replace powers of prime ideals with ``primary'' ideals. The primary decomposition theorem does not depend on having unique factorization of elements, but simply on having the ascending chain condition on ideals, so it remains true for all Noetherian rings.

** 12.3.1. Proposition.**
Let R be a commutative ring.
The set of all nilpotent elements of R forms an ideal of R.

** 12.3.2. Definition.**
Let R be a commutative ring. The
** nil radical**
of R is defined to be the ideal

N(R) = { x R |
x^{n} = 0 for some
n
**Z**^{+} }.

** 12.3.3. Theorem.**
Let R be a commutative ring.

**(a)**
The nil radical of R / N(R) is zero.

**(b)**
The nil radical of R is the intersection
of all prime ideals of R.

The ideal N(R) is also called the
** prime radical**
of R.
This terminology is justified by the above theorem.
This definition can be extended to noncommutative rings,
where the prime radical of R is defined to be
the intersection of all prime ideals of R.

** 12.3.4. Definition.**
Let R be a commutative ring,
and let I be an ideal of R. The ideal

I =
{ x R |
x^{n}
I for some
n **Z**^{+} }

is called the
** radical**
of I.

We note that if I is an ideal of R, then I is the inverse image in R of the nil radical of R/I, which shows that I is an ideal. It also follows that I is the intersection of all prime ideals of R that contain I.

In any principal ideal domain, our next definitions both reduce to the statement that the ideal in question is generated by a power of an irreducible element. But even for the polynomial ring in two indeterminates over a field or the ring of polynomials with integer coefficients the two concepts are distinct.

** 12.3.5. Definition.**
Let I be an ideal of the commutative ring R.

We say that I is a
** primary**
ideal if for all elements a,b R
we have the following condition:

ab I implies
a I or
b^{n} I,
for some
n **Z**^{+}.

We say that I is an
** irreducible**
ideal if

I = JK implies I=J or I=K,

for all ideals J,K of R with
I J and
I K.

Let D be a principal ideal domain,
let p be an irreducible element D, and let
Q=p^{n}D.
If a,b D with
ab Q,
then p^{n}|ab,
so either p|a or p|b.
If a I, then
pa implies
p^{n}|b,
and hence b Q.
This shows that Q is a primary ideal.

Note that if I,Q are ideals of R with
I Q,
then Q is a primary ideal of R
if and only if Q/I is a primary ideal of R/I.
If F is any field,
we can apply this observation to
R=F[x,y],
Q=<x^{2},y>,
and I=<y>
to show that <x^{2},y>
is a primary ideal of F[x,y].
This provides an example
of a primary ideal that is not a power of a prime ideal.

** 12.3.6. Proposition.**
Let R be a commutative Noetherian ring.
Then every irreducible ideal of R is a primary ideal.

** 12.3.7. Lemma.**
Let R be a commutative ring.

**(a)**
If I is a primary ideal of R, then
I is a prime ideal of R.

**(b)**
If { Q_{i} }
is a collection of primary ideals of R such that
Q_{i}=P
for 1 i n, then
Q_{i}
is a primary ideal of R with
(
Q_{i} ) = P.

** 12.3.8. Definition.**
Let Q be a primary ideal of the commutative ring R.
We say that Q is the
** associated prime ideal**
of Q. If P = Q,
then we say that Q
** belongs to the prime ideal**
P and that Q
** is primary for**
P.

** 12.3.10. Theorem.
[Lasker-Noether Decomposition Theorem]
**
Let R be a commutative, Noetherian ring,
and let I be an ideal of R.

There exist primary ideals
{ Q_{i} } with
I = Q_{i},
such that no Q_{i}
contains the intersection of the other primary ideals, and the ideals
Q_{i}
have distinct associated primes.

Furthermore, in any such representation of I
as an intersection of primary ideals, there must be n ideals,
and the set of their associated prime ideals must be the same.

** Example. ** 12.4.1.
Let R be any commutative ring.
We define the ring R [[x]] of all
** formal power series**
over R as the set of all sequences
a=(a_{0},
a_{1},
a_{2}, . . . ),
with componentwise addition and a multiplication given by defining

(a_{0},
a_{1},
a_{2}, . . . )
^{ . }
(b_{0},
b_{1},
b_{2}, . . . ) =
(c_{0},
c_{1},
c_{2}, . . . ) ,

where c_{i} =
_{i=j+k}
a_{j}
b_{k}.

It can be checked easily that R[[x]] is a ring,
that R can be identified with elements of the form
(a_{0}, 0, 0, . . . ),
and that the polynomial ring R[x]
can be identified with the subset of all sequences
with only finitely many nonzero terms.

** 12.4.1. Theorem.
[Cohen]
**
Let R be a commutative ring.
Then R is Noetherian if and only if
every prime ideal of R is finitely generated.

** 12.4.2. Corollary.**
If R is a commutative Noetherian ring,
then so is the ring R[[x]] of all formal power series over R.

** 12.4.3. Theorem.**
Let R be a commutative Noetherian ring.
Then the nil radical of R is nilpotent.
That is, there exists
m **Z**^{+}
such that N(R)^{m}=(0).

** 12.4.4. Corollary.**
If Q is a primary ideal of the commutative Noetherian ring R,
then there exists a positive integer m such that
(Q)^{m}
Q.

** 12.4.5. Lemma.**
Let J be an ideal of the commutative Noetherian ring R, and let
I =
J^{n}.
Then JI = I.

** 12.4.6. Theorem.
[Krull]
**
Let R be a commutative Noetherian ring.
Then the intersection of powers of the Jacobson radical is zero. That is,
(J(R))^{n}=(0),
where J(R) is the Jacobson radical of R.

We now state several results
whose proofs are beyond the scope
of this brief introduction to Noetherian rings.
Recall that in a principal ideal domain
every nonzero prime ideal is maximal,
so that the only proper chains of prime ideals
have the form P (0).
This result was extended by Krull to Noetherian rings,
and is known as the
principal ideal theorem:
If R is a Noetherian ring,
aR is a proper principal ideal of R,
and P is any prime ideal minimal over aR,
then P contains no chains of primes longer than
P P_{1}.

We say that a chain
P_{0}
P_{1}
. . .
P_{m}
of prime ideals has length m.
An induction argument can be used to generalize
the principal ideal theorem,
and this important result is usually called
Krull's generalized principal ideal theorem.
We will investigate some of its consequences
for the study of polynomial rings.

** 12.4.7. Theorem.
[Generalized principal ideal theorem]
**
Let I be a proper ideal of the Noetherian ring R.
If I is generated by m elements,
and P is any prime ideal minimal over I,
then any chain of primes

P
P_{1}
. . .
has length at most m.

One important consequence of the generalized principal ideal theorem
is that any Noetherian ring
satisfies the descending chain condition for prime ideals.
There may or may not be a uniform bound
on the lengths of chains of prime ideals of a Noetherian ring.
If R is Noetherian,
and has a chain of prime ideals of length n,
but none longer,
then we say that R has
** Krull dimension**
equal to n.

A field F is said to be
** algebraically closed**
if every monic polynomial of positive degree
with coefficients in F has a root in F.
As a second consequence of the generalized principal ideal theorem,
we can compute the Krull dimension of the polynomial ring
R=F[x_{1},
x_{2}, . . . ,
x_{n}],
where F is algebraically closed.
It can be shown that any maximal ideal of R is generated by n elements
x_{1}-a_{1},
x_{2}-a_{2},
. . . ,
x_{n}-a_{n}.
It follows that any chain of prime ideals of R has length at most n.
Of course, there is a chain of length n, given by

< x_{1},
x_{2}, . . . ,
x_{n} >
< x_{2},
. . . , x_{n} >
. . .
< x_{n} >
(0).

Thus R has Krull dimension n.

** 12.4.8. Lemma.**
Let F be an algebraically closed field, and let
f, f_{1}, . . . ,
f_{m} be elements of
the polynomial ring
F[x_{1}, . . . ,
x_{n}].
If the system

f(x_{1},
. . . , x_{n})
0

f_{1}(x_{1},
. . . , x_{n}) = . . . =
f_{m}(x_{1},
. . . , x_{n}) = 0

has a solution in some extension field of F,
then it must have a solution in F.

** 12.4.9. Theorem.
[Hilbert's Nullstellensatz]
**
Let F be an algebraically closed field,
and let I be any ideal of the polynomial ring
F[x_{1},
. . . , x_{n}].
Then I consists of all elements
f F[x_{1},
. . . , x_{n}]
such that
f(a_{1},
a_{2},
. . . , a_{n}) = 0 for all
(a_{1},
a_{2}, . . . ,
a_{n}) such that
g(a_{1},
a_{2}, . . . ,
a_{n}) = 0 for all
g I.