Factorization theory and decompositions of modules · of modules and factorization in integral domains. 2 Rings and modules Throughout this paper Rwill denote a commutative, Noetherian

Factorization theory and decompositions of modules

Nicholas R. Baeth1 and Roger A. Wiegand2

October 4, 2011

Abstract

Let R be a commutative ring with identity. It often happens that M1 ⊕M2 ⊕ · · · ⊕Ms∼= N1 ⊕ N2 ⊕

· · · ⊕ Nt for indecomposable R-modules M1, M2, . . . , Ms and N1, N2, . . . , Nt with s 6= t.This behavior canbe captured by studying the commutative monoid [M ] |M is an indecomposable R-module with operationgiven by [M ] + [N ] = [M ⊕ N ]. In this mostly self-contained exposition, we introduce the reader to theinterplay between the the study of direct-sum decompositions of modules and the study of factorizations incommutative monoids.

1 Introduction

Our goal is to study direct-sum decompositions of finitely generated modules over local rings. Unlike the caseof vector spaces (modules over fields), a module can decompose as a direct sum of indecomposable modules inseveral different ways. For example, it is possible to have pairwise non-isomorphic indecomposable modules A,Band C such that A⊕B ∼= C ⊕C ⊕C, so even the number of indecomposable summands need not be invariant.On the other hand, there are a few rules. For instance, one cannot have A⊕C ∼= B⊕C, or A⊕A ∼= B⊕B⊕B.Information of this sort is encoded in the monoid of isomorphism classes. We will focus mainly on rings ofdimension one, and in that context we will determine exactly which semigroups can be realized as semigroupsof isomorphism classes. A recurring theme in the paper will be the analogy between direct-sum decompositionsof modules and factorization in integral domains.

2 Rings and modules

Throughout this paper R will denote a commutative, Noetherian ring with identity. “Noetherian” means thatthere is no infinite strictly ascending chain I1 ⊂ I2 ⊂ I3 ⊂ . . . of ideals in R. More generally, an R-module isNoetherian provided it has no infinite strictly ascending chain of submodules. Equivalently, every non-empty setof submodules has a maximal element. An R-module M is finitely generated if it has a finite subset m1, . . . ,mtsuch that M = Rm1 + · · ·+ Rmt, that is, every element m ∈ M can be expressed as an R-linear combinationof the mi.

Just as for vector spaces, one can form the direct sum M1 ⊕ · · · ⊕ Mt of a family of modules. This isjust the set of t-tuples (x1, . . . , xt) with xi ∈ Mi, and with the obvious coordinate-wise operations. (We will,however, usually write elements of the direct sum as column vectors, in order to allow matrices to operate onthe left.) Alternatively, starting with a module M , we can seek an “internal decomposition” of M . We writeM = M1 ⊕ · · · ⊕Mt provided

(i) each Mi is a submodule of M ,

(ii) M1 + · · ·+Mt = M , and

(iii) Mi ∩ (M1 + · · ·+Mi−1 +Mi+1 + · · ·+Mt) = 0 for each i.

1Department of Mathematics and Computer Science, University of Central Missouri, Warrensburg, MO 64093, [email protected] of Mathematics, University of Nebraska–Lincoln, Lincoln, NE 68588-0130, [email protected]

1

N. Baeth and R. Wiegand Factorization theory and decompositions of modules

A non-zero R-module M is decomposable provided there exist non-zero R-modules M1,M2 such that M ∼=M1 ⊕M2; otherwise M is indecomposable. (The symbol “∼=” denotes isomorphism. Note that the 0 moduleis not considered to be indecomposable, for the same reason that 1 is not regarded as a prime: it’s uselessas a building block!) Sometimes, e.g., in the proof of Proposition 2.1 below, it’s better to work with internaldecompositions: a module M is decomposable if and only if there exist non-zero submodules M1 and M2 of Msuch that M = M1 ⊕M2.

Let C be a class of finitely generated R-modules. We’ll say that C is closed under direct sums, directsummands, and isomorphism, provided the following holds: Whenever M , M1, and M2 are finitely generatedR-modules with M ∼= M1 ⊕M2, we have M ∈ C ⇐⇒ M1,M2 ∈ C. Given such a class C, we will say that Csatisfies the Krull-Remak-Schmidt-Azumaya theorem (KRSA for short), provided the following holds:

(KRSA) If M1 ⊕ · · · ⊕Mt∼= N1 ⊕ · · · ⊕Nu, where the Mi and Nj are indecomposable modules in C, then

t = u and, after renumbering, Mi∼= Ni for each i.

For example, if K is a field, the class C of finite-dimensional vector spaces clearly satisfies KRSA: up toisomorphism, the one-dimensional vector space K is the only indecomposable vector space, and a vector spacehas dimenson n if and only if it is isomorphic to a direct sum of n copies of K. For a more interesting example,let R be any commutative principal ideal domain. Then the class C of finitely generated R-modules satisfiesKRSA, by the elementary divisor theorem. The indecomposable modules in C are the cyclic modules R andR/(pn), where p is an irreducible element and n ≥ 1.

One usually includes in KRSA the condition that every module in C is a direct sum of indecomposablemodules. The next two propositions guarantee that in our context (finitely generated modules over Noetherianrings) this condition is automatically satisfied.

Proposition 2.1. Let M be a non-zero Noetherian module. Then M is a direct sum of finitely many indecom-posable modules.

Proof. We show first that M has an indecomposable direct summand. Suppose not. Then M is decomposable,say, M = X1⊕ Y1, with both summands non-zero. Now write X1 = X2⊕ Y2, X2 = X3⊕ Y3, X3 = X4⊕ Y4, . . . ,with all of the Xi and Yi non-zero. For each n we have Xn ⊕ Yn ⊕ Yn−1 ⊕ · · · ⊕ Y2 ⊕ Y1 = M . We get a strictlyascending chain Y1 ⊂ Y1 ⊕ Y2 ⊂ Y1 ⊕ Y2 ⊕ Y3 ⊂ . . . of submodules of M , contradicting the assumption that Mis Noetherian.

To complete the proof, choose any indecomposable direct summand Z1 of M , and write M = Z1 ⊕W1. IfW1 = 0, we’re done; otherwise (by the first paragraph applied to W1), W1 has an indecomposable direct sum-mand, say, W1 = Z2⊕W2, with Z2 indecomposable. If W2 6= 0, write W2 = Z3⊕W3, with Z3 indecomposable.The chain Z1 ⊂ Z1 ⊕Z2 ⊂ Z1 ⊕Z2 ⊕Z3 ⊂ . . . has to terminate, and eventually we get M = Z1 ⊕ · · · ⊕Zt.

In this paper all of our modules will be finitely generated and therefore, by the next propsition, Noetherian.

Proposition 2.2. Every finitely generated module over a Noetherian ring is a Noetherian module. Everysubmodule of a Noetherian module is finitely generated.

Proof. Given an R-module M and a submodule N , one checks easily that M is Noetherian if and only ifboth N and M/N are Noetherian. Since (M1 ⊕M2)/(M1 ⊕ 0) ∼= M2, it follows that the direct sum of twoNoetherian modules is again Noetherian, and hence that a free module R(t) is Noetherian if R is a Noetherianring. Since a module generated by t elements is a homomorphic image of R(t), the result follows. For thelast statement, observe that if N were a non-finitely generated submodule of a module M , a doomed butpersistent attempt to find a finite generating set xi for N would yield an infinite strictly ascending chainRx1 ⊂ Rx1 +Rx2 ⊂ Rx1 +Rx2 +Rx3 ⊂ . . . of submodules of M .

A recurring theme in this paper is the analogy between decompositions of modules and factorization inintegral domains. In a Noetherian domain the factorization process for a given element has to terminate.Analogously, the decomposition process of a Noetherian module has to stop in a finite number of steps. Onecannot keep splitting off direct summands ad infinitum. In the ring of integers, for example, every integer n ≥ 2is a finite product of prime numbers. A typical approach to proving this is to show first that n is divisible bysome prime p and then to argue, by induction, that n

p is a product of primes. The proof of Proposition 2.1 isin the same spirit.

Here is a family of examples where KRSA fails.

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Example 2.3. Let R be a commutative integral domain with two non-principal ideals I and J satisfyingI + J = R. (For a concrete example, take R = C[x, y], the polynomial ring in two variables, and let I and J bethe maximal ideals Rx+Ry and Rx+R(y − 1).) Then

I ⊕ J ∼= R⊕ (I ∩ J).

To see this, choose a ∈ I and b ∈ J such that a+ b = 1. Define R-homomorphisms ϕ : I ⊕ J → R⊕ (I ∩ J) andψ : R⊕ (I ∩ J)→ I ⊕ J as follows:

ϕ :[xy

]7→[x+ ybx− ay

]ψ :[rs

]7→[ar + sbr − s

].

One checks that ϕ and ψ are inverses of each other. Since R is a domain, every non-zero ideal is indecomposableas an R-module. Moreover, neither I nor J is isomorphic to R (since, by assumption, neither is a principalideal).

This example indicates that the existence of comaximal proper ideals is likely to lead to failure of KRSA.For this reason we will usually restrict our attention to modules over commutative local rings — commutativerings with just one maximal ideal. Notice that a commutative ring is local if and only if the sum of any twonon-units is again a non-unit. This observation motivates the following definition:

Definition 2.4. A ring Λ (not necessarily commutative) is said to be local provided Λ 6= 0, and the sum of anytwo non-units of Λ is again a non-unit.

(The trivial ring in which 1 = 0 is disqualified, since we want the set of non-units to include 0.) By the way,a unit of Λ is an element that has a two-sided inverse. An element can have a left inverse without being a unit.For example, in the endomorphism ring Λ of an infinite-dimensional vector space V with basis v1, v2, v3, . . . ,consider the unilateral shifts: ϕ : vi 7→ vi+1 and ψ : vi+1 7→ vi (with ψ(x1) = 0). Then ψϕ = 1V , but neither ϕnor ψ is an automorphism of V .

Proposition 2.5. Let Λ be a local ring. Then the set J of non-units of Λ is a two-sided ideal of Λ.

Proof. We begin with three observations. First, Λ\J is closed under multiplication, since the product of twounits is a unit. Second, if an element x has a left inverse and a right inverse, say, yx = 1 and xz = 1, theny = y(xz) = (yx)z = z, so x is a unit. Third, if x ∈ J then x2 ∈ J , for if x2 were a unit then x would have botha left and a right inverse. Now let x ∈ J and λ ∈ Λ. We will show that λx ∈ J ; an appeal to symmetry willthen imply that xλ ∈ J . Suppose λx /∈ J . Then λ ∈ J , else x = λ−1(λx) /∈ J by our first observation. Also,xλ ∈ J , else x would have both a left and a right inverse. In the equation

λx = (x+ λ)2 − x2 − xλ− λ2,

all terms on the right-hand side are in J . Since J is closed under addition, this forces λx ∈ J , contradiction.

The reason we’re discussing non-commutative rings is that we will need to study endomorphism rings ofmodules. Given an R-module M , we denote by EndR(M) the ring of endomorphisms of M , that is, R-homomorphisms from M to M . The addition is pointwise, and the product of f and g is the compositionf g. Endomorphism rings are rarely commutative. For example, for a vector space V over a field F , EndF (V )commutes if and only if dim(V ) ≤ 1. We now observe that the structure of the endomorphism ring of a moduledetermines whether or not the module is indecomposable. We write 1M (or 1 if there is no ambiguity) to denotethe identity map 1M (m) = m for all m ∈M . Similarly, 0M (or 0) denotes the map 0M (m) = 0 for all m ∈M .

Proposition 2.6. Let M be a non-zero R-module. Then M is indecomposable if and only if 0 and 1 are theonly idempotents of EndR(M).

Proof. Suppose e is an idempotent of EndR(M) and e 6= 0, 1. If x ∈ e(M) ∩ Ker(e), write x = e(y). Thenx = e2(y) = e(x) = 0; thus e(M) ∩ Ker(e) = (0). Also, given any z ∈ M , we have z = e(z) + (z − e(z)) ∈e(M) + Ker(e). We have shown that M = e(M) ⊕ Ker(e). Moreover, e(M) 6= (0) since e 6= 0. Choosing anyz ∈M with e(z) 6= z, we have 0 6= z − e(z) ∈ Ker(e), and M is decomposable.

Conversely, if M = M1⊕M2, with both summands non-zero, the projection map x1 +x2 7→ x1 (for xi ∈Mi)is a non-trivial idempotent in EndR(M).

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Indecomposable modules behave, with respect to direct-sum decompositions, like irreducible elements in anintegral domain: They cannot be broken down further. The key to proving unique factorization in a principalideal domain is to show that irreducible elements enjoy a stronger property. Recall that a non-zero non-unit p inan integral domain D is prime provided p | ab =⇒ p | a or p | b. Borrowing notation from factorization theory,we write X | Y , for R-modules X and Y to indicate that there is an R-module Z such that X ⊕ Z ∼= Y . Soonwe will verify KRSA in situations where the indecomposable direct summands all have local endomorphismrings, modeling the proof after unique factorization in Z. The following lemma says that modules with localendomorphism rings behave like prime elements in a domain and, further, that they can be cancelled fromdirect-sum relations.

Lemma 2.7. Let R be a commutative, Noetherian ring, and let M , X, Y , and Z be R-modules. Assume thatE := EndR(M) is a local ring.

(i) If M | X ⊕ Y , then M | X or M | Y (“primelike”).

(ii) If M ⊕ Z ∼= M ⊕ Y , then Z ∼= Y (“cancellation”).

Proof. We’ll prove (i) and (ii) sort of simultaneously. In (i) we have a module Z such that M ⊕Z ∼= X ⊕ Y . Inthe proof of (ii) we set X = M and again get an isomorphism M ⊕Z ∼= X⊕Y . Notice that showing that M | Xamounts to producing homomorphisms α : M → X and π : X →M such that πα = 1M . (See Lemma 3.3.)

Choose reciprocal isomorphisms ϕ : M ⊕ Z → X ⊕ Y and ψ : X ⊕ Y →M ⊕ Z. Write

ϕ =[α βγ δ

]and ψ =

[µ νσ τ

],

where α : M → X, β : Z → X, γ : M → Y , δ : Z → Y , µ : X → M , ν : Y → M , σ : X → Z and τ : Y → Z.Since ψϕ = 1M⊕Z =

[1M 00 1Z

], we have µα+νγ = 1M . Therefore, as EndA(M) is local, either µα or νγ must be

an automorphism of M . Assuming that µα is an automorphism, we let π = (µα)−1µ : X →M . Then πα = 1M ,and so M | X. Similarly, the assumption that νγ is an isomorphism forces M to be a direct summand of Y .This proves (i).

To prove (ii) we assume that X = M . Suppose first that α is a unit of E. We use α to diagonalize ϕ:[1 0

−γα−1 1

] [α βγ δ

] [1 −α−1β0 1

]=[α 00 −γα−1β + δ

]=: ξ.

Since all the matrices on the left are invertible, so must be ξ, and it follows that −γα−1β + δ : Z → Y is anisomorphism.

Suppose, on the other hand, that α ∈ J := non-units of E. Then νγ /∈ J (as µα + νγ = 1M ), and itfollows that α+ νγ /∈ J . We define a new map

ψ′ :=[1M νσ τ

]: M ⊕ Y →M ⊕ Z,

which we claim is an isomorphism. Assuming the claim, we diagonalize ψ′ as we did before to ϕ, obtaining, inthe lower-right corner, an isomorphism from Y onto Z. To prove the claim, we use the equation ψϕ = 1M⊕Zto get

ψ′ϕ =[α+ νγ β + νγ

0 1Z

].

As α + νγ is an automorphism of M , ψ′ϕ is clearly an automorphism of M ⊕ Z. Therefore ψ′ = (ψ′ϕ)ϕ−1 isan isomorphism.

We can now prove KRSA for the class C, provided the indecomposable modules in C have local endomorphismrings. The reader will notice that the structure of the proof is identical to the proof of unique factorization inthe ring of integers. In Section 2.1 this result will be applied to the situation where R is a complete local ring.

Theorem 2.8. Let R be a commutative Noetherian ring, and let C be a class of finitely generated modules,closed under direct sums, isomorphism, and direct summands. Assume that EndR(M) is a local ring for everyindecomposable module M ∈ C. Then KRSA holds in C.

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Proof. Let t and u be positive integers, and suppose M1 ⊕M2 ⊕ · · · ⊕Mt∼= N1 ⊕ N2 ⊕ · · ·Nu, where the Mi

and Nj are indecomposable modules in C. We will use induction on t to verify the conclusion of KRSA. Thecase t = 1 is clear, since M1 is assumed to be indecomposable. Proceeding inductively, assume that t ≥ 2. NowMt | N1 ⊕ · · · ⊕ Nu. Since EndRMt is local, (i) of Lemma 2.7 implies that Mt | Nj for some j. But Nj isindecomposable, so in fact Mt

∼= Nj . After reordering the Nj , we may assume that j = u. Using (ii) of Lemmai, we have M1 ⊕ · · ·Mt−1

∼= N1 ⊕ · · ·Nu−1. The inductive hypothesis now implies that t− 1 = u− 1 (so t = u),and, after reordering, that Mi

∼= Ni for each i.

2.1 Completions and a Krull-Remak-Schmidt-Azumaya Theorem

In this section we will see that KSRA holds over any complete local ring. Then, in Section 3, we will geta handle on direct-sum decompositions over arbitrary local rings by analyzing what happens to modules onpassage to the completion. For the rest of the paper, we let R be a commutative, Noetherian local ring. ThenR has exactly one maximal ideal, and we often denote the ring by the pair (R,m) to signal the important roleplayed by the maximal ideal m. We can make R into a metric space by declaring two elements x and y tobe close to each other provided their difference is in a high power of m. More generally, let M be a finitelygenerated R-module. The Krull Intersection Theorem [Mat86, Theorem 8.10], which we shall not prove here,states that

⋂∞n=1 mnM = 0. Thus, if x 6= y there is an integer n such that x− y ∈ mnM\mn+1M . In this case

we let d(x, y) = 2−n, and we let d(x, x) = 0. Note that R is a metric space with metric d. In fact, the triangleinequality holds, in the stronger form d(x, z) ≤ maxd(x, y), d(y, z) for x, y, z ∈ M . The ring R, respectively,the R-module M , is said to be complete provided every Cauchy sequence in R, respectively M , converges.Whenever we use the terminology “complete local ring”, it is tacitly assumed that the ring is commutative andNoetherian.

Proposition 2.9. If (R,m) is a complete local ring, then every finitely generated R-module is complete.

Proof. Given a generating m1, . . . ,mt for M , we get a surjective R-homomorphism from the free R-moduleR(t) onto M by sending the standard basis elements ei to mi. Now a direct sum of two complete modules iseasily seen to be complete, and it follows by induction that R(r) is complete. Since completeness carries overto homomorphic images, we see that M is complete.

The completion R, respectively M , of (R,m), respectively M , with respect to this metric is called the m-adiccompletion of R (respectively M). The completion M can be built as the collection of equivalence classes ofCauchy sequences in M , and alternatively as the inverse limit of the modules M/mnM , with respect to thenatural surjections πn : M/mn+1M R/mnM . (The inverse limit is the submodule of the direct product∏∞n=1(M/mnM) consisting of sequences (xn) satisfying πn(xn+1) = xn for each n.) Either viewpoint makes it

clear that R is a commutative ring and that M is a finitely generated R-module. It is not hard to show thatR is local with maximal ideal m = mR. In fact, (R, m) is Noetherian, though the proof of this fact is far fromtrivial. We refer the reader to [AM69, Chapter 10] for the basic theory of completions. For those who liketensor products, we mention that M can be naturally identified with R⊗RM .

For a familiar example of completion, consider the polynomial ring F [x] in one variable with coefficientsin a field F . The ring R of rational functions defined at 0 is then a local ring with maximal ideal m := f ∈R | f(0) = 0. The m-adic completion R of R is then the ring F [[x]] of formal power series. Similarly, thecompletion of the ring of rational numbers with denominators prime to p is the ring of p-adic integers.

Our goal in this section is to prove KRSA for finitely generated modules over a complete local ring. Inview of Theorem 2.8 it will be enough to show that indecomposable finitely generated modules have localendomorphism rings. The proof requires a few preliminary results on the Jacobson radical J (Λ) of a ring Λ.This is the intersection of the maximal left ideals of Λ. (One can check that if Λ is a local ring then J (Λ)is exactly the set of non-units of Λ.) We refer the reader to [Lam01, Chapter 2] for a nice treatment of thebasics, which we summarize in the next proposition. Recall that an R-algebra is ring Λ together with a ringhomomorphism ϕ : R → Λ that carries R into the center of Λ (that is, (ϕ(r))λ = λ(ϕ(r)) for each r ∈ Rand each λ ∈ Λ). If Λ is an R-algebra and is finitely generated as an R-module (via the structure given byrλ := ϕ(r)λ), then we call Λ a module-finite R-algebra.

Proposition 2.10. Let Λ be a ring, and let J = J (Λ).

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(i) J is the intersection of the maximal right ideals of Λ and therefore is a two-sided ideal of Λ.

(ii) J (Λ/J) = 0.

(iii) If (R,m) is a commutative local ring and Λ is a module-finite R-algebra, then mΛ ⊆ J .

(iv) If Λ has the descending chain condition on left ideals (in particular, if Λ is a finite-dimensional algebraover a field), then J is nilpotent.

Given an ideal I of a ring Λ, we say that idempotents lift modulo I provided for every idempotent e of Λ/Ithere is an idempotent e of Λ such that e + I = e. Similarly, given a surjective ring homomorphism Λ Γ,we say that idempotents lift from Γ to Λ provided every idempotent of Γ comes from an idempotent of Λ.(Notice, for example, that the idempotent 3 in Z/(6) does not lift to an idempotent of Z.) It is well known thatidempotents lift modulo any nil ideal (an ideal in which every element is idempotent). A typical proof of thisfact actually yields the following more general result:

Proposition 2.11. Let I be a two-sided ideal of a (possibly non-commutative) ring Λ, and let e be an idempotentof Λ/I. Given any positive integer n, there is an element x ∈ Λ such that x+ I = e and x ≡ x2 (mod In).

Proof. Start with an arbitrary element u ∈ Λ such that u+ I = e, and let v = 1−u. In the binomial expansionof (u + v)2n−1, let x be the sum of the first n terms: x = u2n−1 + · · · +

(2n−1n−1

)unvn−1. Putting y = 1 − x

(the other half of the expansion), we see that x − x2 = xy ∈ Λ(uv)nΛ. Since uv = u(1 − u) ∈ I, we havex− x2 ∈ In.

Proposition 2.12. Let (R,m) be a complete local ring, and let Λ be a module-finite R-algebra. Then idempotentslift modulo J (Λ).

Proof. Let F = R/m, the residue field of R, and put Λ = Λ/mΛ. By Proposition 2.10, mΛ ⊆ J := J (Λ),and we see that J (Λ) = J := J/mΛ. Now Λ is a finite-dimensional F -algebra, and thus J is nilpotent, byProposition 2.10. Now Λ/J ∼= (Λ/mΛ)/(J/mΛ) = Λ/J (by the “Third Isomorphism Theorem” [DF04, Sec.10.2]). We have factored our homomorphism Λ Λ/J as the composite Λ Λ Λ/J , thereby dividing theheavy lifting into two stages. Proposition 2.11 takes care of the first stage, from Λ/J to Λ. Therefore it willsuffice to show that every idempotent e of Λ = Λ/mΛ lifts to an idempotent of Λ.

Using Proposition 2.11, we can choose, for each positive integer n, an element xn ∈ Λ such that xn+mΛ = eand xn ≡ x2

n (mod mnΛ). (Of course mnΛ = (mΛ)n.) We claim that (xn) is a Cauchy sequence for the m-adictopology on Λ. To see this, let n be an arbitrary positive integer. Given any m ≥ n, put z = xm +xn− 2xmxn.Then z ≡ z2 (mod mnΛ). Also, since xm ≡ xn (mod mΛ), we see that z ≡ 0 (mod mΛ), so 1 − z is a unit ofΛ. Since z(1− z) ∈ mnΛ, it follows that z ∈ mnΛ. Thus we have

xm + xn ≡ 2xmxn, xm ≡ x2m, xn ≡ x2

n (mod mnΛ) .

Multiplying the first congruence, in turn, by xm and by xn, we learn that xm ≡ xmxn ≡ xn (mod mnΛ). If,now, ` ≥ n and m ≥ n, we see that x` ≡ xm (mod mnΛ). This verifies the claim. Since, by Proposition 2.9, Λis m-adically complete, we let x be the limit of the sequence (xn).

Let’s check that x is an idempotent lifting e. Given any n ≥ 1, choose m ≥ n such that x ≡ xm (mod mnΛ).Then x2 ≡ x2

m (mod mnΛ). By construction, xm − x2m ∈ mmΛ, and since mmΛ ⊆ mnΛ, we see that x ≡ x2

(mod mnΛ). Since n was arbitrary, the distance between x and x2 is 0, that is, x = x2. Taking n = 1 andchoosing m as above, we have x ≡ xm (mod mΛ); thus x+ mΛ = xm + mΛ = e.

Theorem 2.13. Let (R,m) be a complete local ring. Then KSRA holds for the class of finitely generatedR-modules.

Proof. By Theorem 2.8 it will suffice to show that Λ := EndR(M) is local whenever M is an indecompos-able finitely generated R-module. Note that EndR(M) is an R-algebra via the ring homomorphism r 7→(multiplication by r. Moreover, one can show, since R is Noetherian and M is finitely generated, that EndR(M)is module-finite over R. (See Lemma 3.2 for a more general result.)

Since M is indecomposable, Λ has no idempotents other than 0 and 1. Now Proposition 2.12 implies thatΛ/J (Λ) has no interesting idempotents either. By Proposition 2.10, mΛ ⊆ J := J (Λ), so Λ/J is a homomorphic

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image of the finite-dimensional algebra Λ/mΛ over the field F := R/m. Therefore Λ/J has the descending chaincondition on left ideals; and J (Λ/J) = 0 by Proposition 2.10. By the Wedderburn-Artin Theorem [Lam01,Theorem 3.5], Λ/J is isomorphic to a direct product Mn1(D1)×· · ·×Mnr

(Dr) of full matrix rings over divisionrings D1, . . . , Dr. But since 0 and 1 are the only idempotents of Λ/J , we see that r = 1 and n1 = 1, that is,Λ/J is a division ring. It follows that J is exactly the set of non-units of Λ, and hence that Λ is a local ring.

3 Monoids of Modules

For our purposes, a monoid is a commutative, additive semigroup with identity element. In keeping with theadditive notation, we denote the identity element by “0”. We will consider two additional conditions one mightimpose on a monoid:

i) A monoid Λ is cancellative provided a+ c = b+ c =⇒ a = b for all a, b, c ∈ Λ.

ii) A monoid Λ is reduced provided a+ b = 0 =⇒ a = b = 0 for all a, b, c ∈ Λ.

For a Noetherian local ring (R,m), we define M(R) to be the set of isomorphism classes [M ] of finitelygenerated R-modules M , endowed with the monoid structure given by the direct sum: [M ] + [N ] = [M ⊕N ].ObviouslyM(R) is commutative and reduced, and we’ll see below, in Corollary 3.7, thatM(R) is cancellative.In the next definition, we formulate several concepts we will need in the language of monoids.

Definition 3.1. An element x of a monoid is an atom provided (i) x 6= 0, and (ii) x = y + z =⇒ y = 0 orz = 0. A monoid Λ is atomic provided it is reduced and cancellative, and every non-zero element of Λ is a finitesum of atoms. A factorial monoid is an atomic monoid in which the representation as a sum of atoms is uniqueup to order of the summands. In detail: If xi and yj are atoms of Λ and x1 + · · · + xm = y1 + · · · + yn, thenm = n, and, after a permutation of 1, . . . ,m, xi = yi for i = 1, . . . ,m. A monoid homomorphism ϕ : Λ → Γis a divisor homomorphism provided x | y ⇐⇒ ϕ(x) | ϕ(y) for all x, y ∈ Λ. Here, x | y means there is z ∈ Λwith x+ z = y.

In an atomic monoid Λ, let H be the set of atoms. Since every element of Λ is uniquely an N0-linearcombination of elements of H, we see that Λ is a free monoid with basis H. In particular, Λ ∼= N(Ω)

0 , where Ωis an index set of cardnality |H|. In what follows, we will use the terms “factorial monoid” and “free monoid”interchangeably.

For reduced, cancellative monoids, divisor homomorphisms are injective: If ϕ(x) = ϕ(y), then x | y andy | x, say, x+ a = y and y + b = x. Then x+ a+ b = x+ 0, whence a+ b = 0. Therefore a = b = 0, so x = y.

A finitely generated R-module M is indecomposable if and only if [M ] is an atom of M(R). Expressing afinitely generated R-module M as a direct sum of indecomposable modules amounts to writing [M ] as a sumof atoms of M(R). Given a submonoid Λ of M(R) closed under direct summands, finite direct sums, andisomorphism, we see that Λ is factorial if and only if Λ satisfies KRSA. (Remember we are tacitly assuming thatR is Noetherian, so Proposition 2.2 and Proposition 2.1 ensure that M(R) and Λ are atomic.) In particular,M(R) is factorial if (R,m) is a complete local ring. This suggests that we should try to understand the monoidhomomorphism Φ : M(R) → M(R) taking [M ] to [M ]. This approach is used in virtually every area ofmathematics: To understand obstreperous behavior, pass to a structure where the behavior is well understood,and then see how much information is lost in the passage. As it turns out, the homomorphism Φ is a divisorhomomorphism (Theorem 3.6) and, in particular, is injective. What is sometimes lost is indecomposability: itcan happen that M is indecomposable but M is not. This phenomenon is the source of the fun we’ll have whenstudying the monoids M(R).

The proof that Φ is a divisor homomorphism involves several little tricks, the first of which is to describe therelation M | N in terms of homomorphisms. We let HomR(M,N) denote the set of R-homomorphisms from Mto N . This is an abelian group with pointwise operations. In fact, it’s also an R-module: If f ∈ HomR(M,N)and r ∈ R, the product rf is defined by (rf)(x) = rf(x). (Notice that the fact that rf is an R-homomorphismdepends on the fact that R is commutative!)

Lemma 3.2. Let M and N be finitely generated modules over a commutative Noetherian ring A. ThenHomA(M,N) is finitely generated as an A-module.

7


Proof. Let x1, . . . , xt be a set of generators for N , and define an R-homomorphism ϕ : HomA(M,N)→ N (t)

by

ϕ(f) =

f(x1)...

f(xt)

.Then ϕ is injective. Since, by Proposition 2.2, N (t) is Noetherian, HomA(M,N) is finitely generated.

Lemma 3.3. Let M and N be finitely generated R-modules. Then M | N if and only if there are homomor-phisms α ∈ HomR(M,N) and β ∈ HomR(N,M) such that βα = 1M .

Proof. If βα = 1M , then α is injective, so α(M) ∼= M . One checks easily that N = α(M) ⊕ Ker(β). For theconverse, suppose M ⊕X ∼= N . Choose reciprocal isomorphisms

[α σ

]: M ⊕X → N and

[βτ

]: N →M ⊕X.

Then [1M 00 1X

]=[βτ

] [α σ

]=[βα βστα τσ

],

and the top left corner yields the equation we want.

Here is one of the most useful results in all of commutative algebra:

Lemma 3.4 (Nakayama’s Lemma). Let (A,m) be a commutative local ring, M a finitely generated A-module,and N a submodule of M . If N + mM = M , then N = M .

Proof. By passing to M/N , we may assume that N = 0. Suppose, by way of contradiction, that M 6= 0.Let x1, . . . , xt be a minimal generating set; then t ≥ 1. Since M = mM , we can write xt = a1x1 + · · · +at−1xt−1 + atxt. Let u = 1− at. Since u /∈ m, u is a unit of A, and we have xt = u−1(a1x1 + · · ·+ at−1xt−1).(By convention, the right-hand side is 0 if t = 1). It follows that M = Ax1 + · · ·+Axt−1, and this contradictsminimality of the original generating set.

Lemma 3.5. Let M be a Noetherian R-module. Every surjective R-endomorphism of M is an automorphism.

Proof. Let f : M M , and put Ln = Ker(fn) for n ≥ 1. The sequence L1 ⊆ L2 ⊆ L3 ⊆ . . . has to stabilize.Choose any n ≥ 1 such that Ln = L2n, and let x be an arbitrary element of Ln. Since fn is surjective, there isan element y ∈ M with fn(y) = x. Then f2n(y) = fn(x) = 0, so y ∈ L2n = Ln, and hence x = 0. This showsthat Ln = 0, whence L1 = 0.

Theorem 3.6. Let (R,m) be a commutative Noetherian local ring, and let M and N be finitely generated R-modules. Then M | N ⇐⇒ M | N . In other words, the homomorphism Φ : M(R) → M(R), taking [M ] to[M ], is a divisor homomorphism.

Proof. If M | N , say, M ⊕X ∼= N , then M ⊕ X ∼= N , so M | N . For the converse, we choose, using Lemma 3.3,homomorphisms α ∈ Hom

R(M, N) and β ∈ Hom

R(N , M) such that βα = 1

M. Put H = HomR(M,N), a

finitely generated R-module by Lemma 3.2. Every h ∈ H induces an element h ∈ HomR

(M, N). Moreover,

HomR

(M, N) is naturally isomorphic to the completion H [Mat86, Theorems 7.11 and 8.14]. This meansthat α can be approximated to any order by elements of H. Taking a first-order approximation, we obtainf ∈ H such that (f − α)(M) ⊆ mN . Similarly, we find g ∈ HomR(N,M) such that (g − β)(N) ⊆ mM . Nowgf−1

M= gf−βα = g(f−α)+(g−β)α, and it follows that (gf−1

M)(M) ⊆ mM . Therefore M = gf(M)+mM ,

and now Nakayama’s Lemma implies that gf(M) = M . From the view of M as the inverse limit of the modulesM/mt, we see that the endomorphisms of M/mtM induced by the composition gf are surjective for each t.Another application of Nakayama’s Lemma shows that gf is itself surjective and hence, by Lemma 3.5, anautomorphism of M . Letting h = (gf)−1g : N → M , we see that hf = 1M , and now Lemma 3.3 implies thatM | N .

8


The following easy corollary of Theorem 3.6 will be useful as we consider examples in Section 4.2.

Corollary 3.7. Let (R,m) be a commutative Noetherian local ring, and let M and N be finitely generatedR-modules. If M ∼= N , then M ∼= N . The monoid M(R) is reduced and cancellative.

Proof. If M ∼= N , then trivially M | N and N | M . Applying Theorem 3.6, we see M | N and N |M . ThereforeN ∼= M ⊕X and M ∼= N ⊕ Y for some finitely generated R-modules X and Y , and hence N ∼= N ⊕ Y ⊕X.Passing to finite-dimensional vector spaces over the field R/m, we have N/mN ∼= N/mN ⊕ Y/mY ⊕ X/mX,whence Y/mY = X/mX = 0. Now Nakayama’s Lemma implies that X = Y = 0. This shows that M ∼= N .

Obviously M(R) is reduced. To verify that it is cancellative, suppose A, B and C are finitely generatedR-modules with A ⊕ C ∼= B ⊕ C. Passing to the completion, we have A ⊕ C ∼= B ⊕ C. Writing each of theseR-modules as a direct sum of indecomposable modules, and using KRSA (Theorem 2.13), we easily deduce thatA ∼= B, and now the first part of the corollary implies that A ∼= B.

Assumption 3.8. From now on, all of our monoids will be assumed to be reduced and cancellative.

Except in special situations, M(R) is too big and complex to afford a precise description. We will focus onlittle pieces of the monoid, whose descriptions will still give us enough information to determine whether or notKRSA uniqueness holds, or how badly it fails. We fix a finitely generated R-module M and then look at thesmallest submonoid of M(R) that contains [M ] and is closed under direct summands and finite direct sums.

Notation 3.9. For a finitely generated R-module M , add(M) consists of isomorphism classes [N ] of finitelygenerated modules N that are direct summands of direct sums of finitely many copies of M .

Thus [N ] ∈ add(M) if and only if there exist a module V and a positive integer t such that N ⊕ V ∼= M (t).If, for example, we want to understand the direct-sum relations among modules M1, . . . ,Ms, we might takeM = M1 ⊕ · · · ⊕Ms. All of these relations are encoded in the monoid structure of add(M).

Similarly, for an element x in a monoid H, we can define add(x): an element h ∈ H belongs to add(x) ifand only if there exist an element y ∈ H and a positive integer n such that h+ y = nx.

3.1 add(M) as a submonoid of N(t)0

Let M be a non-zero finitely generated module over a commutative, Noetherian local ring (R,m). WriteM = V

(n1)1 ⊕ · · · ⊕ V (nt)

t , where

(i) each Vi is an indecomposable R-module,

(ii) each ni is a positive integer, and

(iii) Vi 6∼= Vj if i 6= j.

We preserve this fixed ordering of the indecomposable direct summands Vi of M . Given any [N ] ∈ add(M), weclearly have [N ] ∈ add(M); therefore N ∼= V

(a1)1 ⊕ · · · ⊕ V (at)

t , where the ai are non-negative integers. Let N0

denote the additive monoid of non-negative integers. We have a monoid homomorphism Ψ : add(M) → N(t)0

taking [N ] to the t-tuple [a1 . . . at]. In fact, this is a divisor homomorphism: If [N1] and [N2] are in add(M)and Ψ([N1]) | Ψ([N2]), then clearly N1 | N2. Now Theorem 3.6 implies that N1 | N2, that is, [N1] | [N2]. Wecan identify add(M) with its image Γ(M) := Ψ(add(M)), a submonoid of N(t)

0 . In the monoid N(t)0 , we have

α | β if and only if α ≤ β with the coordinate-wise partial ordering: [a1 . . . at] ≤ [b1 . . . bt] if and only if ai ≤ bifor each i. After making this identification, we see that

N1 | N2 if and only if [N1] ≤ [N2] with respect to the coordinate-wise partial ordering on N(t)0 . (3.1)

Definition 3.10. A full submonoid of a monoid Γ is a submonoid Γ′ such that the inclusion Γ′ → Γ is a divisorhomomorphism.

In summary, we have proved the following:

9


Theorem 3.11. Let M be a finitely generated module over a commutative, Noetherian local ring (R,m). Let tbe the number of non-isomorphic indecomposable R-modules in the (unique) decomposition of of M as a directsum of indecomposables. Then Ψ provides an isomorphism between add(M) and the full submonoid Γ(M) ofN(t)

0 . The module M is indecomposable if and only if Ψ([M ]) is a minimal element of Γ(M)− [0 0 . . . 0].

Using this perspective, we can explain a comment in the introduction: One cannot have finitely generatedmodules A and B over a local ring such that A is indecomposable and A ⊕ A ∼= B ⊕ B ⊕ B. Suppose thecontrary, and let M = A ⊕ B. Note that a := [A] and b := [B] are non-zero elements of add(M). We have2a = 3b, so b ≤ a. But then b | a by (3.1). Since a is indecomposable and b 6= 0, we have b = a. But then therelation 2a = 3a forces a = 0, a contradiction.

Monoids admitting a full embedding into a free monoid N(t)0 are the subject of intense study and go by several

different names. People working in the factorization theory of monoids refer to them as “finitely generatedKrull monoids” or “Diophantine monoids” (more on that later). They also figure prominently in applicationsof commutative algebra to simplicial topology. See, e.g., the book by Bruns and Herzog [BH93], where they arecalled “positive normal affine monoids”.

3.2 One-dimensional rings and Diophantine monoids

The dimension of a commutative ringA is the supremum of integers r for whichA has a chain P0 $ P1 $ · · · $ Prof prime ideals. It can be shown [BH93] that for a local Noetherian ring (R,m), the ring R and its completion Rhave the same dimension. Soon we will restrict our attention to one-dimensional local rings (R,m). With thisrestriction we will be able to describe exactly how M(R) sits inside M(R), that is, exactly which R-modulescome from R-modules. (We say, informally, that a finitely generated R-module N comes from an R-moduleprovided there is a finitely generated R-module M such that M ∼= N . We will soon make this concept formal.)

A Diophantine monoid is a monoid isomorphic to the monoid of non-negative integer solutions to a finitesystem of homogeneous linear equations with integer coefficients. Thus H is Diophantine if and only if thereexist positive integers s and t and an s × t integer matrix ϕ such that H ∼= (Ker(ϕ)) ∩ N(t)

0 . In this case wesee that H is a full submonoid of N(t)

0 . In fact the converse holds: If H is a full submonoid of some N(r)0 then

in fact H is Diophantine (though one might have to take the integer t in the definition strictly larger than r).A proof of this fact, attributed to M. Hochster, is outlined in Exercise 6.1.10 on page 263 of [BH93]. The bigtheorem we are heading toward is a realization theorem, which says that given any Diophantine monoid H thereis a one-dimensional local Noetherian integral domain (R,m) and a finitely generated R-module M such thatΓ(M) ∼= H.

In dimension one, the rank of an R-module N , defined in terms of the localizations at the minimal primeideals, is the key to determining whether or not N comes from an R-module. Let A be any commutative ring,not necessarily local. If P is a prime ideal of A, then AP denotes the set of formal fractions a

s where a ∈ A ands ∈ A\P , with the following equivalence relation: a

s = a′

s′ iff there exists t ∈ A\P such that t(as′ = a′s) = 0.The familiar operations a

s + a′

s′ = as′+a′sss′ and a

s ·a′

s′ = aa′

ss′ then make AP into a local ring, called the “localizationof A at P”. (The maximal ideal is the set of fractions whose numerators are in P .) If A is an integral domainand we take P = (0), then AP is the quotient field of A. For another example, let A = R[x], fix a point p ∈ R,and let P be the maximal ideal consisting of polynomials f for which f(p) = 0. Then AP is the ring of rationalfunctions that are defined near p (whence the term “local”). The prime ideals of AP are exactly the idealsQAP , where Q is a prime ideal of A and Q ⊆ P .

We will be particularly interested in localizations at minimal primes. These are the primes ideals P thatcontain no prime ideal properly. More generally, given an ideal I of A, a minimal prime of I is a prime ideal Psuch that I ⊆ P and there is no prime ideal Q with I ⊆ Q $ P . The minimal primes of A are thus the minimalprimes of the ideal (0).

We can localize modules too: For an A-module M and a prime ideal P of A, MP is the set of formal fractionsms where m ∈M and s ∈ A−P (with a similar equivalence relation). The multiplication (a/s)(m/t) = (am)/(st)

makes MP into an AP -module. If M is finitely generated as an A-module, then MP is a finitely generated RP -module: if mt, . . . ,mt generate M over A, then the fractions mi/1 generate MP over AP .

Proposition 3.12. If A is a Noetherian ring, then A has only finitely many minimal primes.

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Proof. The proof is a quintessential example of a process called “Noetherian induction”. We will prove theformally stronger statement that every ideal has only finitely many minimal primes. Suppose this fails. Then,since A is Noetherian, there is an ideal I maximal with respect to having infinitely many minimal primes. ThenI is certainly not a prime ideal, so there are elements x, y ∈ A−I such that xy ∈ I. The ideals K := I+Ax andL := I + Ay each have only finitely many minimal primes, since they contain I properly. Now every minimalprime P of I must contain either K or L and must therefore be either a minimal prime of K or a minimal primeof L, an obvious contradiction.

In a commutative ring A, the set Nil(A) of nilpotent elements is exactly the intersection of the prime idealsof A. (If x /∈ Nil(A), Zorn’s Lemma provides an ideal P maximal with respect to containing no power of x, anda little fiddling shows that P is a prime ideal.) We say that A is reduced provided Nil(A) = (0). (The word“reduced” here has little to do with the property defined above for monoids; it’s just one of those unfortunatecollisions of terminology.)

Lemma 3.13. Let P be a minimal prime ideal in a commutative reduced ring A. Then AP is a field.

Proof. Suppose x is a non-unit of AP ; our goal is to show that x = 0. We have xAP $ AP , and Zorn’s Lemmaimplies that xAP is contained in some maximal ideal m of AP . Since P is a minimal prime, PAP is the uniqueprime ideal of AP and therefore must equal Nil(AP ). An easy computation shows that Nil(AP ) = (0), so (0) isthe unique prime ideal of AP . Since maximal ideals are prime, we have m = (0), whence x = 0.

Assume, now, that (R,m) is reduced (and Noetherian and local as always), and let M be a finitely generatedR-module. For a minimal prime ideal P of R, let rankP (M) denote the dimension of MP as a vector space overthe field RP . If P1, . . . , Ps are all of the minimal primes of R, the rank of M is the s-tuple (r1, . . . , rs), whereri = rankPi(M). If ri = rj for all i, j, we say that M has constant rank.

Suppose, for example, that R has three minimal primes P1, P2, and P3. Then the R-module R has rank(1, 1, 1). The R-module R/P1 has rank (1, 0, 0), and R/(P1 ∩ P2) has rank (1, 1, 0). One can show easily thatR/m has rank (0, 0, 0).

In general, suppose that P1, P2, . . . , Ps are the minimal primes of R, and let E ⊆ 1, 2, . . . , s. Then

rank(

R⋂i∈E Pi

)= (r1, . . . , rs), where ri =

1, i ∈ E0, i 6∈ E

.

It follows that every s-tuple of zeros and ones can be realized as the rank of a cyclic R-module. Moreover, everynon-zero cyclic module R/I over a local ring (R,m) is indecomposable. We’ll refer to s-tuples of zeros and onesas boring s-tuples. Soon we will see how to build indecomposable modules with more interesting ranks. As weshall see in Interlude 3.19, however, even modules with boring ranks sometimes suffice to demonstrate failureof KRSA.

Assumptions 3.14. From now on (R,m) is a local noetherian ring of dimension one, and its minimal primesare P1, . . . , Ps. We assume further that R is reduced (so of course R is reduced as well).

We now return to the question of which R-modules come from R-modules. We’ll also need to know whichR-modules come from indecomposable R-modules. We’ll call a non-zero finitely generated R-module N extendedif N ∼= M for some finitely generated R-module M , and minimally extended if, in addition, no non-zero properdirect summand of M is extended.

Let’s rephrase Theorem 3.11 in these terms.

Corollary 3.15. Let M be a finitely generated R-module. These are equivalent:

(i) M is indecomposable.

(ii) M is minimally extended.

(iii) Ψ([M ]) is a minimal element of Γ(M)− [0 0 . . . 0].

11


For one-dimensional rings, we have the following result due to L. S. Levy and C. Odenthal [LO96, The-orem 6.2], which tells us exactly which R-modules are extended. There is a partial result for certain two-dimensional rings, but there is no such result for rings of dimension greater than two; in fact, the generalquestion of which modules over the completion of a local ring A come from A-modules seems to be extremelydifficult.

Proposition 3.16. Let (R,m) be a one-dimensional Noetherian local ring. Let R denote the m-adic completionof R, and assume R is reduced. Let M be a finitely generated R-module. Then M is extended from an R-moduleif and only if rankP (M) = rankQ(M) whenever P and Q are minimal prime ideals of R with P ∩R = Q∩R

We will not prove this result here, but refer the reader to [LO96, Theorem 6.2]. If R is a domain, then 0 isthe only minimal prime ideal of R, and Proposition 3.16 takes the following form:

Corollary 3.17. Let (R,m) be a one-dimensional Noetherian local integral domain. Let R denote the m-adiccompletion of R, and assume R is reduced. Let M be a finitely generated R-module. Then M is extended froman R-module if and only if M has constant rank.

Given a finitely generated R-module M , we know that add(M) can be viewed as a full submonoid of someN(t)

0 . Using Proposition 3.16, we can, in fact, represent add(M) as a Diophantine monoid. (Although thisfollows purely formally from Theorem 3.11 and the fact [BH93, Exercise 6.1.10] that full submonoids of N(t)

0 arealways Diophantine, it is instructive to see exactly where the defining equations come from.)

Theorem 3.18. Let (R,m) be a one-dimensional local ring with reduced completion R, and let M be a finitelygenerated R-module. Then add(M) is a Diophantine monoid.

Proof. By Theorem 3.11 it will suffice to show that Γ(M) is a Diophantine monoid. Let P1, . . . , Ps be theminimal prime ideals of R. As in Section 3.1, write M = V

(n1)1 ⊕ · · · ⊕ V (nt)

t , where the Vj are non-isomorphicindecomposable R-modules and the nj are positive integers. Let rij = dimRPi

((Vj)Pi), so that (r1j , . . . , rsj) =

rank(Vj).Suppose a = [a1 . . . at] ∈ N(t)

0 , and let N = V(a1)1 ⊕ · · · ⊕ V

(at)t . Then a ∈ Γ(M) if and only if N

comes from an R-module. (We needTheorem 3.6 here to conclude that if N ∼= V , where V is a finitelygenerated R-module, then [V ] ∈ add(M).) Notice that rank(N) = (ρ1, . . . , ρs), where ρi =

∑tj=1 rijaj . Let

S = (i, l) | 1 ≤ i < l ≤ s and Pi ∩ R = Pl ∩ R. By Proposition 3.16 we see that a ∈ Γ(M) if and only ifρi = ρl whenever (i, l) ∈ S. Thus Γ(M) is the solution set of a family of |S| homogeneous linear equations withinteger coefficients. (If S = ∅, that is, if the Pi lie over distinct prime ideals of R, then there are no relations,and Γ(M) is the free monoid N(t)

0 .)

Interlude 3.19. At this point we describe a couple of “warmup” examples, to illustrate the general method wewill use to demonstrate failure of KRSA. Suppose, for example, that (R,m, k) is a one-dimensional Noetherianlocal domain whose completion R has three minimal prime ideals P1, P2, P3. We assume, as always, that R isreduced. For i = 1, 2, 3, let Xi = R/Pi and Yi = R/

⋂j 6=i Pj . Then rank(Xi ⊕ Yi) = (1, 1, 1), so Corollary 3.17

provides a finitely generated R-module Ai such that Ai ∼= Xi ⊕ Yi. Similarly, there are finitely generatedR-modules B and C such that X1 ⊕ X2 ⊕ X3

∼= B and Y1 ⊕ Y2 ⊕ Y3∼= C, since the given R-modules have

constant ranks (1, 1, 1) and (2, 2, 2) respectively. It is easy to see that B, C and the Ai are minimal extended. ByCorollary 3.15, the corresponding R-modules B, C, Ai are indecomposable. Moreover, Corollary 3.7 implies thatA1⊕A2⊕A3 and B⊕C are isomorphic, since their completions are both isomorphic to X1⊕X2⊕X3⊕Y1⊕Y2⊕Y3.

In order to obtain examples such as the one mentioned in the introduction, we will need R-modules withmore interesting rank functions. Still assuming R is a domain, suppose R has two minimal primes P1 and P2.Fix an integer n ≥ 2, and suppose we can build indecomposable finitely generated R-modules E, F and G withrespective ranks (2n+1, n+1), (n+2, n+1) and (0, n+1). Given [a b c] ∈ N(3)

0 , let N(a, b, c) = E(a)⊕F (b)⊕G(c),an R-module of rank ((2n + 1)a + (n + 2)b, (n + 1)(a + b + c)). Using Corollary 3.17, we see that N(a, b, c) isextended an R-module M(a, b, c) if and only (2n+1)a+(n+2)b = (n+1)(a+b+c), that is, if and only if [a b c]belongs to the Diophantine monoid H := [x y z] ∈ N(3)

0 | nx+ y = (n+ 1)z. Noting that [1 1 1] ∈ H, we putC = M(1, 1, 1). Then C ∼= E⊕F⊕G, and add(C) ∼= Γ(C) = H. Clearly [1 1 1] is an atom of H, as are [n+1 0 n]and [0 n+ 1 1]. Letting A = M(n+ 1, 0, n) and B = M(0, n+ 1, 1), we see that A, B and C are indecomposable

12


R-modules. Moreover, we have the relation A ⊕ B ∼= C(n+1). (The example mentioned in the introduction isthe case n = 2.) An innocent bystander analyzing powers of C would not detect the silly direct-sum behavioruntil the n+ 1st step, since C(n) has only one representation as a direct sum of indecomposable modules.

These examples suggest that in order to realize an arbitrary Diophantine monoid as a monoid of the formΓ(M), we need to find examples where, first of all, R has more minimal primes than R does, and second, R hasfinitely generated modules with interesting rank functions. The next theorem [Wie01, (2.3) and (2.4)] fits thebill perfectly. Moreover the requisite ring can always be chosen to be an integral domain, and the module Mto be torsion-free. (Over an integral domain R, a module M is torsion-free provided rm 6= 0 when r and m arenon-zero elements of R and M , respectively.)

Theorem 3.20. Fix an integer s ≥ 2, and let (r1, . . . , rs) be an arbitrary non-trivial s-tuple of non-negativeintegers. Choose real numbers q1 < · · · < qs. Let R be the subring of the field R(x) of rational functions in onevariable consisting of functions f(x) satisfying the following conditions:

(i) f(q1) = · · · = f(qs)

(ii) f (i)(qj) = 0 for j ∈ 1, 2, . . . , s and i ∈ 1, 2, 3. (Here f (i) denotes the ith derivative of f(x).)

Then

(a) R is a one-dimensional local ring,

(b) R is reduced and has exactly s minimal prime ideals, and

(c) there exists a finitely generated torsion-free R-module N with rank(N) = (r1, . . . , rs).

There’s nothing special about the field of real numbers here. All we need is a field with at least s distinctelements q1, . . . , qs. The construction of the module M and the proof of indecomposability are quite technicaland use ideas going back to a 1967 paper of Drozd and Roıter [DR67] on the classification of one-dimensionalrings of finite representation type. While we won’t say anything more about the module M we will say alittle bit about the ring and where the various maximal ideals of R come from. The maximal ideal of R ism := f ∈ R | f(q1) = · · · = f(qs) = 0. The ringR is a subring of the ring S consisting of rational functions thatare defined at each qi. The ring S has exactly s maximal ideals, namely, mi := f ∈ S | f(qi) = 0, i = 1, . . . , s.One can show that S is finitely generated as an R-module, so that S is a finitely generated R-module. The ringS/mS has a family of orthogonal idempotents, one for each maximal ideal miS, and by Propositions 2.10 and2.12 these lift to idempotents of S. These idempotents give a decomposition S = D1 × · · · ×Ds, in which eachDi is a one-dimensional local domain. Then Qi :=

∏j 6=iDj is a minimal prime ideal of S, and the prime ideals

Pi := Qi ∩ R are exactly the minimal prime ideals of R.

Theorem 3.21 (Realization Theorem for Diophantine Monoids). Let H ⊆ N(t)0 be any Diophantine monoid

containing an element α := [a1 . . . at] in which each ai is positive. Then there exist a one-dimensional localdomain R and a finitely generated torsion-free R-module M such that Γ(M) = H, with Ψ([M ]) = α (in thenotation of Theorem 3.11 ).

Proof. Assume H is defined by m homogeneous linear equations. Then we can write H = N(t)0 ∩Ker(Φ), where

Φ : Qt → Qm is a linear transformation. We regard Φ as an m× t matrix [qij ], and we can assume, by clearingdenominators, that the entries qij are integers. Choose a positive integer h such that qij + h ≥ 0 for all i, j.Let R be the ring of Theorem 3.20 with s = m+ 1. (The number of minimal primes of R is one more than thenumber of defining equations of the monoid.)

For j = 1, . . . , t, choose, using Theorem 3.20, an indecomposable, finitely generated, torsion-free R-moduleNj such that

rank(Nj) = (q1j + h, . . . , qmj + h, h), j = 1, . . . , t. (3.2)

Suppose now that β = [b1 . . . bt] ∈ N(t)0 . If N = N

(b1)1 ⊕ · · · ⊕N (bt)

t , then

rank(N) =

t∑j=1

(q1j + h)bj , . . . ,t∑

j=1

(qmj + h)bj ,

t∑j=1

bj

h

. (3.3)

13


By Corollary 3.17, N comes from an R-module if and only if∑tj=1(qij + h)bj =

(∑tj=1 bj

)h for 1 ≤ i ≤ m,

that is, if and only if β ∈ N(t)0 ∩Ker(Φ) = H.

Let M be the finitely generated R-module (unique up to isomorphism) such that M ∼= L(a1)1 ⊕ · · · ⊕ L(at)

t .Then Γ(M) = H, and the isomorphism Ψ of Theorem 3.11 carries [M ] to the element α.

4 Measuring Failure of KRSA

In this section we describe invariants that are useful in determining the extent to which direct-sum decomposi-tions can be non-unique.

4.1 More on Monoids

We briefly leave the world of direct-sum decompositions and recall some basic concepts of factorization incommutative monoids. We will return later to monoids of modules and use these tools to describe how badlyKSRA can fail over Noetherian local rings of dimension one.

Throughout, let H denote a monoid satisfying (i) and (ii) of §3.

Definition 4.1. A Krull monoid is a monoid admitting a divisor homomorphism ϕ : H → F , where F is afree monoid. A divisor theory is a divisor homomorphism ϕ : H → N(Ω)

0 such that every element of N(Ω)0 is the

greatest lower bound (in the usual product partial ordering) of some non-empty finite set of elements in ϕ(H).Note (cf. [HK98, 20.4]) that if ϕ : H → F and ϕ′ : H → F ′ are divisor theories for H, then there is a monoidisomorphism ψ : F → F ′ such that ϕ′ = ψ ϕ.

In fact [GHK06, Theorem 2.4.8], every Krull monoid has a divisor theory. Although we will not prove thishere, we will need this fact when we discuss the divisor class group of a Krull monoid.

Our next task is to show that all Krull monoids are atomic. The key to proving Theorem 4.3, as well asTheorem 4.4, is to appeal to the descending chain condition on N(t)

0 , with respect to the product (coordinate-wise) partial ordering: [a1 . . . at] ≤ [b1 . . . bt] ⇐⇒ ai ≤ bi for each i. A poset X satisfies DCC provided thereis no infinite strictly descending chain x1 > x2 > x3 > . . . in X. One verifies easily that if X and Y both haveDCC then so does X × Y . Since N0 has DCC, an induction argument shows the following:

Lemma 4.2. Let r be any positive integer. Then N(r)0 satisfies DCC.

Proposition 4.3. If H is a Krull monoid, then H is atomic.

Proof. We may assume that H is a full submonoid of N(I)0 for some (possibly infinite) index set I. Let x be a

non-zero element of H. Then x ∈ N(F )0 for some finite subset F of I. One checks easily that add(x) is also a

full submonoid of N(F )0 . (Technically, we should write either “addH(x)” or “addN(I)

0(x)”, but they are the same,

since H is a full submonoid of N(I)0 .) For elements y, z ∈ add(x), we have y | z in add(x) if and only if y ≤ z in

N(F )0 , that is, if and only if y ≤ z in the product ordering on N(I)

0 . Moreover, if h1 + h2 ∈ add(x), with hi ∈ H,then h1, h2 ∈ add(x); it follows that every atom of add(x) is actually an atom of H. Therefore it will suffice toshow that x is a sum of atoms of add(x). By Lemma 4.2, N(F )

0 , and hence H, satisfies DCC.We will show, in fact, that every non-zero element of add(x) is a sum of atoms. If not, the set B of non-zero

elements of add(x) that cannot be expressed as a sum of atoms of add(x) is non-empty. By the descendingchain condition, B has a minimal element b. Cetainly b is not an atom, so b = c+d, where c and d are non-zeroelements of add(x). Then c < b and d < b, so neither c nor d is in B. Writing c and d as sums of atoms makesb a sum of atoms, contradicting the fact that b ∈ B. This contradiction shows that B = ∅, and the proof iscomplete.

Moreover, if H is a full submonoid to N(t)0 for some positive integer t, in particular, if H is a Diophantine

monoid, then H has only finitely many atoms, as we shall see in Corollary 4.5 below. (Recall that the atomsare the minimal elements of H\0.) Given any poset X, we define a clutter to be a subset S of X with noorder relations among its elements. In other words, if x, y ∈ S, then neither x < y nor y < x. (The less colorfulterm “antichain” is often used in the literature.)

14


Theorem 4.4. Let t be a positive integer. Then N(t)0 has no infinite clutters. For any subset Y of N(t)

0 , min(Y )is finite.

Proof. The second statement follows from the first, since min(Y ) is a clutter. To prove the first statement, weuse induction on t, the case t = 1 being trivial. Suppose, now, that t ≥ 2, and let M be a clutter in N(t)

0 . IfM = ∅, then M is very finite, so we assume that M 6= ∅ and fix an element [a1 . . . an] in M .

For each pair of integers i and j with 1 ≤ i ≤ n and 0 ≤ j ≤ ai, define Mij = [x1 x2 . . . xt] ∈M | xi = jand Bij = [x1 x2 . . . xt−1] ∈ Nt−1

0 | [x1 . . . xi−1 j xi . . . xt−1] ∈ Mij. For each i and j, the bijectionMij ↔ Bij , given by [x1 . . . xi−1 j xi+1 . . . xt]↔ [x1 . . . xi−1 xi+1 . . . xt], shows that Bij is a clutter. Theinductive hypothesis now guarantees that each set Bij is finite, and hence that each Mij is finite. Since thereare only finitely many sets Mij , we need only show that M ⊆

⋃Mij . Let [x1 x2 . . . xt] ∈ M . If ai < xi for

all i, then [a1 a2 . . . at] < [x1 x2 . . . xt], contradicting the assumption that M is a clutter. Thus there existsi ∈ 1, 2, . . . , n with xi ≤ ai, and [x1 x2 . . . xn] ∈Mixi

.

Alternatively, one can prove Theorem 4.4 by verifying a more general result: If A and B are posets, eachwith DCC, and each having no infinite clutters, then A×B has no infinite clutters.

Corollary 4.5. Let H be a full submonoid of N(t)0 for some positive integer t. Then H has only finitely many

atoms. In particular, every Diophantine monoid has only finitely many atoms.

Corollary 4.6. Let x be an element of a Krull monoid H. Then x has only finitely many distinct factorizationsas sums of atoms.

Proof. By choosing a suitable divisor homomorphism, we may assume that H is a full submonoid of N(Ω)0 for

some index set Ω. Choose a finite subset F of Ω such that x ∈ N(F )0 . Then add(x) is a full submonoid of N(F )

0 .Now every atom of H that divides x is actually an atom of add(x), which, by Corollary 4.5, has only finitelymany atoms. Thus there are only finitely many atoms p1, . . . , pn that divide x. The decompositions of x areall of the form x = e1p1 + · · · + enpn, for suitable non-negative integers ei, and our task is to show that thereare only finitely many sequences (e1, . . . , en) arising in this way. It will suffice to establish a bound on eachei. Suppose, by way of contradiction, that there is an atom p of H such that ep | x for every positive integere. Then, in the coordinate-wise partial ordering on N(t)

0 , we have ep ≤ x for each e. Write p = [a1 . . . at] andx = [b1 . . . bt], as elements of N(t)

0 . Since p 6= 0 we have aj > 0 for some j. Then bj ≥ eaj for every e, an obviouscontradiction.

Given a Krull monoid H, we let Q(H) denote the quotient group, which consists of formal differences x−y |x, y ∈ H. If ϕ : H → N(Ω)

0 is a divisor theory, we have an induced quotient homomorphism Q(ϕ) : Q(H) →Q(N(Ω)

0

)defined by Q(ϕ)(x−y) = ϕ(x)−ϕ(y). This gives rise to the divisor class group Cl(H), which is defined

to be the cokernel of the map Q(ϕ). Elements of Cl(H) are called divisor classes. Noting that Q(N(Ω)0 ) = Z(Ω),

we see that a divisor class is a coset z+L, where z ∈ Z(Ω) and L is the image of the homomorphism Q(ϕ). Welet π : Z(Ω) Cl(H) be the canonical surjection. An element z ∈ Z(Ω) belongs to Ker(π) if and only if thereare elements x, y ∈ H such that ϕ(x)− ϕ(y) = z. Moreover, we claim that

Ker(π) ∩ N(Ω)0 = ϕ(H). (4.1)

Of course Ker(π) ∩ N(Ω)0 ⊇ ϕ(H). To verify the reverse inclusion, suppose z ∈ Ker(π) ∩ N(Ω)

0 , and writez = ϕ(x) − ϕ(y), with x, y ∈ H. Then z + ϕ(y) = ϕ(x), so ϕ(y) | ϕ(x). Therefore y | x, as ϕ is a divisorhomomorphism, say, y + h = x. Now ϕ(h) = ϕ(x)− ϕ(y) = z. This shows that Ker(π) ∩ N(Ω)

0 ⊆ ϕ(H).The divisor class group gives a crude measure of failure of unique factorization:

Proposition 4.7. Let H be a Krull monoid. Then H is free if and only if Cl(H) is trivial.

Proof. Let ϕ : H → N(Ω)0 be a divisor theory, and let π : Z(Ω) Cl(H) be the canonical surjection.

Suppose Cl(H) is trivial. Then N(Ω)0 ⊆ Ker(π), and now (4.1) implies that ϕ(H) = N(Ω)

0 . Since all divisorhomomorphisms are injective, H ∼= N(Ω)

0 , and thus H is free. Conversely, if H is free, then the identitymap ι : H → H is a divisor theory. The induced map Q(ι) : Q(H) → Q(H) is also the identity map, andCl(H) = Q(H)/Q(H) = 0.

15


Let H denote a Krull monoid with divisor theory ϕ : H → N(Ω)0 . We refer to the standard basis elements

eω ∈ N(Ω)0 as primes, and we say that an element g ∈ Cl(H) contains a prime if g = π(eω) for some prime

eω ∈ N(Ω)0 . (Here π : Z(Ω) Cl(H) is the canonical projection.) As we shall see in Theorem 4.13, if Cl(H)

is infinite and every divisor class contains a prime, then factorization in H is wildly non-unique. The readershould note that the elements eω are exactly the atoms, equivalently, prime elements, of the free monoid N(Ω)

0 .It might be seem more consistent to say that a divisor class contains an atom, but the terminology above isubiquitous in the literature, and we will stick with it.

In the following theorem [BS] we require the matrix A to contain certain columns of zeros and ones andobserve that the natural inclusion Ker(A)∩N(Ω)

0 ⊆ N(Ω)0 is nearly always a divisor theory. In Section 4.2 we will

use the existence of certain boring ranks to acquire these additional columns. We also determine the divisorclass groups of such Diophantine monoids and consider which divisor classes contain primes. More generalresults for finitely generated Diophantine monoids can be found in [CKO02].

Theorem 4.8. Fix an integer q ≥ 1 and let Iq denote the q× q identity matrix. Let D =[D1 |D2 |D3

], where

D1 =[Iq −Iq

],

D2 =

1 −11 −1...

...1 −1

,and D3 is an arbitrary integer matrix with q rows (and possibly infinitely many columns). Let H = Ker(D)∩N(Ω)

0 ,where Ω is the cardinality of the set of columns of D.

(i) The natural inclusion H → N(Ω)0 is a divisor theory.

(ii) Cl(H) ∼= Z(q).

(iii) Each column of D corresponds, via the isomorphism in (ii), to a divisor class that contain a prime.

Proof. We begin by showing that the inclusion H → N(Ω)0 is a divisor homomorphism. Suppose α, γ ∈ H with

α ≤ γ in N(Ω)0 ; that is, there is an element β ∈ N(Ω)

0 with α + β = γ. Since D(α + β) = 0 and Dα = 0,we see that Dβ = 0 as well, and thus β ∈ H. Therefore H → N(Ω)

0 is a divisor homomorphism. For anelement β ∈ N(Ω)

0 , write −Dβ =[d1 · · · dq

]T where each di is an integer, and set M = max0, d1, . . . , dqand m = min0, d1, . . . , dq. For each i ∈ 1, . . . , 2q+ 2, let εi denote the standard basis vector in N(Ω)

0 with 1in the ith coordinate and 0 elsewhere; note that Dεi is precisely the ith column of D. Define

β1 :=q∑i=1

(di −m)εi −mε2q+2 ∈ N(Ω)0 and β2 :=

q∑i=1

(M− di)εq+i + Mε2q+1 ∈ N(Ω)0 .

One checks that β + β1 and β + β2 are in N(Ω)0 ∩ Ker(D) = H and that β is the greatest lower bound of

β + β1, β + β2. Therefore the inclusion H → N(Ω)0 is a divisor theory.

For each i ∈ 1, 2, . . . , q, the standard basis vector ei occurs as a column of D, and thus D : Z(Ω) → Z(q) issurjective. Therefore

Z(Ω)/Ker(D) ∼= Z(q). (4.2)

Clearly Q(H) ⊆ Ker(D), and we now show the reverse inclusion. Let α ∈ Ker(D), and write α = β − γ

for some β, γ ∈ N(Ω)0 . As in the previous paragraph, find β1 ∈ N(Ω)

0 with β + β1 ∈ H. Then γ + β1 =β+β1−α ∈ N(Ω)

0 ∩ Ker(D) = H, and so α = (β+β1)− (γ+β1) ∈ Q(H). Therefore Q(H) = Ker(D), and nowCl(H) = Z(Ω)/Q(H) = Z(Ω)/Ker(D) ∼= Z(q) by(4.2). This proves (ii).

For (iii), we observe that, for an element z ∈ Z(Ω), the isomorphism in (ii) carries the divisor class π(z) tothe matrix product Dz ∈ Z(q). Suppose now that c is the ωth column of D. Then c = Deω, which corresponds,via the isomorphism in (ii), to the divisor class π(eω). Since this divisor class contains the prime eω, the proofis complete.

16


Let H be a Krull monoid and h a non-zero element of H. The set of lengths of h is L(h) := n | h =a1 + · · ·+ an for atoms ai ∈ H. The elasticity of h ∈ H is ρ(h) := sup L(h)/ inf L(h). Since, by Corollarly 4.6,h has only finitely many distinct factorizations, the elasticity ρ(h) is finite. If H is free, then factorization isunique, and hence ρ(h) = 1 for every non-zero element h ∈ H. The converse can fail, as we see in the followingexample.

Example 4.9. Consider the monoid H = [w x y z] ∈ N(4)0 | w + x = y + z. The atoms are α1 = [1 0 1 0],

α2 = [1 0 0 1], α3 = [0 1 1 0], and α4 = [0 1 0 1]. One can then verify that if∑4i=1 aiαi =

∑4i=1 biαi for

nonnegative integers a1, a2, a3, a4, b1, b2, b3, and b4, then a1 + a2 + a3 + a4 = b1 + b2 + b3 + b4, and henceρ(H) = 1. However, since α1 + α4 = α2 + α3, it is clear that H is not factorial.

We say that a monoid H is half-factorial if |L(h)| = 1 for every non-zero element h ∈ H (equivalently, His atomic and any two representations of an element of H as a sum of atoms have the same length). We putρ(H) := supρ(h) | h ∈ H\0, the elasticity of H. The monoid H is fully elastic provided every rationalnumber in the closed interval [1, ρ(H)] occurs as the elasticity of some element of H. (Note that if ρ(H) =∞,we say H is fully elastic if every rational number in [1,∞) occurs as the elasticity of some element in H.)

We now introduce the block monoid of a Krull monoid. This object is often easier to study, yet it carries agreat deal of information about the original monoid.

Definition 4.10 (block monoid). Let G be an abelian group, let P be any subset of G, and let F(P ) be thefree abelian monoid with basis P . Thus F(P ) consists of formal N0-linear combinations of elements in the setP with the obvious binary operation. We express an element b of F(P ) as follows:

b = n1p1 ⊕ · · · ⊕ ntpt,

where the pi are in P and the ni are non-negative integers and ⊕ denotes a formal sum in F(P ). Consider themap

σ : F(P ) −→ Gn1p1 ⊕ · · · ⊕ ntpt 7→ n1p1 + · · ·+ ntpt

taking the formal sum in F(P ) to the actual sum in G. The submonoid B(G,P ) = s ∈ F(P ) : σ(s) = 0 ofF(P ) is called the block monoid of G with respect to P . In other words, the block monoid is the set of formalsums that add up to 0 in the group G. (In the literature, the free monoid F(P ) and the block monoid B(G,P )are usually written multiplicatively, but the additive notation seems more appropriate to our situation.)

We are particularly interested in the following situation:

(i) H is a Krull monoid,

(ii) G = Cl(H), and

(iii) P is the subset of G consisting of divisor classes that contain primes.

In this situation, we refer to B(G,P ) as the block monoid of H and denote it simply by B(H).

Before describing how B(H) is useful in studying factorization in the Krull monoid H, we need to define atype of monoid homomorphism that preserves basic factorization properties.

Definition 4.11 (transfer homomorphism). Let H and K be monoids (reduced and cancellative as always).A homomorphism ϕ : H → K is a transfer homomorphism provided

(i) ϕ is surjective,

(ii) ϕ(z) 6= 0 for each non-zero z ∈ H, and

(iii) whenever ϕ(z) = a+ b in K, there exist x, y ∈ H such that ϕ(x) = a, ϕ(y) = b, and x+ y = z in H.

17


Suppose that ϕ : H → K is a transfer homomorphism. Given z ∈ H, one checks easily that z is an atomof H if and only if ϕ(z) is an atom of K. In fact, ϕ preserves the basic structure of factorizations in H. Inparticular, L(z) = L(ϕ(z)) (and consequently ρ(z) = ρ(ϕ(z)) and ρ(H) = ρ(K)) [Ger88], and we can studysets of lengths and elasticities in H by studying these same invariants in K. We take this approach when His a Krull monoid and K is the block monoid B(H). In the next result (proved in more generality in [Ger88])we establish a transfer homomorphism β : H → B(H). But first we set the stage by building a commutativediagram that shows the various homomorphisms involved.

Let H be a Krull monoid with divisor theory ϕ : H → N(Ω)0 . Let π : Z(Ω) Cl(H) denote the canonical

surjection onto the divisor class group of H, and let P be the set of divisor classes that contain primes. ThusP = π(eω)ω∈Ω where the set eωω∈Ω is the standard basis for the free monoid N(Ω)

0 , and B(H) = B(Cl(H), P ).We define the monoid homomorphism β : N(Ω)

0 → F(P ) on the basis of N(Ω)0 by eω 7→ π(eω). (Warning:

Although the homomorphism β is obviously surjective, it is not necessarily injective.) Recall also the mapσ : F(P )→ Cl(H) from Definition 4.10 and that B(H) = z ∈ F(P ) | σ(z) = 0. For an element x ∈ N(Ω)

0 onechecks, using (4.1) that

x ∈ ϕ(H) ⇐⇒ β(x) ∈ B(H). (4.3)

Thus β induces a surjective monoid homomorphism β : H → B(H) making the following diagram commute:

H ϕ //

β

N(Ω)0

⊂ //

β

Z(Ω)

π

B(H) ⊆ // F(P ) σ // Cl(H)

(4.4)

Theorem 4.12. The map β : H → B(H) is a transfer homomorphism.

Proof. On replacing H by an isomorphic copy, we may assume that H ⊆ N(Ω)0 and that the map ϕ in the (4.4)

is the inclusion map.We have already verified (i) in Definition 4.11. In order to check (ii) and (iii), we establish some notation

to allow for the fact that the homomorphism β may not be injective. Given an element z ∈ N(Ω)0 , write

z =∑ω∈F mωω, where F is a finite subset of Ω and the mω are non-negative integers. Let p1, . . . , ps be the

distinct elements of π(eω)ω∈F . For i = 1, . . . , s, let Fi = ω ∈ F | π(eω) = pi, and put ni =∑ω∈Fi

mω.Then β(z) = β(z) = n1p1 ⊕ · · · ⊕ nsps ∈ F(P ).

Suppose now that z is a non-zero element of H. Choose ω ∈ Ω such that mω > 0, and suppose ω ∈ Fi.Then ni > 0, whence β(z) 6= 0. This proves (ii). For (iii), suppose β(z) = a + b, with a, b ∈ B(H). Sinceβ(z) = n1p1 ⊕ · · · ⊕ nsps and the pi are distinct basis elements of the free monoid F(P ), it follows thata = k1p1 ⊕ · · · ⊕ ksps and b = `1p1 ⊕ · · · ⊕ `sps, where 0 ≤ ki ≤ ni and ki + `i = ni for each i. In order todecompose z in a manner that is compatible with the decomposition of β(z), we fix an index i for the momentand recall that

∑ω∈Fi

mω = ki + `i. We can choose, for each ω ∈ Fi, a non-negative integer uω ≤ mω in sucha way that

∑ω∈Fi

uω = ki. Put vω = mω − uω. Now, letting i vary, we have uω + vω = mω for all ω ∈ F .Putting x =

∑ω∈F uωeω and y =

∑ω∈F vωeω, we see that β(x) = a, β(y) = b, and x + y = z. Finally, (4.3)

implies that x and y are in H, since their images under β are in B(H).

We state without proof the following amazing result on sets of lengths in a Krull monoid with infinite divisorclass group. This theorem shows how dreadful factorization can be in certain monoids.

Theorem 4.13 ([Kai99], Theorem 1). Let H be a Krull monoid with infinite divisor class group, and assumethat every divisor class contains a prime. Then, for any non-empty finite set L ⊆ n ∈ N | n ≥ 2, there existsan element h ∈ H such that L(h) = L.

Thus, for example, there’s an element h ∈ H with the following property: h is a sum n atoms if and only ifn ∈ 7, 33, 9268. An immediate corollary of Kainrath’s theorem is that any monoid satisfying the hypothesesof Theorem 4.13 has infinite elasticity — a result we will apply in Example 4.17.

Corollary 4.14. Let H be a Krull monoid with infinite divisor class group G, and assume that every divisorclass contains a prime. Then ρ(H) =∞, and H is fully elastic.

18


Proof. Given a rational number p ≥ 1, write p = ab with a, b ∈ N. By Theorem 4.13, there is an element h ∈ H

with L(h) = 2b, 2a, and thus ρ(h) = 2a2b = p.

4.2 Back to Direct-sum Decompositions

Suppose R is a one-dimensional reduced commutative Noetherian local ring. By Propositions 2.1 and 4.3,M(R) is an atomic Krull monoid. Also, by Theorem 3.6, the map [M ] 7→ [M ] gives a divisor homomorphismM(R) → M(R). By Theorem 2.13, M(R) is isomorphic to the free monoid N(Ω)

0 , where Ω is the set ofisomorphism classes of indecomposable finitely generated R-modules. In this section we apply the tools fromSection 4.1 to demonstrate spectacular failure of KRSA over one-dimensional Noetherian local rings.

Given the set of ranks of all indecomposable modules over the ring R, and a description of how minimalprimes of R lie over the minimal primes of R, we can completely describe M(R) as a Diophantine monoid.

Let P1, . . . , Ps denote the minimal primes of R, and for each i ∈ 1, . . . , s let Qi,1, . . . , Qi,ti denote theminimal prime ideals of R with Qi,j ∩ R = Pi. Let q = |Spec(R)| − |Spec(R)|. Since the minimal primes in aone-dimensional local ring are the ones different from the maximal ideal, q is the difference between the numberof minimal primes of R and the number of minimal primes of R. Note that q = (t1−1)+(t2−1)+ · · ·+(ts−1) =∑si=1 ti− s. (By the way, we are using the fact [Mat86, Theorem 7.3] that the map Spec(R)→ Spec(R), taking

P to P ∩R, is surjective.) ThenM(R) ∼= Ker(A)∩N(Ω)0 , where Ω is the set of indecomposable finitely generated

R-modules, and where the matrix A is the q × Ω matrix defined by the following scheme:For an indecomposable R-module M , let (r1,1, . . . , r1,t1 , . . . , rs,1, . . . , rs,ts) denote its rank, where ri,j =

rankQi,j(M), the rank of M at Qi,j . The column indexed by the isomorphism class [M ] is the transpose of the

vector [r1,1 − r1,2 · · · r1,1 − r1,t1 r2,1 − r2,2 · · · r2,1 − r2,t2 · · · rs,1 − rs,2 · · · rs,1 − rs,ts

].

If Q1, Q2, . . . , Qt are the minimal primes of R, and I ⊆ 1, 2, . . . , t, then R/ ∩i∈I Qi is an indecomposablefinitely generated R-module of rank (r1, r2, . . . , rt) where ri = 1 if i ∈ I and ri = 0 if i 6∈ I. Thus thematrix A contains, as a submatrix, the matrix [D1|D2] as in Theorem 4.8. Moreover, if α ∈ Cl(M(R)), thenα contains a prime if α = Aeω for some atom eω in Z(Ω). Since Aeω is the ωth column of A, the elements inCl(M(R)) ∼= Z(Ω) that contain primes correspond to the distinct columns of A. Therefore the block monoid isB(M(R)) ∼= Ker(A′)∩N(Ω′)

0 where A′ denotes the matrix formed by eliminating all repeated columns in A. Inthe following examples, we construct the matrix A and apply factorization-theoretic techniques to the study ofdirect-sum decomposition over certain one-dimensional local rings.

We also use this strategy to investigate direct-sum decompositions of restricted classes of modules, e.g., theclass of all finitely generated torsion-free modules. In this context we note that an extended finitely generatedtorsion-free R-module is necessarily extended from a finitely generated torsion-free R-module. In the firstexample we consider a possibly incomplete list of ranks to exhibit failure of KRSA and give bounds on theelasticity of M(R).

Example 4.15. Suppose that R is an integral domain and that its completion R has two minimal primes P1 andP2. Recall that the indecomposable R-module R/P1 has rank (1, 0) and the indecomposable R-module R/P2

has rank (0, 1). Let M and N be indecomposable R-modules with ranks (m1,m2) and (n1, n2) respectively.Suppose further that a := m1 −m2 and b := n2 − n1 are positive, and let c denote the least common multipleof a and b. Since a, b > 0, neither M nor N is extended. However, one checks that M (c/a)⊕N (c/b) is minimallyextended and is thus the completion of an indecomposable R-module Z. We note also that M ⊕ (R/P2)(a),N⊕ (R/P1)(b), and R/P1⊕ R/P2 are also minimally extended and hence are the completions of indecomposableR-modules X, Y , and W respectively. Since(

M (c/a) ⊕N (c/b))⊕ (R/P1 ⊕ R/P2)(c) ∼=

(M ⊕ (R/P2)(a)

)(c/a)

⊕(N ⊕ (R/P1)(b)

)(c/b)

,

it follows that L := Z ⊕ W (c) is isomorphic to X(c/a) ⊕ Y (c/b) as R-modules. This illustrates the failureof KRSA. Note that the R-module L can be expressed both as the direct sum of c + 1 indecomposable R-modules and as the direct sum of (c/a) + (c/b) indecomposable R-modules. Therefore ρ([L]) ≥ c+1

(c/a)+(c/b) in

19


the monoid M(R). If a, b > 1, then c+1(c/a)+(c/b) > 1 and hence ρ(M(R)) > 1. If the sets a | a = rankP1(M)−

rankP2(M) for some indecomposable MR and b | b = rankP2(M)−rankP1(M) for some indecomposable MRare unbounded, then there is no bound on ρ([L] | L is an R-module, and thus ρ(M(R)) =∞.

Giving a complete list of ranks of indecomposable R-modules would allow a more precise calculation ofelasticity. Unfortunately, acquiring this set of ranks is a difficult problem in general. However, in certain cases,the ranks of all indecomposable modules are known. As the construction of indecomposables with various ranksis rather technical, we simply provide the ranks and go from there. In the next example, we describe thenon-uniqueness of direct-sum decompositions of torsion-free modules over a ring which has finite representationtype, i.e., it has, up to isomorphism, only finitely many indecomposable torsion-free modules.

Example 4.16. Fix a positive integer n. Suppose R is a local domain whose m-adic completion R is isomorphicto C[[x, y]]/(x2y − y2n+1). (We know that such a ring exists by a theorem of Lech [Lec86].) Note that R hasexactly three minimal primes, namely yR, (x− yn)R, and (x+ yn)R.

We first consider all finitely generated R and R modules, even those with torsion. Let r be any positiveinteger, and set s = 1+2+· · ·+r. For each i ∈ 1, 2, . . . , r, s, there exist, by [HRKW08], indecomposable finitelygenerated R-modules Mi and Ni of rank (0, 0, i) and (i, i, 0), respectively. (These modules are not necessarilytorsion-free.) Since R is a domain, none of these modules is extended. However, M1 ⊕ N1, . . . ,Mr ⊕ Nr,(⊕r

i=1Mi)⊕Ns, and (⊕r

i=1Ni)⊕Ms are extended. Moreover, these modules are minimally extended and arethus extended from indecomposable R-modules. Since

M =

((r⊕i=1

Mi

)⊕Ns

)⊕((r⊕i=1

Ni

)⊕Ms

)

∼=r⊕i=1

(Mi ⊕Ni)⊕

(Ms ⊕Ns)

we see that 2, r + 1 ∈ L([M ]), and hence the elasticity of the element [M ] ∈ M(R) is at least r+12 . As r was

arbitrary, the elasticity of M(R) is infinite.The direct-sum behavior of finitely generated torsion-free modules over R isn’t so crazy. In fact, R and R

have finite representation type. The following table (cf. [Yos90], [Bae07]) gives a list, up to isomorphism, of allindecomposable finitely generated torsion-free R-modules, along with their ranks at the three minimal primes.(Note that there are 4n+ 5 indecomposable R-modules.)

module rank module rankA (1, 0, 0) E (1, 0, 1)B (0, 1, 0) Fj (0, 1, 1) 1 ≤ j ≤ nC (0, 0, 1) Gj (2, 1, 1) 1 ≤ j ≤ n− 1D (1, 1, 0) Hj (1, 1, 1) 1 ≤ j ≤ 2n+ 1

We now use this information to describe direct-sum behavior of finitely generated torsion-free modules overR. Since we are dealing only with torsion-free modules, we refer to the monoids C(R) and C(R) of finitelygenerated torsion-free modules over R and R, respectively. Since KRSA holds for the class of all finitelygenerated R modules, C(R) ⊆ C(R) ∼= N(4n+5)

0 . If M is any finitely generated torsion-free R-module, we have

M ∼= Aa ⊕Bb ⊕ Cc ⊕Dd ⊕ Ee ⊕

n⊕j=1

Ffj

j

⊕n−1⊕j=1

Ggj

j

⊕2n+1⊕

j=1

Hhj

j

.

Since R is a domain, L is extended if and only if rank(L) is constant; i.e., if and only if

n−1∑j=1

gj + a+ e =n∑j=1

fj + b and b+ d = c+ e.

Thus, represented as a Diophantine monoid, C(R) = (Ker(A) ∩ N2n+40 )⊕ N2n+1

0 where

A =[1 −1 0 0 1 −1 −1 · · · −1 1 · · · 10 1 −1 1 −1 0 0 · · · 0 0 · · · 0

].

20


This is a 2× (2n+ 4) matrix whose columns correspond to the “unknowns” a, b, c, d, e, f1, . . . , fn, g1, . . . , gn−1.The divisor classes containing primes are the images of the standard basis vectors of N2n+2

0 , that is the sixdistinct columns of A. Thus the block monoid of C(R) is

B(C(R)) ∼= Ker[1 −1 0 0 1 −10 1 −1 1 −1 0

]⋂N6

0. (4.5)

The irreducible elements of this monoid are:

h1 = [1 0 0 0 0 1] h2 = [0 1 0 0 1 0] h3 = [0 0 1 1 0 0]h4 = [1 1 1 0 0 0] h5 = [0 0 0 1 1 1]

Note that h1 + h2 + h3 = h4 + h5, and hence this monoid is not half-factorial. To show that the elasticityis exactly 3/2, begin with

∑6i=1 aihi =

∑6i=1 bihi with each ai and bi nonnegative integers. We leave it as an

exercise to show that assuming(∑6

i=1 ai

)/(∑6

i=1 bi

)> 3/2 leads to a contradiction. Alternatively, one can

apply an algorithm from [Kat04] to compute the elasticity of C(R). Even though KRSA fails for the class oftorsion-free modules over R, the failure is fairly mild: If an R-module M can be expressed as the direct sum ofs indecomposable modules and as the direct sum of t indecomposable modules, then the ratio t/s never exceeds3/2. As we will see in further examples, much worse direct-sum behavior can occur.

In the final example, we investigate the highly non-unique direct-sum decompositions that can occur withtorsion-free modules over a ring with infinite representation type. In stark contrast with the ring in Example4.16, the elasticity is infinite.

Example 4.17. Let (R,m) be a one-dimensional Noetherian local domain whose m-adic completion R isisomorphic to the ring

R ∼=C[[x, y]]

y(x3 − y7).

In this example we apply the same tricks as in Example 4.16. The minimal prime ideals of R are (x3 − y7)Rand yR. To illustrate how badly KRSA fails over R, we need only consider direct-sum relations on some of thefinitely generated torsion-free modules. It was shown in [Sac10] and [BS] that, for each positive integer n, thereexist a positive integer t and an indecomposable finitely generated torsion-free R-module Mn of rank (0, n), andalso an indecomposable finitely generated torsion-free R-module Nn of rank (t+ n, n). Therefore

Ker([

0 1 −1 2 −2 3 −3 · · ·])∩ N(ℵ0)

0

is a submonoid of C(R).By Theorem 4.8 Cl(C(R)) ∼= Z. Moreover, every element of Z appears as a column of the defining matrix.

Therefore Theorem 4.13implies that, given any non-empty finite subset S of integers greater than 1, there existsa finitely generated torsion-free R-module M that can be expressed, for every s ∈ S, as the direct sum of sindecomposable finitely generated torsion-free R-modules. By Corollary 4.14, ρ(C(R)) =∞.

Calculations similar to those in Examples 4.16 and 4.17 have been done for all one-dimensional local ringswith finite representation type (cf. [Bae07] and [BL]) and for some one-dimensional local rings with infiniterepresentation type (cf. [BS]).

5 Questions

In this final section, we give a list of questions related to describing the direct-sum decomposition of modulesvia the study of certain commutative monoids. Questions (1) and (2) can certainly be approached by interestedundergraduates, while Questions (3) and (4) are likely to be very difficult problems.

(1) What monoids can be realized as Diophantine monoids whose defining matrices consist only of entries inthe set −1, 0, 1, and what properties to such monoids have? Since it is easy to construct indecomposablemodules corresponding to columns of this form, how can this information be used to describe direct-sumdecompositions?

21


(2) Certain invariants of the monoidM(R), such as elasticity and the divisor class group, have been studied. Inthe study of commutative monoids, other important invariants such as the ω invariant, catenary degree,critical number, and delta sets are used to further refine descriptions of factorization. What do theseinvariants tell us about direct-sum decompositions?

(3) In order to get a complete description of all possible direct-sum decompositions of torsion-free modulesover one-dimensional local rings, one needs to have a description of all ranks that can occur for inde-composable torsion-free modules over complete rings of this type. All ranks have been computed whenR — equivalently R — has finite representation type. At the other extreme, when R/P has infinite rep-resentation type for every minimal prime ideal P , Crabbe and Saccon [CS] have shown that every tupleoccurs as the rank of some indecomposable torsion-free module. Partial results are known in some of theintermediate cases, where R has infinite representation type but R/P has finite representation type for atleast one minimal prime P . Given a one-dimensional local Noetherian ring with reduced completion R,what tuples occur as ranks of indecomposable R-modules?

(4) Much of our focus has been on one-dimensional rings. However, the same tools from factorization theorycan be applied to study direct-sum decompositions over a ring R of dimension larger than one if it candetermine which R-modules are extended from R-modules. This information is known for certain classes oftwo-dimensional local integral domains and, in this setting, the monoidM(R) was considered in [Bae09].Are there other classes of local rings for which we can determine which R-modules are extended fromR-modules? If so, what factorization-theoretic information can be gleaned from this structure?

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Factorization theory and decompositions of modules · of modules and factorization in integral domains. 2 Rings and modules Throughout this paper Rwill denote a commutative, Noetherian

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