May 07, 2025
For a sequence \(S\) over a finite abelian group, let \(MZ(S)\) denote the length of the shortest nonempty zero-sum subsequence of \(S\). We prove that if \(G\) is finite abelian of order \(n\) and \(S\) has length \(n\), then \(MZ(S)\le n-|\mathop{\mathrm{supp}}(S)|+1\). The same bound holds for every sequence of length at least \(|G|\). In cyclic groups we combine this elementary support bound with the Savchev–Chen structure theorem for long zero-sumfree sequences and obtain the sharper estimate \(MZ(S)\le n-t(t-1)/2\), where \(t=|\mathop{\mathrm{supp}}(S)|\), whenever \(S\) has length \(n\) over \(C_n\) and \(MZ(S)-1>n/2\). As a consequence, every length-\(n\) sequence over \(C_n\) with support size \(3\) has a zero-sum subsequence of length at most \(n-3\), and this is sharp for \(n\ge 5\). We also give an arithmetic application to products of prime ideals in a number field, phrased in the standard class-group and block-monoid setting and a corresponding cyclic class-group sharpening.
Zero-sum theory often asks how long a sequence over a finite abelian group must be before it is forced to contain a zero-sum subsequence. The Davenport constant is the classical example, and it remains central both in additive combinatorics and in factorization theory; see, for example, the survey of Gao and Geroldinger [1] and the monograph of Geroldinger and Halter-Koch [2]. In this paper we look at a slightly more local question. Once a sequence is already in front of us, how does the number of distinct group elements appearing in it constrain the length of its shortest zero-sum subsequence?
To make this precise, for a sequence \(S\) over a finite abelian group we write \(MZ(S)\) for the minimum length of a nonempty zero-sum subsequence of \(S\). The support \(\mathop{\mathrm{supp}}(S)\) is the set of elements occurring in \(S\). Our first result is the support-sensitive bound \[\label{eq:thm31} MZ(S)\le |G|-|\mathop{\mathrm{supp}}(S)|+1\tag{1}\] for every length-\(|G|\) sequence over a finite abelian group \(G\). The proof is short: remove one occurrence of many distinct support elements, apply the elementary lower bound \(|\Sigma(P)|\ge |P|\) to the remaining zero-sumfree sequence, and count inside \(G\setminus\{0\}\). The same argument also gives the natural length-at-least-\(|G|\) corollary by passing to a support-preserving subsequence of length \(|G|\).
The point of view is close to the support-sensitive literature on subsequence sums. Eggleton and Erdős [3] gave early lower bounds for the set of subsequence sums of zero-sumfree sequences, and Freeze and Smith [4] proved the sharper estimate \[|\Sigma(P)|\ge |P|+|\mathop{\mathrm{supp}}(P)|-1\] for zero-sumfree sequences. More recent work, such as Wang–Chao–Peng [5], develops this support-sensitive direction further. Our main bound uses only the elementary part of this circle of ideas, but it is naturally read as a sequence-level companion to those estimates: larger support leaves less room for all zero-sums to be long.
The second part of the paper keeps the group cyclic and uses genuinely cyclic structure. Savchev and Chen proved that every sufficiently long zero-sumfree sequence in \(C_n\) can be put, after multiplication by a unit, into a form whose least positive representatives have total sum less than \(n\) [6]. Combining this with the finite-abelian support bound gives \[\label{eq:thm41} MZ(S)\le n-\binom{|\mathop{\mathrm{supp}}(S)|}{2}\tag{2}\] whenever \(S\) has length \(n\) over \(C_n\) and \(MZ(S)-1>n/2\). In particular, the support-three case has the exact extremal value \(n-3\) for \(n\ge 5\), witnessed by the sequence \(g^{[n-2]}(2g)(3g)\).
Finally we record an arithmetic application. We formulate it for products of prime ideals, which is the setting needed for the direct ideal-factorization argument, or equivalently through the usual transfer from a Krull monoid to the block monoid over the class group. If \(K\) is a number field with class group of order \(h\), and if \(m\ge h\) prime ideals occupy \(t\) ideal classes, then some subproduct of at most \(h-t+1\) of them is principal and generated by an irreducible element. When the class group is cyclic, the cyclic zero-sum refinements above give a stronger bound in the same prime-ideal setting. The restriction to prime ideals is essential for the direct ideal-factorization proof.
Throughout, \(G\) is a finite abelian group written additively, and \(C_n\) denotes the cyclic group of order \(n\). A sequence \(S\) over \(G\) is a finite multiset of elements of \(G\), and subsequences are taken as submultisets of occurrences. We write \(T\mid S\) when \(T\) is a subsequence of \(S\), \(|S|\) for the length of \(S\), \(\mathop{\mathrm{supp}}(S)\) for the set of elements appearing in \(S\), and \(h(S)\) for the maximum multiplicity of an element in \(S\). Thus \(|\mathop{\mathrm{supp}}(S)|\) is the support size. A sequence is zero-sumfree if it has no nonempty zero-sum subsequence. For \(g\in G\) and \(k\ge 0\), the notation \(g^{[k]}\) denotes the sequence consisting of \(k\) copies of \(g\), with \(g^{[0]}\) the empty sequence.
Definition 1. Let \(S\) be a sequence over \(G\), and let \(\sigma(T)\) denote the sum of the terms of a subsequence \(T\mid S\). We set \[\Sigma(S)=\{\sigma(T):T\mid S,\;|T|\ge 1\},\] the set of sums of nonempty subsequences of \(S\). For an integer \(\ell\ge 0\), let \[\Sigma_\ell(S)=\{\sigma(T):T\mid S,\;|T|=\ell\}.\] For \(\ell\ge 1\), define \[\Sigma_{\leq \ell}(S)=\bigcup_{1\le j\le \ell}\Sigma_j(S), \qquad \Sigma_{\geq \ell}(S)=\bigcup_{j\ge \ell}\Sigma_j(S).\] In particular, \(\Sigma(S)=\Sigma_{\geq 1}(S)\).
Definition 2. For a sequence \(S\) over \(G\), define \(MZ(S)\) to be the smallest positive integer \(r\) such that \(S\) has a zero-sum subsequence of length \(r\). If no nonempty zero-sum subsequence exists, put \(MZ(S)=\infty\).
Lemma 1 (Zero-sumfree sumset growth). Let \(P\) be a zero-sumfree sequence over an abelian group. Then \[|\Sigma(P)|\ge |P|.\]
Proof. If \(P\) is empty, the assertion is trivial. Otherwise write \[P=g_1\cdots g_k\] in an arbitrary order. For \(1\le j\le k\), put \[s_j=g_1+\cdots+g_j.\] Each \(s_j\) belongs to \(\Sigma(P)\), since it is the sum of the nonempty subsequence \(g_1\cdots g_j\).
We claim that \(s_1,\ldots,s_k\) are pairwise distinct. If \(s_i=s_j\) for some \(i<j\), then \[g_{i+1}+\cdots+g_j=0,\] which is the sum of a nonempty subsequence of \(P\), contradicting zero-sumfreeness. Hence \(\Sigma(P)\) contains at least the \(k\) distinct elements \(s_1,\ldots,s_k\), and so the conclusion follows. ◻
Remark 1. Freeze–Smith prove the stronger support-sensitive estimate \[|\Sigma(P)|\ge |P|+|\mathop{\mathrm{supp}}(P)|-1\] for zero-sumfree sequences. We only use the elementary bound above in the proof of the main theorem.
We begin with the group-theoretic bound. No cyclic ordering, generator, or residue choice is used here; only the size of the ambient group enters the proof.
Theorem 2. Let \(G\) be a finite abelian group of order \(n\), and let \(S\) be a sequence over \(G\) with \(|S|=n\). Then \[MZ(S)\le n-|\mathop{\mathrm{supp}}(S)|+1.\] Equivalently, if \(MZ(S)=n-s\) for some \(s\in\{0,\ldots,n-1\}\), then \(|\mathop{\mathrm{supp}}(S)|\le s+1\).
Proof. Put \(t=|\mathop{\mathrm{supp}}(S)|\). First, \(S\) cannot be zero-sumfree. Indeed, if \(S\) were zero-sumfree, Lemma 1 would give \(|\Sigma(S)|\ge |S|=n\), while zero-sumfreeness and the convention that \(\Sigma(S)\) contains only nonempty subsequence sums would give \(\Sigma(S)\subseteq G\setminus\{0\}\), whose cardinality is \(n-1\). This is impossible. Thus \(MZ(S)\) is finite; write \(r=MZ(S)\), so \(1\le r\le n\).
If \(r=1\), then the desired inequality is \(1\le n-t+1\), which follows from \(t\le n\). If \(r=2\), then \(0\) is not a term of \(S\), for otherwise \(r=1\). Hence \(\mathop{\mathrm{supp}}(S)\subseteq G\setminus\{0\}\), so \(t\le n-1\), and again \(r\le n-t+1\).
Assume now that \(r\ge 3\) and suppose for contradiction that \(t\ge n-r+2\). Choose distinct support elements \(a_1,\ldots,a_{n-r+2}\) occurring in \(S\), and remove one occurrence of each. The remaining sequence \(P\) has length \(r-2\). Since \(|P|<r\), it is zero-sumfree. By Lemma 1, \(|\Sigma(P)|\ge r-2\).
Let \(A=\{-a_1,\ldots,-a_{n-r+2}\}\). Since \(r>1\), zero is not a term of \(S\), and hence \(A\subseteq G\setminus\{0\}\). The map \(x\mapsto -x\) is a bijection on \(G\), so \(|A|=n-r+2\). Also \(\Sigma(P)\subseteq G\setminus\{0\}\). But \[|\Sigma(P)|+|A|\ge (r-2)+(n-r+2)=n>|G\setminus\{0\}|.\] Therefore \(\Sigma(P)\cap A\ne \varnothing\). Choose \(x\in\Sigma(P)\cap A\), say \(x=-a_i\). By the definition of \(\Sigma(P)\), there is a nonempty subsequence \(Q\mid P\) with \(\sigma(Q)=x\). Adding the selected occurrence of \(a_i\), which is not an occurrence of \(P\), gives a zero-sum subsequence of \(S\) of length at most \[|P|+1=r-1,\] contradicting the definition of \(r\). Thus \(t\le n-r+1\), which is equivalent to \(r\le n-t+1\). ◻
We now present a corollary for the special case in which the length of the sequence is at least \(|G|\).
Corollary 1. Let \(G\) be a finite abelian group of order \(n\). If \(S\) is a sequence over \(G\) with \(|S|\ge n\), then \[MZ(S)\le n-|\mathop{\mathrm{supp}}(S)|+1.\]
Proof. Put \(t=|\mathop{\mathrm{supp}}(S)|\). Since \(\mathop{\mathrm{supp}}(S)\subseteq G\), we have \(t\le n\). Choose one occurrence in \(S\) of each element of \(\mathop{\mathrm{supp}}(S)\), and call the resulting subsequence \(U\). Then \(|U|=t\) and \(\mathop{\mathrm{supp}}(U)=\mathop{\mathrm{supp}}(S)\). After removing \(U\) from \(S\), at least \(n-t\) terms remain, since \(|S|\ge n\). Choose any \(n-t\) of those remaining terms, appending no terms if \(n-t=0\), and add them to \(U\). This gives an \(n\)-term subsequence \(T\mid S\) with \(\mathop{\mathrm{supp}}(T)=\mathop{\mathrm{supp}}(S)\).
By Theorem 2, \(T\) has a nonempty zero-sum subsequence of length at most \[n-|\mathop{\mathrm{supp}}(T)|+1=n-|\mathop{\mathrm{supp}}(S)|+1.\] The same subsequence is a subsequence of \(S\), proving the claim. ◻
The finite-abelian theorem is linear in the support size. In cyclic groups, long zero-sumfree sequences have much more structure, and this lets us improve the bound in the range where the shortest zero-sum subsequence is itself long. The finite-abelian theorem is used below only to obtain the preliminary numerical bound on the support; the refinement is cyclic because it relies on Savchev–Chen’s structure theorem and on least positive representatives in \(C_n\).
Theorem 3 (Cyclic long-regime refinement). Let \(S\) be a length-\(n\) sequence over \(C_n\). Put \(r=MZ(S)\) and \(t=|\mathop{\mathrm{supp}}(S)|\). If \(r-1>n/2\), then \[r\le n-\binom{t}{2}.\]
Proof. The cases \(n\le 2\) are vacuous, since Theorem 2 gives \(r\le n\) and hence \(r-1\le n/2\). Assume \(n\ge 3\). Again by Theorem 2, \[t\le n-r+1.\] Since \(r-1>n/2\), we have \(n-r+1<r-1\), and hence \(t\le r-1\).
Choose one occurrence of each support element of \(S\) and extend this choice arbitrarily to a subsequence \(T\mid S\) of length \(r-1\). This is possible because \(t\le r-1\le n\). By construction, \(\mathop{\mathrm{supp}}(T)=\mathop{\mathrm{supp}}(S)\), so \(T\) has exactly \(t\) distinct terms. Moreover, \(T\) is zero-sumfree: otherwise a nonempty zero-sum subsequence of \(T\) would be a nonempty zero-sum subsequence of \(S\) of length at most \(r-1\), contradicting the definition of \(r=MZ(S)\).
Since \(|T|=r-1>n/2\), Savchev–Chen’s theorem [6] applies. Thus there is a unit \(u\) modulo \(n\) such that the sum of the least positive representatives of the terms of \(uT\) is less than \(n\). Multiplication by \(u\) is an automorphism of \(C_n\), so it preserves zero-sumfreeness and the number of distinct terms. Since \(T\) is zero-sumfree, neither \(T\) nor \(uT\) contains \(0\); hence all these least positive representatives lie in \(\{1,\ldots,n-1\}\).
It remains only to estimate a positive integer multiset. A multiset of \(r-1\) positive integers with exactly \(t\) distinct values has total sum at least \[(r-1)+\binom{t}{2}.\] Indeed, if the distinct values are \(1\le v_1<\cdots<v_t\), then \(v_i\ge i\) for every \(i\), and the remaining \(r-1-t\) terms are each at least \(1\). Thus the total sum is at least \[(1+\cdots+t)+(r-1-t)=(r-1)+\binom{t}{2}.\] The Savchev–Chen representative sum is an integer strictly less than \(n\), so it is at most \(n-1\). Therefore \[(r-1)+\binom{t}{2}\le n-1,\] which rearranges to \(r\le n-\binom{t}{2}\). ◻
In the special case in which \(|\mathop{\mathrm{supp}}(S)| =3\), this yields the following corollary.
Corollary 2. Let \(n\ge 5\), and let \(S\) be a length-\(n\) sequence over \(C_n\) with \(|\mathop{\mathrm{supp}}(S)|=3\). Then \[MZ(S)\le n-3.\] This bound is sharp: if \(g\) is a generator of \(C_n\), then \[MZ\bigl(g^{[n-2]}(2g)(3g)\bigr)=n-3.\]
Proof. Put \(r=MZ(S)\). By Theorem 2, \[r\le n-|\mathop{\mathrm{supp}}(S)|+1=n-2.\] Assume first that \(n\ge 7\). If \(r>n-3\), then the preceding inequality forces \(r=n-2\). Hence \[r-1=n-3>n/2,\] so Theorem 3, applied with \(|\mathop{\mathrm{supp}}(S)|=3\), gives \[r\le n-\binom{3}{2}=n-3,\] a contradiction. Thus \(r\le n-3\) for \(n\ge 7\).
It remains to handle \(n=5\) and \(n=6\). For \(n=5\), if \(0\in\mathop{\mathrm{supp}}(S)\), then \(MZ(S)=1\le 2=n-3\). Otherwise \(\mathop{\mathrm{supp}}(S)\) is a three-element subset of the four nonzero elements of \(C_5\). These four elements split into two inverse pairs, so any three-element subset contains one inverse pair. The corresponding two occurrences form a zero-sum subsequence, and \(MZ(S)\le 2=n-3\).
For \(n=6\), write \(C_6=\langle g\rangle\) and identify terms with their coefficients modulo \(6\). If \(0\in\mathop{\mathrm{supp}}(S)\), then \(MZ(S)=1\le 3=n-3\). If \(\mathop{\mathrm{supp}}(S)\) contains one of the inverse pairs \(\{g,5g\}\) or \(\{2g,4g\}\), then \(S\) has a zero-sum subsequence of length \(2\).
We may therefore assume that \(\mathop{\mathrm{supp}}(S)\) contains no zero and no inverse pair. Then it must contain \(3g\) and one element from each of the pairs \(\{g,5g\}\) and \(\{2g,4g\}\). Thus the support, written in coefficients of \(g\), is one of \[\{1,2,3\},\quad \{1,3,4\},\quad \{2,3,5\},\quad \{3,4,5\}.\] Let \(m_a\) denote the multiplicity of \(ag\) in \(S\). If \(m_3\ge 2\), then \(3g+3g=0\), so \(MZ(S)\le 2\). We may assume \(m_3=1\).
For support \(\{1,2,3\}\), the subsequence \(g(2g)(3g)\) has sum \(0\). For support \(\{3,4,5\}\), the subsequence \((3g)(4g)(5g)\) has sum \(0\). For support \(\{1,3,4\}\), we have \(m_1+m_4=5\). If \(m_1\ge 2\), then \(g+g+4g=0\); if \(m_1\le 1\), then \(m_4\ge 4\), and \(4g+4g+4g=0\). For support \(\{2,3,5\}\), we have \(m_2+m_5=5\). If \(m_2\ge 3\), then \(2g+2g+2g=0\); if \(m_2\le 2\), then \(m_5\ge 3\), and \(2g+5g+5g=0\). Hence \(MZ(S)\le 3=n-3\) in every case.
It remains to prove sharpness. Let \[S_n=g^{[n-2]}(2g)(3g)\] with \(g\) a generator of \(C_n\). Since \(n\ge 5\), the elements \(g,2g,3g\) are distinct. The subsequence \(g^{[n-5]}(2g)(3g)\) has length \(n-3\) and sum \(ng=0\), so \(MZ(S_n)\le n-3\).
Now consider any nonempty subsequence of \(S_n\). It has the form \[g^{[a]}(2g)^{[\epsilon_2]}(3g)^{[\epsilon_3]},\] where \(0\le a\le n-2\) and \(\epsilon_2,\epsilon_3\in\{0,1\}\). Its length is \[L=a+\epsilon_2+\epsilon_3,\] and its coefficient of \(g\) is \[c=a+2\epsilon_2+3\epsilon_3=L+\epsilon_2+2\epsilon_3.\] If \(L<n-3\), then \(L\le n-4\). Since the subsequence is nonempty, \[1\le c\le L+3\le n-1.\] Thus \(c\) is not divisible by \(n\), and the subsequence is not zero-sum. Therefore no zero-sum subsequence has length less than \(n-3\), while one of length \(n-3\) exists. Hence \(MZ(S_n)=n-3\). ◻
Given a number field \(K\) with ring of integers \(\mathcal{O}_K\), as mentioned previously in the introduction, it is known that the Davenport constant of the class group \(Cl(K)\) of \(\mathcal{O}_K\) is related to its elasticity (see [7]), i.e. to lengths of distinct irreducible factorizations of nonzero elements in \(\mathcal{O}_K\). Motivated by this connection, we now record the arithmetic form of Corollary 1. The statement is deliberately made for prime ideals. For arbitrary integral ideals the corresponding subproduct assertion need not hold: a principal factor of a product of arbitrary integral ideals need not be a product of a subcollection of the given ideals.
Proposition 4. Let \(K\) be a number field with ring of integers \(\mathcal{O}_K\), and let \(G=\operatorname{Cl}(K)\) have order \(h\). Let \(\mathfrak p_1,\ldots,\mathfrak p_m\) be nonzero prime ideals of \(\mathcal{O}_K\) with \(m\ge h\), and suppose their ideal classes occupy \(t\) distinct classes of \(G\). Then there is a nonempty index set \(I\subseteq\{1,\ldots,m\}\) such that \[|I|\le h-t+1\] and \[\prod_{i\in I}\mathfrak p_i=(\pi)\] for some irreducible element \(\pi\in\mathcal{O}_K\). Repeated prime ideals are counted as distinct indexed occurrences.
Proof. Form the indexed sequence of ideal classes \[S=[\mathfrak p_1]\cdots[\mathfrak p_m]\] over \(G=\operatorname{Cl}(K)\), written additively. This sequence has length \(m\ge h\) and support size \(t\). By Corollary 1, \[MZ(S)\le h-t+1.\] Choose a nonempty index set \(I\subseteq\{1,\ldots,m\}\) with \(|I|=MZ(S)\) and \[\sum_{i\in I}[\mathfrak p_i]=0.\] Then \[\prod_{i\in I}\mathfrak p_i=(\pi)\] for some nonzero \(\pi\in\mathcal{O}_K\). Since \(I\) is nonempty and each \(\mathfrak p_i\) is a proper ideal, this product is a proper ideal, so \(\pi\) is not a unit.
We prove that \(\pi\) is irreducible. Suppose, to the contrary, that \(\pi=ab\) with nonunits \(a,b\in\mathcal{O}_K\). Then \[(a)(b)=(\pi)=\prod_{i\in I}\mathfrak p_i.\] For each nonzero prime ideal \(\mathfrak q\), put \[e_{\mathfrak q}=\#\{i\in I:\mathfrak p_i=\mathfrak q\}.\] Unique factorization of nonzero ideals into prime ideals gives \[v_{\mathfrak q}((a))+v_{\mathfrak q}((b))=e_{\mathfrak q}\] for every \(\mathfrak q\), and no prime outside the indexed multiset \(\{\mathfrak p_i:i\in I\}\) occurs in \((a)\) or \((b)\). Since \(a\) is a nonunit, \((a)\) is a proper nonzero integral ideal, so its prime-ideal factorization is nonempty. Since \(b\) is a nonunit, \((b)\) is also proper, so \((a)\) cannot use all indexed prime-ideal occurrences. Indeed, if \(v_{\mathfrak q}((a))=e_{\mathfrak q}\) for every \(\mathfrak q\), then all valuations of \((b)\) would be zero, so \((b)=\mathcal{O}_K\), contradicting that \(b\) is a nonunit.
Choose, for each \(\mathfrak q\), exactly \(v_{\mathfrak q}((a))\) of the indices \(i\in I\) with \(\mathfrak p_i=\mathfrak q\). This gives a nonempty proper subset \(J\subsetneq I\) such that \[(a)=\prod_{j\in J}\mathfrak p_j.\] Since \((a)\) is principal, we get \[\sum_{j\in J}[\mathfrak p_j]=0\] in \(\operatorname{Cl}(K)\). This is a nonempty zero-sum subsequence of \(S\) of length \(|J|<|I|=MZ(S)\), a contradiction. Hence no such factorization \(\pi=ab\) exists, and \(\pi\) is irreducible. ◻
When the class group is cyclic, Proposition 4 inherits the cyclic zero-sum refinements from the preceding section. The resulting bound is best read as a dichotomy: either the shortest principal subproduct is already no longer than about half the class number, or the long zero-sumfree structure theorem applies.
Proposition 5 (Cyclic class-group sharpening). Let \(K\) be a number field with ring of integers \(\mathcal{O}_K\), and suppose that \(\operatorname{Cl}(K)\) is cyclic of order \(h\). Let \(\mathfrak p_1,\ldots,\mathfrak p_m\) be nonzero prime ideals of \(\mathcal{O}_K\) with \(m\ge h\), and suppose their ideal classes occupy \(t\) distinct classes. Put \[B_h(t)= \min\left\{ h-t+1,\, \max\left\{\left\lfloor\frac{h}{2}\right\rfloor+1,\, h-\binom{t}{2}\right\} \right\}.\] Then there is a nonempty index set \(I\subseteq\{1,\ldots,m\}\) such that \[|I|\le B_h(t)\] and \[\prod_{i\in I}\mathfrak p_i=(\pi)\] for some irreducible element \(\pi\in\mathcal{O}_K\). If, in addition, \(h\ge 5\) and \(t=3\), then one can take \(|I|\le h-3\).
Proof. Form the indexed class sequence \[S=[\mathfrak p_1]\cdots[\mathfrak p_m]\] over \(\operatorname{Cl}(K)\). Choose one occurrence from each of the \(t\) represented ideal classes and then add arbitrary further occurrences until the resulting subsequence \(T\mid S\) has length \(h\). This is possible because \(m\ge h\), and it gives \[|T|=h,\qquad |\mathop{\mathrm{supp}}(T)|=t.\] Identify \(\operatorname{Cl}(K)\) with \(C_h\), and put \(q=MZ(T)\).
By Corollary 1, or by Theorem 2 applied directly to \(T\), \[q\le h-t+1.\] On the other hand, either \(q-1\le h/2\), in which case \[q\le \left\lfloor\frac{h}{2}\right\rfloor+1,\] or \(q-1>h/2\), in which case Theorem 3 gives \[q\le h-\binom{t}{2}.\] Combining the linear bound with this cyclic dichotomy gives \(q\le B_h(t)\). Since \(T\) is a subsequence of the full class sequence \(S\), the full sequence also has a nonempty zero-sum subsequence of length at most \(B_h(t)\).
Choose a globally shortest nonempty zero-sum subsequence of \(S\), and let \(I\) be its index set. Then \(|I|\le B_h(t)\) and \[\sum_{i\in I}[\mathfrak p_i]=0,\] so \[\prod_{i\in I}\mathfrak p_i=(\pi)\] for some nonzero \(\pi\in\mathcal{O}_K\). The same indexed-prime argument used in Proposition 4 shows that \(\pi\) is irreducible: a nontrivial factorization \(\pi=ab\) would make \((a)\) a principal product of a nonempty proper indexed subcollection of the primes in \(I\), giving a shorter nonempty zero-sum subsequence of the class sequence.
Finally assume \(h\ge 5\) and \(t=3\). The support-preserving subsequence \(T\) has length \(h\) and support size \(3\) over \(C_h\), so Corollary 2 gives \(MZ(T)\le h-3\). Repeating the preceding global-minimality and irreducibility argument gives an indexed principal irreducible subproduct of length at most \(h-3\). ◻
In factorization-theoretic language, the selected class sequence in each proposition is an atom of the block monoid over \(\operatorname{Cl}(K)\), and the principal generator is an atom of \(\mathcal{O}_K^\bullet\). This is the ring-of-integers instance of the standard transfer from Krull monoids to monoids of zero-sum sequences over the class group; see [2], [8], [9].
For a finite abelian group \(G\), define \[f_G(t)=\max\{MZ(S): |S|=|G|,\;|\mathop{\mathrm{supp}}(S)|=t\}.\] Theorem 2 gives \(f_G(t)\le |G|-t+1\), while Corollary 2 shows that \(f_{C_n}(3)=n-3\) for \(n\ge 5\). It would be natural to determine \(f_{C_n}(t)\) for larger fixed \(t\), and then to ask which parts of the cyclic picture survive for noncyclic groups. Moreover, it is natural to ask for which values of t equality is attained in the bound above. One can show that the extremal cases arise only when \(t\in \{1,2,n-1,n\}\), occurring in symmetric pairs.
There is also a stability question behind the linear bound. The Freeze–Smith estimate suggests that equality in Theorem 2 should force a large multiplicity in the original sequence. We do not use such a statement here, and we leave a precise reviewed stability theorem for future work.