Chapter 6 - pantherFILE

Transcription

CHAPTER 6
COMMUTATIVE RING THEORY
In this chapter we present some of the basic ideas in the study of commutative rings.
One of the most fruitful connections in modern mathematics has proven to be that between
commutative rings and both number theory and geometry. We explore those connections
in this chapter.
In Section 1, we discuss basic facts about integral domains and polynomials, and
introduce a key condition we will use throughout the chapter, the Noetherian condition.
In this section, we prove the Hilbert Basis Theorem, which shows the Noetherian condition
can be transferred between a ring and its polynomial ring. In Section 2, we review
and generalize the idea of introducing fractions into a ring, and we derive some basic
properties about the rings so obtained. In Section 3, we define the basic concepts related
to factorization into prime elements in an integral domain, and we discuss when unique
factorization holds in a ring. We show for example that unique factorization can be
transferred between a ring and its polynomial ring. In Section 4, we extend the idea of
factoring elements to that of factoring ideals, and we introduce the ideal analog of the
UFD, the Dedekind domain. We connect this factorization property to the “invertibility”
of ideals. In Section 5, we show how to use localization to “promote” a prime ideal so
that it becomes the unique maximal ideal in a ring, and we study the rings obtained from
this construction. In Section 6, we broaden the notion of algebraic extension of fields to
that of integral extension of integral domains, and we show how this is connected to finite
extensions. In Section 7, we study maximal ideals in rings of functions, and we show
that often, the underlying space on which the functions are defined can be determined
from the maximal ideals of the ring of functions. In particular, we show that there is a
bijection between maximal ideals in the ring of continuous functions on a compact space
and the points of the space. In Section 8, we discuss the connection between geometry
and algebra, defining the basic terms, and proving the Nullstellensatz, which yields the
key correspondence between curves, surfaces, etc. and finitely generated algebras. In
particular, we show that points of an algebraic set correspond to maximal ideals of the
ring of polynomial functions on the set. In Section 9, we discuss how to relate geometric
dimension to some algebraic notions of dimension.
1. Integral Domains, Polynomials, and Noetherian Rings
Chapter 4 was devoted to the theory of fields; this chapter is mostly devoted to the
theory of subrings of fields, that is, integral domains. The two theories are of course
closely related, but they are quite different, since the non-existence of inverses adds both
complications and possibilities. The two most basic examples of an integral domain
which is not a field are Z and the polynomial rings F [x1 , . . . , xn ] over fields F . We
begin this section with a brief review of some facts about polynomials, including a review
of factorization and the division algorithm. We then discuss generation of ideals and
169
170
6. COMMUTATIVE RING THEORY
Noetherian rings, and we show that ideals in F [x] can be generated by a single element,
while those in F [x1 , . . . , xn ] can be generated by a finite set of elements.
As noted above, we could define an integral domain to be a subring of a field, but the
standard definition is that an integral domain is a commutative ring in which 0 6= 1
and in which ab = 0 implies a = 0 or b = 0. It is clear that any subring of a field has
these properties; the converse will follow from the construction of the quotient field in
Section 2. The properties of the integral domain Z are familiar to all, and you surely
know that one of the most important of these properties is that integers can be uniquely
factored into products of primes. We will pursue the theme of factorization in part of this
chapter, beginning in a few paragraphs with polynomials over a field.
Let us review the important example of polynomials. If R is a commutative ring, we
have defined the polynomial rings R[x] and R[x1 , . . . , xn ] before. They are, respectively,
the freeP
objects on x and on {x1 , . . . , xn } in the category of commutative R-algebras.
i
If f = ∞
i=0 ri x ∈ R[x], we define the degree of f , denoted deg f , to be the largest
value of n with rn 6= 0; if f = 0, we either define deg f = −∞ or we take deg f to
be undefined. We call the coefficient rn the leading coefficient of f ; let us call 0 the
leading coefficient of 0. We call f monic if its leading coefficient is 1. The following
lemma is an easy consequence of the definitions of the operations.
Lemma 1.1. Let R be an integral domain and let f, g ∈ R[x].
(1) deg f g = deg f + deg g and the leading coefficient of f g is the product of the
leading coefficients of f and g.
(2) deg(f ± g) ≤ max(deg f, deg g), with equality whenever deg f 6= deg g.
Proof. Exercise.
Corollary 1.2. Let R be a ring. Then R[x] is an integral domain if and only if R
is an integral domain.
Proof. Exercise.
Defining degrees of polynomials in R[x1 , . . . , xn ] is trickier: there are several possibilities. The simplest is to take the degree of a monomial xk11 . . . xknn to be k1 + · · · + kn ,
and then define the degree of f to be the largest degree of a monomial with nonzero
coefficient. We cannot then define leading coefficient (consider for example x1 − x2 ), but
otherwise Lemma 1.1 and Corollary 1.2 remain valid. Note that a polynomial has positive
degree if and only if it is non-constant.
In particular, if F is a field, the polynomial ring F [x1 , . . . , xn ] is an integral domain.
We will prove several other properties of this polynomial ring. Let us start with the important property of factorization. We say a polynomial f ∈ F [x1 , . . . , xn ] is irreducible
if f is not constant and if whenever we write f = gh, one of g, h is constant. It is clear
from Lemma 1.1 (and its multi-variable analog) that any non-constant f can be factored
into a product of irreducible polynomials; just as for integers, we would like to know that
this factorization is unique. The general result is the following; we will only be able to give
the proof in the n = 1 case here — the general case follows from the results of Section 3.
Theorem 1.3. Let F be a field.
(1) Every nonzero polynomial in F [x] can be written in the form λf1 . . . fm for some
λ ∈ F and some monic irreducible polynomials f1 , . . . , fm ∈ F [x]. Moreover, λ
is unique and f1 , . . . , fm are unique up to order.
1. INTEGRAL DOMAINS, POLYNOMIALS, AND NOETHERIAN RINGS
171
(2) Every nonconstant polynomial in F [x1 , . . . , xn ] can be written in the form f1 . . . fm
for some irreducible polynomials f1 , . . . , fm ∈ F [x1 , . . . , xn ]. Moreover, f1 , . . . , fm
are unique up to order and constant multiples.
The existence of a factorization into irreducibles follows easily by induction on degree.
In order to prove uniqueness in the n = 1 case, we need to use the division algorithm.
That is the following fact, which is Theorem 1.3 in Chapter 4.
Theorem 1.4. Let F be a field and let f, g be polynomials in F [x] with g 6= 0. Then
there exist unique polynomials q, r ∈ F [x] with f = qg + r and deg r < deg g (where we
take deg 0 = −∞).
Proof. Exercise.
Once we know this, we can give a quick proof of uniqueness as follows. Suppose
f ∈ F [x] has two distinct factorizations, and suppose f has the smallest possible degree
among such counterexamples. Let f = f1 . . . fm = g1 . . . gn be distinct factorizations of
f into monic irreducible polynomials. By re-labelling if necessary, we may assume deg f1
is less than or equal to the degree of any fi or gj . We first note that f1 cannot divide
any gj , for if it did, then the irreducibility of gj and the fact that f1 , gj are monic would
imply that f1 = gj , and so f2 . . . fm = g1 . . . gj−1 gj+1 . . . gn . This would be a polynomial
of degree less than deg f with two distinct factorizations, contradicting our choice of f .
Now write g1 = qf1 +r where deg r < deg f1 . We’ve just seen that r = 0 is impossible.
We have f1 . . . fm = qf1 g2 . . . gn + rg2 . . . gn ; simplifying this yields f1 h = rg2 . . . gn for
some h ∈ F [x]. Now factor h = h1 . . . hs and r = r1 . . . rt as products of irreducibles.
Certainly f1 6= ri for any i and we saw before that f1 6= gj for any j , so the factorizations f1 h = f1 h1 . . . hs = r1 . . . rt g1 . . . gn differ in that the one on the right contains no
occurrences of f1 . Thus f1 h has two distinct factorizations, and deg f1 h < deg f , since
deg r < deg f1 ≤ deg g1 . This contradiction proves that no such f can exist, and so proves
Theorem 1.3 in the case n = 1.
Let us now examine the ideals of a polynomial ring. If I / F [x], then either I = 0
or I contains a nonzero element f of smallest degree. (A remark on notation is in order
here: we will follow custom and generally write 0 in place of {0}.) In the latter case, let
g ∈ I . Then we can write g = qf + r where deg r < deg f . Clearly r = g − qf ∈ I , so
the choice of f implies r = 0. Thus every element of I is a multiple of f , that is I = Rf
is the ideal of R generated by f . We call an ideal principal if it can be generated by a
single element, and we call a ring a principal ideal ring if all of its ideals are principal;
an integral domain which is a principal ideal ring is called a principal ideal domain,
or simply PID. We’ve just shown F [x] is a PID.
Problem 1.A. Let F be a field and let I be the ideal of F [x, y] generated by x and y
(that is, all elements of F [x, y] with constant coefficient 0). Show that I is not a principal
ideal.
We know that F [x1 , . . . , xn ] is an integral domain, but Problem 1.A and its analog for
more variables tell us it is not a PID for n ≥ 2. Is there anything we can say? It turns
out that there is a weaker property that holds and is quite useful.
Recall that the set of ideals of a ring R is a complete lattice, with the greatest lower
bound of a collection of ideals equal to their intersection and the least upper bound
equal to their sum. In particular, the g.l.b. of I and J is I ∩ J , while the l.u.b. is
I + J = {i + j | i ∈ I, j ∈ J}. We will be interested in the order properties of this lattice,
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and we will see shortly that one important property is equivalent to every ideal being
finitely generated.
We will be especially interested in large and small ideals. We say an ideal I of R is
maximal if it is maximal in the poset of proper ideals of R and minimal if it is minimal
in the poset of nonzero ideals of R. Thus an ideal I is maximal if I ( R and there is no
ideal J with I ( J ( R. The following result is an easy application of Zorn’s Lemma.
Proposition 1.5. Every proper ideal in a ring is a contained in a maximal ideal in
the ring.
Proof. Exercise: use Zorn’s Lemma and the fact that the ring R is generated as an
ideal by 1.
Remark. (1) Proposition 1.5 is not true for pseudorings (“rings” for which we do not
assume the existence of an identity). The corresponding result for subrings of rings or
subgroups of groups is not true in general.
(2) Although Zorn’s Lemma is symmetric, it is not the case that every nonzero ideal in a
ring contains a minimal ideal. (Consider the ring Z.)
We now consider maximal and minimal elements of more general classes of ideals.
Recall that a poset satisfies the a.c.c. (the condition that every ascending chain becomes
stationary) if and only if it satisfies the maximum condition (the condition that every
non-empty subset contains a maximal element). If R is a ring and its lattice of ideals
satisfies these equivalent conditions, we say the ring is Noetherian. If the lattice of ideals
satisfies the d.c.c., or equivalently the minimum condition, we say the ring is Artinian.
The Artinian condition is much more restrictive (see Problem 1.C below), and we will not
discuss it for the moment. The Noetherian condition is equivalent to the condition that
all ideals are finitely generated, as we now prove.
Remark 1.6. The above definitions are only standard when R is commutative. When
R is not commutative, one must distinguish between left, right, and two-sided ideals.
Lemma 1.7. A ring is Noetherian if and only if every one of its ideals is finitely
generated.
Proof. (⇒) Let I be an ideal of the Noetherian ring R. We pick a sequence a0 , a1 , . . .
of elements of I as follows. Pick a0 ∈ I arbitrarily, and given a0 , . . . , an , let In be the
ideal generated by a0 , . . . , an . If any In = I , then I is finitely generated, namely by
a0 , . . . , an . If In ⊂ I , then there is an element an+1 ∈ I \ In .
If we continue in this fashion, and I is not finitely generated, we will construct an
infinite strictly ascending chain I0 ⊂ I1 ⊂ . . . of ideals of R, contrary to the a.c.c.
(⇐) Suppose that every ideal of R is finitely generated and let I0 ⊆ I1 ⊆ . . . be
an ascending chain of ideals of R. It is easy to see that I = ∪∞
n=0 In is an ideal of R.
By our hypothesis, I is finitely generated. Thus there is an integer m with every one of
some finite set of generators contained in Im . Hence I = Im , from which it follows that
Im = Im+1 = . . . .
Problem 1.B. Prove Lemma 1.7 using the maximum condition instead of the a.c.c.
Problem 1.C. Show that the rings Z and F [x1 , . . . , xn ] (for n ≥ 1) are not Artinian.
Clearly any PID is Noetherian, so Z and F [x] are Noetherian for any field F . We
now wish to show all polynomial rings F [x1 , . . . , xn ] are Noetherian. A few paragraphs
above we said that we would study the ideals of polynomial rings. However, it is not easy
1. INTEGRAL DOMAINS, POLYNOMIALS, AND NOETHERIAN RINGS
173
to do this in a concrete way; even proving they are finitely generated is difficult. Instead,
we give an indirect proof discovered by Hilbert at the start of the 20th century.
Theorem 1.8 (Hilbert Basis Theorem). Let R be a ring. Then R is Noetherian if
and only if R[x] is Noetherian.
Proof. Since R is a quotient ring of R[x], it must be Noetherian whenever R[x] is.
We now give two proofs of the fact that if R is Noetherian, so is R[x]. Neither one is
easy, but both are pretty. Although it is not necessary, we will for convenience assume R
is commutative.
In both proofs, we utilize the following notation. For an ideal I of R[x], let `(I)
denote the set of leading coefficients of elements of I , and for a nonnegative integer n, let
`n (I) denote the set of leading coefficients of polynomials in I of degree at most n. These
are both subsets of R, and we claim they are ideals. We prove this for `(I), leaving the
case of `n (I) to the reader.
Pn
i
Let
a,
b
∈
`(I)
and
let
r
∈
R.
Then
there
are
polynomials
f
=
i=0 ai x and
Pm
g = i=0 bi xi in R[x] with an = a and bm = b. If ra = 0, then ra ∈ `(I) (since 0 is the
leading coefficient of 0 ∈ I ), and if ra 6= 0, then ra is the leading coefficient of rf ∈ I .
Thus ra ∈ `(I). If a + b = 0, then a + b ∈ `(I), so suppose a + b 6= 0. If n = m,
then a + b is the leading coefficient of f + g, and we may conclude that a + b ∈ `(I).
Unfortunately, we may have n 6= m, say n > m. In that case, xn−m g is in I and has
degree n, with leading term bxn , so f + xn−m g ∈ I has leading coefficient a + b. Thus in
any case a + b ∈ `(I). This proves `(I) is an ideal of R.
First proof. Let I be an ideal of R[x]. By the above, the set `(I) of leading coefficients of
elements of I is an ideal of R and so is finitely generated, say by a1 , . . . , an , where ai is the
leading coefficient of some fi ∈ I with deg fi = di . Let m be the maximum of d1 , . . . , dn ,
and for each i = 0, . . . , m − 1, choose elements g1,i , . . . , gki ,i ∈ I of degree i whose leading
coefficients b1,i , . . . , bki ,i generate `i (I). Then we claim that the polynomials fl along with
the polynomials gj,i generate the ideal I .
Suppose the claim fails and let J ⊆ I be the ideal of R[x] generated by these polynomials, let f ∈ I \ J be a polynomial of minimal
degree, and let c be the
P
P leading coefficient
of f . If degP
f ≥ m, we can write c = ni=1 ri ai . The polynomial ni=1 (ri xm−di )fi is in
J , and f − ni=1 (ri xm−di )fi has degree less than deg f , whence it is in J by choice of f .
This shows f ∈ J after all.
Now suppose that deg f = i < m. Then we may repeat the above argument using
g1,i , . . . , gki ,i in place of f1 , . . . , fn and using `i (I) in place of `(I), and so reduce the
degree of f by subtracting an element of J . This also leads to the conclusion that f ∈ J .
Thus we get a contradiction in either case. This shows we must have I = J .
Second proof. Let I be a collection of ideals in R[x]. We will prove the maximum
principle holds. We begin by considering the collection {`0 (I) | I ∈ I} of ideals of R.
By the maximum principle for R, this collection has a maximal element J0 . We now let
I0 = {I ∈ I | `0 (I) = J0 }. Next, we consider the collection {`1 (I) | I ∈ I0 } of ideals of
R and we choose a maximal element J1 of this set; next we define I1 = {I ∈ I0 | `1 (I) =
J1 } = {I ∈ I | `0 (I) = J0 , `1 (I) = J1 }. We continue in this way, defining subsets In of
the original I with I0 ⊇ I1 ⊇ . . . .
Now suppose I ∈ ∩∞
n=0 In : we claim I is maximal in I . If not, there is a J ∈ I with
J ⊃ I ; clearly `n (J) ⊇ `n (I) for all n, so by choice of I , `n (J) = `n (I) for all n. Suppose
174
f ∈ J \ I and let n = deg f . By the last observation, there is a g ∈ I with deg g = n and
with the same leading coefficient as f . Thus f − g has smaller degree than f and lies in
J . By induction on n, we may assume f − g ∈ I , which forces f ∈ I . This contradiction
shows I = J and hence that I is indeed maximal.
The only thing left to prove is that ∩∞
n=0 In 6= ∅. Note that for any ideal I of R[x], we
have `0 (I) ⊆ `1 (I) ⊆ · · · (since we may always multiply an element of I by a power of
x). Thus J0 ⊆ J1 ⊆ · · · . The Noetherian condition on R implies that Jn = Jn+1 = · · ·
for some n: this implies that In = In+1 = · · · , which in turn proves that the above
intersection is In 6= ∅.
Iterating the theorem shows for example that F [x1 , . . . , xn ] is Noetherian for any n
and any field F , as is Z[x1 , . . . , xn ]. Since any finitely generated commutative R-algebra
is a factor of a polynomial ring over R, we also get the following corollary.
Corollary 1.9. If R is a commutative Noetherian ring, any finitely generated commutative R-algebra is Noetherian. In particular, any finitely generated commutative algebra over a field is Noetherian and any commutative ring that is finitely generated as a
ring is Noetherian.
Proof. Exercise.
2. Localization
The reader is probably familiar with the formation of the rational numbers from the
integers and the formation of a quotient field from an integral domain by the formal
introduction of fractions. In this section we generalize this construction and allow a more
or less arbitrary set of denominators for our fractions.
Recall that a subset C of R is a multiplicative submonoid if it is closed under multiplication and contains 1. To emphasize our goal in this chapter, let us call such a C a
denominator set. We call a denominator set proper if it does not contain 0. We will
construct a ring of fractions RC −1 with elements r/c for r ∈ R, c ∈ C , so the elements
of C becomes units in RC −1 . The first thing to understand is when two fractions r/c
and r0 /c0 should be equal: the obvious answer is when rc0 = r0 c. This works in an integral domain, but in a general commutative ring it is not enough. To see this, suppose
cs = 0 where c and s are nonzero. Then the fraction cs/c should equal both s = s/1
and 0 = 0/1, so we conclude that s/1 = 0/1, even though s · 1 6= 0 · 1. In general, the
correct rule is the following: r/c = r0 /c0 if and only if there is a d ∈ C with rc0 d = r0 cd,
i.e., with (rc0 − r0 c)d = 0.
With that as motivation, we make the following definition. On the set R × C , we
define a relation ∼ by (r, c) ∼ (r0 , c0 ) if rc0 d = r0 cd for some d ∈ C . This is clearly a
reflexive and symmetric relation. We leave it to the reader to show it is transitive. We
denote the equivalence class of (r, c) under ∼ by r/c, and we define RC −1 to be the set of
all equivalence classes r/c with r ∈ R, c ∈ C . We wish to show RC −1 is a ring in which
the classes r/c correspond to fractions in a natural way.
To show this, it is preferable not to deal with equivalence classes until necessary.
Thus we first work with the set R × C , on which we define two binary operations +
and · by (r, c) + (r0 , c0 ) = (rc0 + r0 c, cc0 ) and (r, c) · (r0 , c0 ) = (rr0 , cc0 ). It is easy to
check that these are associative, commutative operations with identities (0, 1) and (1, 1)
respectively. Moreover, · is distributive over + and (r, c) · (0, 1) = (0, 1), so (R × C, +, ·)
2. LOCALIZATION
175
is a commutative semiring. (Recall that a semiring is a “ring without subtraction”.) We
next check that ∼ is a congruence, that is, that it is compatible with + and ·. We give the
details for +; those for · are easier. Let (r, c) ∼ (r0 , c0 ), say rc0 e = r0 ce for some e ∈ C and
let (s, d) ∈ R × C . Then (r, c) + (s, d) = (rd + sc, cd) and (r0 , c0 ) + (s, d) = (r0 d + sc0 , c0 d),
and (rd + sc)c0 de = rc0 ed2 + scc0 de = r0 ced2 + sc0 cde = (r0 d + sc0 )cde. This proves that
(r, c) + (s, d) ∼ (r0 , c0 ) + (s, d).
This shows ∼ is a congruence, and so RC −1 = R × C/ ∼ becomes a commutative
semiring under the induced operations. Once we pass to equivalence classes we actually get
a ring, since (r, c)+(−r, c) = (rc−rc, c2 ) = (0, c2 ) ∼ (0, 1), whence (r/c)+(−r/c) = 0RC −1 .
Theorem 2.1. Suppose C is a denominator set in a commutative ring R. Then
(1) The localization RC −1 is a ring with the above operations.
(2) The map r 7→ r/1 is a ring homomorphism with kernel {r ∈ R | cr = 0 for some c ∈
C}. In particular, if R is an integral domain and C is proper, then this map is
an embedding.
Proof. (1) Fill in the details as an exercise.
(2) Define φ(r) = r/1: it is easy to see that φ is a ring homomorphism. If cr = 0
for some c ∈ C , then r1c = 01c. Thus φ(r) = r/1 = 0/1 = 0RC −1 . Conversely, if
φ(r) = 0RC −1 , then r/1 = 0/1, so rc = 0 for some c ∈ C .
Example 2.2.
(1) If R is an integral domain and we take C = R \ {0}, then
C is multiplicatively closed, so RC −1 exists. It is not hard to see that RC −1
is a field ((r/c)−1 = c/r), and the theorem shows R embeds in RC −1 . We call
RC −1 the quotient field of R: it is the smallest field into which R embeds. For
example, the quotient field of Z is (can be naturally identified with) Q and the
quotient field of the polynomial ring F [x1 , . . . , xn ] is (can be naturally identified
with) the rational function field F (x1 , . . . , xn ).
(2) If R = F [x] and C = {1, x, x2 , . . . }, then RC −1 is the Laurent polynomial ring
F [x, x−1 ] where we allow negative powers of x (but only finitely many positive
or negative powers).
(3) If R = Z and C is the set of all odd integers, then RC −1 can be naturally
identified with the following subring of Q: { ab | a, b ∈ Z, b odd}.
Problem 2.A. Let R be a commutative ring and C be a denominator set. Show that
RC −1 is not the zero ring if and only if C is proper.
It is natural to ask if there is some sort of “universal” description of localization. The
next problem provides one.
Problem 2.B. If C is a subset of a ring R and φ : R → S is a ring homomorphism, we
will say φ is C -inverting if φ(c) is a unit (has a multiplicative inverse) for every c ∈ C .
If C is a denominator set in a commutative ring R, show that RC −1 together with the
natural map ι : R → RC −1 defined by ι(r) = r/1 has the following universal property.
The map ι is C -inverting, and for any C -inverting ring homomorphism φ : R → S (for
any commutative ring S ), there is a unique ring homomorphism ψ : RC −1 → S with
ψ ◦ ι = φ. See the diagram below.
176
R
ι - −1
RC
@
ψ
@φ
@
?
@
RS
@
Just as we ask what happens to properties when we pass to a polynomial ring, we
desire to know what happens to them when we pass to a localization. In many ways, a
localization is “simpler” than the original ring (as a field is simpler in structure than an
integral domain), and so many properties should transfer. We comment on what happens
to ideals now; in Problem 2.D below we discuss the Noetherian condition.
If I is an ideal of R, then we define the expansion of I to be I e = { i/c ∈ RC −1 |
i ∈ I, c ∈ C }; this is also denoted IC −1 or RC −1 I . It is easy to see that I e is always
an ideal of RC −1 . Conversely, if J is an ideal of RC −1 , we define the contraction of
J to be J c = { r ∈ R | r/1 ∈ J }. When R is an integral domain, J c is often denoted
J ∩ R, since R embeds naturally in RC −1 . It is easy to see that J c is an ideal of R. It is
also easy to see that contraction and expansion are order-preserving, so they define maps
between the lattices of ideals in R and RC −1 . What is the relation between contraction
and expansion? Are they inverses? Unfortunately they are not. We will have more to say
about this in Section 5.
Proposition 2.3. Let R be a ring and C a denominator set in R.
(1) If J is an ideal of RC −1 , then J c is an ideal of R and J = (J c )e .
(2) If I is an ideal of R, then I e is an ideal of RC −1 and (I e )c ⊇ I . Equality holds
if and only if the following condition is satisfied: for any r ∈ R, c ∈ C , if cr ∈ I ,
then r ∈ I .
Proof. Exercise.
Problem 2.C. Find an example of an integral domain R and a denominator set C
such that not every ideal of R is of the form J c for some ideal J / RC −1 .
Problem 2.D. Show that any localization of a commutative Noetherian ring is Noetherian.
3. Factorization
One of the key motivating ideas in commutative ring theory has been the notion of
factorization into primes. In this section we explore this notion. We show for example
that factorization into primes holds in an integral domain if and only if it holds in the
polynomial ring over it. We begin by introducing the basic concepts involved in the study
of factorization, and we discuss what is necessary for the existence of irreducible factorizations; this involves the connection between factorization and ideals. We then study the
more delicate topic of uniqueness of factorizations and introduce UFD’s. UFD’s include
PID’s and rings with a division algorithm. We then discuss localization, polynomials, and
factorization.
Throughout this section R denotes an integral domain. Recall that r ∈ R is a unit if
there exists s ∈ R with rs = 1, i.e., if r has a multiplicative inverse. We denote the set
of units of R by any of GL1 (R), R× , or U(R): it is a multiplicative group. If a, b ∈ R,
we say a divides b or a is a factor of b or b is a multiple of a, if there exists r ∈ R
3. FACTORIZATION
177
with ra = b, and we write a | b. The relation | is a reflexive, transitive relation, that is,
it is a preorder. This means we can define an equivalence relation ≈ on R by a ≈ b if
a | b and b | a: we say a and b are associates if a ≈ b. Now if b = ra and a = sb, then
b = rsb. If b 6= 0, we have rs = 1, whence r and s are units. The case when b = 0 is
easily dealt with, as is the converse, so one can show a ≈ b if and only if there is a unit
u ∈ R with b = ua.
On the set of equivalence classes R/ ≈, the relation | induces a partial order. How
can we describe these equivalence classes? In Z they consist of a number and its additive
inverse, while in F [x1 , . . . , xn ] they consist of all nonzero scalar multiples of some polynomial, together with {0}. In general there is a very nice description of ≈ and of the
induced partial order in terms of ideals.
Lemma 3.1. Let R be an integral domain and let a, b ∈ R.
(1) a | b if and only if Ra ⊇ Rb.
(2) a ≈ b if and only if Ra = Rb.
Proof. Exercise.
Thus divisibility is intimately related to the poset of principal ideals of R. We have
no reason to believe this is a sublattice of the lattice of ideals of R: it may not contain
upper or lower bounds. There are cases where the poset of principal ideals does not form
a lattice, and cases where it forms a lattice that is not a sublattice of the lattice of all
ideals.
Speaking of lattices, we may ask whether R is the pre-ordered equivalent of a lattice
under | . Given a, b ∈ R, a greatest lower bound for a, b under | , that is, an element
r ∈ R with r | a, r | b such that for any s ∈ R with s | a, s | b we have s | r, is called a
greatest common divisor (or highest common factor) of a and b. Since | is only
a preorder, the g.c.d. need not be unique: if r0 ≈ r, then r0 is a g.c.d as well. When 1 is
a g.c.d. of a and b, we say a and b are relatively prime. Likewise, a least upper bound
of a and b relative to | is called a least common multiple of a and b.
We say an element a of R is irreducible (also called atomic) if a is not a unit and
whenever a = bc for b, c ∈ R, one of b, c is a unit in R (equivalently, either b ≈ a or
c ≈ a). Phrased in terms of divisibility, this says that whenever b | a, either b is a unit
or b ≈ a. If we wish to phrase this in terms of ideals, we see that a is irreducible if and
only if the principal ideal Ra is maximal among proper principal ideals.
A complete factorization of r ∈ R is an expression r = ua1 . . . an where u is a unit
in R and a1 , . . . , an are irreducible elements of R. We say two complete factorizations
r = ua1 . . . an = vb1 . . . bm are equivalent if n = m and there is a σ ∈ Sn such that
bi ≈ aσ(i) for all i = 1, . . . , n. We say R is a unique factorization domain (or factorial
domain), UFD for short, if every nonzero element of R has a complete factorization which
is unique up to equivalence.
Problem 3.A. Show that in a UFD, any two elements have a g.c.d. and a l.c.m. Can
you extend this to arbitrary sets of elements (possibly infinite)? That is, can you define
g.c.d. and l.c.m. for arbitrary subsets of a UFD, and prove existence (and uniqueness)?
The existence and uniqueness of complete factorizations are independent of each other,
but in a great many situations, the existence is automatic, as the next lemma shows.
Lemma 3.2. Let R be an integral domain. If the maximum condition holds for the poset
of principal ideals of R, then every nonzero element of R has a complete factorization.
178
Conversely, if R is a UFD, then the maximum condition holds for the poset of principal
ideals of R.
Proof. Suppose that R satisfies the maximum condition on principal ideals but that
the lemma is false for R. Then we may choose a nonzero r ∈ R without a complete
factorization, and such that Rr is maximal among principal ideals generated by elements
without a complete factorization. Certainly r cannot be a unit and cannot be irreducible,
so we may write r = ab where a, b ∈ R are not units. Then it is easy to see that Rr ⊂ Ra
and Rr ⊂ Rb. The maximality of Rr implies that a, b have complete factorizations. But
this implies r = ab does as well, yielding a contradiction. This proves the lemma.
Suppose that R is a UFD. Given a collection I of principal ideals, let Ra ∈ I be such
that a has a complete factorization that is as short as possible. One can show that Ra is
maximal in I (verify! ).
Corollary 3.3. Let R be a Noetherian integral domain. Then every nonzero element
of R has a complete factorization.
Proof. Exercise.
To obtain uniqueness of factorizations, we need to introduce another concept. We say
an element a ∈ R is prime if a 6= 0, a is not a unit, and whenever a | rs, either a | r or
a | s. There is also an interesting ideal-theoretic equivalent of this condition, outlined in
the next problem.
Problem 3.B. Let R be an integral domain and let a ∈ R. Prove that a is prime if
and only if a 6= 0 and the ring R/Ra is an integral domain.
Lemma 3.4. Let R be an integral domain.
(1) Every prime element of R is irreducible.
(2) R is a UFD if and only if every nonzero element of R has a complete factorization
and every irreducible element of R is prime.
Proof. Exercise.
Corollary 3.5. A Noetherian integral domain is a UFD if and only if every irreducible element is prime.
Proof. Exercise.
We can apply this in the following result.
Proposition 3.6. Let R be a PID. Then R is a UFD.
Proof. Since a PID is Noetherian, it is enough to show every irreducible element
a ∈ R is prime. To show this, suppose a | rs for r, s ∈ R and suppose that a 6 | r. This
implies that the ideal Ra + Rr properly contains Ra (since it contains r and Ra does
not). But since R is a PID, Ra + Rr is a principal ideal, and the irreducibility of a thus
implies Ra + Rr = R (see the remarks after the definition of irreducible). This means
there exist x, y ∈ R with xa + yr = 1. Thus xas + yrs = s. Both terms on the left hand
side are divisible by a, whence it follows that a | s, as required.
Remark 3.7. Let us note a very important fact, a special case of which turned up in
the last proof. If R is a PID and r, s ∈ R have d as a g.c.d., then there exist a, b ∈ R
with ar + bs = d. This is because the ideal Rr + Rs equals the ideal Rd.
3. FACTORIZATION
179
This of course yields another proof that Z and F [x] for F a field are UFD’s. We can
generalize these examples. Since the key in both cases was the division algorithm, we make
the following definition. We say an integral domain R has division with remainder
(or sometimes has a division algorithm) if there exists a poset (Γ, ≤) and a function
δ : R → Γ (δ is called the size function) such that the following properties hold: (i)
Γ satisfies the minimum condition; (ii) δ(0) < δ(x) for any nonzero x ∈ R; and (iii) for
any a, b ∈ R with a 6= 0, there exist q, r ∈ R with b = qa + r and δ(r) < δ(a). The two
most obvious examples are R = Z with Γ being the nonnegative integers and δ(n) = |n|
and R = F [x] with Γ = {−∞, 0, 1, 2, . . . } and δ(f ) = deg f . When we may take
Γ ⊆ {−∞, 0, 1, 2, . . . }, we usually say R is a Euclidean domain, although sometimes
extra conditions are imposed on δ, such as δ(a) ≤ δ(ab) for all nonzero a, b ∈ R.
Proposition 3.8. Let R be an integral domain having division with remainder. Then
R is a PID and hence a UFD.
Proof. Let I / R. If I = 0, then certainly I is principal. If not, choose a nonzero
x ∈ I with δ(x) minimal (among such x’s). Let i ∈ I : then i = qx + r for some q, r ∈ R
with δ(r) < δ(x). Since r = i − qx ∈ I , this can only happen if r = 0, i.e., i = qx ∈ Rx.
This proves I = Rx, so I is principal.
We now consider the relationship between unique factorization and localization and
polynomial rings. We start with localization, and we begin with a lemma. Note that in
part (3) of the lemma, we must make a strong assumption about C ; even with this, the
converse may fail.
Lemma 3.9. Let R be an integral domain, let C be a proper denominator set in R,
and let a ∈ R.
(1) a/1 is a unit in RC −1 if and only if a | c for some c ∈ C .
(2) a/1 is prime in RC −1 if it is not a unit in RC −1 and a is prime in R.
(3) Suppose all elements of C are products of units and primes. Then if a is irreducible in R, either a/1 is irreducible in RC −1 or a/1 is a unit in RC −1 and a
is prime in R.
Proof. (1) and (2) Exercise.
(3) If a/1 is a unit, then a | c in R for some c ∈ C by (1). If c is a unit, then a is as
well; otherwise, c is a product of primes p1 , . . . , pn . Write as = c = p1 . . . pn for s ∈ R. If
any pi | a, then since a is irreducible, we have a ≈ pi , which implies a is prime. If pi does
not divide a, then pi | s by primality. Thus if no pi divides a, we may cancel the pi ’s one
at a time and conclude as0 = 1 for some s0 ∈ R. Thus either a is a unit or a is prime.
Suppose that a/1 is not a unit and a/1 = xy for some non-units x = r/c and y = s/d
in RC −1 (where r, s ∈ R, c, d ∈ C ). Then acd = rs. Using the technique in the last
paragraph (do it! ), we may cancel off cd and obtain a = r0 s0 for some r0 , s0 ∈ R, with
r = r0 r00 , s = s0 s00 , r00 s00 = cd ∈ C . Since a is irreducible, one of r0 , s0 is a unit, say r0 is.
Since r00 | cd, r00 is a unit in RC −1 . Consequently, x = r/c = (r0 /1)(r00 /1)(1/c) is a unit
in RC −1 .
Proposition 3.10. Let R be an integral domain and let C be a proper denominator
set in R.
(1) If R is a UFD, then so is RC −1 .
180
(2) Suppose that every element of C is a product of units and primes and suppose
that every nonzero element of R has a complete factorization. Then R is a UFD
if and only if RC −1 is a UFD.
Proof. (1) By Lemma 3.4, every element of R has a factorization into prime factors. By Lemma 3.9(2), this is also true for elements of RC −1 . Another application of
Lemma 3.4(2) shows RC −1 is a UFD.
(2) Suppose RC −1 is a UFD: to show R is a UFD, we need to show every irreducible
element in R is prime. Let r ∈ R be irreducible: Lemma 3.9(3) says that either r is
prime or r/1 is irreducible. In the latter case, r/1 is prime (since RC −1 is a UFD), and
so r is prime in R by Lemma 3.9(2).
We now turn to polynomial rings, and again, we start with a lemma.
Lemma 3.11. Let R be an integral domain and let a ∈ R.
(1) a is a unit in R if and only if a is a unit in R[x].
(2) a is irreducible in R if and only if a is irreducible in R[x].
(3) a is prime in R if and only if a is prime in R[x].
Proof. (1) Exercise.
(2) (⇒) Suppose that a = f g where f, g ∈ R[x]. Then we must have deg f = deg g =
0, so f, g ∈ R, and the result follows from (1).
(⇐) Exercise.
(3) Let π : R → R/Ra be the canonical projection. There is a unique extension of
π to a homomorphism π̃ : R[x] → (R/Ra)[x] with π̃|R = π|R and π̃(x) = x (why? ),
obtained by applying π to the coefficients of polynomials. Clearly ker π̃ = R[x]a, so
(R/Ra)[x] ∼
= R[x]/(R[x]a). Thus by Corollary 1.2, R[x]/(R[x]a) is an integral domain if
and only if R/Ra is an integral domain. By Problem 3.B, this implies that a is prime in
R[x] if and only if a is prime in R.
Theorem 3.12. R is a UFD if and only if R[x] is a UFD.
Proof. (⇒) Suppose f ∈ R[x] \ {0} does not have a complete factorization, and let
deg f be minimal among such polynomials. Let a be the leading coefficient of f and let
a = up1 . . . pn be a complete factorization of a in R. If p1 divides every coefficient of f ,
then write f = p1 f1 ; otherwise set f1 = f . If p2 divides every coefficient of f1 , then write
f1 = p2 f2 ; otherwise set f2 = f1 . Continuing in this fashion, we obtain fn with f = sfn
where s is a product of primes from R, and hence from R[x], and fn has the property
that no prime in R divides all of its coefficients. If fn is irreducible, then f = sfn yields
a complete factorization of f . Otherwise we may write fn = gh where neither g nor h
is a unit of R[x]. It follows from the above that neither g nor h can lie in R, so both
have degree less than deg fn = deg f . Our choice of f ensures that both g and h have
complete factorizations. But this implies f = sgh does as well, contradicting our choice
of f . This proves every nonzero element of R[x] has a complete factorization.
Let C be the set of nonzero elements in R: by Lemma 3.11(3) and the fact that R is
a UFD, every element of C is a product of units and primes. Thus by Proposition 3.10,
R[x] is a UFD if and only if R[x]C −1 is a UFD. But it is easy to see (check! ) that
R[x]C −1 ∼
= RC −1 [x]. The latter is a polynomial ring in one variable over the field RC −1
and so is a UFD. This proves R[x] is a UFD.
4. IDEALS AND DEDEKIND DOMAINS
181
(⇐) Every nonzero element of R can be written as a product of a unit and some
irreducible elements of R[x]. But all these factors must have degree 0, i.e., must lie in
R, so by Lemma 3.11, these factors are a unit and irreducibles in R. This proves every
nonzero element of R has a complete factorization. We may again apply Lemma 3.11 and
the fact that R[x] is a UFD to conclude that all irreducibles in R are prime.
Thus Z[x1 , . . . , xn ] and F [x1 , . . . , xn ] for any field F are UFDs.
4. Ideals and Dedekind Domains
Factorization into primes may fail, but Dedekind saw that in many cases one could
obtain another kind of unique factorization, into prime ideals. In this section we discuss
some properties of ideals, and then discuss Dedekind domains, that is, integral domains
in which ideals may be factored into a product of prime ideals. We will see Dedekind
domains turn up later both in this chapter and the next, indicating their importance in
algebra.
In the ring Z of integers, we know unique factorization of elements into primes is
possible. This is also the case in the ring Z[i] of Gaussian integers,
since it is a Euclidean
√
domain. But let √
us consider√the integral domain R = Z[ −5]. Here we note that
6 = 2 · 3√= (1 + −5)(1 − −5), and it can be verified as follows that the numbers
2, 3, 1 ± −5
in R. To see this, note that if z = xy, then
|z|2 = |x|2 |y|2
√ are2 irreducible
√
and |a + b −5| = a2 + 5b2 . Moreover, if a2 + 5b2 = 1, then a + b −5 = ±1 is a
unit, so if we could write z = xy in R in a nontrivial way, we would have to be able
to write |z|2 =√ AB where A, B > 1 and A, B both have the form a2 + 5b2 . Now for
z = 2, 3, 1 ± −5, we have |z|2 = 4, 6, 9, so we can conclude these values of z are
irreducible in R if we can show a2 + 5b2 = 2 and a2 + 5b2√= 3 have√no solutions. This
is obvious. Since it is clear (verify! ) that none of 2, 3, 1 + −5, 1 − −5 are associates,
we thus√have two distinct factorizations of 6 into irreducibles, and so R is not a UFD. In
fact, Z[ −d] is not a UFD anytime d is an integer greater than or equal to 3.
This seems a bit unfortunate, as R certainly seems like a useful ring, and even our
brief discussion above shows it is related to Diophantine problems involving a2 + 5b2 .
Can we recapture some version of uniqueness of factorization? This question also arose
in attempts to prove Fermat’s last theorem. Kummer decided that what was needed
was to introduce√more numbers, “ideal numbers”, which could serve as common factors
of 2, 3 and 1 ± −5, and enable us to factor 6 further into “prime factors”. Dedekind
proposed an equivalent but more rigorous idea, the approach we presently use. Note that
when dealing with ordinary integers, many statements about numbers can be translated
into statements about the principal ideals they generate. Thus 6 = 2 · 3 is equivalent to
6Z = 2Z · 3Z, and 2 | 6 is equivalent to 6Z ⊆ 2Z. These equivalences are valid in any
integral domain. Dedekind’s idea was to study
√ ideals instead of numbers.
To see how this works in our ring R = Z[ −5], we need to make another observation.
In Z, a number p is prime if and only if the ideal pZ is a maximal ideal; this statement
is not valid in all integral domains — we will discuss this below. But let us use this
idea for now.
The ideal R2 is not maximal; to see this, define a map from R to Z2
√
by a + b −5 7→ [a + b] ∈ Z2 . We leave it to the reader to check
√ that this is a ring
homomorphism; denote its kernel by M . It is clear that
+ b is even },
√ M = { a + b −5 | a √
and it is not hard to see that M = R2 + R(1 + √
−5) = R2 + R(1 − −5). In a
similar way, we may map R to Z3 by sending a + b −5 7→ [a + b] ∈ Z3 . This time,
182
√
the kernel is N = R3 + R(1 − −5). Since both maps are onto fields, the ideals M and
N are
√ maximal ideals of R. In the case of Z3 , there is another map available, namely
a + b −5 7→ [a − b] ∈√Z3 (for Z2 , this would be the same as the original map). Its kernel
is N 0 = R3 + R(1 + −5). We have the ideal R6 contained in all of M, N, N 0 .
Recall that the product of two ideals I and J is the ideal IJ generated by the products
of individual elements ij with i ∈ I and j ∈ J . That is,
n
X
il jl | n a positive integer, il ∈ I, jl ∈ J }.
IJ = {
l=1
√
√
√
√
Thus N N 0 is generated by 3·3 = 9, 3(1+ −5), 3(1− −5), (1+ −5)(1− −5) = 6; this
ideal contains 9−6 = 3, so it is in fact R3. Likewise M 2 = R2, so R6 = R2·R3 = M 2 N N 0
is the complete factorization of R6 into maximal ideals. Such a factorization is possible
for any nonzero proper ideal in R, and so in honor of Dedekind, we say R is a Dedekind
domain.
Now let us formalize and generalize the above discussion; in the course of our work
we will see several equivalent conditions for a ring to be a Dedekind domain. We defined
above the product of two ideals; we also saw that divisibility for elements corresponds
to containment for the ideals they generate. This leads us to define a proper ideal to be
prime if whenever it contains the product of two ideals, it contains one of those ideals.
By induction, this immediately generalizes to any finite product of ideals. We begin by
giving several conditions equivalent to being prime.
Lemma 4.1. Let R be a commutative ring and let P be a proper ideal of R. Then the
following conditions are equivalent.
(1) P is prime.
(2) If A, B are ideals of R with A ⊃ P and B ⊃ P , then AB 6⊆ P .
(3) If a, b ∈ R and ab ∈ P , then a ∈ P or b ∈ P .
(4) R/P is an integral domain.
Proof. (1) ⇒ (2) This is trivial.
(3) ⇐⇒ (4) This is also immediate.
(1) ⇒ (3). Suppose a, b ∈ R and ab ∈ P . Set A = Ra, B = Rb. These are ideals
of R, and it AB = Rab ⊆ P . Since P is prime, this implies either a ∈ Ra ⊆ P or
b ∈ Rb ⊆ P .
(3) ⇒ (2). Suppose (3) holds and let A, B be ideals of R with A ⊃ P , B ⊃ P . Then
there exist elements a ∈ A \ P , b ∈ B \ P . By (3), ab ∈
/ P , so AB 6⊆ P .
Lemma 4.2. Let R be a commutative ring and let M be an ideal of R. Then M is a
maximal ideal if and only if R/M is a field.
Proof. Exercise.
Either condition (2) or condition (4) in Lemma 4.1 now immediately gives us the
following corollary.
Corollary 4.3. Every maximal ideal is a prime ideal.
Proof. Exercise.
One very important connection for us is that between localization and prime ideals.
183
Lemma 4.4. Let R be a commutative ring, let C be a denominator set in R, let P be
an ideal of R disjoint from C , and let J be an ideal in RC −1 .
(1) If P is prime in R, then P e is prime in RC −1 and (P e )c = P .
(2) If P is maximal in R, then P e is maximal in RC −1 .
(3) J is prime in RC −1 if and only if J c is prime in R.
Proof. (1) First we show (P e )c = P . Clearly P ⊆ (P e )c . Suppose r ∈ (P e )c , say
r/1 = p/c for some p ∈ P, c ∈ C . Then we have rcd = pd ∈ P for some d ∈ C . Since P
is prime, this implies r ∈ P or c ∈ P or d ∈ P . The latter two inclusions are not possible
since C ∩ P = ∅, so we must have r ∈ P . This prove (P e )c ⊆ P , which completes the
proof of the equality.
Suppose a/c, b/d ∈ RC −1 (where c, d ∈ C ) and (a/b)(c/d) ∈ P e . Then ab/1 ∈ P e , so
ab ∈ (P e )c = P . Since P is prime, either a ∈ P or b ∈ P , and this in turn implies either
a/c ∈ P e or b/d ∈ P e . This proves P e is prime.
(2),(3) Exercise.
Problem 4.A. What is the noun associated to the condition of being prime — “primeness”, “primality”, . . . ?
We say an integral domain R is a Dedekind domain if every nonzero proper ideal can
be written uniquely (up to order) as a product of maximal ideals. In a PID, an element
a is irreducible if and only if the ideal Ra is maximal. Since every nonzero element has
a unique factorization into irreducibles, every nonzero ideal has a unique factorization as
a product of maximal ideals.
√ Thus every PID is a Dedekind domain. The converse is not
true, as the example of Z[ −5] shows (once we verify it is a Dedekind domain!). We will
see below in Proposition 4.10 that an integral domain is both a UFD and a Dedekind
domain if and only if it is a PID.
In order to study Dedekind domains we need to introduce several tools. One important
tool is the quotient field Q of the integral domain R. (Note there are many quotient fields,
all isomorphic — we fix a particular one and assume R ⊆ Q.) We define an R-ideal of Q
to be a nonempty subset I of Q which is closed under addition and which has the property
that ri ∈ I for every r ∈ R, i ∈ I : this is just an R-submodule of Q. If emphasis is
necessary, we call an ordinary nonzero ideal of R an integral ideal.
We define the product of R-ideals of Q just as we define the product of ordinary
ideals: the R-ideal generated by the products of elements — this product is given by
the usual sum-of-products. It is not hard to see that the R-ideals of Q form a monoid
with identity R under this multiplication. We call an R-ideal I invertible if there is
an R-ideal J ⊆ Q with IJ = R. Clearly the set of invertible ideals forms the largest
subgroup of the monoid of R-ideals. Because of this, the inverse to I is unique when it
exists, and so we may denote it by I −1 .
Problem 4.B.
(1) Show that in any integral domain, any nonzero principal ideal
is invertible.
(2) Show that the ideal generated by x, y in the polynomial ring F [x, y] is not invertible.
(3) Show that if R 6= Q, then Q is an R-ideal that is not invertible.
Problem 4.C. Show that if I is invertible, then I −1 = { q ∈ Q | qI ⊆ R }.
We will show that all integral ideals are invertible precisely when R is a Dedekind
domain. To do this, we need some technical results.
184
Lemma 4.5. Let I be an invertible ideal of the integral domain R. Then I is finitely
generated as an R-module. If I is an integral ideal of R, then I is finitely generated as
an ideal.
Proof. Let J be
, . . . , in ∈ I and
Pan R-ideal of Q with IJ = R. Then there exist i1P
j1 , . . . , jn ∈ J with nl=1 il jl = 1. Thus any i ∈ I can be written i = nl=1 (ijl )il , and
each ijl ∈ R, so i1 , . . . , in generate I as an R-module.
Corollary 4.6. If all nonzero ideals of the integral domain R are invertible, then R
is Noetherian.
We also need the following result. This lemma and a later variant (Lemma 6.3) will
be used several times in this chapter.
Lemma 4.7. Let R be a commutative ring, let M be a finitely generated R-module,
and let J be an ideal of R with JM = M . Then there exists j ∈ J with (1 − j)M = 0.
Proof.
Let m1 , . . . , mn be a generating set for M , and let aik be chosen in J so
P
that nk=1 aik mk = mi . Then in matrix terms, we have (I − A)m = 0, where A is the
matrix (aik ) and m is the column vector (m1 , . . . , mn )T . If we let adj(I − A) denote the
usual adjugate (cofactor) matrix, we know that adj(I − A) · (I − A) = det(I − A)I (see
Proposition 2.10 and Remark 2.12 in the appendix). Thus if we multiply the equation (I −
A)m = 0 on both sides by adj(I −A), we obtain det(I −A)m = 0. Thus det(I −A)mi = 0
for all i, whence det(I − A) annihilates M . It is easy to see that det(I − A) = 1 − j for
some j ∈ J . (For example, use the complete expansion of the determinant.)
Corollary 4.8. If R is a Noetherian integral domain and I, J are ideals of R with
JI = I , then J = R or I = 0.
Proof. By Lemma 4.7, there is an element j ∈ J with (1 − j)I = 0. As R is an
integral domain, this can only happen if I = 0 or j = 1. In the latter case, we have
J = R.
We are now ready for our first major result on Dedekind domains.
Theorem 4.9. Let R be an integral domain. Then the following conditions are equivalent.
(1) R is a Dedekind domain.
(2) Every nonzero proper ideal of R can be written as a product of prime ideals.
(3) Every nonzero ideal of R is invertible.
(4) R is Noetherian and every maximal ideal of R is invertible.
Moreover, when these conditions hold, every nonzero prime ideal in R is a maximal ideal.
Proof. (1) ⇒ (2) is obvious, as is (3) ⇒ (4) (given Corollary 4.6).
(4) ⇒ (1) Suppose there is a nonzero proper ideal I of R which is not a product
of maximal ideals; since R is Noetherian, we may choose I maximal with respect to
this property. There is a maximal ideal M containing I ; thus M −1 I is an ideal of R
containing I . By Corollary 4.8, we know that M I ⊂ I , and this implies I ⊂ M −1 I . Thus
by maximality, M −1 I is a product of maximal ideals. But this implies I = M (M −1 I) is
as well.
This proves existence; now we show uniqueness. Suppose P1 , . . . , Pn , Q1 , . . . , Qm are
invertible prime ideals of R with P1 . . . Pn = Q1 . . . Qm , and let us suppose they are
185
numbered so that Pi ⊂ Pj implies that i < j . Then since Q1 . . . Qm ⊆ P1 , we have
Qi ⊆ P1 for some i. Likewise Pj ⊆ Qi for some j . Thus Pj ⊆ P1 , and by our ordering,
this forces Pj = P1 , so P1 = Qi ; let us re-number on the right so that Qi = Q1 .
Multiplying both sides by P1−1 , we conclude that P2 . . . Pn = Q2 . . . Qm , and we may
assume by induction that n = m and after re-numbering, each Pi = Qj .
(2) ⇒ (3) To prove this, we must show every nonzero prime ideal in R is invertible.
We begin by showing that under the hypothesis of (2), every invertible prime ideal of
R is a maximal ideal. To see this, let P be a nonzero invertible prime ideal and let
a ∈ R \ P . If P + Ra ⊂ R, then there are prime ideals P1 , . . . , Pn and Q1 , . . . , Qm with
P + Ra = P1 . . . Pn and P + Ra2 = Q1 . . . Qm : note each Pi and each Qj contains P .
Let R be the ring R/P , and use ¯ to denote images in this ring. Since P is prime, R
is an integral domain, and each ideal Pi , Qj is prime in R. Moreover, Rā = P1 . . . Pn
2
2
and Rā2 = Q1 . . . Qm , so P1 . . . Pn = Q1 . . . Qm . It is clear that if a product of ideals
is invertible, each of the factors is invertible, and it is also clear that a nonzero principal
ideal is invertible ((Rx)−1 = Rx−1 ), so each ideal Pi and Qj is an invertible ideal of R. It
now follows as in the proof of (4) ⇒ (1) that the two factorizations into prime ideals must
be the same, so the ideals Pi can be paired up with the ideals Qj (with multiplicity two).
As every ideal Pi and Qj contains P , these equalities are valid without the ¯s. It follows
in particular that (P1 . . . Pn )2 = Q1 . . . Qm , so P ⊆ P + Ra2 = (P + Ra)2 ⊆ P 2 + Ra.
Let p ∈ P . If p = x + ra for x ∈ P 2 and r ∈ R, then ra = p − x ∈ P . As a ∈
/ P
2
and P is prime, we see r ∈ P . This shows P = P + P a = P (P + Ra). Multiplying both
sides by the ideal P −1 , we obtain R = P + Ra, contrary to our assumption. This proves
P is a maximal ideal.
Now suppose P is any nonzero prime ideal of R and suppose a ∈ P \ {0}. Then
by hypothesis we can write Ra = P1 . . . Pn for prime ideals P1 , . . . , Pn . As above, Ra is
invertible and so each Pi is invertible. Thus by the last paragraph, each Pi is maximal.
Now P1 . . . Pn ⊆ P , so since P is prime, we have Pi ⊆ P for some i. But as Pi is maximal,
this implies P = Pi , and so P is invertible.
We can now prove the result mentioned before on PIDs and Dedekind domains.
Proposition 4.10. A commutative ring is a PID if and only if it is both a UFD and
a Dedekind domain.
Proof. (⇒) We’ve seen this above.
(⇐) Let R be both a UFD and a Dedekind domain. If R is a field, then R is a PID.
Otherwise, every maximal ideal M of R is nonzero; let a ∈ M be nonzero. Then a is
not a unit, so a = p1 . . . pn for primes p1 , . . . , pn . Since M is prime, some pi ∈ M . As
Rpi is a prime ideal, it is maximal by Theorem 4.9. Thus M = Rpi . Now let I be any
nonzero ideal in R: then I = M1 . . . Mk for some maximal ideals Mi . We’ve just seen
each Mi = Rpi for some pi , so I = Rp1 . . . pk is principal.
Of course if the only examples of Dedekind domains were PID’s, it would not be a
terribly useful concept. It turns out that a great many rings that arise in number theory
and algebraic geometry are Dedekind domains, but this is not so easy to prove. Here are
some examples — without proof. (See Theorem 6.12). If ω is a primitive nth root of
unity, then the ring Z[ω] is always a Dedekind domain, but it is only a PID√for 29 values
of n. Similarly, if d is a squarefree integer with d ≡ 2, 3 (mod 4), then Z[ d] is always
186
a Dedekind domain, but it is only a PID for d < 0 if d = −2, −1, and I believe it is not
known when it is a PID for d > 0.
We close this section by noting that localizations of Dedekind domains are Dedekind.
Proposition 4.11. Let R be a Dedekind domain and let C be a proper denominator
set in R. Then RC −1 is a Dedekind domain.
Proof. We know from Proposition 2.3 that the proper ideals of RC −1 have the form
I e for ideals I of R. Since R is a Dedekind domain, we can write I = P1 . . . Pn for some
prime ideals Pi of R. Clearly I e = P1e . . . Pne , and Lemma 4.4 implies that each Pie is
either prime or all of RC −1 : thus I e is a product of prime ideals of RC −1 . It now follows
from Theorem 4.9(2) that RC −1 is a Dedekind domain.
5. Localization at Prime Ideals
In this section, we study the localization of a ring at a prime ideal. This yields a ring
in which the original prime ideal becomes the unique maximal ideal; such a ring is called
local. We show that any integral domain is the intersection of the local rings obtained
by localizing at all maximal ideals, and we show that a domain is Dedekind if and only
if each of these local rings is Dedekind. In doing so, we introduce the idea of a principal
valuation ring, a UFD with just one prime, and the more general notion of a valuation
ring.
An ideal P is prime if and only if its complement R\P is a denominator set. Thus when
P is prime, we may form the localization R(R \ P )−1 . We call this ring the localization
of R at P , and we denote it RP . We know from Lemma 4.4 that the prime ideals of this
ring correspond to the prime ideals of R disjoint from R \ P , that is, to the prime ideals
of R contained in P . This is an order-preserving correspondence, so P RP = P e contains
all other prime ideals of RP . But every maximal ideal is prime, so P RP contains every
maximal ideal, and hence every proper ideal. A commutative ring with a unique maximal
ideal is called a local ring. (Some authors also require the ring to be Noetherian, but we
do not.) Thus RP is a local ring; it is sometimes called the local ring of R at P .
Example 5.1. Z2Z = { ab | a, b ∈ Z, b odd }. This is a local ring with unique maximal
ideal consisting of all fractions a/b with a even and b odd. Note that any element of this
ring can be written uniquely in the from 2n u where n is a nonnegative integer and u is
a unit of the ring.
We leave it to the reader to perform the analogous construction with 2 replaced by
any prime p.
Many questions about commutative rings can be reduced to questions about the local
rings at the various prime or even just the maximal ideals. One indication of this is
the following result. First note that if R is an integral domain, then any localization
RC −1 can be regarded as a subring of the quotient field Q, so it makes sense to form the
intersection or union of various localizations.
Proposition 5.2. Let R be an integral domain. Then R equals the intersection of
the local rings RM as M ranges over all maximal ideals of R.
Proof. Clearly R ⊆ RM ⊆ Q for any M . Now let q ∈ Q lie in RM for all maximal
ideals M and let I = {r ∈ R | qr ∈ R}. It is easy to see that I is an ideal of R (verify! ).
If q ∈
/ R, then I is a proper ideal of R, so there is a maximal ideal M ⊇ I . Since q ∈ RM
5. LOCALIZATION AT PRIME IDEALS
187
for this M , we can write q = a/b for a ∈ R, b ∈ R \ M . But then qb = a ∈ R, so
b ∈ I ⊆ M . This contradiction shows we must have q ∈ R.
For local rings, different conditions often become the same. Let’s give an example of
this. Suppose a PID is local: then it contains a unique maximal ideal, which must have
the form Rp for some prime p. If a ∈ R, then Ra ⊆ Rp, so p | a. This shows p is the
unique prime in R (up to association), so every element of R can be written uniquely
in the form upn for some nonnegative integer n and some unit u. This implies that for
any a, b ∈ R, either a | b or b | a. An integral domain with this last property is called a
valuation ring. Clearly any valuation ring which is a UFD contains a single prime (up
to association); from this it follows that the ring is a local PID. We call a valuation ring
which is a PID but not a field (equivalently, a PID with a single prime up to association)
a principal valuation ring (or a discrete valuation ring).
Problem 5.A.
(1) Let R be an integral domain with quotient field Q. Show
that the following conditions are equivalent.
(a) R is a valuation ring
(b) For any q ∈ Q, either q ∈ R or q −1 ∈ R.
(c) The ideals of R are totally ordered by inclusion.
(2) Show that any valuation ring is a local ring.
Proposition 5.3. Let R be an integral domain. Then the following statements are
equivalent.
(1) R is a principal valuation ring.
(2) R is a Noetherian valuation ring which is not a field.
(3) R is a UFD with a single prime (up to association).
(4) R a local Dedekind domain which is not a field.
Proof. (1) ⇒ (2), (3), (4) These implications are clear from the above discussion and
previous results.
(2) ⇒ (1) It is easy to see that any finitely generated ideal in a valuation ring is
principal. Thus a Noetherian valuation ring is a PID.
(3) ⇒ (1) Let p be a prime in R: then every element of R has the form upn for some
unit u. If I is a nonzero ideal of R, choose an element upn ∈ I , for a unit u, with n as
small as possible. Then pn ∈ I and pn | i for all i ∈ I , so I = hpn i. Thus R is a PID
with a single prime, i.e., R is a principal valuation ring.
(4) ⇒ (1) Let M be the unique maximal ideal: note M 6= 0, so M 2 6= M . Let
x ∈ M \ M 2 . Then Rx = M n for some n (since R is Dedekind); but x ∈
/ M 2 , so we must
have Rx = M . But any nonzero ideal I = M k = Rxk for some k, so any nonzero ideal is
principal.
It follows from the above results and from Proposition 4.11 that if R is a Dedekind
domain, then every localization RM at a maximal ideal is a principal valuation ring. We
now show the converse is valid if R is Noetherian.
Proposition 5.4. Let R be a Noetherian integral domain. Then R is a Dedekind
domain if and only if RM is a Dedekind domain for each maximal ideal M of R.
Proof. We need to show that every maximal ideal M of R is invertible. Let Q be
the quotient field of R; of course Q is also the quotient field of RM . Set I = { q ∈ Q |
188
qM ⊆ R } and I 0 = { q ∈ Q | qM RM ⊆ RM }. Clearly M ⊆ IM ⊆ R and by hypothesis
I 0 M RM = RM . We need to show IM = R.
What is the relationship between I and I 0 ? Clearly I ⊆ I 0 . Suppose q ∈ Q. Since R
is Noetherian, M is finitely generated, say by m1 , . . . , mn , and it is easy to see that M RM
is generated by m1 /1, . . . , mn /1. Thus q ∈ I 0 if and only if qmi ∈ RM for all i. Since we
can always get a common denominator for a finite set of fractions by multiplying their
denominators, we see that q ∈ I 0 if and only if there exist a1 , . . . , an ∈ R and b ∈ R \ M
with qmi = ai /b, i.e., with bqmi = ai . If this holds, then bqM ⊆ R, so bq ∈ I . In fact,
this shows I 0 = { q ∈ Q | bq ∈ I for some b ∈ R \ M }.
P
Since I 0 M RM = RM , there exist q1 , . . . , qk ∈ I 0 , x1 , . . . , xk ∈ M RM with ki=1 qi xi =
1. We can find a common denominator for the xi and so assume each xi = ai /c for some
ai ∈ M and c ∈ R \ M , and by the remarks in the last paragraph, we can find bi ∈ R \ M
P
with bi qi ∈ I . Set b = b1 . . . bk and note b ∈ R \ M . Now ki=1 (bqi )ai = bc ∈ R \ M .
Thus IM 6⊆ M , which forces IM = R (since M ⊆ IM ⊆ R and M is maximal). This
shows M is invertible.
Problem 5.B. Did we need to assume R is Noetherian in Proposition 5.4?
6. Integral Extensions
The notion of algebraic extension was crucial in studying fields; the corresponding
notion for integral domains is that of an integral extension. In this section, we develop
the basic properties of such extensions. For example, finite dimensionality is intimately
related to algebraic extensions for fields, and we show that finite generation of certain
modules is intimately related to integral extensions.
In Chapter 4 we defined the notion of an algebraic extension of fields; the same definition can be used to define an algebraic extension of integral domains. However, it turns
out to be useful to give a more restrictive definition. For example, every rational number
is algebraic over Z, since a/b is a root of bx − a, but there are many problems where
we only wish to deal with integral solutions to equations. This leads to the following
definition. Let R be an integral domain and let S be an extension ring of R: this means
S is a commutative ring and R is a subring of S — in fact, we will always assume S is an
integral domain. Then we say s ∈ S is integral over R if it is the root of a monic polynomial in R[x], that is, if there exist r0 , . . . , rn−1 ∈ R with sn + rn−1 sn−1 + · · · + r1 s + r0 = 0.
We say S is an integral extension of R if every element of S is integral over R. Note
that when R is a field, “integral” is synonymous with “algebraic”. In general, we have
the following connection.
Lemma 6.1. Let R ⊆ S be integral domains and let s ∈ S . Then s is algebraic over
R if and only if there is a nonzero c ∈ R with cs integral over R.
P
P
Proof. (⇒) Let ni=0 ri si = 0, and set c = rn . Then 0 = ni=0 ri rnn−1 si = (cs)n +
Pn−1 n−1−i
(cs)i ; this shows cs is integral over R.
i=0 ri rn
(⇐) Exercise.
The following result gives another connection between integrality and fields.
Lemma 6.2. Let R ⊆ S be integral domains and suppose S is an integral extension of
R. Then R is a field if and only if S is a field.
Proof. Exercise.
6. INTEGRAL EXTENSIONS
189
In general, the definition of integral is supposed to restrict us to “integer-like” numbers;
for example, a complex number a + bi with a, b ∈ Q is integral over
√ Z if and only if both
a, b ∈ Z. However, there are some surprises; for example, − 21 + 21 −3 is integral over Z,
since it is a root of the polynomial x3 − 1.
Several of the properties of integral elements parallel those of algebraic elements in
field extensions. For example, we will explore the connection between integrality and
finiteness, and we will show that the integral elements form a subring of S . We begin
with a lemma which is a variant of the earlier Lemma 4.7.
Lemma 6.3. Let S be a commutative ring with subring R, let M be a finitely generated
R-submodule of S , and let a ∈ S satisfy aM ⊆ M . Then there is a monic polynomial
f ∈ R[x] with f (a)M = 0.
Proof. Let m1 , P
. . . , mn be a generating set for M as an R-module, and let bik be
chosen in R so that nk=1 bik mk = ami . Then in matrix terms, we have (aI − B)m = 0,
where B is the matrix (bik ) and m is the column vector (m1 , . . . , mn )T . Just as in the
proof of Lemma 4.7, this implies det(aI − B)M = 0. Now f (x) = det(xI − B) is a monic
polynomial in R[x], and f (a) = det(aI − B). This proves the lemma.
Proposition 6.4. Let R be a subring of the commutative ring S and let s ∈ S . Then
the following conditions are equivalent.
(1) s is integral over R.
(2) The subring R[s] generated by R and s is finitely generated as an R-module.
(3) There is a subring S 0 of S containing both R and s which is finitely generated
as an R-module.
Proof. (1) ⇒ (2) The ring R[s] consists of all polynomials in s with coefficients from
r, that is, all R-linear combinations of powers of s. The fact that s is integral over R tells
us that sn is an R-linear combination of 1, s, . . . , sn−1 for some n. It is now easy to show
by induction (Do it! ) that all powers sn+k are R-linear combinations of 1, s, . . . , sn−1 .
Thus R[s] is generated as an R-module by 1, s, . . . , sn−1 .
(2) ⇒ (3) Take S 0 = R[s].
(3) ⇒ (1) Since S 0 is a finitely generated R-module and sS 0 ⊆ S 0 , Lemma 6.3 implies
there is a monic polynomial f ∈ R[x] with f (s)S 0 = 0. But S 0 contains 1, so this forces
f (s) = 0.
Remark 6.5. If R ⊆ S are integral domains, then we may give an even weaker
condition than (3) above. Suppose s ∈ S and there is a nonzero finitely generated Rsubmodule M of S with sM ⊆ M (certainly conditions (2) and (3) each imply this).
Then by Lemma 6.3, there is a monic polynomial f ∈ R[x] with f (s)M = 0. Since
M ⊆ S and S is an integral domain, this implies f (s) = 0.
Corollary 6.6. Let S be an integral domain and let R be a subring of S . Then the
elements of S which are integral over R form a subring of S containing R.
Proof. The elements of R are clearly integral over R. Let a, b ∈ S be integral over
R. Then R[a] is finitely generated as an R-module. Also, b is integral over R[a] since it
is integral over R, so R[a, b] = R[a][b] is finitely generated as an R[a]-module. It follows
(copy the spanning part of the proof of Theorem 2.1 in Chapter 4) that R[a, b] is finitely
generated as an R-module. Since a ± b, ab ∈ R[a, b], Proposition 6.4(3) implies a ± b, ab
are integral over R.
190
We refer to the subring defined in the corollary as the integral closure of R in S .
Problem 6.A. Let T be an integral domain and let R, S be subrings of T with
R ⊆ S ⊆ T . Show that T is an integral extension of R if and only if S is an integral
extension of R and T is an integral extension of S .
As noted above, the elements of R are integral over R; if there are no other elements
of S which are integral over R (that is, if R is the integral closure of R in S ), we say R
is integrally closed in S . When R is integrally closed in its quotient field, we simply
say that R is integrally closed. Thus for example Z is integrally closed, since no nonintegral rational number can satisfy a monic polynomial. This can be greatly generalized
as follows.
Proposition 6.7. If R is a UFD, then R is integrally closed.
Proof. Let q = a/b in the quotient field of R be integral overPR. Since R is a UFD,
i
we may assume a and b are relatively prime (Why? ). Let q n + n−1
i=0 ri q = 0 for some
n
n−1
r0 , . . . , rn−1 ∈ R. Then clearing denominators, we see a = −rn−1 a b − · · · − r1 abn−1 −
r0 bn . Every term on the right contains a b, so we have b | an . But as a and b are relatively
prime, this can only happen if b is a unit in R, which implies q ∈ R.
√
This shows that √Z[ d] cannot be a UFD if d is a squarefree integer with √
d ≡ 1
∈
Z[x]
that
does
not
lie
in
Z[
(mod 4), since (1 + d)/2 is a root of x2 − x − d−1
d].
4
The notion of “integrally closed” is quite important. We now show how it interacts
with localization.
Proposition 6.8. Let R be an integral domain.
(1) If C is a proper denominator set in R and R is integrally closed, then RC −1 is
integrally closed.
(2) R is integrally closed if and only if RM is integrally closed for every maximal
ideal M of R.
Proof. Let Q be the quotient field of R (and hence of RC −1 and RM for any proper
denominator set C and maximal ideal M ).
Pn−1
(1) Let q ∈ Q be integral over RC −1 , so that q n = i=0
xi q i for some x0 , . . . , xn−1 ∈
RC −1 . If we let c ∈ C be a common denominator for the xi , then
xi cj ∈ R for
Pcertainly
n
n−i
n
n
any j ≥ 1. Multiply the above equation by c to obtain (qc) = i=0 (xi c )(qc)i . This
shows qc is integral over R, so since R is integrally closed, we have qc ∈ R. This shows
q ∈ RC −1 , and so RC −1 is integrally closed.
(2) Suppose each RM is integrally closed and let q ∈ Q be integral over R. Then q is
integral over RM ⊇ R, so by integral closure q ∈ RM for all M . By Proposition 5.2, this
implies q ∈ R.
We can use integral closure to give yet another characterization of Dedekind domains.
Theorem 6.9. Let R be an integral domain. Then R is a Dedekind domain if and only
if R is Noetherian and integrally closed and all nonzero prime ideals in R are maximal.
Proof. (⇒) Theorem 4.9 tells us that R is Noetherian and that all nonzero prime
ideals in R are maximal. By Proposition 4.11 and Proposition 5.3 we know each RM
is a UFD, and so is integrally closed by Proposition 6.7. Thus by Proposition 6.8, R is
integrally closed.
6. INTEGRAL EXTENSIONS
191
(⇐) By Proposition 5.4, it suffices to show RM is Dedekind for each maximal ideal
M . By Proposition 6.8, each localization RM is integrally closed, and by Lemma 4.4
and Problem 2.D, each RM is Noetherian and has a unique nonzero prime ideal, namely
M RM . Thus we may assume R is an integrally closed local Noetherian integral domain
with a unique nonzero prime ideal M and with quotient field Q, and we must show R is
Dedekind.
To show this, it of course suffices to show that the maximal ideal M is invertible. For
any ideal J of R, set J 0 = { q ∈ Q | qJ ⊆ R }. This is an R-ideal of Q containing R,
so J ⊆ J 0 J ⊆ R. Thus we have only two possibilities: M 0 M = M or M 0 M = R. In the
former case, Remark 6.5 shows that the elements of M 0 are integral over R, and so by
integral closure, we conclude that M 0 = R. We now show that this is not the case.
We know that if a ∈ R is not zero and not a unit, then Ra is an invertible ideal with
inverse (Ra)0 = Ra−1 ⊃ R. Let P be an ideal of R which is maximal among ideals with
the property that P 0 ⊃ R. We will show P is prime, which implies P = M . To see this,
suppose P is not prime: then there are ideals I, J ⊃ P with IJ ⊆ P . Let q ∈ P 0 \ R.
Then (qI)J ⊆ qP ⊆ R, so qI ⊆ J 0 = R (by maximality of P ). But then q ∈ I 0 = R,
yielding a contradiction. Thus P is prime.
This shows M 0 6= R, and so by the above, M 0 M = R. This proves M is invertible.
We can now give a large number of examples of Dedekind domains, although some
work remains to be done below to verify that they are Dedekind domains. Recall that
an algebraic integer is a complex number which is the root of a monic polynomial
with integer coefficients, that is, which is integral over Z. If F is any subfield of C,
the set of algebraic integers in F forms an integrally closed subring by Corollary 6.6 and
Problem 6.A. We call a field that is finite dimensional over Q a number field and we call
the subring of algebraic integers of a number field a number ring. (This terminology is
the most common, but it is not universal. Some authors, such as Artin, call any subfield
of C a number field.) We wish to show every number ring is a Dedekind domain; we first
prove some preliminary results. One result we will use a couple of times is the following:
every subgroup of Zn (for any positive integer n) is finitely generated. We have not
proven this yet; it will follow from the results of Section 4 in Chapter 7 (although the
reader may already know it from previous study of Abelian groups).
Lemma 6.10. Let R be the ring of algebraic integers in the finite dimensional field
extension E of Q (so R is an arbitrary number ring).
(1) Every element of E has the form r/n for some r ∈ R and some positive integer
n.
(2) Every nonzero ideal of R contains a positive integer.
Proof. (1) Let e ∈ E , whence e is algebraic over Q and hence over Z. By Lemma 6.1,
we have ne integral over Z for some positive integer n. Thus ne ∈ R and e = ne/n.
(2) Let L be the Galois closure of E , that is, Q ⊆ E ⊆ L ⊆ C and L/Q is Galois
— L/Q exists and is finite dimensional by Exercise 26 from Chapter 4 — and let G =
Gal(L/Q).
Q
If I 6= 0 is an ideal of R, let i ∈ I be nonzero and set l = σ∈G σ(i). It is clear
that σ(l) = l for every σ ∈ G. Since L/Q is Galois, this implies l ∈ Q. Each σ(i) is an
algebraic integer, whence l is as well, and so l ∈ Z. As i 6= 0, we have l 6= 0. Now we
192
have l = ii0 where l ∈ Z, i ∈ I ⊆ R, so i0 ∈ E . Since i0 is a product of algebraic integers,
it is an algebraic integer, and so it is in R. Thus l ∈ I .
Lemma 6.11. An integral domain is isomorphic to a number ring if and only if (i)
it is of characteristic 0, (ii) it is integrally closed, and (iii) it is finitely generated as an
Abelian group.
Proof. Let R be the integral domain in question.
(⇒) Suppose R is a number ring. Conditions (i) and (ii) are obvious. To prove
(iii), adopt the notation of Lemma 6.10 and its proof. It will be convenient to write
G = {σ1 , . . . , σn }. We may choose a basis l1 , . . . , ln for L over Q such that l1 , . . . , lm is
a basis for E (for some m ≤ n). By Lemma 6.1, there is a ci ∈ Z \ {0} with ci li an
algebraic integer. It does no harm to replace li by ci li , so we may assume each li is an
algebraic integer, and hence li ∈ R for i ≤ m. P
n
Suppose r ∈ R: then we may write r =
q1 , . . . , qn ∈ Q, with
j=1 qj lj for some P
n
qj = 0 for j > m. If we apply σi to this equation, we obtain
j=1 qj σi (lj ) = σi (r)
for i = 1, . . . , n. We may regard this as a system of n equations in the n unknowns
q1 , . . . , qn , in which case we may solve it by Cramer’s rule, provided that the determinant
c of the n × n matrix (σi (lj )) is nonzero. If c = 0, then theProws of this matrix are
n
linearly dependent, that is, there exist λ1 , . . . , λP
n ∈ L with
i=1 λi σi (lj ) = 0 for all
n
j = 1, . . . , n. Since the lj span L, this means
i=1 λi σi = 0, contrary to Dedekind’s
Lemma (Lemma 10.4 in Chapter 4). Thus c 6= 0 and the system has a unique solution
qj = bj /c where the bj are given by determinants involving the σi (lj ) and σi (r). As these
numbers are all algebraic integers, bj and c are algebraic integers.
Moreover, if we apply σ ∈ G to c, we see that σ(c) is a determinant of the same
form as c, but with the rows permuted. Thus σ(c) = ±c, and so if we set d = c2 , we
have σ(d) = d for all σ ∈ G. This shows d ∈ Q, and as d is an algebraic integer, we
actually have d ∈ Z. Moreover, if we set b0j = bj c, then b0j is an algebraic integer, and
qj = b0j /d, so b0j ∈ Q and hence b0j ∈ Z. This shows that dqj ∈ Z, which proves that
for any r ∈ R, dr ∈ ⊕ni=1 Zli . Since d is independent of r, we can define an Abelian
group homomorphism r 7→ dr from R into ⊕ni=1 Zli . This is clearly one-to-one, so R is
isomorphic to a subgroup of a finitely generated Abelian group, and so R itself is finitely
generated as an Abelian group. (As we remarked above, this is a fact that we have not
yet proved.)
(⇐) Suppose (i),(ii),(iii) hold for R. As R has characteristic 0, it contains a copy of Z,
and its quotient field E contains a copy of Q. Let E 0 = { r/c | r ∈ R, c is a positive integer }:
then E 0 is of course a subring of E . Using the fact that R is finitely generated as an
Abelian group, it is easy to see that E 0 is a finite dimensional vector space over Q, from
which it follows that E 0 is a field. (This is a good exercise; it also follows from Lemma 6.2).
Thus we must have E 0 = E , so E is a finite dimensional extension field of Q. Thus E is
algebraic over Q; we leave it to the reader to show that R consists of all algebraic integers
in E .
Theorem 6.12. Every number ring is a Dedekind domain.
Proof. Let R be a number ring. We need to show R is Noetherian and integrally
closed, and that all nonzero prime ideals are maximal. We know R is integrally closed
(say by Problem 6.A or Lemma 6.11). By Lemma 6.11, R, and hence every ideal of R, is
isomorphic to a subgroup of some Zn , and so each ideal of R is finitely generated as an
7. MAXIMAL IDEALS IN FUNCTION RINGS
193
Abelian group. This certainly implies every ideal of R is finitely generated as an ideal;
thus R is Noetherian. Finally, let P be a nonzero prime ideal of R. By Lemma 6.10,
P contains a positive integer n. Thus R/P is a module over Z/nZ, and it is finitely
generated since R is finitely generated as a Z-module. It follows that R/P is a finite
integral domain, and hence must be a field (Exercise). This proves P is maximal.
√
Problem 6.B. Let d ∈ Z be squarefree (d 6= 1). Show that Z[ d] is a Dedekind
domain if and only if d ≡ 2, 3 (mod 4).
7. Maximal Ideals in Function Rings
We have seen before that a great deal can be learned about a ring simply from its
maximal ideals. For example, a Noetherian integral domain is Dedekind if and only if
each of its maximal ideals is invertible, and any integral domain is the intersection of all
its localizations at maximal ideals. In this section, we show that if R is a ring of functions
from a space X to a field F , we can frequently recover X just from a knowledge of the
maximal ideals of R. We begin by describing maximal ideals in Fun(X, F ) = F X for any
X and F . We then describe maximal ideals in the ring of continuous functions from a
topological space X to R or C. In the next section, we will consider the ring of polynomial
functions from X to F and make some explicit connections with geometry.
Let X be a set and F be a field, so that Fun(X, F ) is a commutative ring with
pointwise operations. We begin with a construction that is useful for all function rings. Fix
x ∈ X and define Φx : Fun(X, F ) → F by Φx (f ) = f (x). That is, Φx is evaluation at x.
This is clearly a ring homomorphism, and it is onto since Fun(X, F ) contains the constant
functions. Set Mx = ker Φx = { f : X → F | f (x) = 0 }. Then Fun(X, F )/Mx ∼
= F
(why? ), so Mx is a maximal ideal. Are these all the maximal ideals of X ? The answer is
“yes” if and only if X is finite. We will give a brief description of the answer in general.
We first recall some terminology from Section 8 in Chapter 1. A filter on X is
a collection F of subsets of X containing X and closed under finite intersections and
supersets (i.e., if Y ⊆ Z ⊆ X and Y ∈ F , then Z ∈ F ). One example of a filter is
the power set P(X). Any other filter is called proper, and a maximal proper filter is
called an ultrafilter. A standard Zorn’s Lemma argument shows that every proper filter
is contained in an ultrafilter.
For example, if Y ⊆ X , then { Z | Z ⊇ Y } is a filter, called the principal filter
generated by Y . If Y is a singleton (i.e., |Y | = 1), then this is an ultrafilter. Any other
ultrafilter is called a non-principal ultrafilter.
Consider F = { Z ⊆ X | X \ Z is finite }. If X is infinite, this is a proper filter that
is not contained in any principal ultrafilter. Thus F is contained in some (in fact all)
non-principal ultrafilters on X .
We need to introduce some additional notation. For f : X → F , the support of f is
supp f = { x ∈ X | f (x) 6= 0 } and the zero set of f is Z(f ) = { x ∈ X | f (x) = 0 }.
(The zero set is also called the vanishing set and denoted V(f ).) If Y ⊆ X , the
characteristic function of Y is the function χY : X → F defined by χY (x) = 1 for
x ∈ Y and χY (x) = 0 for x ∈ X \ Y . Note that supp χY = Y and Z(χY ) = X \ Y .
Theorem 7.1. Let X be a set and F be a field. There is a bijective, order-preserving
correspondence between ideals of Fun(X, F ) and filters on X given by I 7→ { Z(f ) | f ∈
I }, with inverse given by F 7→ { f : X → F | Z(f ) ∈ F }.
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Under this correspondence, the maximal ideals of Fun(X, F ) correspond to ultrafilters
on X .
Proof. If f, g ∈ Fun(X, F ), then Z(f g) = Z(f )∪Z(g) and Z(f +g) ⊇ Z(f )∩Z(g).
Furthermore, Z(0) = X , and Z(−f ) = Z(f ).
Suppose F is a filter on X . The statements in the preceding paragraph immediately
imply that { f : X → F | Z(f ) ∈ F } is an ideal of Fun(X, F ).
Conversely, let I be an ideal and set F = { Z(f ) | f ∈ I }. Then X = Z(0) ∈ F .
Suppose W ⊆ Y ⊆ X and W ∈ F . Then W = Z(f ) for some f . Set g = f · χY ∈ I . It
is clear that Z(g) = Y , and so Y ∈ F .
Finally, let A = Z(f ), B = Z(g) be in F , for f, g ∈ I , and let h ∈ Fun(X, F ) be the
characteristic function of Z(f ). That is, h(x) = 1 if f (x) = 0 and h(x) = 0 if f (x) 6= 0.
Clearly f + gh ∈ I : we claim Z(f + gh) = A ∩ B . Obviously Z(f + gh) ⊇ A ∩ B . If
(f + gh)(x) = 0 and x ∈ Z(f ), then we have f (x) = 0, so g(x) = (gh)(x) = 0 as well.
Thus in this case, x ∈ A ∩ B . If x ∈
/ Z(f ), then h(x) = 0, so (f + gh)(x) = f (x) 6= 0.
This proves that Z(f + gh) = A ∩ B , and so A ∩ B ∈ F .
This shows that F is a filter.
We leave the proof of the claim about ultrafilters and maximal ideals to the reader.
We can recover the space X from a knowledge of the maximal ideals of Fun(X, F )
as follows. The maximal ideal Mx is generated by χX\{x} ; indeed, f ∈ Mx if and only
if f = f · χX\{x} . On the other hand, the other maximal ideals, which correspond to
non-principal ultrafilters, are not finitely generated (exercise! ). Thus there is a natural
bijection between X and the collection of principal (or finitely generated) maximal ideals
of Fun(X, F ). (There is also a natural bijection between X and certain idempotents –
called “primitive” – of Fun(X, F ).)
As we have noted before, we are generally interested in proper subrings of Fun(X, F ).
Suppose X is a topological space and F is R or C. Then we may wish to investigate the
ring C(X, F ) of continuous functions from X to F . The evaluation map Φx : C(X, F ) →
F still makes sense, and its kernel Mx is still a maximal ideal. If X is the closed interval
[0, 1], the Mx are all the maximal ideals, but not if X is the open interval (0, 1). To see this
latter fact, let I = { f ∈ C(X, F ) | there is an > 0 such that f (a) = 0 for all a < }.
It is not hard to see that I is a proper ideal of C(X, F ) that is not contained in any Mx .
The difference between [0, 1] and (0, 1) is that the former space is compact. In general,
it is difficult to describe all maximal ideals of C(X, F ), but when X is compact, we have
the following beautiful result.
Theorem 7.2. Let X be a compact metric space or more generally any compact Hausdorff topological space. Then there is a bijective correspondence between the points of X
and the maximal ideals of C(X, R) or C(X, C) given by x ↔ Mx .
Proof. We will do the complex case. We need to show that if I is a proper ideal of
C(X, C), then there is an x ∈ X with f (x) = 0 for all x ∈ X . Suppose this is not true.
Then for each x ∈ X , there is an fx ∈ I with fx (x) 6= 0. Since fx is continuous,
there is an open set Ux containing x such that fx (y) 6= 0 for all y ∈ Ux . Now the various
sets Ux for x ∈ X cover X (that is, ∪x∈X Ux = X ). Since X is compact, there is a finite
P
subcollection Ux1 , . . . , Uxn that covers X . Set g = ni=1 fxi · fxi (where f (x) = f (x)).
8. ALGEBRAIC GEOMETRY AND THE NULLSTELLENSATZ
195
P
Then g ∈ I and for any x ∈ X , we have g(x) = ni=1 |fxi (x)|2 6= 0, since some fxi (x) 6= 0.
Thus g1 exists and is continuous. This implies 1 = g1 · g ∈ I , so I cannot be proper.
The final point required is to show that Mx 6= My if x 6= y, which amounts to showing
that for any distinct points x, y ∈ X , there is an f ∈ C(X, C) with f (x) = 0, f (y) 6= 0.
This follows from Urysohn’s lemma in the general case. If X is a metric space, we may
explicitly define f (t) = d(x, t).
Remark 7.3. Thus X can be recovered from the maximal ideals of the ring C(X) if
X is a compact Hausdorff space. There is also a correspondence between ideals of C(X)
and filters on X , although it is more complicated.
In algebra, we are most interested in the rings Poly(X, F ) of polynomial functions
from X to F , where X ⊆ F n . We will see that for some sets X , the set X is determined
by the maximal ideals of Poly(X, F ).
Let us explore a simple example, namely X = F . Here if F is infinite, Poly(X, F ) can
be identified with F [x], so we are interested in the maximal ideals of the polynomial ring
F [x]. We know that these maximal ideals are in bijective correspondence with the monic
irreducible polynomials, and they all have the form F [x]p for such p. If F is algebraically
closed, p = x−α for α ∈ F , so maximal ideals are in bijective correspondence with points
of F = X . But if F is not algebraically closed, things are not so simple. For example,
in R[x], the maximal ideal R[x](x2 + 1) corresponds to no point of R, but rather to the
pair −i, i of points in the extension field C. One can show that the maximal ideals of
R[x] are in bijective correspondence with the points of the closed upper half plane (that
is, complex numbers with non-negative imaginary part).
This caution tells us we should stick to algebraically closed F , at least at first. We
will show in the next section that in this case the maximal ideals in F [x1 , . . . , xn ] =
Poly(F n , F ) correspond bijectively to the points of F n . However, this is not true for
arbitrary X ⊆ F n . For example, consider X = F \ {0} ⊆ F , or indeed, any proper
infinite subset of F . Then Poly(F, F ) can be naturally identified with Poly(X, F ) via the
map φ 7→ φ|X : if two polynomials agree on X , they must be equal. It follows that there
is an “extra” maximal ideal of Poly(X, F ) corresponding to the point 0 ∈ F \ X . In the
next section, we will discuss for which subsets X the correspondence works.
8. Algebraic Geometry and the Nullstellensatz
As noted at the beginning of the chapter, one of the most fruitful connections in
mathematics is that between commutative algebra and geometry. We introduce the basics
of this connection in this section. One of our main results is the Nullstellensatz, which
says that over an algebraically closed field F , there is a perfect correspondence between
certain graphs in F n and certain ideals in the polynomial ring F [x1 , . . . , xn ]. Note that
we state our results for the polynomial ring, but of course this ring is identified with the
ring of polynomial functions on F n .
Given a polynomial f ∈ F [x], one thing that we’re interested in is its roots. If
F is algebraically
closed, then we know all roots r1 , . . . , rn lie in F , and we can write
Q
f = λ ni=1 (x − ri ) where λ is the leading coefficient, so the roots essentially determine
the polynomial. What happens when we have more variables? For example, let f (x, y) ∈
F [x, y]. The set of points where f vanishes is a curve in the plane F 2 . We can no longer
write f as a product in terms of these points (after all, there are infinitely many points
196
on the curve), but it might still be the case that the curve determines the polynomial f
up to scalar multiple. This is true under some extra hypotheses. Even in one variable
the roots do not entirely determine the polynomial; we need their multiplicities as well.
So too in any number of variables, f = 0 and f 2 = 0 have the same graph. However,
every polynomial can be factored as a product of irreducible polynomials, so we may ask
whether if f and g are irreducible and have the same graph, they are scalar multiples of
each other. The answer to this question is yes, although this may not be immediately
obvious. This fact will follow from our results in this section.
The discussion in the last paragraph reminds us of the connection of polynomials and
of algebra in general to geometry. Ultimately, we must study the simultaneous zeroes
of several polynomials. For example, a line in F 3 is determined by the simultaneous
vanishing of two linear functions, such as x + y − 1 and y − z − 2.
In order to make sure as many solutions as possible are there, we will assume throughout that the field F is algebraically closed. If F is a subset of F [x1 , . . . , xn ], we define
the affine algebraic set determined by F to be the set of points (a1 , . . . , an ) in F n such
that f (a1 , . . . , an ) = 0 for all f ∈ F . This is also sometimes called the affine variety
determined by F , and it is denoted V(F). Many authors reserve the word “variety” for
a special kind of algebraic set. We will henceforth drop the word affine.
Thus for example, the variety V(x2 + y 2 + z 2 − 1) ⊆ F 3 is the unit sphere in F 3
(although if F = C for example, this sphere looks quite different from the unit sphere in
R3 ). Any set of the form V(F) for some F ⊆ F [x1 , . . . , xn ] is called an algebraic subset
of F n .
Problem 8.A. Show that ∅ and F n are algebraic sets; that any intersection of algebraic sets is an algebraic set; and that any finite union of algebraic sets is an algebraic
set. (If you get stuck, see below.) These facts are precisely the facts required to show
that the collection of algebraic subsets of F n form the closed sets in a topology on F n .
We call this topology the Zariski topology on F n . It is quite different from the usual
topology on Cn or Rn .
The reverse mapping is also of interest. If X ⊆ F n , we define the set of polynomials
vanishing on X to be
I(X) = { f ∈ F [x1 , . . . , xn ] | f (a1 , . . . , an ) = 0 for all (a1 , . . . , an ) ∈ X }.
Clearly F ⊆ I(V(F)) for all F ⊆ F [x1 , . . . , xn ] and X ⊆ V(I(X)) for all X ⊆ F n : does
equality hold? It is clear that it does not always, since for example, I(X) is always an
ideal, or put another way, V(F) = V(I) where I is the ideal of F [x1 , . . . , xn ] generated
by F . Conversely, V(F) is always an algebraic set, so we cannot have X = V(I(X)) if X
is not an algebraic set. These remarks raise many questions. For example, is every subset
of F n an algebraic set, and if not, how can we tell which sets are algebraic? Is every ideal
precisely the set of polynomials vanishing on some algebraic set, and if not, how can we
tell which ideals are? The answer to the latter question is given by the Nullstellensatz,
which we will prove later in this section. The answer to the former is not so clear; let us
consider one example.
Example 8.1. We may as well start with the simplest case, F 1 . Here we claim the
algebraic sets are the finite sets and F . Indeed, if F is a collection of polynomials, let
f be their g.c.d.: clearly V(F) = V(f ) and this is a finite set unless f = 0. Since an
197
algebraically closed field is infinite, this shows that most subsets of F are not algebraic
sets.
Before we answer the question of when equality holds above, let us make some simple
connections between algebra and geometry. We list some elementary properties of V and
I.
Lemma 8.2. Let X, Y ⊆ F n and F, G ⊆ F [x1 , . . . , xn ].
(1) If X ⊆ Y , then I(X) ⊇ I(Y ).
(2) If F ⊆ G , then V(F) ⊇ V(G).
(3) X is an algebraic set if and only if V(I(X)) = X .
(4) F is the set of all polynomials vanishing on some subset of F n if and only if
F = I(V(F)).
(5) I(X ∩ Y ) ⊇ I(X) + I(Y )
(6) V(F ∪ G) = V(F) ∩ V(G)
(7) I(X ∪ Y ) = I(X) ∩ I(Y ).
(8) V(FG)
P = V(F) ∪ V(G). (Here, FG can be taken to be either {f g | f ∈ F, g ∈ G}
or { ni=1 fi gi | fi ∈ F, gi ∈ G }.)
Proof. (1) and (2) follow immediately from the definitions of I, V .
(3) If X = V(I(X)), then X is an algebraic set by definition.
For any X , it is clear that X ⊆ V(I(X). If X is an algebraic set, say X = V(F) for
some F ⊆ F [x1 , . . . , xn ], then certainly F ⊆ I(X). Thus X ⊆ V(I(X)) ⊆ V(F) = X .
Thus equality holds throughout, and X = V(I(X).
(4) Exercise.
(5) Since X ∩ Y ⊆ X and X ∩ Y ⊆ Y , this follows immediately from (1).
(6) Since F, G are subsets of F ∪ G , we have from (2) that V(F ∪ G) ⊆ V(F) ∩ V(G).
If v ∈ V(F) ∩ V(G), then h(v) = 0 for every h ∈ F and every h ∈ G . Thus v ∈ V(F ∪ G).
This proves the desired equality.
(7),(8) Exercise.
The next result shows how the Hilbert Basis theorem can be applied. (In fact, this
result was the original motivation for the Hilbert Basis theorem).
Proposition 8.3. Let X ⊆ F n be an algebraic set. Then there are finitely many
polynomials f1 , . . . , fm ∈ F [x1 , . . . , xn ] such that X = V(f1 , . . . , fm ).
Proof. Choose f1 , . . . , fm to be generators of I(X).
Note that a polynomial in F [x] is irreducible if and only if it has only one root (of
multiplicity one). We seek an analogous geometric idea. Let us call a non-empty algebraic
subset X of F n irreducible if we cannot write X = X1 ∪ X2 for algebraic sets X1 , X2
properly contained in X ; otherwise we say X is reducible. (Irreducibility resembles
connectedness in topology, but is much stronger, as we do not require X1 ∩ X2 = ∅.)
Proposition 8.4. Let X be an algebraic subset of F n . Then X is irreducible if and
only if I(X) is a prime ideal.
Proof. Set I = I(X).
(⇒) Suppose I is not prime; then there are ideals J, K ⊃ I with JK ⊆ I . Since
J ⊃ I , there must be an f ∈ J and an (a1 , . . . , an ) ∈ X with f (a1 , . . . , an ) 6= 0. Thus
V(J) ⊂ X ; likewise, V(K) ⊂ X . By Lemma 8.2(8), we have V(J) ∪ V(K) = V(JK) ⊇
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V(I) = V(I(X)) = X . By the above inclusions, equality holds throughout, and so X is
not irreducible.
(⇐) Suppose X is reducible, say X = X1 ∪ X2 where the Xi are algebraic sets
properly contained in X , and set Ji = I(Xi ). By Lemma 8.2(1,3), we have Ji ⊃ I and
J1 J2 ⊆ J1 ∩ J2 = I(X1 ∪ X2 ) = I(X) = I . This shows I is not prime.
There is another way to characterize I(X) which is related to functions on X . We
have defined before the notion of a polynomial function on F and generalized this to
a polynomial function on F n , namely a function φ : F n → F which has the form
φ(a1 , . . . , an ) = f (a1 , . . . , an ) for some fixed f ∈ F [x1 , . . . , xn ]. Since F is infinite, the
natural map from F [x1 , . . . , xn ] to the ring of polynomial functions from F n to F is
one-to-one (Exercise! ), and we usually identify the two rings. Let X ⊆ F n . We have
defined a polynomial function on X (also called a regular function on X ) to be the
restriction of a polynomial function on F n to X , that is, to be a function φ : X → F
such that there is a polynomial f ∈ F [x1 , . . . , xn ] with φ(a1 , . . . , an ) = f (a1 , . . . , an ) for
all (a1 , . . . , an ) ∈ X . We have denoted the F -algebra of polynomial functions on X by
Poly(X, F ). We will now also use the (more common) notation O(X), and we will call
this ring the coordinate ring of X or the ring of regular functions on X . The restriction of polynomial functions from F n to X defines an F -algebra homomorphism from
F [x1 , . . . , xn ] to O(X), and the kernel is plainly I(X). Thus O(X) ∼
= F [x1 , . . . , xn ]/I(X);
some authors make this the definition of O(X).
We call an F -algebra affine if it is isomorphic to the coordinate ring of some affine
algebraic set. Since such an algebra is a factor of some F [x1 , . . . , xn ], it is a finitely
generated commutative F -algebra, and hence is Noetherian by the Hilbert Basis theorem.
We will see that such an algebra need only satisfy one more requirement to be affine,
namely that if the power of an element is 0, the element must be 0. (We should warn
the reader that “affine” is sometimes defined differently, especially for non-commutative
rings.)
One reason for introducing the coordinate ring is to give some kind of invariant description of a variety. Thus the line F 1 and the x-axis in F 2 , as well as the line x = y, z = 0
in F 3 , all are the same geometrically, but they lie in different spaces. However, they all
have isomorphic coordinate rings, namely the polynomial ring F [t].
We now begin our description of the vanishing ideals of algebraic sets. This comes
in two parts; first we concentrate on points, and then we consider the general case. We
begin with an important lemma.
Lemma 8.5 (Zariski). Let E/F be an extension of fields. Then E is finitely generated
as an F -algebra if and only if E is finite dimensional over F .
Proof. (⇐) This is trivial.
(⇒) Let E = F [α1 , . . . , αn ]. It is sufficient to show each αi is algebraic over F . We
will proceed by induction on the number n of generators. When n = 1, the result is true,
since if α1 is not algebraic over F , then F [α1 ] ∼
= F [x], which is not a field.
Suppose n > 1 and suppose some αi is not algebraic over F , say αn is transcendental
over F . Set R = F [αn ] and K = F (αn ); by induction E/K is finite dimensional, so
each αi is algebraic over K . We can choose an element c ∈ R so that the coefficients
of the (monic) minimal polynomials of the αi over K all have the form r/c for some
199
r ∈ R. If we do this, then it is plain that each αi is integral over the ring R[c−1 ] = RC −1 ,
where C = {1, c, c2 , . . . }. It follows from Corollary 6.6 that E is integral over R[c−1 ]. By
Lemma 6.2, this implies R[c−1 ] is a field. But if we let p be a prime in R not dividing c
(such a p exists by Exercise 18), then we cannot write p−1 = rc−n for any n. Thus R[c−1 ]
cannot be a field. This contradiction shows every αi must be algebraic over F .
Note that if X consists of just the point a = (a1 , . . . , an ), then the polynomials xi − ai
all lie in I(X). Now let Ma be the ideal of F [x1 , . . . , xn ] generated by x1 −a1 , . . . , xn −an .
Then Ma is the kernel of the unique F -algebra homomorphism from F [x1 , . . . , xn ] to F
that sends xi to ai for each i, just as in Section 7. Since this map is plainly onto (constants
go to themselves), Ma is a maximal ideal of F [x1 , . . . , xn ]. As I(X) is proper, it follows
that Ma = I(X). Our first result tells us that all maximal ideals have this form.
Theorem 8.6 (Weak Nullstellensatz). Let F be an algebraically closed field and let
M be a maximal ideal of F [x1 , . . . , xn ]. Then M = Ma for some a ∈ F n .
Proof. The F -algebra F [x1 , . . . , xn ]/M is finitely generated and it is a field since M
is maximal. Thus by Lemma 8.5 it is finite dimensional and hence algebraic over F . As
F is algebraically closed, this field must be F . This means that for each i there is an
ai ∈ F with xi + M = ai + M , that is, xi − ai ∈ M . If we set a = (a1 , . . . , an ), this means
that Ma ⊆ M . Since both are maximal ideals, they must be equal.
Corollary 8.7. Let F be an algebraically closed field and let X be an algebraic subset
of F n . Then there is a bijective correspondence between the points of X and the maximal
ideals of Poly(X, F ) = O(X) given by x 7→ Mx .
Proof. Identify O(X) with F [x1 , . . . , xn ]/I(X). The maximal ideals of O(X) correspond to the maximal ideals of F [x1 , . . . , xn ] containing I(X), that is, the maximal
ideals of O(X) correspond to the ideals Ma ⊇ I(X). Since Ma ⊇ I(X) if and only if
a ∈ X , this completes the proof.
Corollary 8.8 (Weak Nullstellensatz). Suppose F is an algebraically closed field. If
I is a proper ideal of F [x1 , . . . , xn ], then V(I) 6= ∅.
Proof. Exercise.
We can now use a very nice idea (known as the “Rabinowitsch trick”) to extend this
result to all ideals. Before doing this, though, we need another definition. We say an ideal
I is semiprime if whenever J is an ideal with J n ⊆ I for some positive integer n, we
have J ⊆ I . By induction it is clear that it is enough to check this for n = 2, and just as
in the proof of Lemma 4.1, we only need to check this for J ⊃ I . Again as in the proof
of Lemma 4.1, I is semiprime if and only if whenever rn ∈ I , we have r ∈ I (that is, the
ring R/I has no non-zero nilpotent elements), and it is enough to check this for n = 2.
A semiprime ideal is also called a radical ideal for this reason.
Given any ideal I in a commutative ring R, we define the radical of I to be
√
I = { r ∈ R | rn ∈ I for some positive integer n }.
It is not hard to see that this is an ideal (Exercise! ) containing I . We will see in Chapter 7
that it is precisely the intersection of all prime ideals of R containing I , or equivalently,
the smallest semiprime ideal of R containing I .
200
If X is any subset of F n and f m ∈ I(X), then (f (a1 , . . . , an ))m = 0 for any
(a1 , . . . , an ) ∈ X . As F is a field, this implies f (a1 , . .√. , an ) = 0, and so f ∈ I(X). Thus
I(X) is always a semiprime ideal, and so I(V(I)) ⊆ I for any ideal I of F [x1 , . . . , xn ].
Theorem 8.9 (Nullstellensatz). Let √F be an algebraically closed field and let I be an
ideal of F [x1 , . . . , xn ]. Then I(V(I)) = I .
Proof. Set X = V(I) and let f ∈ I(X). We consider the ring R0 = F [x0 , x1 , . . . , xn ]
and we let I 0 be the ideal of R0 generated by the elements of I and by 1−x0 f (x1 , . . . , xn ),
that is I 0 = R0 I + R0 (1 − x0 f ). Suppose (a0 , a1 , . . . , an ) ∈ V(I 0 ): then since I ⊆ I 0 , we
have (a1 , . . . , an ) ∈ X . But then (1 − x0 f )(a0 , a1 , . . . , an ) = 1 − a0 f (a1 , . . . , an ) = 1 6= 0.
This shows V(I 0 ) = ∅, and so by the weak Nullstellensatz, we must have I 0 = R0 .
Since R0 is generated by x0 in addition to R, the fact that I 0 = R0 , i.e., that 1 ∈ I 0 ,
can be stated as
1=
k
X
j=0
xj0 (ij +hj (1−x0 f )) = i0 +h0 +(i1 +h1 −h0 f )x0 +· · ·+(ik +hk −hk−1 f )xk0 −hk f xk+1
0
for some i0 , . . . , ik ∈ I and h0 , . . . , hk ∈ F [x1 , . . . , xn ]. Since the ij and hj do not involve
x0 , the only way this equation can hold is to have i0 + h0 = 1, i1 + h1 − h0 f = 0, . . . ,
ik + hk − hk−1 f = 0, and hk f = 0. This means h0 = 1 − i0 , h1 = f − (i1 + i0 f ), . . . ,
hk = f k − (ik + · · · + i0 f k ), hk f = 0. Now if f = 0, certainly f ∈ I . If f 6= 0, then we
must have hk = 0, which implies f k = ik + · · · + i0 f k ∈ I .
There is another way to think of the close of the proof in the previous paragraph:
0
I’ll
P sketch it and let the reader fill in the details. Since 1 ∈ I , we can write 1 =
l gl il + h(1 − x0 f ) for some il ∈ I and gl , h ∈ F [x0 , x1 , . . . , xn ]. Regard this as an
equation
P in F [x1 , . . . , xn , 1/f ][x0 ] and substitute 1/f for x0 . This yields the equation
1 = l g̃l il , where g̃l is obtained from gl by replacing each occurrence of x0 by 1/f . If k
is the largest power of x0 that occurs in
Pany gl , then we can clear fractions by multiplying
through by f k . We then obtain f k = l ḡl il ∈ I , since each ḡl = f k g̃l ∈ F [x1 , . . . , xn ].
Corollary 8.10. Let F be an algebraically closed field. Then I 7→ V(I) and X 7→
I(X) provide inverse bijective correspondences between the set of radical ideals of F [x1 , . . . , xn ]
and the set of algebraic subsets of F n .
Proof. Exercise.
9. Dimension Theory
In the last section we saw a connection between algebra and geometry. In this section
we pursue that connection by showing how to determine the dimension of a geometric
object (is it a curve, surface, 3-dimensional space, etc.) from the corresponding algebraic
object. The algebraic dimension can be measured either by chains of prime ideals, or by
transcendence degree over the base field (that is, the size of the largest set in the ring
which is algebraically independent over the base field).
We begin with the geometric definition. Suppose for example that X is an irreducible
curve. Then the only irreducible algebraic subsets of X are single points. If X is an
irreducible surface, then the the only irreducible algebraic subsets are curves and points.
This motivates the following definition. If X is an algebraic set, we define the dimension
10. HOMEWORK PROBLEMS
201
of X to be the supremum of the positive integers n for which there is a chain X0 ⊂ X1 ⊂
· · · ⊂ Xn of irreducible algebraic subsets of X . This supremum is actually a maximum.
Problem 9.A. Show that if F is an algebraically closed field, then the dimension of
the algebraic set F n is n.
We can immediately connect this with one algebraic definition. Recall that an irreducible algebraic subset of X corresponds to a prime ideal of O(X). We define the Krull
dimension of a commutative ring R to be the supremum of the positive integers n for
which there is a chain P0 ⊂ P1 ⊂ · · · ⊂ Pn of prime ideals of R. We then have the
following.
Proposition 9.1. Let F be an algebraically closed field and let X ⊆ F m be an
algebraic set. Then the dimension of X equals the Krull dimension of O(X), and this is
at most m.
Proof. Exercise.
Even if F is not algebraically closed, the Krull dimension of F [x1 , . . . , xn ] is n.
There are many other ways to obtain the dimension of X or R, at least in special
cases. We will mention just one here.
Theorem 9.2. Let F be an algebraically closed field and let X ⊆ F m be an algebraic
set. Then the dimension of X is the largest positive integer n such that O(X) contains
a copy of the polynomial ring F [x1 , . . . , xn ].
Proof. Difficult exercise.
We close by making an observation about Dedekind domains.
Theorem 9.3. Let F be an algebraically closed field.
(1) If X is an algebraic set and O(X) is a Dedekind domain, then dim X = 1.
(2) Let f ∈ F [x, y]. Then F [x, y]/F [x, y]f is a Dedekind domain if and only if f is
irreducible and Of = (∂f /∂x, ∂f /∂y) does not vanish on V(f ).
Proof. Difficult exercise.
This theorem is a special case of a more general result. It marks the start of the study
of singularities of algebraic subsets of F n .
10. Homework Problems
All rings in these problems are assumed to be commutative.
(1) Let R be a commutative
ring. Recall that the ring R[[x]] consists of all formal
P
i
power series ∞
r
x
for
r0 , r1 , · · · ∈ R, with the same operations as for polyi
i=0
nomials. When dealing with power series, we cannot define degree, so we define
the order to be the smallest positive integer n with rn 6= 0, and the coefficient
rn takes the place of the leading coefficient.
(a) State and prove the properties of order in R[[x]] that are analagous to the
standard properties of degree in R[x]. P
i
(b) Show that the power series f = ∞
i=0 ri x ∈ R[[x]] is a unit if and only if
r0 is a unit in R.
[Hint for (b): It is possible to do this directly, or as follows. Recall (1 − x)(1 +
202
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
x + x2 + . . . ) = 1; show that you can replace x by any power series g divisible by
x, that is, any g with constant coefficient 0. Make sure everything makes sense.]
Let R be a commutative ring. Show that the following conditions are equivalent.
(a) R is an integral domain.
(b) R[x] is an integral domain.
(c) R[[x]] is an integral domain.
Suppose R is an integral domain, m, n are relatively prime positive integers in
R, and am = bm and an = bn for a, b ∈ R. Prove that a = b.
Let R be a commutative ring. Show that R is a field if and only if R[x] is a PID.
(a) Let R be a Noetherian ring with I / R. Show that R/I is Noetherian.
(b) Let R be the subring of the polynomial ring Z[x] consisting of all polynomials a0 + b1 x + · · · + bn xn with a0 , b1 , . . . , bn ∈ Z and b1 , . . . , bn even. Show
that R is not Noetherian.
[Since Z[x] is Noetherian, this shows that a subring of a Noetherian ring need
not be Noetherian.]
Prove that an integral domain is Artinian if and only if it is a field.
Let R be a commutative ring and C a denominator set in R. Show that RC −1
is the zero ring if and only if 0 ∈ C .
Let R ⊆ R0 be integral domains with quotient fields Q and Q0 respectively.
(a) Show that Q embeds in Q0 in a natural way.
(b) Show that if R0 is finitely generated as an R-module, then Q0 is finite
dimensional over Q.
[Hint: show Q0 = R0 (R \ {0})−1 .]
Let C be a denominator set of R and let I / R. Show that (I e )c = { r ∈ R | rc ∈
I for some c ∈ C }.
Prove Proposition 2.3.
Let C ⊆ D be denominator sets in R. Show that RD−1 ∼
= (RC −1 ){d/1 | d ∈
−1
D} .
Let R be a commutative ring and C a denominator set in R.
(a) Show that if R is Noetherian, then RC −1 is Noetherian.
(b) Show that RC −1 can be Noetherian even if R is not.
If C is a subset of a ring R and φ : R → S is a ring homomorphism, we will
say φ is C -inverting if φ(c) is a unit for every c ∈ C . If C is a denominator
set in a commutative ring R, show that RC −1 together with the natural map
ι : R → RC −1 defined by ι(r) = r/1 has the following universal property. The
map ι is C -inverting, and for any C -inverting ring homomorphism φ : R → S (for
any commutative ring S ), there is a unique ring homomorphism ψ : RC −1 → S
with ψ ◦ ι = φ.
The following is a picture of the universal property of localization.
R
ι - −1
RC
@
@φ
@
(14) Prove Lemma 3.1.
(15) Prove Lemma 3.4 and Corollary 3.5.
ψ
@
RS
@
?
203
(16) Let R be a UFD. Show that the set of principal ideals of R forms a complete
lattice (under ⊆).
Note that this need not be a sublattice of the lattice of all ideals of R; the greatest
lower bound is obtained from the g.c.d. and the least upper bound from the l.c.m.
(17) Let R be a subring of the integral domain S , and let a, b, d ∈ R.
(a) If d is a g.c.d. of a, b in S , is d a g.c.d. of a, b in R?
(b) If d is a g.c.d. of a, b in R, is d a g.c.d. of a, b in S ?
(c) Do your answers to (a) or (b) change if we assume R is a PID? If we
assume S is a PID? If we assume both are PIDs?
(18) Show that F [x1 , . . . , xn ] has infinitely many non-associate irreducible polynomials
for any field F and any positive integer n.
Can you replace F by any Noetherian integral domain?
(19) Prove Lemma 3.9. (Supply all missing details.)
(20) Let R be a UFD and let C be a proper denominator set in R.
(a) Show that RC −1 is a field if and only if every prime in R divides some
element of C .
(b) Conclude that F [x1 , . . . , xn ]C −1 cannot be a field if C is a denominator
set in R which is finitely generated as a multiplicative monoid.
(21) Show that R is a UFD if and only if R[[x]] is a UFD.
(22) Prove Corollary 4.3.
(24) Let R be an integral domain with quotient field Q, let I be a nonzero R-ideal
of Q, and let I 0 = { q ∈ Q | qI ⊆ R }.
(a) Show that I 0 is an R-ideal of Q.
(b) Show that there is an R-ideal J of Q with IJ = R if and only if II 0 = R
(in which case J must equal I 0 ).
(25) Show that a Dedekind domain with only finitely many maximal ideals is a PID.
[Hint: You can modify the proof of Proposition 5.3(4) ⇒ (1). Let M, P1 , . . . , Pn
be all maximal ideals and choose m0 ∈ M \ M 2 . Show M 2 + P1 . . . Pn = R,
so there are x ∈ M 2 , y ∈ P1 . . . Pn with x + y = 1. Set m = x + m0 y, show
m ∈ M \ (M 2 ∪ P1 ∪ · · · ∪ Pn ), and conclude Rm = M .]
(26) Prove Proposition 4.11 by focussing on invertible ideals. That is, show that if
I / R is invertible, then I e is an invertible ideal of RC −1 .
(27) Let R be a commutative ring and let I / R. In this problem some facts will be
stated, and you will be asked to use them to draw conclusions. (Hint: use Zorn’s
Lemma.) If you wish to try to prove the facts, you may, but they are not easy.
(See Problem 14 in Chapter 7.)
(a) If I is maximal among non-finitely-generated ideals of R, then I is prime.
Conclude that R is Noetherian if and only if every prime ideal in R is finitely
generated.
(b) If I is maximal among non-principal ideals of R, then I is prime. Conclude that R is a principal ideal ring if and only if every prime ideal in R is
principal.
(c) Suppose that R is an integral domain. If I is maximal among noninvertible ideals of R, then I is prime. Conclude that R is a Dedekind domain
if and only if all nonzero prime ideals in R are invertible.
204
(28) Show that if P is a prime ideal in the commutative ring R, then RP /P RP is
isomorphic to the quotient field of R/P .
(29) Let R be an integral domain with quotient field Q. Show that the following
conditions are equivalent.
(a) R is a valuation ring
(b) For any q ∈ Q, either q ∈ R or q −1 ∈ R.
(c) The ideals of R are totally ordered by inclusion.
(30) Show that any valuation ring is a local ring.
(31) Show that any maximal proper subring of a field is a valuation ring. (Such a ring
also has the property that for any a, b in the ring, there is a positive integer n
with a | bn .)
(32) Show that if R is a subring of the integral domain S and S is integral over R,
and S is finitely generated as an algebra over R, then S is finitely generated as
an R-module.
(33) Let R ⊆ S be integral domains and let r ∈ R. Show that if r−1 exists in S and
is integral over R, then r−1 ∈ R.
(35) Show that any valuation ring is integrally closed.
[One can show that an integral domain is integrally closed if and only if it equals
the intersection of all valuation rings contained between it and its quotient field,
but this is not easy.]
(36) Let R be a UFD with quotient field Q, let E be an extension field of Q, and let
e ∈ E . Show that e is integral over R if and only if e is algebraic over F and the
(monic) minimal polynomial of e in F [x] lies in R[x]. Do you need to assume R
is a UFD?
(37) Let R ⊆ S be commutative rings.
(a) Show that if P is a prime ideal of S , then P ∩ R is a prime ideal of R.
(b) Suppose that Q is a prime ideal of R, and let P be an ideal of S which is
maximal with respect to the property that P ∩ R ⊆ Q. Show that P is a prime
ideal.
(c) Taking R = Z, S = Q, show that given a prime ideal Q of R, there need
not be a prime ideal P of S with P ∩ R = Q. (When such a P exists for every
Q, we say lying over holds for the extension S/R.)
(38) Let R ⊆ S be commutative rings and let I / S . Then R/(I ∩ R) embeds in S/I
in a natural way. Decide whether if S is integral over R, we can conclude that
S/I is integral over R/(I ∩ R).
(39) Let R ⊆ S be commutative rings. We say lying over holds for this extension of
rings if for any prime ideal Q of R, there is a prime ideal P of S with P ∩R = Q.
Prove that if S is integral over R, lying over holds.
[One approach: Using Problem 37 and Zorn’s Lemma, our candidate for P might
be an ideal maximal with respect to P ∩ R ⊆ Q. By Problem 38, we may assume
P = 0 = P ∩ R, so R and S are integral domains and we need to show Q = 0.
Let C = R \ Q and localize R and S at C to pass to RC −1 ⊆ SC −1 . Since
RC −1 = RQ is a local ring with maximal ideal QC −1 , show that SC −1 must be a
field (else some nonzero I / S has I ∩ R ⊆ Q). By Lemma 6.2, this shows RC −1
is a field and so Q = 0.]
205
(40) Let R ⊆ S be integral domains such that S is integral over R and let I be a
nonzero ideal of S . Show that I ∩ R 6= 0.
√
(41) Let d be a squarefree integer. Determine the integral closure of Z[ d].
(42) Let F be a field. A monomial subalgebra of the polynomial ring F [x] is
a subalgebra spanned as a vector space over F by powers of x (for example,
F [x2 , x3 ] is spanned by all powers xn with n = 0 or n ≥ 2). Show that if R is a
monomial subalgebra of F [x] other than F , then the quotient field of R is F (xl )
where l is the g.c.d. of the powers of x that occur in R. Determine whether F [xl ]
is the integral closure of R in its quotient field.
(43) Let R be an integral domain with quotient field Q. We say R is completely
integrally closed if and only if whenever q ∈ Q and there is a q 0 ∈ Q \ {0} with
q 0 q n ∈ R for all positive integers n, then q ∈ R.
(a) Show any completely integrally closed integral domain is integrally closed.
(b) Is every PID completely integrally closed? Every UFD? Every Dedekind
domain?
(44) Supply all the details in the proof of Theorem 7.1.
(45) Let F be a field and X a set, and let F = { Y ⊆ X | X \ Y is finite }. Call an
element of Fun(X, F ) almost constant if it takes on a single value on all but
finitely many points of X . If we set I = { f ∈ Fun(X, F ) | Z(f ) ∈ F }, show
that Fun(X, F )/I ∼
= { f ∈ Fun(X, F ) | f is almost constant }.
(46) Describe the correspondence claimed to exist in Remark 7.3.
(47) Let F be a field. Determine whether the ideal generated by x + y and xn + y n
is prime in F [x, y], where n ≥ 2.
[Hint: this may depend on the field and on n.]
(48) Let F be a field, let a = (a1 , . . . , an ) ∈ F n , and let φ : F [x1 , . . . , xn ] → F
be the unique F -algebra homomorphism with φ(xi ) = ai for all i. Prove that
ker φ = Ma is the ideal generated by x1 − a1 , . . . , xn − an .
(49) Complete the proof of Lemma 8.2.
(50) Let F be an algebraically closed field. Show that there can be no infinite chain
X1 ⊃ X2 ⊃ X3 ⊃ . . . of algebraic sets in F n . (Hence the poset of algebraic
subsets of F n satisfies the minimum condition.)
(51) Prove that every algebraic set can be written as a finite union of irreducible
subsets. Can you say anything about uniqueness?
[Hint: Use Problem 50 to find a “minimal counterexample”.]
(52) Let F be an algebraically closed field. Show that every proper algebraic set in
F 2 is the union of finitely many points and finitely many irreducible algebraic
curves (that is, curves defined by f (x, y) = 0 for some irreducible polynomial f ).
(53) Let F an algebraically closed field and let X be an algebraic set in F n . Find
a natural bijection between the points of X and the set of F -algebra homomorphisms from the coordinate ring O(X) to F .
(54) Let F be an algebraically closed field and let E/F be a field extension. Show that
if F is a set of polynomials in F [x1 , . . . , xn ] and F has a simultaneous solution
in E , then it has a simultaneous solution in F .
(55) Let F be an algebraically closed field and let R be a finitely generated commutative F -algebra. Show that every prime ideal of R is the intersection of some
collection of maximal ideals.
206
(56) Let F be an algebraically closed field. Use Corollary 8.10 to prove that a proper
algebraic subset of F 1 is finite.

Chapter 6 - pantherFILE

Transcription

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