Torsionless Modules and Minimal Generating Sets for Ideals of Integral
Domains
Except where reference is made to the work of others, the work described in this thesis is
my own or was done in collaboration with my advisory committee. This thesis does not
include proprietary or classifled information.
Wesley R. Brown
Certiflcate of Approval:
Overtoun Jenda
Professor
Mathematics and Statistics
H. Pat Goeters, Chair
Professor
Mathematics and Statistics
Peter Nylen
Professor
Mathematics and Statistics
William Ullery
Professor
Mathematics and Statistics
Stephen L. McFarland
Acting Dean
Graduate School
Torsionless Modules and Minimal Generating Sets for Ideals of Integral
Domains
Wesley R. Brown
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
August 7, 2006
Torsionless Modules and Minimal Generating Sets for Ideals of Integral
Domains
Wesley R. Brown
Permission is granted to Auburn University to make copies of this thesis at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Wesley Robert Brown was born July 17th, 1982 in Birmingham, Alabama. In 2000, he
graduated from Thompson High School in Alabaster, Alabama where he was Valedictorian
of his senior class and captain of the football team. He earned a Bachelor?s degree in
Mathematics with a minor in Computer Science from Birmingham-Southern College in
2004. While at BSC, he served as president of Kappa Mu Epsilon Mathematics Honor
Society, an editor of The Hilltop News, and was a founding member of Tri-Epsilon. He
began his graduate studies at Auburn University in 2004.
iv
Thesis Abstract
Torsionless Modules and Minimal Generating Sets for Ideals of Integral
Domains
Wesley R. Brown
Master of Science, August 7, 2006
(B.S., Birmingham-Southern College, 2004)
33 Typed Pages
Directed by H. Pat Goeters
This is a treatise of relationships between the number of elements necessary to gen-
erate the ideals of a domain and the torsionless modules of that domain. Three types of
domains are identifled according to natural decompositions of their torsionless modules.
The descriptions of the domains follow the historical approach of Dedekind by focusing on
the ideals of the domains.
v
Acknowledgments
The author would like to thank Dr. Pat Goeters for his generous contributions and
patience. The author would also like to thank his family and friends for their love, support,
and encouragement.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speciflcally LATEX)
together with the departmental style-flle aums.sty.
vii
Table of Contents
1 Introduction 1
2 Definitions 2
3 Principal Ideal Domains 4
4 Dedekind Domains 6
5 Noetherian Domains 14
Bibliography 25
viii
Chapter 1
Introduction
We will attempt to develop a strong relation between the number or elements necessary
to generate the ideals of a domain and properties of the torsionless modules of that domain.
We will consider three types of domains: principal ideal domains, the class of which is
properly contained in the class of Dedekind domains, which in turn is properly contained
in the class of Noetherian domains whose ideals are generated by two elements.
By deflnition, the ideals of a principal ideal domain are generated by a single element.
For these domains, every torsionless module is a direct sum of copies of R, or free. For
Dedekind domains, every ideal is generated by two elements, one of which can be arbitrarily
chosen. The terminology is that ideals are 112-generated. For Dedekind domains, every
torsionless module is projective. Finally, for the broadest class, Noetherian domains whose
ideals are 2-generated, every torsionless module is a direct sum of ideals of R:
Throughout this paper, R always represents an integral domain with quotient fleld Q.
1
Chapter 2
Definitions
The following deflnitions are basic staples of ring and module theory but are included
here due to their fundamental importance throughout this thesis.
Deflnition 1. Let X be a subset of a module A over R. The intersection of all submodules
of A containing X is called the submodule generated by X. If X is a flnite set, then the
submodule generated by X is said to be flnitely generated.
If a module A is generated by the set fx1;x2;:::;xng then we write A = (x1;x2;:::;xn):
Deflnition 2. A module M over R is torsion-free if for all a 2 M and r 2 R, ra = 0
implies that either r = 0 or a = 0.
Deflnition 3. A module M over R is free if M is isomorphic to a direct sum of copies of
R.
Deflnition4. A subset X of a torsion-free R-module M is said to be linearly independent
if for distinct x1;x2;:::;xn 2 X and r1;r2;:::;rn 2 R; r1x1 + r2x2 + ::: + rnxn = 0 implies
that ri = 0 for all i:
Deflnition 5. The rank of a torsion-free R-module M is the cardinality of a maximal
linearly independent subset of M.
Deflnition 6. A module M over R is torsionless if M is isomorphic to a submodule of a
flnite-rank free module.
Deflnition 7. A module M over R is projective if for every R-module homomorphism
A ! B ! 0
2
and any R-module homomorphism ? : M ! B; then there is an R-module homomorphism
? : M ?! A such that ?? = .
It is apparent that free modules are projective. However there are domains for which
projective modules are not necessarily free as the example in Chapter 4 shows.
Deflnition 8. Let P be a prime ideal of R. Let
P P1 P2 ::: Pn
be a descending chain of unique prime ideals of R such that for any other proper chain
P P1 P2 ::: Pm;
m ? n. Then the rank of the prime ideal P is n.
Deflnition 9. The Krull dimension of R is the supremum of the ranks of the prime
ideals of R.
Deflnition 10. An ideal I of R is invertible if II?1 = R where I?1 = fq 2 Q : qI Rg.
3
Chapter 3
Principal Ideal Domains
We will flrst consider principal ideal domains
Deflnition 11. A domain R is a principal ideal domain, or a PID for short, if every
ideal of R can be generated by a single element.
In the case of a PID, every torsionless R-module is free as stated below in the single
result of this section.
Theorem 3.1. Every torsionless R-module is free if and only if R is a PID.
Proof. ()) Assume every torsionless R-module is free. Let I be an ideal of R. We will
show that I is principal. Since I is an ideal, it is an R-module. R is a free R-module of
rank 1 so I is a torsionless R-module and thus is free. Therefore, I ?= 'nR for some n. Let
a;b 2 I. Note that b(a)+(?a)b = 0 and thus any two elements of I are linearly dependent.
However, (1;0;:::;0);(0;1;0;:::;0) 2 'nR are linearly independent as long as n > 1. Thus,
I ?= R and thus I is principal.
(() Assume R is a PID. Let K be a torsionless R-module. Then,
0 6= K ?= K0 ? F ?= 'nR;
where F is some free R-module. Without loss of generality, we may identify K with K0:
Now, let ? : F ?! R be a projection map such that ?(K) = I 6= 0, for some ideal I of
R. Since I is principal, we can write I = aR. Let x 2 K such that ?(x) = a. Now, deflne
? : I ?! K by ?(ar) = rx. Since ?? = 1I, K ?= Im?'K0, where K0 = Ker(?jK). Note
4
that Im? = xR ?= R. Also,
Ker? = R'R':::'0':::'R;
where 0 corresponds to the copy of R being projected. Since K0 ? Ker?, induction applies
to show that K0, and thus K, is a direct sum of copies of R and is thus free.
We can now state the following corollary, which creates a symmetry with the flrst result
of the next section.
Corollary 3.1. Every torsionless R-module is free if and only if every ideal is free.
Since free modules are projective, the proof of this corollary follows in the exact manner
of the flrst proof of the next section.
5
Chapter 4
Dedekind Domains
We will begin this section by characterizing domains whose torsionless modules are
projective. We will then relate this to the class of Dedekind domains deflned below.
Theorem 4.1. Every torsionless R-module is projective if and only if every ideal is projec-
tive.
Proof. ()) Assume every torsionless R-module is projective. Let I be an ideal of R. Then
I is an R-module. Also, since R is a free R-module of rank 1, I is torsionless. Thus, I is
projective.
(() Assume every ideal of R is projective. Let K be a torsionless R-module. Then
0 6= K ?= K0 ? F ?= 'nR. As in the proof of Theorem 3.1, we will identify K with K0:
Since F is free, it is projective. Let ? : F ! R be a projection map such that ?(K) = I 6= 0.
I is an ideal of R and is thus projective. Consider the sequence:
0 ! Ker(?jK) ! K ?! I ! 0:
Since I is projective, we know there exists ? : I ! K such that ?? = 1I. So, K ?=
Im? ' Ker(?jK). Im? is isomorphic to an ideal of R and is thus projective. As in the
proof of Theorem 3.1, inductively Ker(?jK) is a direct sum of ideals and so is a direct sum
of projective modules and is thus projective. Thus, K is a direct sum of projective modules
and is projective.
6
Deflnition 12. A domain R is called a Dedekind domain if every proper ideal of R is
the product of a flnite number of prime ideals of R.
Every principal ideal domain is a Dedekind domain. In fact, every nonzero element
a 2 R can be factored as a = p1p2:::pm with each pj a prime element of R. Thus, (a) =
(p1)(p2):::(pn) There are, however, Dedekind domains that are not principal. An example
is
Z[p10] = fa+bp10 : a;b 2 Zg
as discussed in [1].
We now relate the result from the theorem proved above to Dedekind domains.
Theorem 4.2. The following are equivalent for a domain R:
(a) R is a Dedekind Domain.
(b) Every proper ideal of R is uniquely the product of a flnite number of prime ideals.
(c) Every nonzero (fractional) ideal in R is invertible.
(d) Every (fractional) ideal of R is projective.
(e) R is Noetherian, integrally closed, and every nonzero prime ideal is maximal.
(f) R is Noetherian and for every nonzero prime ideal P of R the localization RP is a
principal ideal domain.
The proof of this theorem is rather sprawling. We will give a sketch here. A complete
proof can be found in [1]. A related proof can be found in [4]. The equivalence of (c) and
(d) will be discussed again in Lemma 5.7.
7
Sketch of Proof. (a) ) (b) We begin with I, a proper ideal of R such that I = P1P2:::Pm =
Q1Q2:::Qn for primes Pi and Qj. Since I ? Pi;Qj, any Pi contains some Qj. Once it is
shown that every nonzero prime ideal of a Dedekind domain is invertible and maximal, the
proof follows by cancellation and induction.
(b) ) (c) This follows from the fact that a flnite product of ideals I1I2???In of a
domain is invertible if and only if each IJ is invertible and the previously stated fact that
every nonzero prime ideal of a Dedekind domain is invertible.
(c) ) (d)IfI isanonzeroidealofR, thenII?1 = Rimpliesthattherearea1;a2;:::;an 2
I and b1;b2;:::;bn 2 I?1 such that a1b1 + a2b2 + ::: + anbn = 1. The epimorphism
` : 'nR ! I given by `(r1;r2;:::;rn) = r1a1 + r2a2 + ::: + rnan is split by f : I ! 'nR
where f(a) = (ab1;ab2;:::;abn). Thus, I is projective.
(d) ) (c) If I is a nonzero ideal of R, then there is a split epimorphism ` : 'nR ! I !
0. So, there exists bi 2 I?1 for i = 1;2;:::n such that for ai = `(ei), where ei is the standard
basis vector in 'nR, a1b1 +a2b2 +:::+anbn = 1. Thus II?1 = R and I is invertible.
(c) ) (e) Since every invertible fractional ideal of R is a flnitely generated R-module,
and since a commutative ring is Noetherian if and only if every ideal of R is flnitely gener-
ated, we can conclude that R is Noetherian.
If u 2 K, the quotient fleld of R, then Ru is a flnitely generated R-module and is thus
a fractional ideal of R. So, R[u] is invertible. Since R[u]R[u] = R[u],
R[u] = RR[u] = (R[u]?1R[u])R[u] = R[u]?1R[u] = R
which implies that u 2 R and thus R is integrally closed.
Next, we will show that any nonzero prime ideal is maximal. Let P be a nonzero prime
ideal of R. There is some maximal ideal M containing P. M is invertible and thus M?1P
8
is a fractional ideal of R and M?1P ? M?1M = R which means M?1P is an ideal of R.
Also, since M(M?1P) = RP = P and P is prime either M ? P or M?1P ? P. In the
latter case,
R ? M?1 = M?1R = M?1PP?1 ? PP?1 ? R
which means that M?1 = R. But then R = MM?1 = MR = M, a contradiction since M
is maximal. So, it must be that M ? P. Therefore M = P and P is maximal.
(e) ) (f) This implication is established using several results in [1]. First, we know
that RP is an integrally closed integral domain. Also, every ideal of RP is of the form IP
where I is an ideal of R and furthermore every nonzero prime ideal of RP is of the form IP
where I is a nonzero prime ideal of R that is contained in P. By assumption, every nonzero
prime ideal of R is maximal and thus PP is the unique nonzero prime ideal of RP. This
implies that RP is a principal ideal domain.
(f) ) (a) First, we will show that any ideal I of R is invertible. II?1 is an ideal of R
and thus if II?1 6= R then there is some maximal ideal M containing I. M is prime so by
assumption, IM is principal. Let r=q be the element that generates IM, where r 2 I and
q =2 M. R is Noetherian and thus I is flnitely generated. Let fb1;b2;:::;bng be a generating
set for I. For each i, bi=1R = (ri=qi)(r=q) 2 IM where each ri 2 I and each qi =2 M. Thus
qiqbi = rir 2 I. Let t = qq1q2:::qn =2 M. Then for all i, we have (t=r)bi = tbi=r 2 R and
thus (t=r) 2 I?1. Finally, we have that t = (t=r)r 2 II?1 ? M. But t =2 M and so it must
be that II?1 = R. This shows that I is invertible.
Then for each I, we can choose a maximal ideal M 6= R that contains I. The invertibil-
ity of M and I implies that I must be properly contained in the ideal IM?1 since otherwise
9
we have
R = RR = (II?1)(MM?1) = I?1(IM?1)M = I?1IM = RM = M;
a contradiction since M is maximal.
For any proper ideal J of R, we can now set up a chain of ascending ideals containing
J that must terminate since R is Noetherian. But, the ideals are of the form discussed in
the last paragraph and thus the only way the chain can terminate is for the terminal link
to be R itself. This allows us to express J as a product of maximal and thus prime ideals.
This shows that R is Dedekind.
Stated here, Nakayama?s lemma is a well known result of ring theory.
Nakayama?s Lemma 4.3. Let S be a domain. If J(S) is the intersection of all maximal
ideals of S, K is a flnitely-generated S-module, and L is a submodule of K, then
K = L+J(S)K
implies that K = L:
Since our main theme is to relate torsionless modules to the magnitude of maximal
generating sets of ideals, we now introduce the following concept.
Deflnition 13. A nonzero ideal I of a domain R is said to be 112-generated if for any
a 2 I, a 6= 0, there exists b 2 I such that I is generated by fa,bg.
Theorem 4.4. A domain R is Dedekind if and only if every nonzero ideal I of R is one-
and-a-half generated.
10
Proof. (() We know that R being Dedekind is equivalent to the statement R is Noetherian
and for all maximal ideals M of R, the localization RP is a PID. Assume that every ideal
I of R can be generated by fa,bg as above. Thus, R is clearly Noetherian.
Now, let P be a maximal ideal of R. Let J be an ideal of RP. So, J = RPI for some
ideal I of R. Choose a 2 PI and write I = Ra+Rb. Then,
J = RP(Ra+Rb) = RPa+RPb;
and a 2 PI PJ. Thus
RPb+PJ = J;
so by Nakayama?s Lemma, J = RPb. Thus, J is principal and RP is a PID. By Theorem
4.2, R is Dedekind.
()) Let I be a nonzero ideal of R. Let a 2 I be nonzero. R is Dedekind so the ideal
Ra is a product of maximal ideals, say Ra = P1P2:::Pm for maximal ideals P1;P2;:::Pm.
Then, aR P implies Pi P for some i and therefore these are the only maximal ideals
that contain a. By Theorem 4.2, if R is Dedekind, then R is Noetherian and for every
maximal ideal P, RP is a PID. So, RPi is a PID for i = 1;2;:::m and so the ideal IRPi is
a principal ideal of RPi. Thus, for each i there is some bi 2 I such that IRPi = RPibi. We
claim that there is a b 2 I such that IRPi = RPib for all i. Order P1;P2;:::Pm such that
P1;P2;:::Pk are distinct and for j > k, Pj is equal to one of P1;P2;:::Pk. Now, Qkj=1;j6=i Pj
is not contained in Pi for any i so
Pi +
kY
j=1;j6=i
Pj = R:
11
Then clearly,
PiI +(
kY
j=1;j6=i
Pj)I = I
and for each i = 1;2;:::m, we can write bi = ci +di where ci 2 Pi and di 2 (Qkj=1;j6=i Pj)I.
Note that bi is congruent to di modulo PiI for all i. Let b = d1 + d2 + ::: + dm. Then for
any i, dj is congruent to zero modulo PiI if j 6= i. So, b ? bi (mod Pi) for all i. Then
IRPi = RPibi = RPibi +PiIRPi:
But since b ? bi (mod Pi), this is equal to RPib + PiIRPi. Moreover, as in the reverse
implication, the only maximal ideal of RPi is PiRPi and since
IRPi = RPib+PiIRPi = RPib+(IRPi)(PiRPi);
we can apply Nakayama?s Lemma and arrive at the equality: IRPi = RPib for all i.
Let J = Ra + Rb. Note that J I. Also note that IRP = IP = JP = JRP for any
maximal ideal P of R. We can see this by flrst considering the case where P is not one of
our P1;P2;:::Pm. In this case,
RP = (Ra)P JP IP RP:
If P is one of our P1;P2;:::Pm then
IRP = RPb JRP IRP:
12
To complete the proof, we will show that I = J. Since J I, we can consider
I?1J R. Now, note that
RP(I?1J) = (RPI?1)(RPJ) = (RPI?1)(RPI) = RP(I?1I) = RP
for any maximal ideal P. Assume I?1J 6= R. I?1J is an ideal and so it is contained in
some maximal ideal P. But then
RP = RP(I?1J) RPP ? RP;
while RPP 6= RP since P is a proper ideal, a contradiction. Thus, I?1J = R and flnally
I = J.
So, we can follow the logical chain laid out above and summarize the ideas of this
section in the following theorem that follows the form of the other results in this paper. As
we have seen, these equivalent conditions are met precisely when R is a Dedekind domain.
Theorem 4.5. Every torsionless R-module is projective (hence a direct sum of rank-1
projective ideals) if and only if every ideal of R is one-and-a-half generated.
13
Chapter 5
Noetherian Domains
In this section, we will work within the class of Noetherian domains (domains that
satisfy the Ascending Chain Condition for ideals).
Proposition 5.1. A domain R is Noetherian if and only if every prime ideal of R is
flnitely generated.
We will rely upon the following standard in commutative ring theory, as well as some
well-known results whose proofs are beyond the scope of this thesis.
Krull Principal Ideal Theorem. Let R be a Noetherian domain. If P is a minimal
prime over some nonzero principal ideal aR, then P is a maximal ideal.
Krull Intersection Theorem. If I is an ideal in a Noetherian domain R, then \nIn = 0.
The proof of the Krull Principal Ideal Theorem. is contained in [3]. We now state a
necessary deflnition.
Deflnition 14. An R-module M is re exive if Hom(Hom(M;R);R) ?= M: A domain R
is re exive if every torsionless R-module is re exive.
The following is the main result of this section.
Theorem 5.2. The following are equivalent for a Noetherian domain R:
(a) Every ideal of R is 2-generated.
(b) Every fractional overring of R is re exive.
(c) Every nonzero ideal of R is projective over its ring of endomorphisms.
14
(d) Every torsionless module is isomorphic to a direct sum of ideals of R.
Note that the equivalence of items (a) and (d) is the primary result of this chapter.
Before we prove this result, we will establish a necessary foundation through a series of
lemmas.
Lemma 5.3. If R is Noetherian and M and A torsionless modules, then for any prime
ideal P of R,
Hom(M;A)P = Hom(MP;AP):
Proof. SinceHom(MP;AP) = f` 2 Hom(QM;QA) j `(MP) APg, thenclearlyHom(M;A)P
Hom(MP;AP):
Let ` 2 Hom(MP;AP): M is torsionless and thus flnitely-generated. Let x1;x1;:::;xn
be a flnite generating set for M; then there exists an r 2 RnP such that r`(xj) 2 A for all
j. But then
` = (1=r)?r` 2 Hom(M;A)P:
Lemma 5.4. A Noetherian domain R is re exive if and only if R has Krull dimension 1
and M?1 can be generated by two elements for each maximal ideal M of R.
We will give a sketch of this proof. For more detail see [2].
Sketch of Proof. ()) If R is re exive, then every ideal of R is re exive. Let P be a
nonzero minimal prime ideal of R. By results in [2], there exist elements a;b 2 R such
that P = fr 2 Rjra 2 Rbg. Let q = a=b and x = q + R. Let M be a maximal ideal of R
containing P. As in the proof of Lemma 5.10 below, Rx \ (M?1=R) 6= 0 and thus there
must be some r 2 R such that rq 2 M?1 n R. Therefore, r is not an element of P, but
15
Mr ? fs 2 R : sa 2 Rbg = P implying that M = P. Thus, every nonzero prime ideal is
maximal and R has Krull dimension 1.
Let M be a maximal ideal of R. First, we claim that M?1=R ?= R=M. This is because
there is a one-to-one correspondence between the set of ideals between M and R and the
set of ideals between M?1 and R. Thus, M?1=R is simple and the result follows. Our next
claim is that this implies that M?1 is generated by two elements. This is clear since if we
take some u 2 M?1nR, then M?1 can be generated by 1 and u.
(() Once it is shown that the R-module Q=R is injective, the proof proceeds as follows.
Let P be a rank 1 prime ideal of R. It can be shown that P must then be re exive and thus
P?1=R 6= 0. It is su?cient to show that P is maximal (this is done by showing that R=P
is a fleld) and thus Q=R is a universal injective R-module, meaning that it is injective and
contains every simple R-module. It can then be shown that this is equivalent to R being
re exive.
The following lemma is a a weaker version of one direction of our main result of this
section. It requires the domain to be local.
Lemma 5.5. If R is a local domain such that every flnitely-generated, torsion-free module
is a direct sum of modules of rank 1, then every flnitely-generated, torsion-free R-module of
rank 1 is 2-generated.
Sketch of Proof. Let I be a nonzero, flnitely-generated R module. It will be su?cient to
show that I is 2-generated. Let a1;a2;:::;an be a minimal generating set for I. Assume that
n is at least 2. Let x = (a1;a2;:::;an) 2'nR = F. Let B be the submodule of F generated
by x. B is torsion-free of rank 1 and F=B is flnitely-generated of rank n?1. Then by the
hypothesis, F=B = C1 'C2 ':::'Cn?1 where each Ci is torsion-free of rank 1.
16
It can be shown that there is a decomposition F = F1 'F2 ':::'Fn?1 such that if
Bi = Fi\B then B = B1'B2':::'Bn?1 and Ci ?= Fi=Bi. Since B is torsion-free of rank
1, it is indecomposable. So, we can assume it is contained in one of the summands of F,
say B F1. From here we can use the fact that the coordinates of x relative to any basis
for F are a generating set for I and the minimality of such a generating set to conclude
that B can not be properly contained in any proper summand of F and thus F = F1. So
n?1 = 1 and n = 2.
Lemma 5.6. If every nonzero ideal of a ring R is contained in only a flnite number of
maximal ideals of R, then any ideal I of R can be generated by by maxf2;kg elements, where
k is the supremum over the maximal ideals M of R of the number of elements necessary to
generate IM as an RM-module.
More thorough treatments of the previous three lemmas appear in [2].
Lemma 5.7. Let I be a flnitely generated ideal of a domain R. Then the following are
equivalent:
(a) I is projective.
(b) I is invertible.
(c) IP is principal for all maximal ideals P of R.
Proof. (a ) b) Assume I is projective. Then for any sequence:
R'R':::'R `! I ! 0;
17
there is f : I ! R'R':::'R that splits `. Deflne ` by
`(r1;r2;:::;rn) = r1a1 +r2a2;:::+rnan;
where fa1;a2;:::;ang is a flnite generating set for I, and let f : I ! R'R':::'R be such
that f(x) = (f1x;f2x;:::;fnx). Then fj 2 I?1 for all j. By deflnition, `f(x) = x. But then
`(f1x;f2x;:::;fnx) = (f1a1x+f2a2x+:::+fnanx) = x(f1a1 +f2a2 +:::+fnan) = x:
This implies that f1a1 +f2a2 +:::+fnan = 1 and since each fj 2 I?1 and each aj 2 I, this
means that f1a1 +f2a2 +:::+fnan 2 I?1I and so 1 2 I?1I. Therefore, I?1I = R and I is
invertible.
(b ) c) Assume II?1 = R. Then IPI?1P = RP for any maximal ideal P of R. Thus
1 2 IPI?1P . So, assume a1b1 + ::: + anbn = 1. Thus, it must be that for some i, aibi = u,a
unit. Therefore aibiu?1 = (u?1bi)ai = 1. Let c 2 IP. c = c(u?1bi)ai = (cu?1bi)a: This
shows that IP = RPa and thus IP is principal.
(c ) b) Assume IP is principal for all maximal P. Furthermore, assume that I is not
invertible. Then II?1 6= R. II?1 is an ideal of R so it must be contained in some maximal
P. Then IP(I?1)P PP. It follows from Lemma 5.3 that (IP)?1 = (I?1)P where (IP)?1
is the inverse of IP over RP. So, IP is not invertible over RP. But this is a contradiction
since over the local ring RP, principal and invertible are equivalent. So, it must be that I
is invertible over R.
(b ) a) Assume II?1 = R. So, there must be some a1b1 + a2b2 + ::: + anbn 2
II?1 that is equal to 1. Thus I can be generated by fa1;a2;:::;ang since for any c 2 I,
c = ((cb1)a1 + ::: + (cbn)an). Since this set generates I, we only need to consider a map
18
` : R ' R ' ::: ' R ! I such that `(r1;r2;:::;rn) = r1a1 + r2a2 + ::: + r3a3. Then if we
consider f : I ! R'R':::'R such that f(x) = (b1x;b2x;:::;bnx), then
`f(x) = `(b1x;b2x;:::;bnx) = x(b1a1 +b2a2 +:::+bnan) = x:
Thus, I is projective and the proof is complete.
Lemma 5.8. If R is Noetherian and satisfles (a), (b), (c), or (d), then R has Krull dimen-
sion 1.
Proof. (a) Assume every ideal of R is 2-generated. Without loss of generality, we can assume
that R is a local domain with maximal ideal M. We will then assume that this maximal
ideal has rank greater than 1.
Let x and y be the elements that generate M. Consider R=Rx, which we will call
?R. Note that this is a local ring with maximal ideal ?M = M=Mx. Note that this ideal is
generated by y + Rx and is thus principal. By the Krull Intersection Theorem, we have
that Tn Mn = 0 and so Tn ?Mn = 0 since ?Mn = (Mn + Rx)=Rx. We claim that either ?R
is an integral domain or ?Mn = 0 for some n.
Assume ?Mn 6= 0 for all n. Let r;s 2 R and assume (r + Rx)(s + Rx) = rs + Rx = 0
but r and s are not elements of Rx. Since Tn ?Mn = 0, there are m and n such that
r + Rx 2 ?Mn n ?Mn+1 and s + Rx 2 ?Mm n ?Mm+1. Now, ?Mn = fayn + Rx : a 2 Rg and
?Mm = fbym + Rx : b 2 Rg. So, we can write r + Rx = ayn + Rx where a 2 R nM and
s+Rx = byn +Rx where b 2 RnM. Then (r +Rx)(s+Rx) = abyn+m +Rx. Since ab is
not in M, it is a unit and there exists c 2 RnM such that cab = 1R. So,
c(r +Rx)(s+Rx) = yn+m +Rx = 0:
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Therefore, yn+m 2 Rx and thus
?Mn+m = fdyn+m +Rx : d 2 Rg = 0;
a contradiction. Thus, if ?Mn 6= 0 for all n, then ?R is an integral domain.
Assume ?Mn = 0 for some positive n. But then M Rx and by the Krull Principal
Ideal Theorem, the rank of M is no greater than 1. So, ?R is an integral domain. Thus Rx
is a prime ideal of R since if (r + Rx)(s + Rx) = 0 then either (r + Rx) or (s + Rx) must
be zero and thus either r or s is in Rx.
Since the rank of M is greater than one, M is not principal and thus x and y are
distinct elements and they are linearly independent modulo M2. But, M2 can be generated
by two elements and it must be that those two elements are among x2, y2, and xy. So,
one of these must be generated by the other two. If x2 is generated by y2 and xy, then
M2 ? Rx and by the Krull Principal Ideal Theorem, the rank of M is at most 1. Similarly,
y2 cannot be generated by the other two. So, it must be that xy = ax2+by2 2 Rx for some
a;b 2 M. Now, by2 2 Rx and Rx is a prime ideal. Thus, either b 2 Rx or y2 2 Rx. But Ry
is not contained in Rx and so y2 must not be in Rx. So, b 2 Rx. Let b = cx with c 2 R.
Then xy = ax2 +cxy2 and y = ax+cy2. But this means y 2 M2, a contradiction. So, the
rank of M must be 1 and thus R has Krull dimension 1.
(b) By Lemma 5.4, R has Krull dimension 1.
(c) Let P be a nonzero prime ideal of R. By Lemma 5.3, given an ideal I of R,
End(I)P = End(IP) and so (c) holds for the domain RP. It is enough to show that no
prime ideal properly contains another prime ideal and therefore, without loss of generality,
we may assume that R is local with maximal ideal P. Note that P is invertible over its ring
S of endomorphisms. Let P0 be a maximal ideal of S. Then P0 contains P and there is
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no other prime ideal P00 containing P because S is integral over R. Thus, P0 is a minimal
prime of S containing the principal ideal P and so by the Krull Principal Ideal Theorem,
P0 must also be a minimal prime. But S is integral over R, and so there can be no proper
prime ideals of R contained in P since if there were there would have to be a proper prime
ideal of S contained in P0. Thus, R has Krull dimension 1.
(d) As in part (a), assume that R is local with maximal ideal M. Then by Lemma 5.5,
R can be generated by two elements. We can then retrace our steps from (a).
Lemma 5.9. If R is Noetherian of Krull dimension 1 and M is a re exive R-module, then
for I = traceR(M) = Hom(M;R)M, one has
(i) I?1 is a ring, and
(ii) M is an I?1-module.
Proof. Let M be any module and let a 2 I?1 = Hom(I;R). If f : M ! R, then f(M) I
and so af(M) aI R implying that af is again in Hom(M;R). When M = I?1,
aI?1 I?1 and so I?1I?1 I?1. Furthermore, I R implies R I?1. Hence I?1 is a ring
and Hom(M;R) is an I?1-module. This latter fact implies that M ?= Hom(Hom(M;R);R)
is also an I?1-module.
Lemma 5.10. If R is a re exive, local domain with maximal ideal P, then every R-
submodule, L, of Q containing R properly contains P?1.
Proof. Let x 2 LnR. Let L0 = Rx + R. Consider L?10 , an ideal of R. If L?10 = R, then
L0 = (L?10 )?1 = R, a contradiction since x 2 L n R. Thus, L?10 P. This implies that
P?1 (L?10 )?1 = L0 L.
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Lemma 5.11. If M is a torsionless module of rank ? 2 over a Noetherian domain R of
Krull dimension 1 such that traceR(M) = R, then M ?= R'M0 for some module M0.
The proof of the above lemma appears in [5]. We can now begin the proof of the main
result of this section.
Proof. Note that by Lemma 5.8, under any of the conditions, R is Noetherian of Krull
dimension 1 so we will assume this as a matter of course.
(a) ) (b) If S is a fractional overring of R, and J is an ideal of S, then rJ R for
some 0 6= r 2 R, and so rJ = Ra+Rb for some a;b 2 rJ. Then rJ = Sa+Sb from which
it follows that J = S(a=r) + S(b=r) is 2-generated over S. Therefore, it su?ces to show
that a domain whose ideals are 2-generated is re exive, and it is enough to argue that R is
re exive. But, by Lemma 5.4, R being re exive is equivalent to R having Krull dimension
1 and M?1 being 2-generated. We know that R has Krull dimension 1 and also that M?1
is isomorphic to an ideal of R and is thus 2-generated. Therefore, R is re exive as desired.
(b) ) (c) An ideal I of R is projective over its endomorphism ring S if and only if
Hom(I;S)I = S, which in turn holds if and only if SP = Hom(I;S)PIP = Hom(IP;SP)IP
for every prime ideal P of R. We know by Lemma 5.3 that the second equality holds.
Therefore, we may assume that R is local with maximal ideal P.
P?1P is an ideal of R, so either P?1P = R or P?1P = P since P is the maximal ideal
of R. If P?1P = R, then P is invertible. Let I P be an ideal of R. It can be seen that
either I is principal or I TPn = 0. This shows that, in this case, R is a PID. By Lemma
5.7, any ideal I is then projective. So, assume that P?1P P. But then,
P?1P?1P P?1P P;
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and thus, P?1P?1 P?1. So P?1 is an overring of R. Moreover, if S is any fractional
overring of R, then S?1 R implies that S?1 P. Thus, R1 = P?1 (S?1)?1 = S since
R is re exive.
Let I be an ideal of R. If I?1I P, then R1I?1I R1P = P and thus R1I?1 I?1.
Consequently, I = Hom(I?1;R) is also an R1-module. Thus, either I is principal (as when
P?1P = R) or I is a module over R1.
We will argue that either R1 is a pid, or, R1 is a re exive local domain whose maximal
ideal P1 properly contains P. Since P has endomorphism ring R1 and R1 is re exive, P is
invertible over R1 by the argument in the last paragraph. Write P = R1a. Recall that
0 ! R=P ! R1=P ! R1=R ! 0
exact, implies that R1=P has length 2 as an R (or R=P) module. If P1;P2 are distinct
maximal ideals of R1, then
R1=P1P2 ?= R1=P1 'R1=P2;
by the Chinese Remainder Theorem. Since P P1P2, we must have P = P1P2. Thus P1
and P2 are invertible since P?11 = P2P?1 and P?12 = P1P?1 Thus, they are also principal,
and R is a pid. Otherwise, R1 is local with maximal ideal P1. If P = P1, then R1 is a pid.
Let S be an overring of R and let I be an ideal with endomorphism ring S. If S = R,
then I is principal by the same argument used earlier in this proof. Otherwise, R1 S by
Lemma 5.10. In this case, either S = R1, or R2 = P?11 S where P1 is the maximal ideal
of R1. In the flrst case, again, I is principal, and in the second case, either S = R2 or S
contains P?12 where P2 is the maximal ideal of R2. We claim that this ascending chain of
23
ideals must terminate at some point and thus leave us with the conclusion that I is principal
and thus invertible over S and flnally projective over S. To see that this chain terminates,
recall that R is Noetherian and note that S is flnitely generated as an R-module. Thus S is
a Noetherian R-module and thus satisfles the ascending chain condition on ideals and thus
our ascending chain must terminate, leaving us with the conclusion that I is projective over
S.
(c) ) (b) This proof, while not excessively long, is beyond the scope of this paper. It
can be found in [6].
(b) ) (d) Let I = traceR(M) and let S be the endomorphism ring of I. Since (b) ) (c),
M is a re exive R-module and so by Lemma 5.9, I?1 is a ring and M is an I?1-module.
But I is also an I?1-module and so I?1I I which implies I?1 S. Clearly S I?1
and so S = I?1. Thus, M is an S-module. Because I is projective over S, it follows that
traceS(M) = S. Hence by Lemma 5.11, M = M0'S, and induction applies to M0 to show
that M is a direct sum of fractional ideals of R.
(d) ) (a) For any maximal ideal M of R, every ideal of the local ring RM can be
generated by two elements by Lemma 5.5. We know that R has Krull dimension 1 and thus
any nonzero prime ideal of R is contained in only one maximal ideal of R and any element
of R is contained in only a flnite number of maximal ideals of R: So, we can apply Lemma
5.6 and conclude that every ideal of R can be generated by two elements.
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Bibliography
[1] Hungerford, Thomas W., Algebra, Springer-Verlag, 1974.
[2] Matlis, Eben, Torsion-Free Modules, University of Chicago Press, 1972.
[3] Matsumura, H., Commutative Ring Theory, Cambridge University Press, 1980.
[4] Ribenboim, Paolo Algebraic Numbers, John Wiley and Sons, Inc, 1972.
[5] Rush,DavidE., Ringswithtwo-generatedideals, J. of Pure and Applied Alg.73, (1991),
257-275.
[6] Sally, J.D., Vasconcelos, W.V., Stable Rings, J. of Pure and Applied Alg. 4 (1974),
319-336.
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