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Arithmetic of Krull monoids
Alfred Geroldinger
Institute of Mathematics and Scienti�c Computing
University of Graz
Additive Combinatoricsin Paris, 2012
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Zero-Sum Sequences form a semigroup
Let G = (G ,+) be an additively written abelian group.
• A sequence S = (g1, . . . , g`) over G : �nite, unorderedsequence of terms from G , repetition allowed.
• S has sum zero if σ(S) = g1 + . . .+ g` = 0.
• The set of sequences forms a semigroup with concatenation ofsequences as the operation: (F(G ), ·). Indeed
if T = (h1, . . . , hk), then ST = (h1, . . . , hk , g1, . . . , g`) .
• The set of zero-sum sequencesB(G ) = {S ∈ F(G ) | σ(S) = 0} is a subsemigroup.
• Unit element in B(G ): empty sequence.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
• Divisibility in F(G ): T | S ⇐⇒ T is a subsequence of S⇐⇒ there exists an T ′ such that S = TT ′.
• If S and T have sum zero, then T ′ has sum zero: in otherwords,
T | S in B(G ) if and only if T | S in F(G ) .
PROBLEM: Factorize a zero-sum sequence S into minimalzero-sum sequences, say
S = S1 · . . . · Sk ,
and study the set L(S) of all possible k .
WHY ?
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
De�nition of Krull monoids
A monoid H is a Krull monoid if one of the following equivalentconditions is satis�ed:
• There is a homomorphism ϕ : H → F(P) such that, for alla, b ∈ H, we have
a | b in H if and only if ϕ(a) |ϕ(b) in F(P) .
(ϕ is called a divisor homomorphism).
• H is completely integrally closed and satis�es theascending chain condition on divisorial ideals.
Then
• N = F(P); Every monoid F(P) (factorial monoid) is Krull.
• B(G ) ↪→ F(G ) is Krull.
• R Dedekind ⇐⇒ Ideals = F(spec•(R)); R Dedekind impliesϕ : R• → Ideals : a 7→ aR is a divisor homomorphism.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
More Examples: Ring theory
Domains: Let R be a domain. Then (R•, ·) is a monoid.
• R is a Krull domain if and only if R• is a Krull monoid.
• The monoid algebra R[H] is Krull if and only if R is Krull andH is Krull (for a reduced monoid H; Chouinard 1981)).
• Integrally closed noetherian domains are Krull.• One-dimensional Krull: Dedekind domains• Higher dimensional Krull: a�ne K -algebras, rings of invariants
Submonoids of Domains: Regular congruence monoids in Krulldomains are Krull.Example: Let R be a non-principal order in a Dedekind domain R
with conductor f = (R :R). Then
H = {a ∈ R• | aR + f = R}
is a Krull monoid.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
More Examples: Module Theory
Let R be a commutative ring, C a class of R-modules closed underisomorphisms, �nite direct sums and direct summands.For a module M, let [M] denote its isomorphism class.Then
H = {[M] | M ∈ C}is an additive semigroup where addition is de�ned as
[M] + [N] = [M ⊕ N] .
If C is the class of noetherian R-modules, then every M ∈ C is a�nite direct sum of indecomposable R-modules.
Theorem
• (Krull-Schmidt 1930s) If EndR(M) is local for all M ∈ C(e.g., all M have �nite length), then H is factorial.
• (Wiegand, Facchini, 2000s) If EndR(M) is semilocal for all
M ∈ C, then H is a Krull monoid.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Class groups I
There are equivalent:
• H is a Krull monoid.
• H has a divisor theory. This is a divisor homomorphismϕ : H → D = F(P) such that, for every p ∈ P , there is a �nitesubset X ⊂ H such that p = gcd
(ϕ(X )
).
Divisor theories are unique (up to isomorphism).If ϕ = (H ↪→ D), then
G = q(D)/q(H) = {aq(H) = [a] | a ∈ D}
is called the class group of G , and
GP = {[p] | p ∈ P} ⊂ G
denotes the set of classes containing prime divisors.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Class Groups II
FACT 1: A reduced Krull monoid is uniquely determined by itsclass group and by the number of prime divisors in the classes.
Note: If R is a ring of integers in an algebraic number �eld. Thenevery class contains in�nitely many prime ideals. Thus Fact 1justi�es the
Classical Philosophy: The class group determines the arithmetic.
FACT 2: Let G = (G ,+) be an abelian group with |G | 6= 2. ThenB(G ) is the unique Krull monoid with class group (isomorphic to )G where every class contains precisely one prime divisor.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Sets of lengths: First observations
Let H be a Krull monoid. Every nonunit a ∈ H can be written as
a = u1 · . . . · uk where u1, . . . , uk ∈ A(H) .
• A(H) is the set of atoms (irreducible elements).
• k is called the length of the factorization.
•LH(a) = {k | a has a factorization of length k} ⊂ N
is the set of lengths of a.
• All sets of lengths are �nite: if H ⊂ F(P) anda = p1 · . . . · p` ∈ H, then max L(a) ≤ `.
• L(H) = {L(a) | a ∈ H} is the system of all sets of lengths.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Sets of lengths: Classical results
Let H be a Krull monoid with class group G , and suppose thatevery class has a prime divisor. We have
• (19th Century) H is factorial i� |G | = 1.
• (Carlitz 1960) |G | ≤ 2 if and only if |L(a)| = 1 for all a ∈ H.
• (�liwa 1982) If |G | ≥ 3, then for every m ∈ N, there is anam ∈ H with |L(am)| = m.
Note: If a = u1 · . . . · uk = v1 · . . . · v` with ui , vj ∈ A(H), then
am = (u1 · . . . · uk)i (v1 · . . . · v`)m−i for all i ∈ [0,m]
and henceL(am) ⊃ {`m + i(k − `) | i ∈ [0,m]} .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Narkiewicz 1979: There is a Transfer Homomorphism
from a general Krull monoid to B(G )
Let H be a Krull monoid, ϕ = (H ↪→ F(P)) its divisor theory, G itsclass group, and GP = G . Consider
H −−−−→ F(P)
β
y yβ̃
B(G ) −−−−→ F(G )
Then β = β̃ | H is a transfer hom., where β̃ maps an element
a = p1 · . . . · pl ∈ F(P) to β̃(a) = [p1] · . . . · [pl ] ∈ F(G ) ,
and we have
• a ∈ H ⇐⇒ β̃(a) is a zero-sum sequence.
• LH(a) = LB(G)(β(a))
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Main Question
Let H be a Krull monoid with class group G such that every classcontains a prime divisor.
FACT : Sets of lengths in H can be studied in theassociated monoid of zero-sum sequences B(G ).
HOW DO SETS OF LENGTHS LOOK LIKE ?
How do we get started ?We have
UV = W1 · . . . ·Wm
(g1, . . . , gk)(h1, . . . , hl ) = (g1, g5, h3)(g2, h4, h7, h7) . . . . . . ,
where U,V ,Wj are atoms in B(G ).In other words, they are minimal zero-sum sequences over G .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Let G0 ⊂ (G ,+) and S = g1 · . . . · gl ∈ F(G0) a sequence over G0.
• |S | = l is the length of S ,
• σ(S) = g1 + . . .+ gl ∈ G is the sum of S ,
• Σk(S) = {∑
i∈I gi | ∅ 6= I ⊂ [1, l ] with |I | = k} ⊂ G isthe set of k-term subsums of S ,
• Σ(S) =∑
k≥1 Σk(S) is the set of (all) subsums of S .
Furthermore,
• S is called a zero-sum sequence of σ(S) = 0,
• S is called zero-sum free if 0 /∈ Σ(S),
• B(G0) ↪→ F(G0) monoid of zero-sum sequences over G0,
• A(G0) := A(B(G0)
)the set of atoms over G0,
in other words, the minimal zero-sum sequences over G0,
• D(G0) = max{|U| | U ∈ A(G0)} ∈ N isDavenport constant of G0. Equivalently, we have
• D(G ) is the smallest integer ` such that every sequence oflength at least ` has a nontrivial zero-sum subsequence.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Methods: Group algebras
Let R be a domain. Let d(G ,R) denote the largest integer l ∈ Nhaving the following property:
There is some sequence S = g1 · . . . · g` such that
(a1−X g1)·. . .·(al−X g`) 6= 0 ∈ R[G ] for all a1, . . . , al ∈ R• .
If S is zero-sum free, then all these expressions are nonzero(the coe�cient of X 0 is nonzero), and hence
D(G )− 1 ≤ d(G ,R) .
Note: Let G ,G ′ be two �nite abelian groups.
• If K is a splitting �eld, then K [G ] ∼= K |G | ∼= K [G ′], but wemay have D(G ) 6= D(G ′). HOWEVER,
• (Higman) Z[G ] ∼= Z[G ′] implies that G ∼= G ′.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Methods: Alon's Combinatorial Nullstellensatz 1999
Let R be a domain, A1, . . . ,An ⊂ R �nite nonempty subsets,f ∈ R[X ], and gi =
∏a∈Ai
(Xi − a) ∈ R[Xi ] for all i ∈ [1, n].Then the following statements are are equivalent:
• f (a1, . . . , an) = 0 for all (a1, . . . , an) ∈ A1 × . . .× An.
• There are h1, . . . , hn ∈ R[X ] with deg(gi ) + deg(hi ) ≤ deg(f )such that
f =n∑
i=1
gihi .
Many applications; many variations and extensions (Ball-Serra,Kouba, Heinig, Michalek,..)
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Methods: Addition Theorems
Theorem
Let A,B ⊂ G be �nite nonempty subsets, H = Stab(A + B) be the
stabilizer of the sumset A + B, and ΦH : G → G/H.
(a) (Kneser) |ΦH(A) + ΦH(B)| ≥ |ΦH(A)|+ |ΦH(B)| − 1.
(b) (Kemperman-Scherk)
|A + B| ≥ |A|+ |B| −min{ rA,B(g) | g ∈ A + B} .(c) (Grynkiewicz 2005) Partition Theorem.
(d) (DeVos-Goddyn-Mohar 2009) A generalization of Kneser's ....
Corollary (to (b))
If S = S1S2 is zero-sum free, then |Σ(S)| ≥ |Σ(S1)|+ |Σ(S2)|.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Methods: Inductive Method I
Corollary (to Grynkiewicz or DeVos-Goddyn-Mohar)
Let S ∈ F(G ), n ∈ [1, |S |], and H = Stab(Σn(S)). Then
|Σn(S)| ≥( ∑g∈G/H
min{n, vg(φH(S)
)} − n + 1
)|H| .
To study a given sequence S = g1 · . . . · g` proceed as follows:
• Find a suitable subgroup K ⊂ G , consider the naturalepimorphism ϕ : G → G/K , and ϕ(S) = ϕ(g1) · . . . · ϕ(g`).
• Consider a factorization S = S0S1 · . . . · Sk such that |Si | issmall and ϕ(Si ) ∈ B(G/K ) for all i ∈ [1, k].
• Investigate the sequences T = σ(S1) · . . . · σ(Sk) ∈ F(K ) andS0T ∈ F(G ). Clearly, if S is zero-sum free, then S0T iszero-sum free too.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Methods: Inductive Method II
Let η(G ) resp. s(G ) denote the smallest integer l ∈ N with thefollowing property:
• Every sequence S over G of length |S | ≥ l has azero-sum subsequence T of length |T | ∈ [1, exp(G )] resp.zero-sum subsequence T of length |T | = exp(G )
For the Erd®s-Ginzburg-Ziv constant s(G ) we have
s(G ) ≥ η(G ) + exp(G )− 1 . (∗)
Theorem
• (Gao) η(G ) ≤ |G | and s(G ) ≤ |G |+ exp(G )− 1.If exp(G ) ≤ 4 or exp(G ) is large, then equality in (∗)
• If G = Cn1 ⊕ Cn2 , then s(G ) = η(G ) + n2 − 1 = 2n1 + 2n2 − 3(n1 = 1: EGZ-Theorem; n1 = n2: Reiher).
• Groups of higher rank: remember Bhowmik's talk (yesterday).
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
The Davenport constant again
Let G = Cn1 ⊕ . . .⊕ Cnrwhere 1 < n1 | . . . | nr . Then
D∗(G ) := 1 +r∑
i=1
(ni − 1) ≤ D(G ) .
• (Olson, Kruyswijk 1960s): Equality for p-groups and rank 2groups.
• Very limited progress for groups of small rank and groups closeto p-groups: Bhowmik, Gao, Schlage-Puchta, Schmid,....
• Conjecture: Equality for rank three groups and for G = C rn .
• (G.+Schneider) For every r ≥ 4 there are in�nitely manygroups G of rank r for which inequality holds.
• (Liebmann-G.-Philipp, 2012) Inequality for G = C2 ⊕ C r2n with
n odd, r ≥ 4.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Structure of minimal zero-sum sequences: Cyclic groups
Theorem ( Savchev-Chen 2007)
Let G be a cyclic group of order |G | = n.
1. Let S ∈ F(G ) be zero-sum free of length l ≥ n+12 .
Then there are g ∈ G with G = 〈g〉 and 1 = n1 ≤ . . . ≤ n`with m = n1 + . . .+ n` < |G | such that
S = (n1g)(n2g) · . . . · (nlg) and Σ(S) = {g , 2g , . . . ,mg} .
2. Let U ∈ B(G ) be a minimal zero-sum sequence of length
l ≥⌊n2
⌋+ 2. Then there is some g ∈ G with G = 〈g〉 such
that, with 1 = n1 ≤ n2 ≤ . . . ≤ nl ,
S = (n1g)(n2g) · . . . · (nlg) and n1 + . . .+ nl = n .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Structure of minimal zero-sum sequences: Rank two groups
Let G = Cm ⊕ Cmn with m, n ∈ N and m ≥ 2. A sequence S overG of length D(G ) = m + mn − 1 is a minimal zero-sum sequence ifand only if it has one of the following two forms :
•
S = eord(e1)−11
ord(e2)∏ν=1
(xνe1 + e2) , where
{e1, e2} is a basis of G , x1, . . . , xord(e2) ∈ [0, ord(e1)− 1], andx1 + . . .+ xord(e2) ≡ 1 mod ord(e1).
•
S = g sm−11 g(n−s)m+12
m−1∏ν=1
(−xνg1 + g2) , where
{g1, g2} is a generating set of G with ord(g2) = mn, s ∈ [1, n],x1, . . . , xm−1 ∈ [0,m − 1], x1 + . . .+ xm−1 = m − 1, and(s = 1 or mg1 = mg2
).
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Proof of the Structural Result
The result was proved in four papers:
• The prime case G = Cp ⊕ Cp: Christian Reiher,to appear in Journal of the London Math. Soc.
• Multiplicity by two: If the Structural Result holds for Cn ⊕ Cn,then it holds for C2n ⊕ C2n.W. Gao and A. G.: Integers 3 (2003)
• Multiplicity by odd numbers:
W. Gao and A.G. and D.J. Grynkiewicz:Acta Arith. 141 (2010), 103 � 152
• General Groups of Rank Two: W.A. Schmid:Acta Arith. 143 (2010), 333 � 343.
Partial results and special cases: Bhowmik, Halupczok,Schlage-Puchta; Fang Chen, Svetoslav Savchev.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Set of distances: De�nition
• For a �nite subset L = {a1, . . . , at} ⊂ Z with a1 < . . . < at let
∆(L) = {aν+1 − aν | ν ∈ [1, t − 1]} ⊂ N
denote the set of (successive) distances of L.
•∆(H) =
⋃a∈H
∆(L(a)
)⊂ N
denotes the set of distances of H.
By de�nition we have
• ∆(H) = ∅ ⇐⇒ all sets of lengths are singletons,
• ∆(H) = {d} ⇐⇒ All sets of lengths are arithmeticalprogressions with di�erence d .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Set of distances: Properties
For a subset G0 ⊂ (G ,+) we set
∆(G0) = ∆(B(G0)
).
• For every monoid H, we have min∆(H) = gcd∆(H).
• max∆(G0) ≤ D(G0)− 2.
• (Carlitz 1960) If |G | ≤ 2, then ∆(G ) = ∅.
• If 2 < |G | <∞, then ∆(G ) is an interval with 1 ∈ ∆(G ).
• If G is in�nite, then ∆(G ) = N.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Re�ned elasticities
For a monoid H and k ∈ N, let
ρk(H) = max{` | there exists an equation u1 · . . . · uk = v1 · . . . · v`}
A simple counting argument shows that for ρk(G ) := ρk
(B(G )
)• ρ2k(G ) = kD(G ).
• 1 + kD(G ) ≤ ρ2k+1(G ) ≤ kD(G ) +⌊D(G)2
⌋.
The upper bound can be realized for "many" groups withD∗(G ) = D(G ). However, using Savchev-Chen
Theorem (Gao+G.)
Let G be a cyclic group. Then for every k ∈ N, we have
ρ2k+1(G ) = kD(G ) + 1 .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
AAMPs: Almost Arithmetical Multiprogressions
Let d ∈ N, M ∈ N0 and {0, d} ⊂ D ⊂ [0, d ].A subset L ⊂ Z is called an
AAMP with di�erence d , period D, and bound M,if
L = y + (L′ ∪ L∗ ∪ L′′) ⊂ y +D + dZ
where
• y ∈ Z is a shift parameter,
• L′ ⊂ [−M,−1] and L′′ ⊂ max L∗ + [1,M], are the (short)beginning and end parts of L,
• L∗ is (the ) �nite nonempty (important and long middle part)with min L∗ = 0 and
L∗ = (D + dZ) ∩ [0,max L∗] .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Structure Theorem - Realization Theorem
Theorem (Freiman, G., Halter-Koch,...)
Let H be a Krull monoid with �nite class group G.
Then there is an M = M(G ) ∈ N0 such that every set of lengths is
an AAMP with di�erence d ∈ ∆(G ) and bound M.
Theorem (Schmid 2009)
Let M ∈ N0 and ∆ ⊂ N be a �nite nonempty set. Then there
exists a Krull monoid H with �nite class group such that :
For every AAMP L with di�erence d ∈ ∆ and bound M there is
some yH,L ∈ N such that
y + L ∈ L(H) for all y ≥ yH,L.
Indeed, there exists an algebraic number �eld such that its ring of
integers has this property.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
An interlude
Theorem (Kainrath)
Suppose that H is a Krull monoid with in�nite class group, and
that every class contains a prime divisor.
Then every �nite subset L ⊂ N≥2 can be realized as a set of
lengths of H.
Theorem (Frisch)
Suppose that H = Int(Z) is the ring of integer valued polynomials.
Then every �nite subset L ⊂ N≥2 can be realized as a set of
lengths of H.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
On the parameters M and ∆
The bound M: Only very rough upper bounds are known.
On the set of di�erences ∆ in long AAMPs:By the proof of the Structure Theorem, this set ∆ is equal to
∆∗(G ) = {min∆(G0) | G0 ⊂ G ,∆(G0) 6= ∅} ⊂ ∆(G ) .
There is some work on ∆∗(G ). Recall that |G | ≥ 3.
SIMPLE FACTS:
• 1 ∈ ∆∗(G ).
• If g ∈ G with ord(g) ≥ 3, then ord(g)− 2 ∈ ∆∗(G ).
• If r(G ) ≥ 2, then [1, r(G )− 1] ⊂ ∆∗(G ).
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Cross numbers and ∆(G0)
For a sequence S = g1 · . . . · gl let
k(S) =∑̀i=1
1ord(gi )
denote the cross number of S ,
and de�ne (Krause 1984)
K(G ) = max{k(S) | S ∈ A(G )} denote the cross number of G ,
For p-groups, the precise value of K(G ) is known !!There are equivalent: (Skula, Zaks 1970s)
• ∆(G0) = ∅.• k(U) = 1 for all U ∈ A(G0).
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
On ∆∗(G )
Theorem
• (Schmid) max∆∗(G ) = max{exp(G )− 2,m(G )} and
m(G ) ≤ max{r∗(G )− 1,K(G )− 1} .
If G is a p-group, then K(G ) < r∗(G ), m(G ) = r(G )− 1and thus max∆∗(G ) = max{exp(G )− 2, r(G )− 1}.
• There are results for groups with large rank and for groups
with large exponent.
• (G. + Hamidoune 2002)
max(∆∗(Cn) \ {n − 2}
)=⌊n2
⌋− 1 .
• (Plagne + Schmid 2012) Detailed study of ∆∗(Cn).
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Arithmetical Characterizations of Class Groups
Let R be a ring of integers with class group G .
Classical Philosophy: The class group determines the arithmetic.
• R is factorial i� |G | = 1.
• Carlitz 1960: |L(a)| = 1 for all non-zero a ∈ R i� |G | ≤ 2.
Narkiewicz 1970s:
• What about the converse:Do arithmetical phenomena characterize the class group ?
• Give arithmetical characterizations of the class group.
Positive answers by Kaczorowski, Halter-Koch, Rush, and others.Let
L(G ) := L(B(G )
)= {L(A) | A is a zero-sum sequence over G}
denote the system of sets of lengths over G .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Systems of Sets of Lengths
PROBLEM: Given two �nite abelian groups G ,G ′ such thatL(G ) = L(G ′). Does it follow that G ∼= G ′?
Apart from
L(C1) = L(C2) and L(C3) = L(C2 ⊕ C2) ,
we get, by using ALL what we had so far,
Theorem
Let G and G ′ be �nite abelian groups with D(G ) ≥ 4, and suppose
that L(G ) = L(G ′).
1. If G is cyclic or an elementary 2-group, then G ∼= G ′.
2. (Schmid) If G = Cn ⊕ Cn with n ≥ 3, then G ∼= G ′.
3. (Baginski-G.-Grynkiewicz-Philipp) If G has rank two and
D(G ′) = D∗(G ′), then G ∼= G ′.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Quantitative aspects of non-unique factorizations
Let R be the ring of integers in an algebraic number �eld, G itsideal class group and |G | ≥ 3.
Narkiewicz 1960s: Study the asymptotic behaviour of the followingcounting functions:
Fk(x) = #{aR | a ∈ R•, (R :aR) ≤ x and |Z(a)| ≤ k}Gk(x) = #{aR | a ∈ R•, (R :aR) ≤ x and |L(a)| ≤ k}Mk(x) = #{aR | a ∈ R•, (R :aR) ≤ x and max L(a) ≤ k}
Using the Tauberian Theorem of Ikehara and Delange, one canshow that that all these quantities behave for x →∞asymptotically like
x(log x)−A(log log x)B
with exponents A, B ∈ R>0 depending only on G .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
FURTHER DIRECTIONS:
1. Better asymptotics: Recent progress by Kaczorowski, Perelli,Radziejewski.
2. Determine the exponents A,B .
3. Non-principal orders.
In particular, we have
Fk(x) ∼ x(log x)−1+1/|G |(log log x)Nk(G) , where
Nk(G ) is the maximal length of an ordered zero-sum sequencehaving at most k distinct factorizations. We have
•∑r
i=1 ni ≤ N1(G ) ≤ N2(G ) ≤ .....• Conjecture(Narkiwiecz-Sliwa 1982):
∑ri=1 ni = N1(G )
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Theorem (Gao et al., 2011,2012,201?)
(a) If k is small with respect to n, then Nk(Cn) = n
(use Savchev-Chen).
(b) N1(Cn ⊕ Cn) = 2n.
(c) If D(C 3n1
) ≤ 3n1 − 1, then N1(Cn1 ⊕ Cn2) = n1 + n2.
Method for (b):
• Use an addition theorem and group algebras to show thatN1(Cp ⊕ Cp) = 2p.
• Show Multiplicity:
N1(Cm⊕Cm) = 2m + N1(Cn⊕Cn) = 2n⇒ N1(Cmn⊕Cmn) = 2mn .
To do so, use the structural description of minimal zero-sumsequences of maximal length over groups of rank two.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
How does the typical set of lengths look like?
Theorem (G. + Halter-Koch 2006)
(a) Let S = g1 · . . . · g` be a zero-sum sequence over G
such that
{g1, . . . , g`} ∪ {0} ⊂ G is a subgroup .
Then the set of lengths L(S) is an arithmetical progression
with di�erence 1.
(b) Let R be a ring of integers with class group G.
limx→∞
#{aR | (R :aR) ≤ x and L(a) is an AP as above}#{aR | (R :aR) ≤ x}
= 1 .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Distance between factorizations
Let b ∈ H, and let z , z ′ ∈ Z(b) be two factorizations, say
z = w1 · . . . · wn u1 · . . . · uk , z ′ = w1 · . . . · wn v1 · . . . · v`
where all ui , vj ,wk are atoms and ui , vj are pairwise non-associated.
Then
d(z , z ′) = max{k , `} ∈ {0}∪N≥2 the distance between z and z ′.
Note: Let m ∈ N, a = u1 · . . . · uk = v1 · . . . · v`, and study am:
d(
(u1 · . . . · uk)m, (v1 · . . . · v`)m)
= max{km, `m} ≥ 2m ,
but
Z(am) ⊃ {zi = (u1 · . . . · uk)i (v1 · . . . · v`)m−i | i ∈ [0,m]}
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Catenary degree
For a ∈ H, let c(a) ∈ N0 ∪ {∞} denote the smallestC ∈ N0 ∪ {∞} with the following property:
For any two factorizations z , z ′ ∈ Z(a), there exists a �nitesequence of factorizations of a
z = z0, z1, . . . , zm = z ′ concatenating z and z ′ in Z(a)
such that
d(zi−1, zi ) ≤ C for all i ∈ [1,m] .
Then
c(H) = sup{c(a) | a ∈ H} is the catenary degree of H .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Arithmetical Properties of the catenary degree
By de�nition, we have• c(H) = 0 ⇐⇒ H is factorial.• If H is not factorial, then 2 + sup∆(H) ≤ c(H). In particular,c(H) = 3 =⇒ all sets of lengths are AP with di�erence 1.
Setc(G ) := c
(B(G )
).
Theorem
• c(G ) ≤ D(G ), and equality holds if and only if
G is either cyclic or an elementary 2-group.
• (G.+Grynkiewicz+Schmid) If D∗(G ) = D(G ), then
k(G ) = 2 + max∆(G ) = c(G ) , where k(G ) =
max{k | u1u2 = v1 · . . . · vk , no lengths between 2 and k}.
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Outline
De�nitions and Examples
Problems
Monoids of Zero-sum sequences
Sets of Lengths
Quantitative Aspects
Beyond Sets of Lengths
Beyond Krull monoids
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
On not integrally closed noetherian domains
Let R be a noetherian domain with quotient �eld q(R) = K , andR ⊂ K the integral closure of R .
FACTS:
• Every non-zero non-unit of R can be written as a product ofatoms.
• R is Krull with dim(R) = dim(R), andR is Krull if and only if R = R .
• R is a �nitely generated R-module if and only ifthe conductor f = (R :R) = {a ∈ R | aR ⊂ R} 6= {0}.
• Example: Non-principal orders in algebraic number �elds:so in quadratic number �elds
R = Z[f ω] ⊂ R = Z[ω] ⊂ K = Q(ω) .
De�nitions Problems Zero-Sums Sets of Lengths Quantitative Aspects Beyond I Beyond II
Arithmetic of not integrally closed noetherian domains
Let R be a noetherian domain with f = (R :R) 6= {0}.Suppose that the class group C(R) and R/f are �nite. Then
• The catenary degree c(R) and the set of distances ∆(R) are�nite.
• There is an M ∈ N0 such that every set of lengths is anAAMP with di�erence d ∈ ∆(R) and bound M.
• There is a �niteness criterion for the re�ned elasticities.For non-principal orders it runs as follows:
• One (equivalently, all) invariants ρk(R) are �nite.• For every nonzero p / R there is precisely one P / R with
P ∩ R = p.
METHOD: Study a transfer homomorphism θ : R → B , where Bis a monoid having similar algebraic properties as R , but which issimpler from a combinatorial point of view.