\documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern} \usepackage[margin=1.1in]{geometry} \usepackage{microtype} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{enumitem} \usepackage{hyperref} \usepackage{xcolor} \hypersetup{colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black,pdfauthor={The Clankers},pdftitle={Secondary Note on Split-Zero Globalization}} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newcommand{\B}{\mathbf B} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\im}{\operatorname{Im}} \newcommand{\id}{\operatorname{id}} \newcommand{\cO}{\mathcal O} \newcommand{\cP}{\mathcal P} \newcommand{\N}{\mathbf N} \newcommand{\Z}{\mathbf Z} \newcommand{\Q}{\mathbf Q} \newcommand{\F}{\mathbf F} \newcommand{\C}{\mathbf C} \newcommand{\rad}{\operatorname{rad}} \newcommand{\Cl}{\operatorname{Cl}} \newcommand{\Diag}{\mathsf{Diag}} \newcommand{\SMod}{\mathbf{SMod}} \newcommand{\Mod}{\mathbf{Mod}} \newcommand{\one}{\mathbf 1} \title[Secondary Note on Split-Zero Globalization]{A Secondary Note on Split-Zero Globalization\\[4pt]\large Semimodule Classification, Multiplicative Reconstruction, and All-Orders Zeta Deformation} \author{The Clankers} \date{\today} \begin{document} \begin{abstract} We extend the baseline globalization theory of the split-zero semiring \[ S=G(R)=R\sqcup\{\tau\}, \qquad r\oplus s=r+s, \quad r\oplus\tau=r, \quad \tau\oplus\tau=\tau, \] \[ rs=rs, \qquad r\tau=\tau r=\tau, \qquad \tau^2=\tau, \] for a nonzero commutative ring $R$ with identity. Throughout, we assume the results of the companion note: the ring reflection $p_R$, the support character $\chi_R$, the classification of ideals, prime ideals, congruences, and localizations, and the supported-zero projector $\nu_R(x)=0_Rx$. The main results are the exact classification of $G(R)$-semimodules as join-semilattice-indexed diagrams of $R$-modules; the identification of the cancellative subcategory with $R$-Mod; the Boolean-cubical form of free semimodules; the product theorem $G(A\times B)\cong G(A)\times_{\B}G(B)$; reconstruction from the multiplicative monoid together with the missing additive relations; the infinite-dimensionality of the multiplicative-monoid spectrum in the UFD case; the contracted monoid-algebra decomposition; conservative ideal/class/factorization results; and the full all-orders deformation theory of the shifted quotient-size zeta sheet, including universal jets, cumulants, arithmetic kernels, non-Eulerianity, Faulhaber interpolation, factorial interpolation, and the exact endpoint defect controlled by a two-point Stone model. \end{abstract} \maketitle \tableofcontents \section{Standing background and notation} Throughout $R$ denotes a nonzero commutative ring with identity and \[ S:=G(R)=R\sqcup\{\tau\}, \qquad e:=0_R\in R\subset S. \] We assume the notation and the proved results of the baseline globalization note, namely: \begin{enumerate}[label=(\roman*)] \item the ring reflection \[ p_R:S\to R, \qquad p_R(\tau)=0_R, \quad p_R(r)=r; \] \item the support character \[ \chi_R:S\to\B, \qquad \chi_R(\tau)=0, \quad \chi_R(r)=1; \] \item the classification of ideals, prime ideals, congruences, and localizations away from nilpotent supported elements; \item the supported-zero projector \[ \nu_R:S\to S, \qquad \nu_R(x)=ex, \] with image $\{\tau,e\}\cong\B$. \end{enumerate} No proof from that baseline package is repeated below. The only external analytic facts used are standard properties of the Hurwitz zeta function and of Dedekind zeta functions; see the references at the end. \section{Algebraic refinements beyond the baseline package} \subsection{Algebraic normal form and universal quotients} \begin{proposition}\label{prop:normal-form} There is a canonical semiring isomorphism \[ G(R)\xrightarrow{\sim}D_R:=\{(r,b)\in R\times\B: b=0\Rightarrow r=0_R\} =\{(0_R,0)\}\cup (R\times\{1\}). \] \end{proposition} \begin{proof} Define \[ \iota:G(R)\to D_R, \qquad \iota(\tau)=(0_R,0), \qquad \iota(r)=(r,1) \quad(r\in R). \] This is bijective. On $D_R$, componentwise addition and multiplication restrict to \[ (r,1)+(s,1)=(r+s,1), \qquad (r,1)+(0,0)=(r,1), \qquad (0,0)+(0,0)=(0,0), \] \[ (r,1)(s,1)=(rs,1), \qquad (r,1)(0,0)=(0,0), \qquad (0,0)^2=(0,0), \] which are exactly the defining operations of $G(R)$ transported through $\iota$. \end{proof} \begin{theorem}\label{thm:ring-quotient} For every commutative ring $A$, composition with $p_R$ induces a bijection \[ \Hom_{\mathrm{Ring}}(R,A) \xrightarrow{\sim} \Hom_{\mathrm{SemiRing}}(G(R),A). \] In particular, $R$ is the universal ring quotient of $G(R)$. \end{theorem} \begin{proof} Let $f:G(R)\to A$ be a semiring homomorphism. Since $\tau=0_S$, one has \[ f(\tau)=0_A. \] Also, \[ f(0_R)=f(0_R\oplus 0_R)=f(0_R)+f(0_R), \] so subtraction in the ring $A$ gives $f(0_R)=0_A$. Restrict $f$ to $R$ and call the restriction $g$. For $r,s\in R$, \[ g(r+s)=f(r\oplus s)=f(r)+f(s)=g(r)+g(s), \] \[ g(rs)=f(rs)=f(r)f(s)=g(r)g(s), \qquad g(1_R)=f(1_R)=1_A. \] Hence $g:R\to A$ is a ring homomorphism. Since $p_R(\tau)=0_R$ and $p_R(r)=r$, one has \[ (g\circ p_R)(\tau)=g(0_R)=0_A=f(\tau), \qquad (g\circ p_R)(r)=g(r)=f(r). \] Thus $f=g\circ p_R$. Uniqueness follows from surjectivity of $p_R$. \end{proof} \begin{corollary}\label{cor:no-splitting} There is no semiring homomorphism $s:R\to G(R)$ such that \[ p_R\circ s=\id_R. \] \end{corollary} \begin{proof} If such $s$ existed, then $s(0_R)=\tau$ because $s$ preserves the semiring zero, and $s(1_R)=1_R$ because $p_R\circ s=\id_R$. Also $p_R(s(-1_R))=-1_R\neq0_R$, so $s(-1_R)$ is supported. Hence \[ \tau=s(0_R)=s(1_R+(-1_R))=s(1_R)\oplus s(-1_R)=1_R\oplus s(-1_R), \] but the right-hand side is supported, contradiction. \end{proof} \begin{theorem}\label{thm:boolean-character-unique} The support character \[ \chi_R:G(R)\to\B \] is the unique semiring homomorphism from $G(R)$ to $\B$. \end{theorem} \begin{proof} Let $u:G(R)\to\B$ be a semiring homomorphism. Then $u(\tau)=0$ and $u(1_R)=1$. Now \[ u(0_R)=u(1_R\oplus (-1_R))=u(1_R)+u(-1_R)=1+u(-1_R)=1 \] in $\B$. If $u(r)=0$ for some $r\in R$, then \[ 1=u(0_R)=u(r0_R)=u(r)u(0_R)=0, \] contradiction. Hence $u(r)=1$ for every $r\in R$. Therefore $u=\chi_R$. \end{proof} \begin{corollary}\label{cor:no-ring-to-bool} There is no semiring homomorphism $R\to\B$. \end{corollary} \begin{proof} If $v:R\to\B$ were such a homomorphism, then \[ 0=v(0_R)=v(1_R+(-1_R))=v(1_R)+v(-1_R)=1+v(-1_R)=1, \] contradiction. \end{proof} \begin{corollary}\label{cor:no-unital-bool-section} There is no unit-preserving semiring homomorphism $\B\to G(R)$. \end{corollary} \begin{proof} If $w:\B\to G(R)$ were unit-preserving, then $w(1)=1_R$. Since $1+1=1$ in $\B$, \[ 1_R=w(1)=w(1+1)=w(1)\oplus w(1)=1_R\oplus 1_R=1_R+1_R. \] Subtracting $1_R$ inside the supported copy of $R$ yields $1_R=0_R$, contradiction. \end{proof} \begin{theorem}\label{thm:full-faithfulness} For nonzero commutative rings $R,R'$, restriction induces a bijection \[ \Hom_{\mathrm{SemiRing}}(G(R),G(R')) \xrightarrow{\sim} \Hom_{\mathrm{Ring}}(R,R'). \] Consequently, \[ \End_{\mathrm{SemiRing}}(G(R))\cong \End_{\mathrm{Ring}}(R), \qquad \Aut_{\mathrm{SemiRing}}(G(R))\cong \Aut_{\mathrm{Ring}}(R). \] \end{theorem} \begin{proof} Let $f:G(R)\to G(R')$ be a semiring homomorphism. Since $f$ preserves semiring zero, $f(\tau)=\tau$. By Theorem~\ref{thm:boolean-character-unique}, the composite $\chi_{R'}\circ f$ must equal $\chi_R$. Therefore for every $r\in R$, \[ \chi_{R'}(f(r))=\chi_R(r)=1, \] so $f(r)$ is supported, hence belongs to $R'$. Thus $f$ restricts to a map $R\to R'$. For $r,s\in R$, \[ f(r+s)=f(r\oplus s)=f(r)\oplus f(s)=f(r)+f(s), \] \[ f(rs)=f(r)f(s), \qquad f(1_R)=1_{R'}. \] Hence $f|_R$ is a ring homomorphism. Conversely, if $g:R\to R'$ is a ring homomorphism, define \[ G(g)(\tau)=\tau, \qquad G(g)(r)=g(r) \quad(r\in R). \] A direct check shows that $G(g)$ is a semiring homomorphism. The two constructions are inverse. \end{proof} \subsection{Product structure and multiplicative shadows} \begin{theorem}\label{thm:product-fiber} For nonzero commutative rings $A,B$, \[ G(A\times B)\cong G(A)\times_{\B}G(B), \] where the fiber product is taken with respect to the support characters. \end{theorem} \begin{proof} Define \[ \Phi:G(A\times B)\to G(A)\times_{\B}G(B) \] by \[ \Phi(\tau)=(\tau,\tau), \qquad \Phi((a,b))=(a,b). \] This is well defined because every supported element of $G(A\times B)$ has support character $1$ in both components, while $\tau$ has support character $0$ in both components. Conversely, an element of the fiber product is either $(\tau,\tau)$ or a pair $(x,y)$ with $\chi_A(x)=\chi_B(y)=1$, hence $x\in A$ and $y\in B$. Define \[ \Psi(\tau,\tau)=\tau, \qquad \Psi(x,y)=(x,y)\in A\times B. \] These maps are inverse semiring homomorphisms. \end{proof} \begin{theorem}\label{thm:monoid-reconstruction} Let $M_S=(S,\cdot)$ be the multiplicative monoid of $S$. Let $T(S)$ be the quotient of the free commutative semiring $\mathbb N[M_S]$ by the semiring congruence generated by \[ [a]+[b]\sim [a+b] \qquad (a,b\in R), \] and \[ [x]+[\tau]\sim [x] \qquad (x\in M_S). \] Then \[ T(S)\cong S. \] \end{theorem} \begin{proof} Define \[ \Phi:\mathbb N[M_S]\to S, \qquad \Phi([x])=x. \] The generating relations hold in $S$, because addition of supported elements is ring addition and $x\oplus\tau=x$ for every $x\in M_S$. Hence $\Phi$ factors through a semiring homomorphism \[ \overline\Phi:T(S)\to S. \] Conversely, define \[ \Psi:S\to T(S), \qquad \Psi(x)=[x]. \] This is multiplicative by construction and additive because \[ \Psi(a\oplus b)=\Psi(a+b)=[a+b]=[a]+[b] \qquad(a,b\in R), \] and \[ \Psi(x\oplus\tau)=\Psi(x)=[x]=[x]+[\tau] \qquad(x\in M_S). \] The maps $\overline\Phi$ and $\Psi$ are inverse on generators, hence everywhere. \end{proof} \begin{theorem}\label{thm:monoid-prime-explosion} Assume that $R$ is a UFD with infinitely many pairwise nonassociate irreducibles. Let $\Pi$ be a fixed set of representatives of the irreducible classes. For $\Sigma\subseteq\Pi$, define \[ \mathfrak p_\Sigma:=\{\tau,0_R\}\cup \{x\in R\setminus\{0\}:p\mid x\text{ for some }p\in\Sigma\}. \] Then: \begin{enumerate}[label=(\arabic*)] \item $\mathfrak p_\Sigma$ is a prime ideal of the commutative monoid $M_S$; \item for $\Sigma_1,\Sigma_2\subseteq\Pi$, \[ \mathfrak p_{\Sigma_1}\subseteq \mathfrak p_{\Sigma_2} \iff \Sigma_1\subseteq\Sigma_2; \] \item $\Spec(M_S)$ has infinite Krull dimension. \end{enumerate} \end{theorem} \begin{proof} First, $\mathfrak p_\Sigma$ is an ideal: multiplying $\tau$ by anything gives $\tau$, multiplying $0_R$ by anything gives $0_R$ or $\tau$, and multiplying a supported element divisible by some $p\in\Sigma$ by another supported element preserves divisibility by $p$. It is proper because $1\notin\mathfrak p_\Sigma$. Now suppose $ab\in\mathfrak p_\Sigma$. If $ab=\tau$, one factor is $\tau$. If $ab=0_R$, then $a,b\in R$, and since $R$ is a domain, one factor is $0_R$. If $ab\neq0_R$ and is divisible by some $p\in\Sigma$, then $a,b\in R\setminus\{0\}$ and irreducibles are prime in a UFD, so $p\mid a$ or $p\mid b$. Hence one factor lies in $\mathfrak p_\Sigma$. Thus $\mathfrak p_\Sigma$ is prime. If $\Sigma_1\subseteq\Sigma_2$, then $\mathfrak p_{\Sigma_1}\subseteq\mathfrak p_{\Sigma_2}$ is immediate. Conversely, if $\mathfrak p_{\Sigma_1}\subseteq\mathfrak p_{\Sigma_2}$ and $p\in\Sigma_1$, then $p\in\mathfrak p_{\Sigma_2}$, so some $q\in\Sigma_2$ divides $p$. Since $p$ and $q$ are irreducible, they are associate; by the choice of $\Pi$, one has $q=p$. Thus $p\in\Sigma_2$. Choose pairwise distinct $p_1,p_2,\dots\in\Pi$. Then \[ \{\tau\} \subsetneq \mathfrak p_{\{p_1\}} \subsetneq \mathfrak p_{\{p_1,p_2\}} \subsetneq\cdots \] is a chain of prime ideals of arbitrary finite length. \end{proof} \begin{theorem}\label{thm:contracted-monoid-algebra} Assume that $R$ is an integral domain. Let \[ R^{\bullet}:=R\setminus\{0\}. \] Then \[ \Z_0[M_S]:=\Z[M_S]/([\tau])\cong \Z\times \Z[R^{\bullet}]. \] Consequently, \[ \Spec(\Z_0[M_S])\cong \Spec(\Z)\sqcup \Spec(\Z[R^{\bullet}]). \] \end{theorem} \begin{proof} Killing $[\tau]$ removes the absorbing element, so \[ \Z_0[M_S]\cong \Z[M], \] where $M=(R,\cdot)$. Define \[ \Phi:\Z[M]\to \Z\times \Z[R^{\bullet}] \] on basis elements by \[ \Phi([0_R])=(1,0), \qquad \Phi([r])=(1,[r]) \quad(r\neq0). \] If one factor is $[0_R]$, multiplicativity is immediate. If $r,s\neq0$, then $rs\neq0$ because $R$ is a domain, and \[ \Phi([r])\Phi([s])=(1,[r])(1,[s])=(1,[rs])=\Phi([rs]). \] Hence $\Phi$ is a ring homomorphism. Define \[ \Psi:\Z\times \Z[R^{\bullet}]\to \Z[M] \] by \[ \Psi\!\left(c,\sum_{r\neq0}a_r[r]\right)= \left(c-\sum_{r\neq0}a_r\right)[0_R]+\sum_{r\neq0}a_r[r]. \] A direct computation shows $\Psi\Phi=\id$ and $\Phi\Psi=\id$. The spectrum statement follows from the elementary description of spectra of product rings. \end{proof} \section{Semimodule classification: exact replacement of $R$-linear algebra} \begin{lemma}\label{lem:epsilon-basic} Let $M$ be a left $S$-semimodule and define \[ \varepsilon_M(m):=em. \] Then: \begin{enumerate}[label=(\arabic*)] \item $\varepsilon_M$ is additive; \item $\varepsilon_M^2=\varepsilon_M$; \item for every $m\in M$, \[ m+\varepsilon_M(m)=m. \] \end{enumerate} \end{lemma} \begin{proof} Additivity and idempotence are immediate: \[ \varepsilon_M(m+n)=e(m+n)=em+en, \qquad \varepsilon_M^2(m)=e(em)=e^2m=em. \] Finally, \[ m+\varepsilon_M(m)=1\cdot m+e\cdot m=(1\oplus e)m=m, \] because $1\oplus e=1+0_R=1$ inside the supported copy of $R$. \end{proof} \begin{proposition}\label{prop:support-semilattice} Let \[ L_M:=eM=\im(\varepsilon_M). \] Then $L_M$ is an idempotent commutative monoid under the ambient addition of $M$. In particular, \[ l\vee l':=l+l' \qquad(l,l'\in L_M) \] defines a join-semilattice structure on $L_M$ with least element $0_M$. \end{proposition} \begin{proof} If $l=em$ and $l'=en$, then \[ l+l'=e(m+n)\in L_M. \] Moreover, \[ l+l=em+em=(e\oplus e)m=em=l, \] because $e\oplus e=e$. Thus $L_M$ is additively idempotent. Since $0_M=e0_M$, the least element is $0_M$. \end{proof} \begin{theorem}\label{thm:fiber-modules} For each $l\in L_M$, define the fiber \[ M_l:=\{m\in M:\varepsilon_M(m)=l\}. \] Then: \begin{enumerate}[label=(\arabic*)] \item $M_l$ is canonically an $R$-module; \item the zero element of the module $M_l$ is $l$ itself; \item if $l\le l'$ in $L_M$, then \[ T_{l,l'}:M_l\to M_{l'}, \qquad T_{l,l'}(m):=m+l' \] is a well-defined $R$-linear map; \item one has \[ T_{l,l}=\id_{M_l}, \qquad T_{l',l''}\circ T_{l,l'}=T_{l,l''} \quad(l\le l'\le l''). \] \end{enumerate} \end{theorem} \begin{proof} Fix $l\in L_M$. If $m,n\in M_l$, then \[ \varepsilon_M(m+n)=\varepsilon_M(m)+\varepsilon_M(n)=l+l=l, \] so $M_l$ is closed under addition. Now $l$ is the additive identity on $M_l$: if $m\in M_l$, then by Lemma~\ref{lem:epsilon-basic}, \[ m+l=m+\varepsilon_M(m)=m. \] Also, \[ m+(-1)m=(1\oplus(-1))m=0_Rm=em=l. \] Hence $(-1)m$ is the additive inverse of $m$ in the fiber. Thus $M_l$ is an abelian group. If $r\in R$ and $m\in M_l$, then \[ \varepsilon_M(rm)=e(rm)=(er)m=em=l, \] because $er=0_Rr=0_R=e$. Hence $rm\in M_l$. The module axioms are inherited from the semimodule axioms. Finally, \[ 0_Rm=em=l, \] which is the zero element of the module $M_l$. Thus $M_l$ is an $R$-module. Assume $l\le l'$. If $m\in M_l$, then \[ \varepsilon_M(m+l')=\varepsilon_M(m)+\varepsilon_M(l')=l+l'=l', \] so $m+l'\in M_{l'}$. Therefore $T_{l,l'}$ is well defined. It is additive because \[ T_{l,l'}(m+n)=m+n+l'=(m+l')+(n+l'), \] using $l'+l'=l'$. It preserves the zero element because \[ T_{l,l'}(l)=l+l'=l'. \] It is $R$-linear because if $r\in R$, \[ T_{l,l'}(rm)=rm+l'=rm+rl'=r(m+l'), \] since every supported scalar acts trivially on $L_M$. Finally, \[ T_{l,l}(m)=m+l=m, \] and for $l\le l'\le l''$, \[ T_{l',l''}(T_{l,l'}(m))=(m+l')+l''=m+(l'+l'')=m+l''=T_{l,l''}(m). \] \end{proof} \begin{proposition}\label{prop:addition-via-join-fiber} If $m\in M_l$ and $n\in M_{l'}$, then \[ m+n = T_{l,l\vee l'}(m) +_{M_{l\vee l'}} T_{l',l\vee l'}(n), \] where $+_{M_{l\vee l'}}$ denotes the module addition inside the fiber $M_{l\vee l'}$. \end{proposition} \begin{proof} The right-hand side equals \[ (m+l\vee l')+(n+l\vee l'). \] Since \[ \varepsilon_M(m+n)=\varepsilon_M(m)+\varepsilon_M(n)=l\vee l', \] Lemma~\ref{lem:epsilon-basic} gives \[ (m+n)+(l\vee l')=m+n. \] Using idempotence of $l\vee l'$, the expressions coincide. \end{proof} \begin{definition}\label{def:support-diagram} A \emph{support diagram over $R$} is a triple \[ \mathcal A=\bigl(L,\{A_l\}_{l\in L},\{\rho_{l,l'}\}_{l\le l'}\bigr) \] where: \begin{enumerate}[label=(\arabic*)] \item $L$ is a join-semilattice with least element $0$; \item each $A_l$ is an $R$-module; \item each $\rho_{l,l'}:A_l\to A_{l'}$ is $R$-linear; \item $\rho_{l,l}=\id_{A_l}$ and $\rho_{l',l''}\circ\rho_{l,l'}=\rho_{l,l''}$ whenever $l\le l'\le l''$. \end{enumerate} Morphisms are the obvious join-preserving natural transformations. The resulting category is denoted $\Diag_R$. \end{definition} \begin{theorem}\label{thm:reconstruct-semimodule} Let \[ \mathcal A=\bigl(L,\{A_l\}_{l\in L},\{\rho_{l,l'}\}_{l\le l'}\bigr)\in \Diag_R. \] Define \[ \mathbb M(\mathcal A):=\bigsqcup_{l\in L}A_l. \] For $x\in A_l$ and $y\in A_{l'}$, define \[ x\boxplus y:=\rho_{l,l\vee l'}(x)+\rho_{l',l\vee l'}(y)\in A_{l\vee l'}. \] Let $0$ denote the zero element of $A_0$. For $r\in R$, define \[ r\cdot x:=rx\in A_l, \qquad \tau\cdot x:=0\in A_0. \] Then $\mathbb M(\mathcal A)$ is a left $S$-semimodule. \end{theorem} \begin{proof} Commutativity of $\boxplus$ is immediate. To prove associativity, let \[ x\in A_l, \qquad y\in A_m, \qquad z\in A_n, \qquad u:=l\vee m\vee n. \] Then \[ (x\boxplus y)\boxplus z = \rho_{l\vee m,\nu}\bigl(\rho_{l,l\vee m}(x)+\rho_{m,l\vee m}(y)\bigr)+\rho_{n,\nu}(z). \] By linearity and functoriality of the $\rho$'s this equals \[ \rho_{l,\nu}(x)+\rho_{m,\nu}(y)+\rho_{n,\nu}(z), \] which is symmetric in the three terms. Hence $\boxplus$ is associative. If $x\in A_l$, then \[ x\boxplus 0=\rho_{l,l}(x)+\rho_{0,l}(0)=x+0_{A_l}=x, \] and similarly $0\boxplus x=x$. Now let $r,s\in R$. If $x\in A_l$, then \[ (r+s)\cdot x=(r+s)x=rx+sx=(r\cdot x)\boxplus (s\cdot x), \] because both terms lie in the same fiber $A_l$. Also \[ (rs)\cdot x=r\cdot(s\cdot x), \qquad 1\cdot x=x. \] If $x\in A_l$ and $y\in A_{l'}$, then \[ r\cdot(x\boxplus y) = r\bigl(\rho_{l,l\vee l'}(x)+\rho_{l',l\vee l'}(y)\bigr) = \rho_{l,l\vee l'}(rx)+\rho_{l',l\vee l'}(ry) = (r\cdot x)\boxplus(r\cdot y). \] Finally, \[ \tau\cdot(x\boxplus y)=0=(\tau\cdot x)\boxplus(\tau\cdot y), \] and \[ (r\oplus\tau)\cdot x=r\cdot x=(r\cdot x)\boxplus(\tau\cdot x). \] Hence all semimodule axioms hold. \end{proof} \begin{theorem}\label{thm:equivalence-semimodules-diagrams} The assignments \[ M\longmapsto\bigl(L_M,\{M_l\}_{l\in L_M},\{T_{l,l'}\}_{l\le l'}\bigr) \] and \[ \mathcal A\longmapsto \mathbb M(\mathcal A) \] extend to mutually quasi-inverse equivalences \[ \boxed{ S\text{-}\SMod\simeq \Diag_R. } \] \end{theorem} \begin{proof} The extraction of a support diagram from an $S$-semimodule is justified by Theorem~\ref{thm:fiber-modules} and Proposition~\ref{prop:addition-via-join-fiber}. Conversely, Theorem~\ref{thm:reconstruct-semimodule} constructs an $S$-semimodule from any support diagram. These constructions are functorial. If $f:M\to N$ is $S$-linear, then \[ \varepsilon_N(f(m))=ef(m)=f(em)=f(\varepsilon_M(m)), \] so $f$ carries each fiber $M_l$ into the corresponding fiber of $N$, and the resulting maps commute with transport. If $(\phi,\eta):\mathcal A\to\mathcal A'$ is a morphism of support diagrams, then the map on the disjoint unions is additive because naturality gives \[ \eta_{l\vee l'}(\rho_{l,l\vee l'}(x)+\rho_{l',l\vee l'}(y)) = \rho'_{\phi(l),\phi(l)\vee\phi(l')}(\eta_l(x)) + \rho'_{\phi(l'),\phi(l)\vee\phi(l')}(\eta_{l'}(y)). \] Starting from an $S$-semimodule $M$, the reconstructed semimodule has the same underlying set, because the fibers partition $M$. Proposition~\ref{prop:addition-via-join-fiber} shows that the reconstructed addition agrees with the original one. Scalar compatibility is immediate. Starting from a support diagram $\mathcal A$, the reconstructed semimodule has support semilattice identified with $L$ via the zero elements of the fibers, and the extracted fibers and transport maps are the original ones by construction. \end{proof} \begin{corollary}\label{cor:cancellative-category} Restriction along $p_R$ induces an equivalence \[ R\text{-}\Mod\simeq S\text{-}\SMod^{\mathrm{canc}}, \] where the right-hand side is the full subcategory of cancellative $S$-semimodules. \end{corollary} \begin{proof} If $M$ is cancellative, Lemma~\ref{lem:epsilon-basic} gives \[ m+\varepsilon_M(m)=m. \] Cancellation implies $\varepsilon_M(m)=0_M$ for all $m$. Hence $L_M$ is the one-point semilattice, so Theorem~\ref{thm:equivalence-semimodules-diagrams} identifies $M$ with a single $R$-module. Conversely, any $R$-module $A$ becomes an $S$-semimodule by \[ r\cdot a:=ra, \qquad \tau\cdot a:=0, \] and this semimodule is cancellative. \end{proof} \begin{corollary}\label{cor:adjunctions} The functor \[ M\longmapsto eM \] is left adjoint to restriction along $\chi_R:S\to\B$, and the functor \[ M\longmapsto K_M:=\ker(\varepsilon_M) \] is right adjoint to restriction along $p_R:S\to R$. \end{corollary} \begin{proof} Let $N$ be a $\B$-semimodule, regarded as an $S$-semimodule via $\chi_R$. If $f:M\to N$ is $S$-linear, then for every $m\in M$, \[ f(m)=e\cdot f(m)=f(em), \] because $e$ is supported and so acts as $1$ on $N$. Thus $f$ factors uniquely through $\varepsilon_M:M\to eM$, giving \[ \Hom_S(M,\chi_R^*N)\cong \Hom_{\B}(eM,N). \] Now let $A$ be an $R$-module, viewed as an $S$-semimodule via $p_R$. If $g:A\to M$ is $S$-linear, then \[ eg(a)=g(ea)=g(0)=0_M, \] so $g(A)\subseteq K_M$. Conversely, any $R$-linear map $A\to K_M$ is automatically $S$-linear. Hence \[ \Hom_S(p_R^*A,M)\cong \Hom_R(A,K_M). \] \end{proof} \begin{theorem}\label{thm:free-semimodule-boolean-cube} Let $F_n:=S^n$. Then the support diagram of $F_n$ is the Boolean lattice \[ \cP(\{1,\dots,n\}) \] with fiber over $I\subseteq\{1,\dots,n\}$ canonically equal to $R^I$, and transport along inclusions $I\subseteq J$ given by extension by zero in the coordinates of $J\setminus I$. \end{theorem} \begin{proof} The support of a vector $x=(x_1,\dots,x_n)\in S^n$ is determined exactly by the set of indices $j$ for which $x_j\in R$. Thus \[ L_{F_n}\cong \cP(\{1,\dots,n\}), \] with join equal to union. Fix $I\subseteq\{1,\dots,n\}$. The fiber over $I$ consists of vectors whose coordinates in $I$ are supported and whose coordinates outside $I$ are $\tau$. Projection to the supported coordinates identifies that fiber with $R^I$. If $I\subseteq J$, adding the support idempotent of $J$ inserts zeros in the new coordinates $J\setminus I$. \end{proof} \begin{theorem}\label{thm:subsemimodule-classification} Let $M$ be an $S$-semimodule. A subsemimodule $N\subseteq M$ is equivalent to the data of: \begin{enumerate}[label=(\arabic*)] \item a join-subsemilattice with zero $L'\subseteq L_M$; \item for each $l\in L'$, an $R$-submodule $N_l\subseteq M_l$ containing the zero element $l$ of the fiber; \item compatibility with transport: \[ T_{l,l'}(N_l)\subseteq N_{l'} \qquad(l\le l'\text{ in }L'). \] \end{enumerate} Under this correspondence, \[ N=\bigsqcup_{l\in L'}N_l. \] \end{theorem} \begin{proof} If $N\subseteq M$ is a subsemimodule, then $eN\subseteq N$, so \[ L_N:=eN\subseteq L_M. \] If $l,l'\in L_N$, choose $x,y\in N$ with $ex=l$ and $ey=l'$. Then \[ l\vee l'=e(x+y)\in eN=L_N, \] so $L_N$ is a join-subsemilattice with zero. For $l\in L_N$, let \[ N_l:=N\cap M_l. \] Because $N$ is stable under addition and under the action of $R\subset S$, each $N_l$ is an $R$-submodule of $M_l$. If $l\le l'$ and $x\in N_l$, then \[ T_{l,l'}(x)=x+l'\in N, \] because both $x$ and $l'$ lie in $N$. Hence $T_{l,l'}(N_l)\subseteq N_{l'}$. Conversely, given such data, define \[ N:=\bigsqcup_{l\in L'}N_l. \] Then $0_M\in N$. If $x\in N_l$ and $y\in N_{l'}$, then by Proposition~\ref{prop:addition-via-join-fiber}, \[ x+y = T_{l,l\vee l'}(x)+_{M_{l\vee l'}}T_{l',l\vee l'}(y) \in N_{l\vee l'}. \] Thus $N$ is closed under addition. It is stable under $R$ because each $N_l$ is an $R$-submodule, and under $\tau$ because $\tau x=0_M$. Hence $N$ is an $S$-subsemimodule. \end{proof} \begin{theorem}\label{thm:finite-generation} Assume that $M$ is generated as an $S$-semimodule by $m_1,\dots,m_n$. Set \[ l_i:=\varepsilon_M(m_i)\in L_M. \] Then: \begin{enumerate}[label=(\arabic*)] \item $L_M$ is the join-subsemilattice generated by $l_1,\dots,l_n$; in particular, \[ |L_M|\le 2^n; \] \item for each $l\in L_M$, the fiber $M_l$ is generated as an $R$-module by the transported elements \[ T_{l_i,l}(m_i) \qquad (1\le i\le n,\ l_i\le l). \] \end{enumerate} \end{theorem} \begin{proof} Every element of $M$ is a finite sum of terms $r_jm_{i_j}$ with $r_j\in R$. Since supported scalars preserve support, \[ \varepsilon_M(r_jm_{i_j})=\varepsilon_M(m_{i_j})=l_{i_j}. \] Therefore \[ \varepsilon_M\!\left(\sum_j r_jm_{i_j}\right)=\bigvee_j l_{i_j}. \] Hence every support element is a join of some subset of $\{l_1,\dots,l_n\}$, and every such join occurs as the support of the corresponding sum of generators. This proves (1). Now let $m\in M_l$ and write \[ m=\sum_j r_jm_{i_j}. \] Then \[ l=\varepsilon_M(m)=\bigvee_j l_{i_j}, \] so each $l_{i_j}\le l$. Therefore each $T_{l_{i_j},l}(m_{i_j})$ is defined. Since transport is $R$-linear, \[ \sum_j r_jT_{l_{i_j},l}(m_{i_j}) = \sum_j T_{l_{i_j},l}(r_jm_{i_j}) = \sum_j (r_jm_{i_j}+l) = m+l = m, \] because $l$ is the zero of the module $M_l$. \end{proof} \section{Conservative ideal, class, and factorization phenomena} \begin{theorem}\label{thm:ideal-multiplication-classical} Assume the baseline classification of ideals, \[ I_J:=J\cup\{\tau\} \qquad(J\triangleleft R). \] Then for all ideals $J,K\triangleleft R$, \[ I_JI_K=I_{JK}. \] If $R$ is an integral domain, then the ideal class monoid of the nontrivial ideals of $G(R)$ is canonically isomorphic to the ordinary ideal class monoid of $R$. \end{theorem} \begin{proof} Every product of an element of $I_J$ with an element of $I_K$ is either $\tau$ or an element of $JK$, so $I_JI_K\subseteq I_{JK}$. Conversely, $\tau\in I_JI_K$, and every element of $JK$ is a finite sum of products $ab$ with $a\in J$ and $b\in K$, hence belongs to $I_JI_K$. Thus $I_JI_K=I_{JK}$. Now assume $R$ is a domain. For nonzero ideals $J_1,J_2\subset R$, classical ideal-class equivalence is \[ J_1\sim_R J_2 \iff \exists a,b\in R\setminus\{0\}:aJ_1=bJ_2. \] For the corresponding $G(R)$-ideals, \[ I_{J_1}\sim_G I_{J_2} \iff \exists a,b\in R\setminus\{0\}:aI_{J_1}=bI_{J_2}. \] But \[ aI_J=aJ\cup\{\tau\}. \] Therefore \[ aI_{J_1}=bI_{J_2} \iff aJ_1=bJ_2. \] Hence the class monoids coincide. \end{proof} \begin{proposition}\label{prop:separated-weight-nonmultiplicative} Assume that every nonzero quotient $R/J$ is finite, and write \[ N(J):=|R/J|. \] Then the separated-sheet weight \[ w(J):=N(J)+1 \] is not multiplicative on nonzero ideals. More precisely, \[ w(JK)=N(J)N(K)+1, \qquad w(J)w(K)=N(J)N(K)+N(J)+N(K)+1. \] Consequently the separated congruence zeta sheet \[ \zeta^{\mathrm{sep}}_{G(R)}(s):=\sum_{0\neq J}(N(J)+1)^{-s} \] is not an Euler product indexed by prime ideals. \end{proposition} \begin{proof} The displayed identities are immediate from the definition. Their difference is $N(J)+N(K)$, which is strictly positive. Therefore $w$ is not multiplicative. \end{proof} \begin{proposition}\label{prop:same-abscissa} Under the same finiteness hypothesis, the classical ideal zeta sheet \[ \zeta^{\mathrm{idl}}_{G(R)}(s):=\sum_{0\neq J}N(J)^{-s} \] and the separated sheet $\zeta^{\mathrm{sep}}_{G(R)}(s)$ have the same abscissa of absolute convergence. \end{proposition} \begin{proof} For every nonzero ideal $J$ and every $\sigma>0$, \[ N(J)\le N(J)+1\le 2N(J), \] whence \[ 2^{-\sigma}N(J)^{-\sigma}\le (N(J)+1)^{-\sigma}\le N(J)^{-\sigma}. \] The conclusion follows by comparison. \end{proof} \begin{proposition}\label{prop:units-irreducibles} Assume that $R$ is an integral domain. Then: \begin{enumerate}[label=(\arabic*)] \item the units of $G(R)$ are exactly the units of $R$; \item every nonzero supported irreducible of $G(R)$ is exactly an irreducible of $R$, and conversely. \end{enumerate} \end{proposition} \begin{proof} If $u\in G(R)$ is a unit, then $u\neq \tau$ because $\tau x=\tau\neq1$. Hence $u\in R$, and its inverse is also supported, so $u$ is a unit of $R$. The converse is immediate. Let $x\in R\setminus\{0\}$. If $x=ab$ in $G(R)$, then neither factor can be $\tau$, since otherwise $ab=\tau$. Hence $a,b\in R$, so factorization of $x$ in $G(R)$ is exactly factorization in $R$. \end{proof} \begin{theorem}\label{thm:supported-zero-prime-reducible} Assume that $R$ is an integral domain. The supported zero $e=0_R\in G(R)$ is reducible and prime in the divisibility sense. \end{theorem} \begin{proof} It is reducible because \[ e=e\cdot e, \] and $e$ is not a unit. Now suppose $e\mid xy$. Then \[ xy=e\cdot c \] for some $c\in G(R)$. If $c=\tau$, then $xy=\tau$. If $c\in R$, then $xy=e$. Hence \[ xy\in\{e,\tau\}. \] If both $x$ and $y$ are supported nonzero elements, then $xy$ is again a supported nonzero element of $R$, impossible. Hence one of $x,y$ lies in $\{e,\tau\}$, so $e$ divides one of them. \end{proof} \begin{proposition}\label{prop:scalar-norm-collapse} Assume that $R$ is an integral domain with a multiplicative ideal norm $N$ on nonzero ideals, and assume that there exists a proper nonzero ideal $J$ with $N(J)>1$. Let \[ I_{(0)}:=\{e,\tau\}. \] If \[ h:\{\text{ideals of }G(R)\}\to [0,\infty) \] satisfies \[ h(I_1I_2)=h(I_1)h(I_2) \] for all ideals and \[ h(I_J)=N(J) \] for every nonzero ideal $J\triangleleft R$, then \[ h(I_{(0)})=0, \qquad h(\{\tau\})=0. \] \end{proposition} \begin{proof} Since $I_{(0)}I_J=I_{(0)}$, one has \[ h(I_{(0)})=h(I_{(0)})h(I_J)=h(I_{(0)})N(J). \] Because $N(J)\neq1$, it follows that $h(I_{(0)})=0$. Similarly, \[ \{\tau\}I_J=\{\tau\} \] implies \[ h(\{\tau\})=h(\{\tau\})N(J), \] so $h(\{\tau\})=0$. \end{proof} \section{Universal all-orders deformation of the shifted zeta sheet} \begin{definition}\label{def:Lt} Let \[ L(s)=\sum_{n\ge1}a_n n^{-s} \] be a Dirichlet series with abscissa of absolute convergence $\sigma_a$. For $|t|<1$, define \[ L_t(s):=\sum_{n\ge1}a_n(n+t)^{-s}. \] Let \[ (Tf)(s):=f(s+1), \qquad D:=sT. \] \end{definition} \begin{theorem}\label{thm:universal-taylor-deformation} For $|t|<1$ and $\Re(s)>\sigma_a$, \[ \boxed{ L_t(s)=\sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^mL(s+m). } \] Equivalently, \[ \boxed{ L_t=e^{-tD}L. } \] \end{theorem} \begin{proof} For $|u|<1$, \[ (1+u)^{-s}=\sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}u^m. \] Apply this with $u=t/n$. Since $|t|<1$, one has $|t/n|<1$ for every $n\ge1$, so \[ (n+t)^{-s} = n^{-s}\Bigl(1+\frac{t}{n}\Bigr)^{-s} = \sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^m n^{-s-m}. \] Multiplying by $a_n$ and summing over $n$ gives \[ L_t(s) = \sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^m\sum_{n\ge1}a_n n^{-s-m} = \sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^mL(s+m). \] Also, \[ D^mL(s)=(s)_mL(s+m) \] by induction on $m$. Hence \[ e^{-tD}L= \sum_{m\ge0}\frac{(-t)^m}{m!}D^mL = \sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^mL(s+m)=L_t. \] \end{proof} \begin{corollary}\label{cor:jets} For every integer $n\ge0$, \[ \boxed{ \partial_t^{\,n}L_t(s)=(-1)^n(s)_nL_t(s+n). } \] In particular, \[ \boxed{ \partial_t^{\,n}L_t(s)\big|_{t=0}=(-1)^n(s)_nL(s+n). } \] \end{corollary} \begin{proof} Differentiate termwise: \[ \partial_t^{\,n}(m+t)^{-s}=(-1)^n(s)_n(m+t)^{-s-n}. \] Summing over $m$ yields the formula. \end{proof} \begin{corollary}\label{cor:semigroup-law} If $|t_0|<1$ and $|h|<1-|t_0|$, then \[ L_{t_0+h}=e^{-hD}L_{t_0}. \] \end{corollary} \begin{proof} Apply Theorem~\ref{thm:universal-taylor-deformation} to the Dirichlet series \[ s\longmapsto L_{t_0}(s)=\sum_{n\ge1}a_n(n+t_0)^{-s}. \] \end{proof} \begin{theorem}\label{thm:meromorphic-continuation-shifted-tower} Assume that $L$ extends meromorphically to $\C$. Let $N\ge1$ and $|t|<1$. Then on the half-plane $\Re(s)>\sigma_a-N$, \[ \boxed{ L_t(s)= \sum_{m=0}^{N-1}(-1)^m\frac{(s)_m}{m!}t^mL(s+m) + H_{N,t}(s), } \] where $H_{N,t}(s)$ is holomorphic on $\Re(s)>\sigma_a-N$. \end{theorem} \begin{proof} Apply Taylor's formula with integral remainder to \[ f_s(u):=(1+u)^{-s} \] at $u=0$, then substitute $u=t/n$. One obtains \[ (n+t)^{-s} = \sum_{m=0}^{N-1}(-1)^m\frac{(s)_m}{m!}t^m n^{-s-m} + t^N n^{-s-N}R_N(t/n,s), \] where $R_N(t/n,s)$ is holomorphic in $s$ and uniformly bounded on compact subsets of $\Re(s)>\sigma_a-N$. Multiplying by $a_n$ and summing over $n$ gives the stated decomposition. The remainder series converges absolutely and locally uniformly on $\Re(s)>\sigma_a-N$, hence defines a holomorphic function there. \end{proof} \begin{corollary}\label{cor:pole-rigidity} If $L$ has a unique simple pole at $s=1$, then every $L_t$ has the same unique simple pole at $s=1$, with the same residue: \[ \operatorname*{Res}_{s=1}L_t(s)=\operatorname*{Res}_{s=1}L(s). \] \end{corollary} \begin{proof} In Theorem~\ref{thm:meromorphic-continuation-shifted-tower}, the only term that can contribute a pole at $s=1$ is $L(s)$ itself, since $L(s+m)$ is regular there for all $m\ge1$. \end{proof} \begin{definition}\label{def:cumulants} Assume $L_t(s)\neq0$. Define \[ A_n(t;s):=(s)_n\frac{L_t(s+n)}{L_t(s)} \qquad(n\ge1), \] and define the logarithmic cumulants \[ \kappa_n(t;s):=\partial_t^{\,n}\log L_t(s). \] \end{definition} \begin{theorem}\label{thm:cumulant-partition-formula} For every $n\ge1$, \[ \boxed{ \kappa_n(t;s) = (-1)^n \sum_{\pi\in\Pi_n} (-1)^{|\pi|-1}(|\pi|-1)! \prod_{B\in\pi}A_{|B|}(t;s), } \] where $\Pi_n$ is the set of set partitions of $\{1,\dots,n\}$. \end{theorem} \begin{proof} Set \[ m_n(t;s):=\frac{\partial_t^{\,n}L_t(s)}{L_t(s)}. \] By Corollary~\ref{cor:jets}, \[ m_n(t;s)=(-1)^nA_n(t;s). \] Now form the exponential generating series \[ M(z):=\sum_{n\ge0}m_n\frac{z^n}{n!}, \qquad K(z):=\sum_{n\ge1}\kappa_n\frac{z^n}{n!}. \] Taylor expansion yields \[ L_{t+z}(s)=L_t(s)M(z), \] whereas the definition of $\kappa_n$ gives \[ \log L_{t+z}(s)-\log L_t(s)=K(z), \] so \[ L_{t+z}(s)=L_t(s)e^{K(z)}. \] Hence \[ M(z)=e^{K(z)}, \qquad K(z)=\log M(z). \] The standard moment--cumulant formula for exponential generating series implies \[ \kappa_n= \sum_{\pi\in\Pi_n} (-1)^{|\pi|-1}(|\pi|-1)! \prod_{B\in\pi}m_{|B|}. \] Substituting $m_{|B|}=(-1)^{|B|}A_{|B|}$ and using $\sum_{B\in\pi}|B|=n$ yields the displayed formula. \end{proof} \begin{corollary}\label{cor:first-four-cumulants} The first four cumulants are \[ \kappa_1=-A_1, \] \[ \kappa_2=A_2-A_1^2, \] \[ \kappa_3=-A_3+3A_1A_2-2A_1^3, \] \[ \kappa_4=A_4-4A_1A_3-3A_2^2+12A_1^2A_2-6A_1^4. \] \end{corollary} \begin{proof} Expand the partition formula of Theorem~\ref{thm:cumulant-partition-formula} for $n=1,2,3,4$. \end{proof} \begin{theorem}\label{thm:arithmetic-kernels-riemann} For every integer $r\ge1$, the ratio \[ Q_r(s):=\frac{\zeta(s+r)}{\zeta(s)} \] has a Dirichlet expansion \[ Q_r(s)=\sum_{n\ge1}\alpha_r(n)n^{-s}, \] where $\alpha_r$ is multiplicative and \[ \alpha_r(p^k)=-\frac{p^r-1}{p^{rk}} \qquad(k\ge1). \] Equivalently, \[ \boxed{ \alpha_r(n)=(-1)^{\omega(n)}\frac{\prod_{p\mid n}(p^r-1)}{n^r}. } \] \end{theorem} \begin{proof} Using the Euler product for $\zeta$, \[ Q_r(s)=\prod_p\frac{1-p^{-s}}{1-p^{-s-r}}. \] At a fixed prime $p$, set $u=p^{-s}$. Then \[ \frac{1-u}{1-u/p^r} = (1-u)\sum_{k\ge0}p^{-rk}u^k = 1+\sum_{k\ge1}(p^{-rk}-p^{-r(k-1)})u^k = 1-\sum_{k\ge1}\frac{p^r-1}{p^{rk}}u^k. \] Hence \[ \alpha_r(p^0)=1, \qquad \alpha_r(p^k)=-\frac{p^r-1}{p^{rk}} \quad(k\ge1). \] Multiplicativity follows from the Euler product, and the explicit formula for general $n$ is the product over the primes dividing $n$. \end{proof} \begin{corollary}\label{cor:first-log-jet-explicit} One has \[ \partial_t\log\zeta_t(s)\Big|_{t=0} = -s\sum_{n\ge1}\alpha_1(n)n^{-s}, \] where \[ \alpha_1(n)=(-1)^{\omega(n)}\frac{\varphi(\rad(n))}{n}. \] Moreover, \[ \partial_t^2\log\zeta_t(s)\Big|_{t=0} = s(s+1)\sum_{n\ge1}\alpha_2(n)n^{-s} - s^2\left(\sum_{n\ge1}\alpha_1(n)n^{-s}\right)^2. \] Hence the second logarithmic jet is already not primewise local. \end{corollary} \begin{proof} By Theorem~\ref{thm:arithmetic-kernels-riemann}, \[ \frac{\zeta(s+1)}{\zeta(s)}=\sum_{n\ge1}\alpha_1(n)n^{-s}, \qquad \frac{\zeta(s+2)}{\zeta(s)}=\sum_{n\ge1}\alpha_2(n)n^{-s}. \] Now Corollary~\ref{cor:first-four-cumulants} evaluated at $t=0$ gives \[ \partial_t\log\zeta_t(s)\Big|_{t=0}=-A_1(0;s)=-s\frac{\zeta(s+1)}{\zeta(s)}, \] and \[ \partial_t^2\log\zeta_t(s)\Big|_{t=0}=A_2(0;s)-A_1(0;s)^2. \] Substituting the explicit Dirichlet expansions yields the formulas. The identity \[ \prod_{p\mid n}(p-1)=\varphi(\rad(n)) \] gives the expression for $\alpha_1(n)$. The square term involves Dirichlet convolution and therefore mixes prime supports. \end{proof} \begin{theorem}\label{thm:non-eulerian-nonintegral-t} Let \[ L(s)=\sum_{n\ge1}a_n n^{-s} \] with $a_1\neq0$. If $01$, then $b(m)=b(n)=b(mn)=1$, so the defect is $0$. If $m=n=1$, then the defect is $0$. If exactly one of $m,n$ equals $1$, the defect is $1$. Hence the displayed formula holds, and summing over coprime pairs gives \[ \sum_{n\ge2}n^{-t}+\sum_{m\ge2}m^{-s}=\zeta(s)+\zeta(t)-2. \] The formulas for $\mathcal D(\mathbf u)$ and $\mathcal D(\mathbf v)$ are immediate. The comultiplication formulas are obtained by evaluation on the four pairs $(\xi_1,\xi_1)$, $(\xi_1,\xi_\ast)$, $(\xi_\ast,\xi_1)$, $(\xi_\ast,\xi_\ast)$. Subtracting gives the final identity. \end{proof} \begin{theorem}\label{thm:two-sheet-defect} For \[ \mathcal Z_{1,u}(s):=(1+u)\zeta(s)-u, \qquad \lambda:=1+u, \] the coefficient function is \[ c_u(1)=1, \qquad c_u(n)=\lambda \quad(n\ge2). \] Hence the coprime multiplicativity defect \[ \Delta_u(m,n):=c_u(mn)-c_u(m)c_u(n) \qquad((m,n)=1) \] is given by \[ \Delta_u(m,n)= \begin{cases} 0,&\min(m,n)=1,\\[1mm] \lambda-\lambda^2=-u(1+u),&m,n>1. \end{cases} \] Equivalently, if \[ a_\lambda:=\mathbf u+\lambda\mathbf v, \] then \[ \Delta(a_\lambda)-a_\lambda\otimes a_\lambda = (\lambda-\lambda^2)\,\mathbf v\otimes\mathbf v. \] Thus the scalar \[ \delta(\lambda):=\lambda-\lambda^2=-u(1+u) \] is the exact idempotence defect of the non-unit Stone component. \end{theorem} \begin{proof} The coefficient formula for $\mathcal Z_{1,u}$ is immediate from the definition. If one of $m,n$ equals $1$, then multiplicativity holds because $c_u(1)=1$. If $m,n>1$, then $mn>1$ and \[ \Delta_u(m,n)=\lambda-\lambda^2. \] This proves the first formula. For the Stone identity, note that \[ \Delta(a_\lambda)=\Delta(\mathbf u)+\lambda\Delta(\mathbf v) =\mathbf u\otimes\mathbf u+ \lambda(\mathbf v\otimes\mathbf u+\mathbf u\otimes\mathbf v+\mathbf v\otimes\mathbf v), \] while \[ a_\lambda\otimes a_\lambda =\mathbf u\otimes\mathbf u+ \lambda(\mathbf v\otimes\mathbf u+\mathbf u\otimes\mathbf v)+ \lambda^2\mathbf v\otimes\mathbf v. \] Subtracting yields the result. \end{proof} \begin{theorem}\label{thm:coprime-kernel} Define the interior coprime kernel by \[ C(s,t):=\sum_{\substack{(m,n)=1\\ m,n>1}}m^{-s}n^{-t}. \] Then \[ C(s,t)=\frac{\zeta(s)\zeta(t)}{\zeta(s+t)}-\zeta(s)-\zeta(t)+1. \] Consequently, \[ \mathfrak D_u(s,t):= \sum_{\substack{(m,n)=1}}\Delta_u(m,n)m^{-s}n^{-t} = -u(1+u) \left( \frac{\zeta(s)\zeta(t)}{\zeta(s+t)}-\zeta(s)-\zeta(t)+1 \right). \] In particular, at the doubled endpoint $u=1$, \[ \mathfrak D_1(s,t)= -2 \left( \frac{\zeta(s)\zeta(t)}{\zeta(s+t)}-\zeta(s)-\zeta(t)+1 \right). \] \end{theorem} \begin{proof} The full coprime sum is \[ \sum_{(m,n)=1}m^{-s}n^{-t} = \prod_p\left(1+\sum_{a\ge1}p^{-as}+\sum_{b\ge1}p^{-bt}\right) = \prod_p\frac{1-p^{-s-t}}{(1-p^{-s})(1-p^{-t})} = \frac{\zeta(s)\zeta(t)}{\zeta(s+t)}. \] Subtracting the cases $m=1$ or $n=1$ gives the formula for $C(s,t)$. Now Theorem~\ref{thm:two-sheet-defect} shows that the coefficient-level defect equals $-u(1+u)$ exactly on the interior sector $m,n>1$ and vanishes otherwise. Hence \[ \mathfrak D_u(s,t)=-u(1+u)C(s,t), \] which is the asserted formula. The case $u=1$ is immediate. \end{proof} \begin{corollary}\label{cor:origin-values} One has \[ F_1(0)=-\frac32, \qquad \mathcal Z_{1,1}(0)=-2, \qquad \mathfrak D_1(0,0)=-3. \] Moreover, \[ \mathfrak D_u(0,t)=u(1+u)F_1(t), \qquad \mathfrak D_1(0,t)=2F_1(t). \] \end{corollary} \begin{proof} Since $F_1(s)=\zeta(s)-1$, \[ F_1(0)=\zeta(0)-1=-\frac32. \] Also, \[ \mathcal Z_{1,1}(0)=2\zeta(0)-1=-2. \] Now \[ C(0,0)=\frac{\zeta(0)\zeta(0)}{\zeta(0)}-\zeta(0)-\zeta(0)+1 =-\frac12+\frac12+\frac12+1 =\frac32, \] so \[ \mathfrak D_1(0,0)=-2\cdot\frac32=-3. \] Further, \[ C(0,t)=\frac{\zeta(0)\zeta(t)}{\zeta(t)}-\zeta(0)-\zeta(t)+1=1-\zeta(t), \] hence by Theorem~\ref{thm:coprime-kernel}, \[ \mathfrak D_u(0,t)=-u(1+u)(1-\zeta(t))=u(1+u)(\zeta(t)-1)=u(1+u)F_1(t). \] The case $u=1$ follows immediately. \end{proof} \section{Number-field extension} \begin{definition}\label{def:number-field-shift} Let $K$ be a number field, let $a_K(n)$ be the number of nonzero ideals of $\cO_K$ of norm $n$, and define \[ \zeta_K(s)=\sum_{n\ge1}a_K(n)n^{-s}. \] For $|t|<1$, define \[ \zeta_{K,t}(s):=\sum_{n\ge1}a_K(n)(n+t)^{-s} = \sum_{0\neq J\triangleleft \cO_K}(N(J)+t)^{-s}. \] \end{definition} \begin{theorem}\label{thm:number-field-deformation} For $|t|<1$, \[ \zeta_{K,t}(s)=\sum_{m\ge0}(-1)^m\frac{(s)_m}{m!}t^m\zeta_K(s+m), \] for $\Re(s)>1$, and this identity determines a meromorphic continuation of $\zeta_{K,t}$ to the whole plane. Moreover, $\zeta_{K,t}$ has the same unique simple pole at $s=1$ as $\zeta_K$, with the same residue. \end{theorem} \begin{proof} Apply Theorem~\ref{thm:universal-taylor-deformation} and Corollary~\ref{cor:pole-rigidity} to the Dirichlet series $L=\zeta_K$, using the standard meromorphic continuation and pole structure of Dedekind zeta functions. \end{proof} \begin{theorem}\label{thm:number-field-arithmetic-kernels} For every integer $r\ge1$, the ratio \[ Q_{K,r}(s):=\frac{\zeta_K(s+r)}{\zeta_K(s)} \] has an ideal-Dirichlet expansion \[ Q_{K,r}(s)=\sum_{\mathfrak a\neq0}\alpha_{K,r}(\mathfrak a)N(\mathfrak a)^{-s}, \] where $\alpha_{K,r}$ is multiplicative and \[ \alpha_{K,r}(\mathfrak p^k)=-\frac{N(\mathfrak p)^r-1}{N(\mathfrak p)^{rk}} \qquad(k\ge1). \] Equivalently, \[ \boxed{ \alpha_{K,r}(\mathfrak a)=(-1)^{\omega(\mathfrak a)} \frac{\prod_{\mathfrak p\mid \mathfrak a}(N(\mathfrak p)^r-1)}{N(\mathfrak a)^r}. } \] \end{theorem} \begin{proof} Use the Euler product for $\zeta_K$: \[ \zeta_K(s)=\prod_{\mathfrak p}(1-N(\mathfrak p)^{-s})^{-1}. \] Hence \[ Q_{K,r}(s)=\prod_{\mathfrak p}\frac{1-N(\mathfrak p)^{-s}}{1-N(\mathfrak p)^{-s-r}}. \] The local computation is identical to the proof of Theorem~\ref{thm:arithmetic-kernels-riemann}, with $p$ replaced by $N(\mathfrak p)$. Multiplicativity follows from the Euler product. \end{proof} \begin{corollary}\label{cor:number-field-non-eulerian} If $0