January 01, 1970
We introduce the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\), extending a positive information algebra \(A\) by a join-semilattice \(L\), where elements of \(L\) dominate \(A\) and arithmetic in \(L\) reduces to the join. This models saturation or overflow in computational systems and generalizes the transition from finite to infinite cardinal arithmetic. We characterize the idempotent elements of \(S\) and \(S[X]\), fully classify idempotent power series over cardinal numbers, describe the structure of prime and maximal ideals, compute the Krull dimension of \(S\) (\(\dim S = \dim A + |L|\) for well-ordered finite \(L\)), and establish Noetherian and Artinian criteria.
Inspired by the arithmetic of the semiring of cardinal numbers, we introduce a novel extension of a positive information algebra by means of a join-semilattice. More precisely, given a positive information algebra \(A\) and a join-semilattice \(L\), we construct a new positive information algebra \[S = A \oplus_{\mathop{\mathrm{ord}}} L\] containing \(A\) as a subsemiring and \(L\) as an upper semilattice component. Conceptually, this construction abstracts the passage from natural number arithmetic to cardinal arithmetic: within the lower component \(A\), operations follow the original algebraic structure, whereas once computations enter the upper component \(L\), the behavior becomes order-dominated, resembling the arithmetic of infinite cardinals.
Beyond its intrinsic algebraic interest, this construction also admits a natural interpretation in engineering and computational systems. In many practical settings, ordinary arithmetic governs system behavior only within a safe operational range, while saturation, overflow, fault escalation, or resource exhaustion push the system into a different computational regime. The semilattice component \(L\) models such exceptional states, where arithmetic no longer behaves classically but is instead governed by ordered propagation rules (see Example 4).
Since the terminology of semiring theory is not completely standardized [1], we begin by fixing the notation and terminology used throughout the paper. We also briefly summarize the main results obtained herein.
A bimagma \((H,+,\cdot)\) is called a hemiring if \((H,+,0)\) is a commutative monoid, \((H,\cdot)\) is a semigroup, multiplication distributes over addition from both sides, and \(0\) is multiplicatively absorbing; that is, \(h0 = 0h = 0\) for all \(h \in H\). A hemiring \(H\) is called commutative if \(ab = ba\) for all \(a,b \in H\). A hemiring \(S\) is called a semiring if it possesses a nonzero multiplicative identity element \(1\) [2]. Throughout this paper, all semirings are assumed to be commutative.
A semiring \(S\) is called zerosumfree if \(a+b=0\) implies \(a=b=0\) for all \(a,b \in S\), and entire if \(ab=0\) implies \(a=0\) or \(b=0\) for all \(a,b \in S\). A semiring that is both zerosumfree and entire is called an information algebra [2]. We say that a semiring \(S\) is conical if \(ab=1\) implies \(a=1\) (and hence \(b=1\); see Definition 3). Finally, a semiring \(S\) is called positive if it is equipped with a partial order \(\leq\) compatible with both addition and multiplication, and such that \(0\) is the least element [3].
Let \((A, +, \cdot, \leq_A)\) be a positive information algebra and let \((L, \leq_L)\) be a join-semilattice, with \(A\) and \(L\) disjoint. Set \(S = A \cup L\). We extend the orders on \(A\) and \(L\) to an order on \(S\) by declaring \[a \leq \ell \quad\text{for all } a \in A,\;\ell \in L,\] while \(\ell \leq a\) never holds for \(\ell \in L\) and \(a \in A\). Addition and multiplication on \(S\) are defined for \(x,y \in S\) by \[x + y = \begin{cases} x +_A y & \text{if } x,y \in A,\\ \sup\{x,y\} & \text{otherwise}, \end{cases} \qquad x \cdot y = \begin{cases} x \cdot_A y & \text{if } x,y \in A\setminus\{0\},\\ 0 & \text{if } x=0 \text{ or } y=0,\\ \sup\{x,y\} & \text{otherwise}. \end{cases}\] The upper semilattice component \(L\) may be interpreted as an overflow regime, fault layer, or saturation zone in which the ordinary arithmetic of the base semiring \(A\) is replaced by order-dominated behavior (see Definition 1). In §2, we prove that the resulting structure \[S = A \cup L\] forms a positive information algebra (Theorem 3). Motivated by its interpretation as a transition from ordinary arithmetic to an overflow regime, we call \[S = A \oplus_{\mathop{\mathrm{ord}}} L\] the overflow semiring associated to \(A\) and \(L\) (see Definition 2). We also show that for every \(s \in L\), the lower sets \(S^{<s}\) and \(S^{\leq s}\) are subsemirings of \(S\) (cf. Proposition 5), yielding a directed filtration \(\{S^{\leq s}\}_{s \in L}\) of \(S\) (cf. Corollary 1).
After introducing overflow semirings, in §3 we study their distinguished elements. For instance, Proposition 6 and Proposition 10 characterize idempotent and von Neumann regular elements of \(S\), respectively. We then determine families of multiplicatively idempotent polynomials and formal power series over overflow semirings (see Propositions 11, 13, and 15). Special attention is given to formal power series over the semiring of cardinal numbers, for which Corollary 2 provides a complete characterization of the idempotent elements. Recall that an element \(a\) of a semiring \(S\) is called small if \(b \in S \setminus U(S)\) implies \(a+b \in S \setminus U(S)\) [2]. We denote the set of all small elements of \(S\) by \(\mathop{\mathrm{sml}}(S)\). In Theorem 9, we show that if \(S=A\oplus_{\mathop{\mathrm{ord}}}L\) is the overflow semiring, then \[\mathop{\mathrm{sml}}(S) = \mathop{\mathrm{sml}}(A)\cup L.\]
A nonempty subset \(I\) of \(S\) is called an ideal, denoted \(I \trianglelefteq S\), if \(a+b \in I\) and \(sa \in I\) for all \(a,b \in I\) and \(s \in S\) [2]. The set of all ideals of a semiring \(S\) is denoted by \(\operatorname{Id}(S)\). An ideal is finitely generated if it is generated by a finite subset of \(S\), and principal if it is generated by a single element \(a \in S\), denoted \((a)\) [2]. Since \(S\) is commutative, \[(a)=aS=\{as:s\in S\}.\] An ideal \(I\) is proper if \(I \neq S\), and subtractive if \(x+y \in I\) and \(x \in I\) imply \(y \in I\) for all \(x,y \in S\) [2]. A proper ideal \(P\) is called prime if \(ab \in P\) implies \(a \in P\) or \(b \in P\) (see Corollary 7.6 in [2]).
The final section investigates the ideal theory of overflow semirings. We first show that \(S\) is always austere, meaning its only subtractive ideals are trivial (Proposition 20), and examine extension and contraction along the inclusion \(A \hookrightarrow S\) (Proposition 21). After characterizing the ideals, primes, and maximals of \(S\) (Theorems 24 and 22), we establish the Krull dimension formula: \[\dim S = \dim A + \lvert L \rvert\] (see Theorem 28). Finally, assuming \(L\) is well-ordered, we prove that \(S\) is Noetherian if and only if \(A\) is Noetherian (Theorem 30), and Artinian if and only if \(A\) is Artinian and \(L\) is a finite chain (Theorem 31).
Definition 1. Let \((A, +, \cdot, \leq_A)\) be a positive information algebra and let \((L, \leq_L)\) be a join-semilattice, with \(A\) and \(L\) disjoint. Set \(S = A \cup L\). Define an order on \(S\) by:
preserving the original orders on \(A\) and \(L\),
\(a \leq \ell\) for all \(a \in A\), \(\ell \in L\),
\(\ell \leq a\) never holds for \(\ell \in L\), \(a \in A\).
Define addition and multiplication on \(S\) for \(x, y \in S\) by \[x + y = \begin{cases} x +_A y & \text{if } x, y \in A, \\ \sup\{x, y\} & \text{otherwise}, \end{cases} \qquad x \cdot y = \begin{cases} x \cdot_A y & \text{if } x, y \in A\setminus\{0\}, \\ 0 & \text{if } x = 0 \text{ or } y = 0, \\ \sup\{x, y\} & \text{otherwise}. \end{cases}\]
Lemma 1. The structure \((S,+,0)\) in Definition 1 is a commutative monoid.
Proof. Commutativity of \(+\) is clear. For associativity \[(x+y)+z = x+(y+z),\] we consider the location of the arguments. If all three are in \(A\), it holds in the semiring. If all three are in \(L\), it holds in the semilattice. If one, say, \(x\) is in \(A\), and others are in \(L\), then the value of both sides is \(\sup\{y,z\}\). And if two of them are in \(A\), say, \(x\) and \(y\), and \(z\) is in \(L\), then the value of both sides is \(z\) (by commutativity, other permutations follow). The element \(0 \in A\) is the identity: for \(x \in A\), \(x+0 = x\) by the semiring; for \(x \in L\), \(x+0 = \sup\{x,0\} = x\) because \(0 < x\). Hence \((S,+,0)\) is a commutative monoid. ◻
Lemma 2. The structure \((S,\cdot,1)\) in Definition 1 is a commutative monoid and \(0\) is its absorbing element.
Proof. First, \(0\) is absorbing: for any \(x \in S\), if \(x \in A\) then \(0 \cdot x = 0\) by the semiring property; if \(x \in L\) then \(0 \cdot x = 0\) by definition (the second case of multiplication). Hence \(0 \cdot x = 0 = x \cdot 0\) for all \(x \in S\). Commutativity of \(\cdot\) is clear. For associativity \[(x \cdot y) \cdot z = x \cdot (y \cdot z),\] we consider the location of the arguments. If any of \(x,y,z\) is \(0\), both sides are \(0\) by the absorbing property of \(0\) just proved. If all three are in \(A\), it holds in the semiring. If all three are in \(L\), it holds in the semilattice. If one is in \(A\) (non-zero) and the others in \(L\), both sides equal \(\sup\{y,z\}\). If two are in \(A\) (non-zero) and one in \(L\), both sides equal the element in \(L\). The element \(1 \in A\) is the identity: for \(x \in A\), \(x \cdot 1 = x\) by the semiring; for \(x \in L\), \(x \cdot 1 = \sup\{x,1\} = x\) because \(1 < x\). Hence \((S,\cdot,1)\) is a commutative monoid and \(0\) is its absorbing element. ◻
Proposition 1. The structure \((S,+,\cdot)\) in Definition 1 is a semiring.
Proof. By Lemma 1, \((S,+,0)\) is a commutative monoid. By Lemma 2, \((S,\cdot,1)\) is a commutative monoid with \(0\) absorbing. First we verify distributivity: \[x \cdot (y + z) = (x \cdot y) + (x \cdot z), \qquad\forall\, x,y,z \in S.\] The verification proceeds by cases on \(x\):
Case \(x = 0\): Then \(x \cdot (y+z) = 0\) and \((x \cdot y)+(x \cdot z) = 0+0 = 0\).
Case \(x \in A \setminus \{0\}\):
If \(y,z = 0\): both sides are \(0\).
If \(y = 0\), \(z \in A \setminus \{0\}\): LHS \(= xz\), RHS \(= 0 + xz = xz\).
If \(y = 0\), \(z \in L\): LHS \(= x \cdot z = z\), RHS \(= 0 + z = z\).
If \(y \in A \setminus \{0\}\), \(z = 0\): symmetric.
If \(y,z \in A \setminus \{0\}\): This case holds by distributivity in \(A\).
If \(y \in A \setminus \{0\}\), \(z \in L\): LHS \(= x \cdot (y + z) = x \cdot z = z\), RHS \(= xy + xz = xy + z = z\) (since \(z \in L\) dominates).
If \(y \in L\), \(z = 0\): symmetric.
If \(y \in L\), \(z \in A \setminus \{0\}\): symmetric.
If \(y,z \in L\): LHS \(= x \cdot (y \vee z) = y \vee z\), RHS \(= y + z = y \vee z\).
Case \(x \in L\):
If \(y,z = 0\): both sides are \(0\).
If \(y = 0\), \(z \in A \setminus \{0\}\): LHS \(= x \cdot z = x\), RHS \(= x \cdot 0 + x \cdot z = 0 + x = x\).
If \(y = 0\), \(z \in L\): LHS \(= x \cdot z = x \vee z\), RHS \(= 0 + (x \vee z) = x \vee z\).
If \(y \in A \setminus \{0\}\), \(z = 0\): symmetric.
If \(y,z \in A \setminus \{0\}\): by zerosumfree property for \(A\), \(y+z \in A \setminus \{0\}\), so LHS \(= x \cdot (y+z) = x\), RHS \(= x + x = x\).
If \(y \in A \setminus \{0\}\), \(z \in L\): LHS \(= x \cdot z = x \vee z\), RHS \(= x + (x \vee z) = x \vee z\).
If \(y \in L\), \(z = 0\): symmetric.
If \(y \in L\), \(z \in A \setminus \{0\}\): symmetric.
If \(y,z \in L\): LHS \(= x \cdot (y \vee z) = x \vee y \vee z\), RHS \(= (x \vee y) + (x \vee z) = x \vee y \vee z\).
Hence distributivity holds. Thus \((S,+,\cdot)\) is a semiring, completing the proof. ◻
Remark 2. In Definition 1, requiring \(A\) to be an information algebra (that is, zerosumfree and entire) is essential in order for \(S\) to satisfy the semiring axioms. If \(A\) is not zerosumfree, choose \(x \in L\) and \(y, z \in A \setminus \{0\}\) with \(y+z=0\). Then \[x \cdot (y+z) = x \cdot 0 = 0 \neq x = (x \cdot y)+(x \cdot z),\] violating distributivity. If \(A\) is not entire, choose \(x \in L\) and \(y, z \in A \setminus \{0\}\) with \(y \cdot z = 0\). Then \[(x \cdot y) \cdot z = x \cdot z = x \neq 0 = x \cdot 0 = x \cdot (y \cdot z),\] violating associativity of multiplication. Thus both zerosumfreeness and entireness of \(A\) are necessary for \(S\) to be a semiring.
Lemma 3. The extended relation \(\leq\) on \(S = A \cup L\) in Definition 1 is a partial order and the least element of \(S\) is \(0\).
Proof. We verify the three axioms:
The relation \(\leq\) on \(S\) is reflexive because \(\leq_A\) and \(\leq_L\) are both reflexive.
For antisymmetry condition of \(\leq\), suppose \(x \leq y\) and \(y \leq x\) for \(x,y \in S\).
If \(x,y \in A\) or \(x,y \in L\), then antisymmetry follows from the antisymmetry condition of \(\leq_A\) or \(\leq_L\).
If \(x \in A\) and \(y \in L\), then \(x \leq y\) holds by definition, but \(y \leq x\) never holds. Hence this case cannot occur. Similarly, \(x \in L\) and \(y \in A\) cannot occur with both inequalities.
Thus \(x = y\) in all possible cases.
For transitivity, suppose \(x \leq y\) and \(y \leq z\) for \(x,y,z \in S\).
If all three are in \(A\) or in \(L\), transitivity follows from the transitivity of \(\leq_A\) or \(\leq_L\).
If \(x \in A\) and \(z \in L\), then \(x \leq z\) no matter \(y \in A\) or \(y \in L\) because every element in \(A\) is smaller than every element in \(L\).
The other cases never happen because any element in \(L\) cannot be smaller than or equal to any element in \(A\).
Hence \(x \leq z\) in all cases.
Therefore \(\leq\) is a partial order on \(S\). It is clear that \(0 \le x\) for all \(x \in S\). ◻
Theorem 3. The structure \((S,+,\cdot,\leq)\) in Definition 1 is a positive information algebra.
Proof. We first verify the two compatibility conditions:
For any \(x,y,z \in S\) with \(x \leq y\), we show \(x+z \leq y+z\).
If \(x,y \in A\) and \(z \in A\): follows from \(A\) being an ordered semiring.
If \(x,y \in A\) and \(z \in L\): then \(x+z = z\) and \(y+z = z\), so equality holds.
If \(x \in A\), \(y \in L\) (so \(x \leq y\) automatically), and \(z \in A\): then \(x+z \in A\), \(y+z = y\) (since \(y \in L\) dominates), and \(x+z \leq y\) because every element in \(A\) is below every element in \(L\).
If \(x \in A\), \(y \in L\), and \(z \in L\): then \(x+z = z\), \(y+z = y \vee z\), and \(z \leq y \vee z\).
If \(x,y \in L\): then \(x \leq y\) implies \(x \vee z \leq y \vee z\) for any \(z \in L\); if \(z \in A\), then \(x+z = x\), \(y+z = y\), and \(x \leq y\).
For any \(x,y,z \in S\) with \(x \leq y\) and \(z \geq 0\), we show \(x \cdot z \leq y \cdot z\).
If \(z = 0\): both sides are \(0\), so equality holds.
If \(z \in A \setminus \{0\}\):
If \(x,y \in A\): then \(x \cdot z \leq y \cdot z\) because \(A\) is an ordered semiring.
If \(x,y \in L\): then \(xz = x\) and \(yz = y\), and \(x \leq y\) implies \(x \cdot z \leq y \cdot z\).
If \(x \in A\) and \(y \in L\): then \(x \leq y\) automatically. Here \(xz \in A\) (since \(x,z \in A\)) and \(yz = y \in L\). Since every element of \(A\) is below every element of \(L\), we have \(xz \leq y\), i.e., \(x \cdot z \leq y \cdot z\).
If \(x \in L\) and \(y \in A\): this cannot happen because \(x \leq y\) would imply an \(L\)-element is below an \(A\)-element, contradicting \(A < L\).
If \(z \in L\): then for any \(w \in S\), we have \(w \cdot z = z\) if \(w \neq 0\), and \(0 \cdot z = 0\). Thus:
If \(x = 0\), then \(x \cdot z = 0\). Since \(x \leq y\), we have \(0 \leq y\), so \(y\) could be \(0\) or nonzero. If \(y = 0\), then \(y \cdot z = 0\) and \(0 \leq 0\) holds. If \(y \neq 0\), then either \(y \in A\), and so \(y \cdot z = z\) and \(0 \leq z\) holds because \(z \in L\); or \(y \in L\), and so \(y \cdot z = y \vee z\) which is obviously \(\geq 0\).
Let \(x \neq 0\). Since \(x \leq y\) and \(x \neq 0\), we must have \(y \neq 0\) (otherwise \(x \leq 0\) would force \(x = 0\), a contradiction).
If \(x,y \in A\), then \(xz = z\) and \(yz = z\), so equality holds.
If \(x \in A\) and \(y \in L\), then \(xz = z\) while \(yz = y \vee z\), and the inequality \(z \leq y \vee z\) holds clearly.
The case \(x \in L\) and \(y \in A\) cannot happen because \(x \leq y\) would imply an element in \(L\) is below an element in \(A\), a contradiction: The assumption \(x \leq y\) for \(x \in L\) and \(y \in A\) would imply \(x = y\), contradicting the disjointness of \(A\) and \(L\).
If \(x,y \in L\), then \(xz = x \vee z\) and \(yz = y \vee z\), and \(x \leq y\) implies \(x \vee z \leq y \vee z\) in the semilattice.
Therefore in all cases, \(x \cdot z \leq y \cdot z\).
Thus, \((S,+,\cdot,\leq)\) is a positive semiring. We then show \(S\) is zerosumfree by showing that if \(0 < x\), then \(0 < x+y\) for any \(y \geq 0\). Consider cases:
\(x,y \in A\): then \(x+y \in A\) and \(x+y > 0\) because \(A\) is zerosumfree.
\(x \in A\), \(y \in L\): then \(x+y = y \in L\), and since every element of \(A\) is smaller than every element of \(L\), we have \(y > 0\), so \(x+y > 0\).
\(x \in L\), \(y \in A\): then \(x+y = x \in L\), and \(x > 0\) by assumption.
\(x,y \in L\): then \(x+y = x \vee y \in L\), hence \(x+y > 0\).
Thus \(0 < x+y\) in all cases. Similarly one may prove that \(S\) is entire by showing that the conditions \(0 < x\) and \(0 < y\) imply \(0 < xy\). Hence \(S\) is a positive information algebra. ◻
Definition 2. We call the positive information algebra \((S,+,\cdot,\leq)\) obtained by extending a positive information algebra \(A\) with a join-semilattice \(L\) (see Theorem 3) the overflow semiring, denoted by \(A \oplus_{\mathop{\mathrm{ord}}} L\).
Examples 4. To illustrate the construction in Definition 2, we present the following examples.
In Zermelo–Fraenkel set theory with the Axiom of Choice, consider the positive information algebra \(\mathbb{N}_0\) together with the join-semilattice \(\mathcal{L}\) of all infinite cardinal numbers. The overflow semiring \(\mathscr{C} = \mathbb{N}_0 \oplus_{\mathop{\mathrm{ord}}} \mathcal{L}\), equipped with cardinal addition and multiplication, forms a class-semiring satisfying the axioms of a positive information algebra and is totally ordered by the usual ordering of cardinals (see Chapter 4 of [4]).
It is straightforward to verify that the restricted max-plus algebra \(A=[-\infty,0]\) (with addition \(\oplus=\max\) and multiplication \(\odot=+\)) is a positive information algebra. Let \(L=\mathbb{N}\) denote the set of positive integers equipped with the usual order. In the overflow semiring \(S=A\oplus_{\mathop{\mathrm{ord}}}L\), addition is simply given everywhere by \(\max\). Multiplication is defined by \[x\odot y= \begin{cases} x+y, & \text{if } x,y\in A\setminus\{-\infty\},\\[4pt] -\infty, & \text{if } x=-\infty \text{ or } y=-\infty,\\[4pt] \max\{x,y\}, & \text{otherwise}. \end{cases}\] The resulting algebraic structure exhibits a surprisingly hybrid behavior between tropical arithmetic and ordered overflow dynamics.
The set \(A=[0,1]\), equipped with \(\max\) as addition, \(\min\) as multiplication, and the usual order on the real numbers, forms a positive information algebra. Extending this structure by the singleton join-semilattice \(L=\{\ell\}\) yields the overflow semiring \(S=A\oplus_{\mathop{\mathrm{ord}}}\{\ell\}\), whose operations are given by \[x\oplus y= \begin{cases} \max(x,y), & x,y\in[0,1],\\ \ell, & \text{otherwise}, \end{cases} \qquad x\odot y= \begin{cases} \min(x,y), & x,y\in[0,1]\setminus\{0\},\\ 0, & x=0 \text{ or } y=0,\\ \ell, & \text{otherwise}. \end{cases}\] This construction admits a natural interpretation in resource allocation:
\([0,1]\) represents normalized resource levels (\(0\) meaning no resources and \(1\) full capacity);
\(\ell\) represents an oversaturated state beyond system capacity;
\(\max\) models competitive aggregation by selecting the larger demand;
\(\min\) models cooperative interaction through bottleneck behavior;
once the system becomes oversaturated, it remains oversaturated unless reset by multiplication with \(0\).
Thus, \(S\) models a resource system in which exceeding capacity irreversibly triggers a failure state, while multiplication by \(0\) acts as a complete reset mechanism.
The Viterbi semiring \(([0,1], \max, \cdot, \leq)\) (see p. 3 in [3]) is a positive information algebra widely used in dynamic programming. For a join-semilattice \((L,\leq_L)\), the overflow semiring \[V = [0,1] \oplus_{\text{ord}} L\] is also a positive information algebra. It is additively idempotent since addition is the \(\sup\) operation. This construction extends standard Viterbi optimization to scenarios with multi-tiered or infinite utility states by gluing \(L\) above \([0,1]\) to represent degrees of non-standard certainty. For instance, if \(L = \{+\infty\}\), then \[V = [0,1] \cup \{+\infty\}\] where \(+\infty\) acts as a “Super-Pass” token in path-finding. If a path possesses this token, no amount of scaling by fractional probabilities can degrade it, unless it hits a dead end (\(0\)).
The Boolean semiring \((\mathbb{B}=\{0,1\},+,\cdot, \leq)\) forms a positive information algebra. Extending this structure by a singleton join-semilattice \(\{m\}\) yields the overflow semiring \(S=\mathbb{B}\oplus_{\mathop{\mathrm{ord}}}\{m\}\). Structurally, the additive behavior of \(S\) closely resembles Bochvar’s internal three-valued logic, which introduces a third truth value \(m\) (meaningless) governed by a contagion principle satisfying \(\neg m=m\), and for which an implication \(x\to y\) evaluates to \(m\) whenever either operand equals \(m\) [5]. Interpreting semiring addition through the classical equivalence \(a+b \equiv \neg a \to b\), one obtains an additive Cayley table isomorphic to the additive structure of \(S\). A fundamental structural divergence nevertheless appears in the multiplicative behavior. Bochvar’s logic enforces complete logical contagion, requiring \(0\cdot m=m\), whereas the overflow semiring preserves the algebraic annihilator law \(0\cdot m=0\). From a computational perspective, this models short-circuit evaluation and local fault isolation: once a controlling gate is closed (\(0\)), a downstream exceptional state (\(m\)) is never propagated or executed. By privileging algebraic annihilation over universal logical contagion, the semiring \(S\) ensures that a masked or inactive computational channel cannot be corrupted by a latent runtime fault.
Proposition 5. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. For any \(s \in L\), the sets \[S^{< s} = \{ x \in S : x < s \} \quad \text{and} \quad S^{\leq s} = \{ x \in S : x \leq s \}\] are subsemirings of \(S\).
Proof. Since \(s \in L\) and \(A < L\) by definition, every element \(a \in A\) satisfies \(a < s\). Thus, the additive and multiplicative identities satisfy \(0, 1 \in S^{<s}\). Now, let \(x, y \in S^{<s}\). By Definition 1, their sum \(x + y\) and multiplication \(x \cdot y\) either evaluate within the information algebra \(A\) (yielding an element in \(A\), which is strictly less than \(s\)) or reduce to a supremum bounded above by \(\max\{\sup\{x,y\}, \sup\{x,y\}\} = \sup\{x,y\} < s\). Thus, \(S^{< s}\) is closed under addition and multiplication, making it a subsemiring of \(S\). Replacing the strict inequality with \(\leq\) yields identical closure bounds, completing the proof for \(S^{\leq s}\). ◻
Corollary 1 (Directed Filtration by \(L\)). Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. The family of subsemirings \(\{S^{\leq s}\}_{s \in L}\) forms an increasing directed filtration of \(S\), such that \[S = \bigcup_{s \in L} S^{\leq s}.\]
Proof. Let \(s, t \in L\) with \(s \leq t\). If \(x \in S^{\leq s}\), then \(x \leq s \leq t\) implies \(x \in S^{\leq t}\). Thus, \(S^{\leq s} \subseteq S^{\leq t}\), establishing that the family is tracking the poset order. Now, let \(x \in S\). If \(x \in A\), then \(x \leq s\) for all \(s \in L\), placing \(x \in \bigcup_{s \in L} S^{\leq s}\). If \(x \in L\), then by reflexivity \(x \leq x\), placing \(x \in S^{\leq x}\). Hence, the identity \(S = \bigcup_{s \in L} S^{\leq s}\) holds, completing the proof. ◻
Proposition 6. Let \(I^+(S)\) and \(I^\times(S)\) denote the additively and multiplicatively idempotent elements of \(S\), respectively. Then for the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\), \[I^+(S) = I^+(A) \cup L \qquad \text{and} \qquad I^\times(S) = I^\times(A) \cup L.\]
Proof. If \(x \in L\), then \(x+x = \sup\{x,x\} = x\) and \(x \cdot x = \sup\{x,x\} = x\), so \(L \subseteq I^+(S) \cap I^\times(S)\). If \(x \in A\), additive or multiplicative idempotency follows from that in \(A\). Conversely, no element outside \(I^+(A) \cup L\) can be additively idempotent, and similarly for multiplication. Hence the equalities hold. ◻
An element \(a\) of a semiring \(S\) is called multiplicatively invertible (or a unit) if \(ab = 1\) for some \(b \in S\). The group of all units of \(S\) is denoted by \(U(S)\).
Proposition 7. For the overflow semiring \(S = A \oplus_{\operatorname{ord}} L\), the group of units satisfies \(U(S) = U(A)\).
Proof. If \(\ell \in L\), then \(\ell y \in L \cup \{0\}\) for every \(y \in S\), so \(\ell y \neq 1\). Hence no element of \(L\) is a unit, and therefore \(U(S)\subseteq A\). Now multiplication on \(A\) is unchanged in \(S\). Thus every \(a\in U(A)\) remains invertible in \(S\), so \(U(A)\subseteq U(S)\). Conversely, if \(a\in A\) is invertible in \(S\), its inverse must lie in \(A\) since elements of \(L\) are not units. Hence \(a\in U(A)\). Hence, \(U(S)=U(A)\) as claimed. ◻
Recall that a multiplicative monoid \(M\) is called conical if \(ab = 1\) implies \(a = b = 1\) for all \(a,b \in M\) [6]. We extend this notion to semirings as follows.
Definition 3. We call a semiring \(S\) conical if \(ab = 1\) implies \(a = b = 1\) for all \(a,b \in S\).
Proposition 8. If \(A\) is conical, then so is the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) and the polynomial semiring \(S[X]\).
Proof. Suppose \(x \cdot y = 1\) in \(S\). Since \(1 \in A\) and any multiplication involving an element of \(L\) lies in \(L\) (or is \(0\) if the other factor is \(0\)), we must have \(x, y \in A\) and \(x \cdot y = 1\). Conicality of \(A\) then yields \(x = y = 1\). Now let \(f, g \in S[X]\) with \(fg = 1\). Since \(S[X]\) is entire, \[\deg(f) + \deg(g) = \deg(fg) = \deg(1) = 0,\] so \(\deg(f) = \deg(g) = 0\). Thus \(f\) and \(g\) are constant polynomials, i.e., \(f = a\), \(g = b\) for some \(a, b \in S\). Then \(ab = 1\) in \(S\), so by the first part \(a = b = 1\). Hence \(f = g = 1\), proving \(S[X]\) is conical. ◻
Example 1. The following are examples of conical positive information algebras:
The class-semiring of cardinal numbers \(\mathscr{C}\) and the semiring \(S\) obtained from the restricted max-plus algebra in Examples 4 are both conical. However, they are not isomorphic because \(S\) is additively idempotent (\(\max(x,x) = x\)), whereas \(\mathscr{C}\) is not (since \(\kappa + \kappa = \kappa\) holds for infinite cardinals but fails for finite nonzero cardinals).
Every bounded distributive lattice \((L,\vee,\wedge,0,1)\) is a conical positive information algebra.
The Viterbi positive information algebra \(([0,1], \max, \cdot, \leq)\) illustrated in Examples 4 is conical.
Definition 4. An element \(a\) of a semiring \(S\) is called small if \[b \in S \setminus U(S) \implies a+b \in S \setminus U(S).\] The set of all small elements of \(S\) is denoted by \(\mathop{\mathrm{sml}}(S)\).
Theorem 9. Let \(S=A\oplus_{\mathop{\mathrm{ord}}}L\) be the overflow semiring. Then \[\mathop{\mathrm{sml}}(S) = \mathop{\mathrm{sml}}(A)\cup L.\]
Proof. First let \(\ell\in L\) and let \(b \notin U(S)\). Since \(L\) is absorbing under the overflow addition, \(\ell+b\in L\), and hence \(\ell+b\notin U(S)\). Thus \(\ell\in \mathop{\mathrm{sml}}(S)\), so \(L\subseteq \mathop{\mathrm{sml}}(S)\). Now let \(a\in \mathop{\mathrm{sml}}(A)\) and let \(b\notin U(S)\). Since \(U(S)=U(A)\) (Proposition 7), either \(b\in A \setminus U(A)\) or \(b\in L\).
If \(b\in A \setminus U(A)\), then \(a+b\in A \setminus U(A)\) because \(a\in \mathop{\mathrm{sml}}(A)\), and so, \(a+b \notin U(S)\).
If \(b\in L\), then \(a+b\in L\), so again \(a+b\notin U(S)\).
Thus \(a\in \mathop{\mathrm{sml}}(S)\), proving \(\mathop{\mathrm{sml}}(A)\cup L \subseteq \mathop{\mathrm{sml}}(S)\).
Conversely, let \(a\in \mathop{\mathrm{sml}}(S)\) and assume \(a\notin L\). Then \(a\in A\). We show that \(a\in \mathop{\mathrm{sml}}(A)\). Let \(b\in A\setminus U(A)\). Since \(U(A)=U(S)\), we have \(b\notin U(S)\). Because \(a\in \mathop{\mathrm{sml}}(S)\), \[a+b \notin U(S).\] Moreover, since \(a,b\in A\), the operations of \(S\) restrict to those of \(A\), hence the sum \(a+b\) computed in \(S\) lies in \(A\) and coincides with the sum in \(A\). Therefore, \[a+b \in A \setminus U(A),\] so \(a\in \mathop{\mathrm{sml}}(A)\). Thus \(\mathop{\mathrm{sml}}(S)\subseteq \mathop{\mathrm{sml}}(A)\cup L\). ◻
Definition 5. An element \(x\) in a semiring is called additively (resp., multiplicatively) (von Neumann) regular if there exists an element \(y\) such that \(x + y + x = x\) (resp., \(x \cdot y \cdot x = x\)). We denote the sets of these additively and multiplicatively regular elements by \(\operatorname{VNR}^+(S)\) and \(\operatorname{VNR}^\times(S)\), respectively.
Proposition 10. For the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\), the sets of additively and multiplicatively regular elements satisfy \[\operatorname{VNR}^+(S) = \operatorname{VNR}^+(A)\cup L, \qquad \operatorname{VNR}^\times(S) = \operatorname{VNR}^\times(A)\cup L.\]
Proof. For every \(x\in L\), taking \(y=x\) gives \[x+x+x=x, \qquad x\cdot x\cdot x=x,\] since both operations on \(L\) are given by the semilattice join. Hence \[L\subseteq \operatorname{VNR}^+(S)\cap \operatorname{VNR}^\times(S).\] If \(x\in \operatorname{VNR}^+(A)\) or \(x\in \operatorname{VNR}^\times(A)\), the corresponding regularity equation in \(A\) remains valid in \(S\), since the operations agree on \(A\). Thus \[\operatorname{VNR}^+(A)\cup L \subseteq \operatorname{VNR}^+(S), \qquad \operatorname{VNR}^\times(A)\cup L \subseteq \operatorname{VNR}^\times(S).\] Conversely, let \(x\in A\). If \(x\cdot y\cdot x=x\) for some \(y\in S\) and \(y\in L\), then \[x\cdot y\cdot x = \sup\{x,y\} = y,\] since every element of \(L\) dominates every element of \(A\). Hence \(y=x\), contradicting \(A\cap L=\emptyset\). Therefore \(y\in A\), so \[x\cdot_A y\cdot_A x=x,\] and consequently \(x\in \operatorname{VNR}^\times(A)\). Similarly, if \(x+y+x=x\) for some \(y\in S\) and \(y\in L\), then \[x+y+x=\sup\{x,y\}=y,\] again a contradiction. Hence \(y\in A\), implying \[x+_A y+_A x=x,\] so \(x\in \operatorname{VNR}^+(A)\). Thus \[\operatorname{VNR}^+(S) = \operatorname{VNR}^+(A)\cup L, \qquad \operatorname{VNR}^\times(S) = \operatorname{VNR}^\times(A)\cup L,\] completing the proof. ◻
Proposition 11. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. Then the multiplicatively idempotent elements of \(S[X]\) are the constant polynomials in \(I^\times(A) \cup L\).
Proof. If \(f^2 = f\), then \(2\deg(f) = \deg(f)\) because \(S\) is entire. Hence \(\deg(f) = 0\), so \(f\) is a constant polynomial, say \(f = c \in S\). Then \(c^2 = c\), i.e., \(c \in I^\times(A) \cup L\) by Proposition 6. Conversely, any constant polynomial \(c\) with \(c \in I^\times(A) \cup L\) satisfies \(c^2 = c\). ◻
Proposition 12. Let \(S\) be any semiring. An element \(f = \sum_{n=0}^{\infty} a_n X^n\) with \(a_0 = 0\) is multiplicatively idempotent in the formal power series semiring \(S[[X]]\) if and only if \(f = 0\).
Proof. If \(f = 0\), then clearly \(f^2 = f\). Conversely, suppose \(f^2 = f\) and \(a_0 = 0\). Then for all \(n \ge 0\), \[a_n = \sum_{k=0}^n a_k a_{n-k}.\] For \(n = 0\), we have \(a_0 = a_0^2 = 0\), which holds. Assume inductively that \(a_1 = \cdots = a_{n-1} = 0\). Then for \(n \ge 1\), \[a_n = a_0 a_n + a_n a_0 + \sum_{k=1}^{n-1} a_k a_{n-k} = 0.\] Thus by induction, \(a_n = 0\) for all \(n \ge 1\), so \(f = 0\). ◻
Proposition 13. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring and let \(f = \sum_{n=0}^{+\infty} a_n X^n\) with \(a_0 = 1\) be an element of the formal power series semiring \(S[[X]]\) satisfying the following conditions:
\(a_1 = a_2 = \dots = a_m = 0\),
\(a_j\) are arbitrary elements of \(L\) for \(m+1 \leq j \leq 2m+1\),
\(a_j \geq \sup\{a_{m+1}, a_{m+2}, \dots, a_{j-m-1}\}\) for all \(j \geq 2m+2\).
Then \(f\) is an idempotent element of \(S[[X]]\).
Proof. Let \(b_n\) be the coefficients of \(f^2 = f \cdot f\). We prove \(b_n = a_n\) for all \(n \geq 0\) by considering the following cases:
Case \(n=0\): \(b_0 = a_0 a_0 = 1 \cdot 1 = 1 = a_0\).
Case \(1 \leq n \leq m\): Observe that since \(a_1 = \dots = a_m = 0\), every term in \(b_n = \sum_{k=0}^n a_k a_{n-k}\) involves at least one factor from this zero range (or \(k=0\) and \(n-k=n \leq m\), yielding \(1 \cdot 0 = 0\)). Hence, \(b_n = 0 = a_n\).
Case \(m+1 \leq n \leq 2m+1\): For such \(n\), the only nonzero contributions occur when \(k = 0\) or \(k = n\) (because \(a_k = 0\) for \(1 \leq k \leq m\)). Thus, \[b_n = a_0 a_n + a_n a_0 = 1 \cdot a_n + a_n \cdot 1 = a_n + a_n = a_n,\] since \(a_n \in L\) (by hypothesis 2) and addition in \(L\) is idempotent.
Case \(n \geq 2m+2\): First, we establish that \(a_n \in L\). By hypothesis (3), \(a_n \geq a_{m+1}\). By hypothesis (2), \(a_{m+1} \in L\). Because every element of \(A\) is strictly below every element of \(L\) in the ordered semiring \(S\), the inequality \(a_n \geq a_{m+1}\) structurally forces \(a_n \in L\). Consequently, for any \(k \geq m+1\), we are guaranteed that \(a_k \in L\).
Now, write \(b_n = \sum_{k=0}^n a_k a_{n-k}\). Observe:
\(k = 0\) and \(k = n\) each contribute \(a_0 a_n = a_n a_0 = a_n\).
For \(1 \leq k \leq m\): \(a_k = 0\), so \(a_k a_{n-k} = 0\).
For \(n-m \leq k \leq n-1\): then \(n-k \leq m\), so \(a_{n-k} = 0\), giving \(a_k a_{n-k} = 0\).
For \(m+1 \leq k \leq n-m-1\): both \(k\) and \(n-k\) are \(\geq m+1\). As shown above, this implies \(a_k, a_{n-k} \in L\). Thus, their multiplication reduces to the semilattice join: \(a_k a_{n-k} = \sup\{a_k, a_{n-k}\}\).
Thus, \[b_n = a_n + a_n + \sum_{k=m+1}^{n-m-1} \sup\{a_k, a_{n-k}\} = a_n + \sup \{a_k\}_{k=m+1}^{n-m-1},\] where we used \(a_n + a_n = a_n\) (idempotence in \(L\)) and \[\sup\{ \sup\{a_k, a_{n-k}\}\}_k = \sup \{a_k\}_k.\] By hypothesis (3), \(a_n \geq \sup\{a_{m+1}, \dots, a_{n-m-1}\}\). Because \(a_n \in L\) dominates this supremum under addition, the expression simplifies cleanly to \(b_n = a_n\).
Thus \(b_n = a_n\) for all \(n\), proving \(f^2 = f\). ◻
Theorem 14. An element \(f = \sum_{n=0}^{+\infty} a_n X^n\) with \(a_0 = 1\) is multiplicatively idempotent in the formal power series semiring \(\mathscr{C}[[X]]\) if and only if either \(f = 1\) or coefficients of \(f\) satisfy the following conditions:
\(a_1 = a_2 = \dots = a_m = 0\), and
\(a_j\) are arbitrary infinite cardinals for \(m+1 \leq j \leq 2m+1\), and
\(a_j \geq \max\{a_{m+1}, a_{m+2}, \dots, a_{j-m-1}\}\) for all \(j \geq 2m+2\).
Proof. \((\Rightarrow)\): Suppose \(f^2 = f\) and \(f\) is not constant. Let \(m\) be the smallest index such that \(a_{m+1} \neq 0\). Then \(a_1 = a_2 = \dots = a_m = 0\).
Step 1. For \(m+1 \le n \le 2m+1\), the only possible nonzero terms in \(b_n = \sum_{k=0}^n a_k a_{n-k}\) are \(k=0\) and \(k=n\) (any other \(k\) would put one index \(\le m\)). Hence \(b_n = a_0 a_n + a_n a_0 = 2a_n\). Since \(f^2 = f\), \(b_n = a_n\), so \(a_n = 2a_n\). In cardinal arithmetic, \(\kappa = 2\kappa\) implies \(\kappa = 0\) or infinite. By minimality of \(m\), \(a_{m+1} \neq 0\), thus \(a_{m+1}\) is infinite. For \(n = m+2,\dots,2m+1\), if \(a_n \neq 0\) it must be infinite; the statement allows arbitrary infinite cardinals here. Hence \(a_{m+1},\dots,a_{2m+1}\) are arbitrary infinite cardinals.
Step 2. Using strong induction we prove that \(a_j\) is infinite for all \(j \ge m+1\).
Base: \(j = m+1,\dots,2m+1\) are infinite by Step 1.
Inductive step: Assume for some \(t \ge 2m+2\) that \(a_{m+1},\dots,a_{t-1}\) are all infinite. From \(f^2 = f\): \[a_t = b_t = \sum_{k=0}^t a_k a_{t-k}.\] Separate the terms: \[a_t = 2a_t + \sum_{k=m+1}^{t-m-1} a_k a_{t-k},\] because \(k=0\) and \(k=t\) give \(2a_t\), and other nonzero contributions require \(k \ge m+1\) and \(t-k \ge m+1\), i.e. \(k \in [m+1, t-m-1]\).
Now \(k \le t-m-1 < t\) and \(t-k \le t-m-1 < t\), so by the induction hypothesis \(a_k\) and \(a_{t-k}\) are infinite. Hence each multiplication \(a_k a_{t-k}\) is infinite, so the sum \[S_t := \sum_{k=m+1}^{t-m-1} a_k a_{t-k}\] is infinite (any sum containing an infinite cardinal is infinite). Then \(2a_t + S_t\) is infinite. Thus \(a_t\) must be infinite (otherwise LHS finite, RHS infinite). This completes the induction: all \(a_j\) for \(j \ge m+1\) are infinite.
Step 3. Now we proceed to derive the inequality for \(j \ge 2m+2\). Since all \(a_j\) (\(j \ge m+1\)) are infinite, we have \(2a_j = a_j\) for such \(j\). For \(j \ge 2m+2\): \[a_j = 2a_j + \sum_{k=m+1}^{j-m-1} a_k a_{j-k} = a_j + S_j,\] where \(S_j = \sum_{k=m+1}^{j-m-1} a_k a_{j-k}\). In \(\mathscr{C}\), for infinite cardinals \(a_k, a_{j-k}\): \[a_k a_{j-k} = \max(a_k, a_{j-k}),\] and the sum of infinite cardinals equals their maximum. Hence \[S_j = \max_{k=m+1}^{j-m-1} \max(a_k, a_{j-k}) = \max_{r=m+1}^{j-m-1} a_r.\] The equation \(a_j = a_j + S_j\) in cardinal arithmetic implies \(a_j \ge S_j\). Therefore: \[a_j \ge \max_{r=m+1}^{j-m-1} a_r \quad \text{for all } j \ge 2m+2.\] That is exactly \[a_j \ge \max\{a_{m+1}, a_{m+2}, \dots, a_{j-m-1}\}.\] Thus the three conditions hold, completing the proof of \(\Rightarrow\).
\((\Leftarrow)\): This direction holds by Proposition 13. ◻
Proposition 15. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring and let \(f = \sum_{n=0}^{+\infty} a_n X^n\) be an element of the formal power series semiring \(S[[X]]\) satisfying the following conditions for some \(\lambda \in L\) and \(m \ge 1\):
\(a_0 = \lambda\), and \(a_1 = a_2 = \dots = a_m = 0\),
\(a_j \in L\) and \(a_j \geq \lambda\) for all \(m+1 \leq j \leq 2m+1\),
\(a_j \geq \sup\{a_{m+1}, a_{m+2}, \dots, a_{j-m-1}\}\) for all \(j \geq 2m+2\).
Then \(f\) is a multiplicatively idempotent element of \(S[[X]]\).
Proof. Let \(b_n = \sum_{k=0}^n a_k a_{n-k}\) denote the coefficients of the squared series \(f^2 = f \cdot f\). We verify that \(b_n = a_n\) for all \(n \geq 0\) by matching the case architecture of Proposition 13:
Case \(n = 0\): The constant term evaluates to \(b_0 = a_0 a_0 = \lambda \cdot \lambda\). Because \(\lambda \in L\) and multiplication in \(L\) reduces to the semilattice join, we have \(\lambda \cdot \lambda = \sup\{\lambda, \lambda\} = \lambda = a_0\).
Case \(1 \leq n \leq m\): Since \(a_1 = \dots = a_m = 0\), every internal multiplication \(a_k a_{n-k}\) contains at least one factor equal to \(0\). For the boundary terms where \(k=0\) or \(k=n\), we have \(a_0 a_n = \lambda \cdot 0 = 0\) and \(a_n a_0 = 0 \cdot \lambda = 0\), owing to the multiplicative absorption of \(0\) across \(S\). Thus, every term vanishes, yielding \(b_n = 0 = a_n\).
Case \(m+1 \leq n \leq 2m+1\): For indices in this window, the interior terms where \(1 \leq k \leq m\) or \(n-m \leq k \leq n-1\) vanish due to the zero-coefficients. The sum collapses entirely to the boundary terms: \[b_n = a_0 a_n + a_n a_0 = \lambda a_n + a_n \lambda.\] By hypothesis, \(a_n \in L\) and \(a_n \geq \lambda\). Under the multiplication rules of the extended algebra, the multiplication reduces to the join: \(\lambda a_n = a_n \lambda = \sup\{\lambda, a_n\} = a_n\). Substituting this back into the expression yields \(b_n = a_n + a_n = a_n\), via additive idempotence in \(L\).
Case \(n \geq 2m+2\): By hypothesis, \(a_n \geq a_{m+1} \geq \lambda\). Since \(a_{m+1} \in L\), this transitively forces \(a_n \in L\). Splitting the convolution sum yields: \[b_n = a_0 a_n + a_n a_0 + \sum_{k=m+1}^{n-m-1} a_k a_{n-k} = \lambda a_n + a_n \lambda + \sum_{k=m+1}^{n-m-1} \sup\{a_k, a_{n-k}\}.\] Because \(a_n \geq \lambda\), the boundary terms evaluate to \(\lambda a_n = a_n \lambda = a_n\). The summation simplifies exactly as in Proposition 13: \[b_n = a_n + a_n + \sup\{a_k\}_{k=m+1}^{n-m-1} = a_n + \sup\{a_k\}_{k=m+1}^{n-m-1}.\] Since \(a_n \geq \sup\{a_{m+1}, \dots, a_{n-m-1}\}\), the element \(a_n\) absorbs the entire supremum under addition, leaving \(b_n = a_n\).
Thus, \(b_n = a_n\) holds globally for all index regions, confirming that \(f^2 = f\). ◻
Theorem 16. Let \(\lambda\) be an infinite cardinal. An element \(f = \sum_{n=0}^{+\infty} a_n X^n\) with \(a_0 = \lambda\) is multiplicatively idempotent in the formal power series semiring \(\mathscr{C}[[X]]\) if and only if either \(f = \lambda\) or coefficients of \(f\) satisfy the following conditions:
\(a_1 = a_2 = \dots = a_m = 0\), and
\(a_j \geq \lambda\) are arbitrary infinite cardinals for \(m+1 \leq j \leq 2m+1\), and
\(a_j \geq \max\{a_{m+1}, a_{m+2}, \dots, a_{j-m-1}\}\) for all \(j \geq 2m+2\).
Proof. The proof follows the same strategy as that of Theorem 14, and is therefore omitted. ◻
Corollary 2. The multiplicatively idempotent elements of \(\mathscr{C}[[X]]\) are precisely those characterized by their constant term \(a_0 \in \{0,1\} \cup \{\text{infinite cardinals}\}\) under the conditions of Proposition 12, Theorem 14, and Theorem 16.
Proof. If \(f^2 = f\), equating the constant terms yields \(a_0^2 = a_0\), forcing \(a_0 \in \{0,1\} \cup \{\lambda \in \mathscr{C} : \lambda \text{ is infinite}\}\). The full classification follows by exhaustion:
If \(a_0 = 0\), then \(f = 0\) by Proposition 12.
If \(a_0 = 1\), the coefficients must satisfy Theorem 14.
If \(a_0 = \lambda\), the coefficients must satisfy Theorem 16.
Conversely, sufficiency in each case is guaranteed by the respective referenced results. ◻
Recall that an element \(a\) of a semiring \(A\) is called multiplicatively subidempotent if \(a+a^2 = a\). A semiring is called multiplicatively subidempotent if each of its elements is multiplicatively subidempotent [3].
Proposition 17. Let \(A\) be a multiplicatively subidempotent positive information algebra. Then the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) is also multiplicatively subidempotent.
Proof. If \(s \in A\), then \(s\) is multiplicatively subidempotent by hypothesis. If \(s = \ell \in L\), then multiplication in \(L\) is join, so \[\ell + \ell^2 = \sup\{\ell, \sup\{\ell,\ell\}\} = \ell.\] Hence every element of \(S\) is multiplicatively subidempotent, as claimed. ◻
Remark 18. Golan describes that the Viterbi semiring \(V\) (see Example 4) is not only additively idempotent and multiplicatively subidempotent, but also satisfies the following condition: \[\begin{align} \label{Viterbirarecondition} a+b \geq ab, \qquad \forall\, a,b \in V = [0,1]. \end{align}\tag{1}\] Note that in the Viterbi semiring, addition is the max operation.
Proposition 19. Let \(A\) be a positive information algebra satisfying the condition 1 . Then the overflow semiring \(S = V \oplus_{\mathop{\mathrm{ord}}} L\) satisfies the same condition.
Proof. Let \(a,b \in S\). If \(a,b \in A\), then \(a+b = a+_A b\) and \(ab = a\cdot_A b\), so \(a+b \ge ab\) by hypothesis. If \(a \neq 0\) and \(b \in L\), then both operations give \(\sup\{a,b\}\), so \(a+b = ab\). If \(a = 0\) and \(b \in L\), then \(a+b = \sup\{0,b\} = b\) and \(ab = 0\). Because \(0 \le b\) in \(S\) (every element of \(A\) is \(\le\) every element of \(L\)), we have \(b \ge 0\), i.e., \(a+b \ge ab\). The cases with \(a \in L\), \(b \in A\) are symmetric. Thus \(a+b \ge ab\) holds for all \(a,b \in S\). ◻
Recall that a semiring \(S\) is called austere if its only subtractive ideals are \(\{0\}\) and \(S\) [2].
Proposition 20. The overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) is austere.
Proof. Let \(I\) be a proper subtractive ideal of \(S\) and assume \(I \neq \{0\}\). Pick \(x \in I\), \(x \neq 0\). If \(x \in A\), then for any \(\ell \in L\), \[x\ell = \sup\{x,\ell\} = \ell \in I,\] so \(L \subseteq I\). Since \(1 + \ell = \ell \in I\) and \(1 \notin I\), subtractivity forces \(1 \in I\), a contradiction. If \(x \in L\), then \(1 + x = x \in I\) with \(1 \notin I\), again contradicting subtractivity. Hence no proper nonzero subtractive ideal exists, so the only subtractive ideals are \(\{0\}\) and \(S\). ◻
Proposition 21. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. Then:
If \(I \trianglelefteq A\), then \(I \cup L\) is an ideal of \(S\), and the map \[I \mapsto I \cup L\] is injective from \(\mathrm{Id}(A)\) into \(\mathrm{Id}(S)\).
If \(J \trianglelefteq S\), then \(J \cap A\) is an ideal of \(A\).
The contraction map \(J \mapsto J \cap A\) is injective on the set of ideals of \(S\) containing \(L\).
Proof. Let \(I \trianglelefteq A\). Then \(I \cup L\) is closed under addition and absorbs multiplication by elements of \(S\), hence is an ideal of \(S\). Injectivity of \(I \mapsto I \cup L\) follows from \((I \cup L) \cap A = I\).
Let \(J \trianglelefteq S\). If \(a,b \in J \cap A\), then \(a+b \in J \cap A\), and if \(a \in J \cap A\) and \(x \in A\), then \(ax \in J \cap A\), so \(J \cap A \trianglelefteq A\).
If \(J_1, J_2 \trianglelefteq S\) with \(L \subseteq J_1, J_2\) and \(J_1 \cap A = J_2 \cap A\), then \(J_i = (J_i \cap A) \cup L\), hence \(J_1 = J_2\). ◻
Theorem 22. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. Let \(J\) be a nonzero ideal of \(S\).
If \(J\) contains a nonzero element of \(A\), then \[J = J_A \cup L,\] where \(J_A = J \cap A\) is an ideal of \(A\). Moreover, \(J\) is a prime ideal of \(S\) if and only if \(J_A\) is a prime ideal of \(A\).
If \(J\) contains no nonzero element of \(A\) and \((L, \leq_L)\) is a well-ordered set, then \[J = \{0\} \cup \{\ell \in L: \ell \geq \ell_0\},\] where \(\ell_0\) is the smallest element of \(J \cap L\). In such a case, \(J\) is a prime ideal of \(S\).
Proof. We treat the two cases separately.
Let \(a \in J \cap A\) be nonzero and \(\ell \in L\). Since \(J\) is an ideal of \(S\), \[\ell \cdot a = \sup\{\ell, a\} = \ell \in J.\] Therefore \(L \subseteq J\), and so \(J_A \cup L \subseteq J\). Conversely, if \(x \in J\) and \(x \notin L\), then \(x \in A\), and thus \(x \in J_A\). Hence \(J = J_A \cup L\). Note that \(J_A\) is a contraction of \(J\) and so it is an ideal of \(A\). Moreover, if \(J\) is prime in \(S\), then its contraction \(J_A = J \cap A\) is prime in \(A\). Conversely, suppose \(J_A\) is prime in \(A\) and let \(x, y \notin J\). Since \(L \subseteq J\), we have \(x, y \in A \setminus J_A\). Because \(J_A\) is prime, \(x \cdot_A y \notin J_A\), hence \(x \cdot y \notin J\). Therefore \(J\) is prime in \(S\).
If \(J\) contains no nonzero element of \(A\), then every nonzero element of \(J\) lies in \(L\). Since \(L\) is well-ordered, \(J \setminus \{0\}\) has a smallest element \(\ell_0\). Then \[J \subseteq \{0\} \cup \{\ell \in L : \ell \geq \ell_0\}.\] For any \(\ell \geq \ell_0\), we have \(\ell = \ell \cdot \ell_0 = \sup\{\ell, \ell_0\} \in J\), so the reverse inclusion holds. Thus \[J = \{0\} \cup \{\ell \in L : \ell \geq \ell_0\}.\] We now show \(J\) is prime. Take \(x, y \in S\) with \(x \cdot y \in J\).
If \(x \cdot y = 0\), then \(x = 0\) or \(y = 0\) because \(S\) is entire, so \(x \in J\) or \(y \in J\).
If \(x \cdot y = \ell \geq \ell_0\) (nonzero), then the multiplication lies in \(L\).
If \(x, y \in L\), then \(x \cdot y = \sup(x,y) \in J\). Hence the larger of \(x,y\) is \(\geq \ell_0\), so it belongs to \(J\).
If one factor is in \(A \setminus \{0\}\) and the other in \(L\), say \(x = a \in A \setminus \{0\}\), \(y = \ell' \in L\), then \(x \cdot y = \sup(a,\ell') = \ell' \in J\), so \(y \in J\).
Thus in all cases \(x \in J\) or \(y \in J\), so \(J\) is prime.
This completes the proof. ◻
Corollary 3. Let \(J\) be a nonzero ideal of the semiring \(\mathscr{C}\) of cardinal numbers. Then exactly one of the following cases happens:
\(J\) contains a nonzero finite cardinal. Then \[J = J_f \cup \{\kappa : \kappa \text{ is an infinite cardinal}\},\] where \(J_f \trianglelefteq \mathbb{N}_0\) is the set of all finite cardinals in \(J\). Furthermore, \(J\) is a prime ideal of \(\mathscr{C}\) if and only if \(J_f\) is a prime ideal of \(\mathbb{N}_0\).
\(J\) contains no nonzero finite cardinal. Then \[J = \{0\} \cup \{\kappa \in \mathscr{C} : \kappa \geq \alpha\},\] where \(\alpha\) is the smallest infinite cardinal in \(J\). In this case, \(J\) is prime.
Proof. The semiring \(\mathbb{N}_0\) is a positive information algebra and the set of infinite cardinals is well-ordered by \(\leq\) [7]. ◻
Remark 23. The contraction map \(J \mapsto J \cap A\) is generally not injective on the full ideal lattice \(\mathrm{Id}(S)\) of the overflow semiring \(S\). Theorem 22 immediately provides counterexamples. For example, let \(L\) be the two-element chain \(\ell_1 < \ell_2\). By Theorem 22(2), choosing \(\ell_0=\ell_1\) and \(\ell_0=\ell_2\) yields two distinct ideals of \(S\) containing no nonzero elements of \(A\): \[J_1=\{0\}\cup\{\ell\in L:\ell\ge \ell_1\}=\{0,\ell_1,\ell_2\},\] \[J_2=\{0\}\cup\{\ell\in L:\ell\ge \ell_2\}=\{0,\ell_2\}.\] Since \[J_1\cap A=\{0\}=J_2\cap A\] while \(J_1\neq J_2\), the contraction map fails to be injective on \(\mathrm{Id}(S)\).
Theorem 24. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. If \(\mathfrak{m}\) is a maximal ideal of \(A\), then \(\mathfrak{m} \cup L\) is a maximal ideal of \(S\). Conversely, if \(\mathfrak{M}\) is a maximal ideal of \(S\), then \(\mathfrak{M} \cap A\) is a maximal ideal of \(A\). In particular, \(A\) is local if and only if \(S\) is local.
Proof. Let \(\mathfrak{m}\) be a maximal ideal of \(A\). Then \(\mathfrak{m} \cup L\) is an ideal of \(S\) by Proposition 21. Suppose \(J\) is an ideal of \(S\) properly containing \(\mathfrak{m} \cup L\). Then \(J\) contains some \(a \notin \mathfrak{m} \cup L\). Since \(A\) and \(L\) are disjoint, \(a \in A \setminus \mathfrak{m}\). Because \(\mathfrak{m}\) is maximal in \(A\), the ideal generated by \(\mathfrak{m}\) and \(a\) in \(A\) is \(A\) itself; hence \(1_A \in A\) belongs to this ideal. Consequently \(1_A \in J\), which forces \(J = S\). Thus \(\mathfrak{m} \cup L\) is maximal in \(S\).
Conversely, let \(\mathfrak{M}\) be a maximal ideal of \(S\). Then \(\mathfrak{M} \cap A\) is the contraction of \(\mathfrak{M}\), hence an ideal of \(A\). If \(\mathfrak{M} \cap A\) were not maximal, there would exist a proper ideal \(\mathfrak{n}\) of \(A\) such that \(\mathfrak{M} \cap A \subset \mathfrak{n} \subset A\). Then \(\mathfrak{n} \cup L\) is an ideal of \(S\) properly containing \(\mathfrak{M}\) because \[\mathfrak{M} = (\mathfrak{M} \cap A) \cup L \subset \mathfrak{n} \cup L\] and is proper because \(1_A \notin \mathfrak{n}\). This contradicts the maximality of \(\mathfrak{M}\). Therefore \(\mathfrak{M} \cap A\) is maximal in \(A\). The final statement follows immediately: \(A\) has a unique maximal ideal if and only if \(S\) does. ◻
Proposition 25. Let \(A\) be a positive information algebra such that \(a \geq 1\) for all nonzero \(a \in A\). Also suppose that if \(a,b > 1\), then \(a+b > 1\). Let \(L\) be a join-semilattice. Then the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) is local and its unique maximal ideal is \(\mathfrak{M} = S \setminus \{1\}\).
Proof. Consider the subset \(\mathfrak{m} = A \setminus \{1\}\) of \(A\). If \(x,y \in \mathfrak{m}\), then \(x,y > 1\) and by assumption \(x+y > 1\), so \(x+y \in \mathfrak{m}\). If \(a \in A\) is arbitrary and \(x \in \mathfrak{m}\), then \(a \geq 1\) and \(x > 1\). It follows that \(ax \geq x > 1\), and hence \(ax \in \mathfrak{m}\). Thus \(\mathfrak{m}\) is an ideal of \(A\). No proper ideal of \(A\) properly contains \(\mathfrak{m}\) because any such ideal would have to contain \(1\). Hence \(\mathfrak{m}\) is a maximal ideal of \(A\). Also, \(\mathfrak{m}\) is the unique maximal ideal since every proper ideal of \(A\) is contained in \(\mathfrak{m}\). By Theorem 24, \[\mathfrak{m} \cup L = S \setminus \{1\} = \mathfrak{M}\] is the unique maximal ideal of \(S\). Therefore \(S\) is local, completing the proof. ◻
Example 2. Observe that \(B = \{0\} \cup [1, +\infty)\) is a subsemiring of \(\mathbb{R}_{\geq 0}\) with the usual operations. Take any subsemiring \(A\) of \(B\), and let \(L\) be the class of infinite cardinal numbers under the usual cardinal ordering. Then \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) satisfies the conditions of Proposition 25.
Following the terminology for PM-rings (cf. [8]), we define PM-semirings as follows:
Definition 6. A semiring \(S\) is called a PM-semiring if each prime ideal of \(S\) is contained in a unique maximal ideal of \(S\).
Proposition 26. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring. Then the following statements are equivalent:
\(A\) is local.
\(S\) is local.
\(S\) is a PM-semiring.
Proof. \((1) \Leftrightarrow (2)\) This follows directly from Theorem 24.
\((2) \Rightarrow (3)\) If \(S\) is local, it possesses exactly one maximal ideal, which must automatically contain every proper ideal (and thus every prime ideal) of \(S\). Hence, \(S\) is a PM-semiring.
\((3) \Rightarrow (1)\) Suppose \(S\) is a PM-semiring. Since \(S\) is an entire semiring by Theorem 3, the zero ideal \(\{0\}\) is a prime ideal of \(S\). By definition of a PM-semiring, \(\{0\}\) must be contained in a unique maximal ideal of \(S\). Since every maximal ideal is proper and thus contains \(\{0\}\), it follows that \(S\) can possess only one unique maximal ideal. Thus, \(S\) is local, implying that \(A\) is local by Theorem 24, which completes the proof. ◻
Proposition 27. Let \(A\) be a positive information algebra such that \(a \leq 1\) for all nonzero \(a \in A\), and assume that \(a,b<1\) implies \(a+b<1\). Let \(L\) be a join-semilattice. Then the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) is local with unique maximal ideal \(\mathfrak{m} = S \setminus \{1\}\).
Proof. The proof is dual to that of Proposition 25, with all inequalities reversed relative to the order structure of \(A\), and is therefore omitted. ◻
Example 3. The restricted max-plus algebra \(A = [-\infty, 0]\), with addition \(x \oplus y = \max(x,y)\) and multiplication \(x \odot y = x + y\), satisfies the conditions of Proposition 27. Indeed, the additive identity is \(0_A = -\infty\) and the multiplicative identity is \(1_A = 0\). Every \(a \in A\) with \(a \neq 0_A\) (i.e., \(a \in (-\infty, 0]\)) satisfies \(a \leq 1_A\). Moreover, if \(a,b < 1_A\), then \(a,b < 0\), and their sum \(a \oplus b = \max(a,b) < 0 = 1_A\).
Definition 7. Let \(S\) be a semiring. The Krull dimension of \(S\), denoted \(\dim S\), is the supremum of all integers \(n \geq 0\) such that there exists a strictly ascending chain \[P_0 \subset P_1 \subset \cdots \subset P_n\] of prime ideals of \(S\). If no such finite supremum exists, we set \(\dim S = \infty\).
Theorem 28. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring, where \(A\) is a positive information algebra and \(L\) is a well-ordered set. Then the Krull dimension of \(S\) is given by \[\dim S = \begin{cases} \dim A, & \text{if } L = \varnothing,\\[4pt] \infty, & \text{if } \dim A = \infty \text{ or } L \text{ is infinite},\\[4pt] |L| + \dim A, & \text{if } L \text{ is finite and nonempty and } \dim A < \infty. \end{cases}\]
Proof. If \(L = \varnothing\), then \(S = A\), so \(\dim S = \dim A\). If \(\dim A = \infty\), then \(S\) contains an infinite chain from \(A\) lifted via \(I \mapsto I \cup L\), so \(\dim S = \infty\). If \(L\) is infinite, then for any \(m \in \mathbb{N}\) choose \(\ell_1 < \cdots < \ell_m\) in \(L\) and a chain \(P_0 \subset \cdots \subset P_n\) in \(A\); then \[I_{\ell_m} \subset \cdots \subset I_{\ell_1} \subset P_0 \cup L \subset \cdots \subset P_n \cup L\] is a chain of length \(m + n\). Since \(m\) is arbitrary, \(\dim S = \infty\).
Now suppose \(L\) is finite and nonempty with \(|L| = k\) and \(\dim A = n < \infty\). Since \(\dim A = n\), there exists a chain \(P_0 \subset \cdots \subset P_n\) of primes in \(A\). Lifting gives \(P_0 \cup L \subset \cdots \subset P_n \cup L\) in \(S\). List \(L\) increasingly: \(\ell_1 < \cdots < \ell_k\). For each \(\ell_i\), the ideal \(I_{\ell_i} = \{0\} \cup \{\ell \in L : \ell \geq \ell_i\}\) is prime (Type II), and \(\ell_i < \ell_{i+1}\) implies \(I_{\ell_{i+1}} \subset I_{\ell_i}\). Moreover, \(I_{\ell_1} \subset P_0 \cup L\). Concatenating, \[I_{\ell_k} \subset \cdots \subset I_{\ell_1} \subset P_0 \cup L \subset \cdots \subset P_n \cup L\] is a chain of length \(k + n\), so \(\dim S \geq k + n\). By Theorem 22, every prime ideal of \(S\) is either Type I (\(P \cup L\)) or Type II (\(I_\ell\)), and any chain can contain at most \(k\) Type II primes (all of them, in reverse order) followed by at most \(n+1\) Type I primes. Therefore \(\dim S \leq k + n\). Hence, \(\dim S = k + n\), as claimed. ◻
Examples 29. We illustrate the Krull dimension of the overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) below:
Let \(A\) be a zerosumfree positive semifield (so \(\dim A = 0\)) and let \(L\) be a finite chain of length \(m\) (i.e., \(|L| = m\)). Then \(\dim S = m\).
The Krull dimension of the semiring \(\mathscr{C}\) of cardinal numbers is \(\infty\) because we have infinitely many cardinal numbers, including \(\beth_0 = \aleph_0\) and \(\beth_k = 2^{\beth_{k-1}}\) for every positive integer \(k\) [7].
Lemma 4. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring, where \(A\) is a positive information algebra and \(L\) is a well-ordered set. Let \(I\) be an ideal of \(S\). Then:
If \(I\) contains no nonzero element of \(A\), then \(I = (\ell_0)\) is principal, where \(\ell_0 = \min(I \cap L)\).
If \(I\) contains a nonzero element of \(A\) (so \(I = I_A \cup L\)), then:
If \(G\) generates \(I_A\) in \(A\), then \(G \cup \{\ell_{\min}\}\) generates \(I\) in \(S\), where \(\ell_{\min} = \min L\).
If \(H\) generates \(I\) in \(S\), then \(H \cap A\) generates \(I_A\) in \(A\).
Proof. The first case follows directly from Theorem 22.
Now assume \(I\) contains a nonzero element of \(A\). For the first statement, \(G\) generates \(I_A\) in \(A\), hence in \(S\). Since every \(\ell \in L\) satisfies \(\ell = \ell \cdot \ell_{\min}\), \(G \cup \{\ell_{\min}\}\) generates \(I\).
For the converse, let \(H\) generate \(I\) in \(S\), and let \(a \in I_A\). Then \(a\) can be expressed as a finite sum \(a = \sum_{i=1}^n s_i h_i\) for some \(s_i \in S\) and \(h_i \in H\). By Definition 1, addition with any element of \(L\) results in an element of \(L\). Since \(a \in A\) and \(A \cap L = \varnothing\), it follows that no individual nonzero term \(s_i h_i\) can belong to \(L\). Because the multiplication of any nonzero element with an element of \(L\) always lands in \(L\), each nonzero term \(s_i h_i \in A\) strictly requires that both \(s_i \in A\) and \(h_i \in H \cap A\). Thus, \(a\) is generated entirely by \(H \cap A\) using \(A\)-operations, proving that \(H \cap A\) generates \(I_A\) in \(A\). ◻
Recall that a semiring \(S\) is Noetherian if and only if every ideal of \(S\) is finitely generated [2].
Theorem 30. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring, where \(A\) is a positive information algebra and \(L\) is a well-ordered set. Then \(S\) is Noetherian if and only if \(A\) is Noetherian.
Proof. \((\Rightarrow)\) Suppose \(S\) is Noetherian. Assume, toward a contradiction, that \(A\) is not Noetherian. Then there exists an ideal \(I_A\) of \(A\) that is not finitely generated. Since the zero ideal \(\{0\}\) is finitely generated, \(I_A\) must contain a nonzero element of \(A\). By Theorem 22, \(I=I_A\cup L\) is an ideal of \(S\). By Lemma 4, \(I\) is finitely generated in \(S\) if and only if \(I_A\) is finitely generated in \(A\). This contradicts the assumption that \(S\) is Noetherian. Hence \(A\) is Noetherian.
\((\Leftarrow)\) Conversely, assume \(A\) is Noetherian, and let \(I\) be an ideal of \(S\). If \(I\) contains no nonzero element of \(A\), then by Theorem 22, \(I=(\ell_0)\), where \(\ell_0=\min(I\cap L)\). Thus \(I\) is principal. Now suppose \(I\) contains a nonzero element of \(A\). Then \(I=I_A\cup L\), where \(I_A=I\cap A\) is an ideal of \(A\). Since \(A\) is Noetherian, \[I_A=(a_1,\dots,a_n)\] for some \(a_1,\dots,a_n\in A\). By Lemma 4, \[I=(a_1,\dots,a_n,\ell_{\min}),\] where \(\ell_{\min}=\min L\). Hence \(I\) is finitely generated. Therefore every ideal of \(S\) is finitely generated, so \(S\) is Noetherian. ◻
Corollary 4. Let \(L\) be a well-ordered set. Then the overflow semiring \(S= \mathbb{N}_0 \oplus_{\mathop{\mathrm{ord}}} L\) is a Noetherian semiring. In particular, the class-semiring of cardinal numbers is Noetherian.
Proof. It is easy to see that \(I\) is an ideal of the semiring \(\mathbb{N}_0\) if and only if \(I\) is a submonoid of \((\mathbb{N}_0,+)\). By Corollary 2.8 in [9], every submonoid of \(\mathbb{N}_0\) is finitely generated. It follows that the semiring \(\mathbb{N}_0\) is Noetherian. Thus by Theorem 30, \(S\) is Noetherian finishing the proof. ◻
Recall that a semiring \(S\) is called Artinian if \(S\) satisfies the descending chain condition on its ideals (see Definition 1.1 in [10]).
Theorem 31. Let \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) be the overflow semiring, where \(A\) is a positive information algebra and \(L\) is a well-ordered set. Then \(S\) is Artinian if and only if:
\(A\) is Artinian;
\(L\) is finite.
Proof. \((\Rightarrow)\) Suppose \(S\) is Artinian. First, we show that \(A\) is Artinian. Let \[J_1 \supseteq J_2 \supseteq J_3 \supseteq \cdots\] be a descending chain of ideals in \(A\). By Proposition 21, \[J_1\cup L \supseteq J_2\cup L \supseteq J_3\cup L \supseteq \cdots\] is a descending chain of ideals in \(S\). Since \(S\) is Artinian, there exists \(N\) such that \[J_n\cup L = J_{n+1}\cup L\] for all \(n\ge N\). Intersecting with \(A\) yields \(J_n=J_{n+1}\) for all \(n\ge N\). Hence \(A\) is Artinian. Next, we prove that \(L\) is finite. Assume, toward a contradiction, that \(L\) is infinite. Since \(L\) is well-ordered, there exists an infinite strictly increasing sequence \[\ell_1<\ell_2<\ell_3<\cdots.\] For each \(n\ge1\), define \[I_n=\{0\}\cup\{\ell\in L:\ell\ge\ell_n\}.\] By Theorem 22, each \(I_n\) is an ideal of \(S\). Since \(\ell_n\in I_n\) but \(\ell_n\notin I_{n+1}\), \[I_1 \supset I_2 \supset I_3 \supset \cdots\] is a strictly descending chain of ideals of \(S\), contradicting the Artinian property. Therefore \(L\) must be finite.
\((\Leftarrow)\) Conversely, suppose \(A\) is Artinian and \(L\) is finite. Let \[I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots\] be a descending chain of ideals in \(S\).
Case 1: Infinitely many ideals \(I_n\) contain a nonzero element of \(A\). Passing to a subsequence if necessary, we may assume every \(I_n\) contains a nonzero element of \(A\). By Theorem 22, \[I_n = J_n \cup L,\] where \(J_n = I_n \cap A\) is an ideal of \(A\). Since \[I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots,\] we obtain a descending chain \[J_1 \supseteq J_2 \supseteq J_3 \supseteq \cdots\] in \(A\). Because \(A\) is Artinian, there exists \(N\) such that \[J_n=J_{n+1}\] for all \(n\ge N\). Consequently, \[I_n=J_n\cup L = J_{n+1}\cup L = I_{n+1},\] so the chain stabilizes.
Case 2: Only finitely many ideals \(I_n\) contain a nonzero element of \(A\). Then there exists \(N\) such that for all \(n\ge N\), the ideal \(I_n\) contains no nonzero element of \(A\). By Theorem 22, \[I_n=\{0\}\cup\{\ell\in L:\ell\ge \ell_n\},\] where \[\ell_n=\min(I_n\cap L).\] Since \(I_n\supseteq I_{n+1}\), we have \[\ell_n\le \ell_{n+1}.\] Thus \((\ell_n)\) is a nondecreasing sequence in the finite set \(L\), and hence eventually constant. Therefore the chain \((I_n)\) stabilizes. In all cases, every descending chain of ideals stabilizes. Hence \(S\) is Artinian. ◻
Example 4. Note that in the semiring \(\mathbb{N}_0\), the ideals generated by \(2^n\) form an infinite strictly descending chain: \[(1) \;\supset\; (2) \;\supset\; (4) \;\supset\; (8) \;\supset\; \cdots\] Hence \(\mathbb{N}_0\) is not Artinian. Therefore by Theorem 31, for any well-ordered set \(L\), the overflow semiring \(S = \mathbb{N}_0 \oplus_{\mathop{\mathrm{ord}}} L\) is not Artinian.
Remark 32. The overflow semiring \(S = A \oplus_{\mathop{\mathrm{ord}}} L\) is structurally analogous to the indigenous semirings \[S_k = \{0, 1, \dots, k, m\}\] introduced in [11]. Both frameworks model computational saturation and boundary capping: while \(S_k\) truncates arithmetic above \(k\) into a monolithic element \(m\), \(S\) expands this saturation zone into an arbitrary join-semilattice \(L\) dominating a general positive information algebra \(A\). Algebraically, both structures are entire and zerosumfree, their upper-layer elements collapse into idempotents, and both are austere. However, whereas \(S_k\) is a finite local semiring with a fixed Krull dimension (\(\dim S_k = 1\)), \(S\) possesses a far richer prime ideal spectrum, dynamic Noetherian and Artinian properties, and a customizable Krull dimension governed by the formula \[\dim S = \dim A + |L|\] for finite chains \(L\).