Conductor-Reduced Defect Decompositions for Rees Algebras and a Fiber-Corrected Bridge to Linear Type

Abstract

Let $(A,\mathfrak m)$ be an excellent normal local domain of dimension $d\ge 2$, and let $I$ be an $\mathfrak m$-primary ideal. Write $$\mathcal R(I)=A[It],\qquad \mathcal R(I)^$$=\bigoplus_{n\ge 0}\overline{I^n}t^n, \qquad \mathcal U(I)=\bigoplus_{n\ge 0}(I^n)_1t^n,R(I)=A[It],R(I)∗=n≥0⨁​Intn,U(I)=n≥0⨁​(In)1​tn, where $(I^n)_1$ is the first coefficient ideal of $I^n$. The quotient $$\mathcal C(I)=\mathcal R(I)^$$/\mathcal U(I) is the pure codimension-one defect controlling Serre's $R_1$ condition for $\mathcal R(I)$, while the quotient $$\mathcal E(I)=J/L$$ between the defining ideal $J$ of $\mathcal R(I)$ and the ideal $L$ of linear relations measures failure of linear type.

This paper records a verified theorem package extracted from the interaction of Puthenpurakal's $R_1$ criterion and the conductor formalism of Gasanova--Herzog--Kling--Moradi. For every nonzero conductor element $r\in \mathfrak m\cap \mathfrak c(I)$ we define the reduced comparison map $$\phi_n^r:\operatorname{Sym}_A^n(I)/r\operatorname{Sym}_A^n(I)\longrightarrow \overline{I^n}/r\overline{I^n},$$ and prove:

1. the exact index formula $$\lambda(\ker\phi_n^r)-\lambda(\operatorname{coker}\phi_n^r)$$

   \lambda((J/L)_n);
2. the exact decomposition $$\nu_r(n)=d_r(n)+\kappa_r(n)-\tau_r(n), \qquad \nu_r(n):=\lambda!\left(\frac{\overline{I^n}}{I^n+r\overline{I^n}}\right),$$ where $d_r$ is the first-coefficient correction, $\kappa_r$ is the conductor-sliced codimension-one defect, and $\tau_r$ is an explicit intersection defect;
3. the asymptotic criterion $$\mathcal R(I)\text{ is }R_1 \iff \deg \nu_r(n)\le d-2,$$ while in the non-$R_1$ case one has $\deg \nu_r(n)=d-1$;
4. a fiber-corrected bridge theorem: if the fiber-cone equation ideal $$J_{\mathrm{fib}}=\ker!\big(\operatorname{Sym}_k(I/\mathfrak mI)\to F(I)\big)$$ vanishes, then there exists a constant $C>0$ such that $$\lambda((J/L)_n)\le C,\lambda((I^n)_1/I^n)\qquad\text{for all }n\ge 1,$$ hence $$\lambda((J/L)_n)=O(n^{d-2}).$$

The raw bridge between equation defects and codimension-one defects fails in general: if $J_{\mathrm{fib}}\neq 0$, then the equation defect has maximal symmetric-fiber dimension. Thus the honest contribution is a corrected bridge theorem, not an unconditional one.

Code availability. The full theorem package has been formalized in Lean 4 (Mathlib) as the ReesDefects package. Source code: https://github.com/Shengrong-Wu/AI-Driven-Math-Research/tree/main/ReesDefects. A section-by-section account of the formal development appears in the appendix.

Introduction

Context and motivation

For an $\mathfrak m$-primary ideal $I$ in a normal local domain, two natural defects govern two different failures of regularity in the blow-up algebra of $I$.

The first is the equation defect $$\mathcal E(I)=J/L,$$ where $J$ is the defining ideal of the Rees algebra $\mathcal R(I)$ inside a polynomial presentation of $\operatorname{Sym}_A(I)$, and $L$ is the ideal of linear relations. The module $\mathcal E(I)$ vanishes exactly when $I$ is of linear type.

The second is the codimension-one normalization defect $$\mathcal C(I)=\mathcal R(I)^*/\mathcal U(I),$$ where $$\mathcal U(I)=\bigoplus_{n\ge 0}(I^n)_1t^n$$ is the first-coefficient Rees algebra. Puthenpurakal proved that $\mathcal C(I)$ detects the $R_1$ property of $\mathcal R(I)$: $$\mathcal R(I)\text{ is }R_1 \iff \mathcal C(I)=0 \iff \overline{I^n}=(I^n)_1\ \text{for all }n\ge 1.$$

These two defects live on different sides of the symmetric--Rees comparison: $$\operatorname{Sym}_A(I)\twoheadrightarrow \mathcal R(I)\subseteq \mathcal R(I)^*.$$ The conductor formalism of Gasanova--Herzog--Kling--Moradi supplies the bridge on the equation side: $$\mathfrak c(I)=L:_A J=\operatorname{Ann}_A(J/L).$$ The guiding question of the research program summarized here was whether one can compare the equation defect and the codimension-one defect quantitatively.

Related work

Two recent results form the base of this paper.

First, Puthenpurakal showed that for an excellent normal local ring $(A,\mathfrak m)$ of dimension at least $2$ and an $\mathfrak m$-primary ideal $I$, the $R_1$ property of the Rees algebra is equivalent to equality between integral closures and first coefficient ideals: $$\mathcal R(I)\text{ is }R_1 \iff \overline{I^n}=(I^n)_1\ \text{for all }n\ge 1.$$

Second, Gasanova--Herzog--Kling--Moradi introduced the conductor $\mathfrak c(I)=L:_A J$, proved that it localizes, and in the domain case obtained the identity $$J=L:r\qquad\text{for every }0\neq r\in \mathfrak c(I).$$ This turns multiplication by a conductor element into an equation detector inside the symmetric algebra.

The present paper does not improve the base $R_1$ theorem itself. It reorganizes the mechanism behind it and records the strongest verified bridge statements obtained in the subsequent research log.

Main result

The strongest theorem package proved in the research log is the following.

For every nonzero conductor element $r\in \mathfrak m\cap \mathfrak c(I)$, the reduced comparison map $$\phi_n^r:\operatorname{Sym}_A^n(I)/r\operatorname{Sym}_A^n(I)\to \overline{I^n}/r\overline{I^n}$$ has finite-length kernel and cokernel, and satisfies the exact index formula $$\lambda(\ker\phi_n^r)-\lambda(\operatorname{coker}\phi_n^r)=\lambda((J/L)$$n).λ(kerϕnr​)−λ(cokerϕnr​)=λ((J/L)n​). Its cokernel is the reduced normalization defect $$\nu_r(n)=\lambda!\left(\frac{\overline{I^n}}{I^n+r\overline{I^n}}\right),$$ and one has the exact decomposition $$\nu_r(n)=d_r(n)+\kappa_r(n)-\tau_r(n),$$ where $$d_r(n)=\lambda!\left(0:$${(I^n)1/I^n}r\right), \qquad \kappa_r(n)=\lambda!\left(0:{\overline{I^n}/(I^n)_1}r\right), and $$\tau_r(n)= \lambda!\left( \frac{(I^n)_1\cap (I^n+r\overline{I^n})}{I^n+r(I^n)_1} \right).$$ The leading term of $\nu_r(n)$ detects $R_1$.

The second main theorem is a corrected bridge to linear type. Let $$F(I)=\bigoplus_{n\ge 0}I^n/\mathfrak m I^n$$ be the fiber cone and $$J_{\mathrm{fib}}=\ker!\big(\operatorname{Sym}$$k(I/\mathfrak m I)\to F(I)\big)Jfib​=ker(Symk​(I/mI)→F(I)) its equation ideal over $k=A/\mathfrak m$. If $J${\mathrm{fib}}=0, then $$\lambda((J/L)_n)\le C,\lambda((I^n)_1/I^n)$$ for a constant $C$ independent of $n$, hence $\lambda((J/L)_n)=O(n^{d-2})$.

What is genuinely new

The paper centers on the mechanism, not on a stronger replacement for Puthenpurakal's theorem.

The genuinely new verified outputs are:

1. the reduced comparison map $\phi_n^r$ and its exact index formula;
2. the exact defect decomposition $$\nu_r=d_r+\kappa_r-\tau_r;$$
3. the identification of $\tau_r$ as an explicit intersection defect;
4. the fiber-corrected bridge theorem, which is the strongest true bridge between equation defect and codimension-one defect obtained in the research log;
5. the obstruction theorem showing why the raw uncorrected bridge cannot hold in general.

Paper outline

Section 2 fixes notation and collects the exact sequences inherited from the two source papers. Section 3 develops the conductor-reduced mechanism and proves the structural identity $\nu_r=d_r+\kappa_r-\tau_r$. Section 4 states the main results. Section 5 proves them. Section 6 discusses the remaining obstruction and the only currently viable route to stronger theorems. Appendix A records the special-fiber obstruction theorem.

Background and Preliminaries

Setup and notation

Throughout the paper:

• $(A,\mathfrak m)$ is an excellent normal local domain;
• $\dim A=d\ge 2$;
• $I$ is an $\mathfrak m$-primary ideal;
• $Q(A)$ is the fraction field of $A$;
• $k=A/\mathfrak m$.

Write $$\mathcal R(I)=\bigoplus_{n\ge 0}I^n t^n\subseteq A[t], \qquad \mathcal R(I)^*=\bigoplus_{n\ge 0}\overline{I^n}t^n\subseteq Q(A)[t].$$

For each $n\ge 1$, let $U_n=(I^n)$1Un​=(In)1​ be the first coefficient ideal of $I^n$, and define $$\mathcal U(I)=\bigoplus$${n\ge 0}U_n t^n. Following the notation of the research log, set $$D_n:=U_n/I^n,\qquad C_n:=\overline{I^n}/U_n,\qquad B_n:=\overline{I^n}/I^n,$$ and globally $$\mathcal D=\bigoplus_{n\ge 1}D_n,\qquad \mathcal C=\bigoplus_{n\ge 1}C_n,\qquad \mathcal B=\bigoplus_{n\ge 1}B_n.$$

On the symmetric side, fix a polynomial presentation $$T=A[y_1,\dots,y_m],\qquad \operatorname{Sym}_A(I)=T/L,\qquad \mathcal R(I)=T/J,$$ where $L\subseteq J$ and $$\mathcal E:=J/L.$$

Standard exact sequences

The inclusions $$\mathcal R(I)\subseteq \mathcal U(I)\subseteq \mathcal R(I)^$$R(I)⊆U(I)⊆R(I)∗ yield exact sequences $$0\to \mathcal R(I)\to \mathcal U(I)\to \mathcal D\to 0, \tag{2.1}$$ $$0\to \mathcal U(I)\to \mathcal R(I)^$$\to \mathcal C\to 0, \tag{2.2} $$0\to \mathcal D\to \mathcal B\to \mathcal C\to 0. \tag{2.3}$$

The symmetric--Rees presentation yields $$0\to \mathcal E\to \operatorname{Sym}_A(I)\to \mathcal R(I)\to 0. \tag{2.4}$$

Splicing (2.4) with $$0\to \mathcal R(I)\to \mathcal R(I)^$$\to \mathcal B\to 00→R(I)→R(I)∗→B→0 gives $$0\to \mathcal E\to \operatorname{Sym}_A(I)\to \mathcal R(I)^$$\to \mathcal B\to 0. \tag{2.5}

Inherited structural facts

We use the following results proved in the source papers and already extracted in the research log.

1. $\mathcal D$ is a finite graded $\mathcal U(I)$-module and $$\dim \mathcal D\le d-1.$$
2. $\mathcal C$ is a finite graded $\mathcal U(I)$-module and $$\mathcal C=0\quad\text{or}\quad \dim \mathcal C=d.$$
3. One has $$\mathcal R(I)\text{ is }R_1 \iff \mathcal C=0 \iff \overline{I^n}=U_n\ \text{for all }n\ge 1.$$
4. The conductor $$\mathfrak c(I):=L:_A J$$ satisfies $$\mathfrak c(I)=\operatorname{Ann}_A(\mathcal E),$$ it localizes, and for every nonzero $r\in \mathfrak c(I)$, $$J=L:r.$$

We also use the elementary finite-length identity $$\lambda(0:_M r)=\lambda(M/rM) \tag{2.6}$$ for any finite-length $A$-module $M$ and any $r\in \mathfrak m$.

Fiber-cone notation

Let $$F(I)=\mathcal R(I)\otimes_A k=\bigoplus_{n\ge 0}I^n/\mathfrak m I^n$$ be the fiber cone. Set $$V:=I/\mathfrak m I, \qquad S_k:=\operatorname{Sym}$$k(V),V:=I/mI,Sk​:=Symk​(V), and define the fiber-cone equation ideal $$J$${\mathrm{fib}}:=\ker(S_k\to F(I)).

Mechanism and Main Structural Identity

Conductor-reduced comparison maps

Fix a nonzero element $$r\in \mathfrak m\cap \mathfrak c(I).$$ For each $n\ge 1$ define $$\phi_n^r:\operatorname{Sym}_A^n(I)/r\operatorname{Sym}_A^n(I)\longrightarrow \overline{I^n}/r\overline{I^n}. \tag{3.1}$$ This factors as $$\operatorname{Sym}_A^n(I)/r\operatorname{Sym}_A^n(I) \xrightarrow{\alpha_n} I^n/rI^n \xrightarrow{\beta_n} U_n/rU_n \xrightarrow{\gamma_n} \overline{I^n}/r\overline{I^n}. \tag{3.2}$$

The conductor identity $J=L:r$ turns multiplication by $r$ into an exact torsion detector in the symmetric algebra: $$0\to \mathcal E\to \operatorname{Sym}_A(I)\xrightarrow{\cdot r}\operatorname{Sym}_A(I)\to \operatorname{Sym}_A(I)/r\operatorname{Sym}_A(I)\to 0. \tag{3.3}$$

Since $\mathcal R(I)\subset A[t]$, multiplication by $r$ is injective on $\mathcal R(I)$ and on $\mathcal U(I)$. Since $\mathcal R(I)^$\subset Q(A)[t]R(I)∗⊂Q(A)[t], it is also injective on $\mathcal R(I)^$. Applying the snake lemma to (2.4), (2.1), and (2.2) yields degreewise exact sequences $$0\to \mathcal E_n\to \operatorname{Sym}_A^n(I)/r\operatorname{Sym}_A^n(I)\xrightarrow{\alpha_n} I^n/rI^n\to 0, \tag{3.4}$$ $$0\to D_n[r]\to I^n/rI^n\xrightarrow{\beta_n} U_n/rU_n\to (D_n/rD_n)\to 0, \tag{3.5}$$ $$0\to C_n[r]\to U_n/rU_n\xrightarrow{\gamma_n} \overline{I^n}/r\overline{I^n}\to (C_n/rC_n)\to 0. \tag{3.6}$$

The four reduced defects

Define $$d_r(n):=\lambda(D_n[r])=\lambda(D_n/rD_n), \tag{3.7}$$ $$\kappa_r(n):=\lambda(C_n[r])=\lambda(C_n/rC_n), \tag{3.8}$$ $$\nu_r(n):=\lambda!\left(\frac{\overline{I^n}}{I^n+r\overline{I^n}}\right), \tag{3.9}$$ and let $$\tau_r(n)$$ be the length of the image of the boundary map $$\partial_n^r:C_n[r]\longrightarrow D_n/rD_n$$ coming from multiplication by $r$ on $$0\to D_n\to B_n\to C_n\to 0. \tag{3.10}$$

By (2.6), the two equalities in (3.7) and (3.8) hold because $D_n$ and $C_n$ have finite length.

The interaction term as an intersection defect

The boundary map $\partial_n^r$ admits an explicit ideal-theoretic description.

Take a class $x+U_n\in C_n[r]$. Then $x\in \overline{I^n}$ and $rx\in U_n$. The connecting map sends $$x+U_n\longmapsto rx+(I^n+rU_n)\in U_n/(I^n+rU_n)\cong D_n/rD_n.$$ Hence $$\operatorname{im}\partial_n^r \cong \frac{U_n\cap (I^n+r\overline{I^n})}{I^n+rU_n}, \tag{3.11}$$ and therefore $$\tau_r(n)= \lambda!\left( \frac{U_n\cap (I^n+r\overline{I^n})}{I^n+rU_n} \right). \tag{3.12}$$

Thus $$\tau_r(n)=0 \iff U_n\cap (I^n+r\overline{I^n})=I^n+rU_n. \tag{3.13}$$ This is the precise intersection-rigidity statement measured by the interaction term.

Main Result

We now state the strongest verified theorem package.

Theorem A (conductor-reduced comparison package)

Assume the setup of Section 2. Let $0\neq r\in \mathfrak m\cap \mathfrak c(I)$.

1. For each $n\ge 1$, the reduced comparison map $\phi_n^r$ has finite-length kernel and cokernel and satisfies $$\lambda(\ker\phi_n^r)-\lambda(\operatorname{coker}\phi_n^r)$$

   \lambda(\mathcal E_n). \tag{4.1}λ(kerϕnr​)−λ(cokerϕnr​)=λ(En​).(4.1)

2. The cokernel of $\phi_n^r$ is the reduced normalization defect: $$\operatorname{coker}\phi_n^r \cong \frac{\overline{I^n}}{I^n+r\overline{I^n}}, \tag{4.2}$$ and its length satisfies the exact decomposition $$\nu_r(n)=d_r(n)+\kappa_r(n)-\tau_r(n). \tag{4.3}$$

3. The interaction term is given by $$\tau_r(n)= \lambda!\left( \frac{U_n\cap (I^n+r\overline{I^n})}{I^n+rU_n} \right). \tag{4.4}$$

4. One has $$\mathcal R(I)\text{ is }R_1 \iff \deg \nu_r(n)\le d-2, \tag{4.5}$$ while $$\mathcal R(I)\text{ is not }R_1 \iff \deg \nu_r(n)=d-1. \tag{4.6}$$

5. If $\mathcal D=0$, equivalently $U_n=I^n$ for all $n$, then $$\tau_r(n)=0,\qquad d_r(n)=0,\qquad \nu_r(n)=\kappa_r(n). \tag{4.7}$$

Theorem B (fiber-corrected bridge to linear type)

Assume the setup of Section 2 and, in addition, $$J_{\mathrm{fib}}=0.$$ Then there exists a constant $C>0$ such that for all $n\ge 1$, $$\lambda(\mathcal E_n)\le C,\lambda(D_n). \tag{4.8}$$ Consequently $$\lambda(\mathcal E_n)=O(n^{d-2}). \tag{4.9}$$ For every nonzero $r\in \mathfrak m\cap \mathfrak c(I)$ one then has $$\lambda(\ker\phi_n^r)=\kappa_r(n)+O(n^{d-2}), \tag{4.10}$$ $$\lambda(\operatorname{coker}\phi_n^r)=\kappa_r(n)+O(n^{d-2}). \tag{4.11}$$

Corollary C

Assume $J_{\mathrm{fib}}=0$ and $\mathcal D=0$. Then $$\mathcal E=0,$$ so $I$ is of linear type.

Proof of the Main Result

Proof of Theorem A

We prove the five statements in order.

Step 1: auxiliary kernel and cokernel sequences

Let $$\psi_n^r:=\gamma_n\circ\beta_n:I^n/rI^n\to \overline{I^n}/r\overline{I^n}.$$ Write $$P_n:=U_n/rU_n,$$ and let $$q_n:P_n\to D_n/rD_n$$ be the natural cokernel map coming from (3.5). Restrict $q_n$ to $C_n[r]\subseteq P_n$ and define $$\theta_n^r:=q_n|_{C_n[r]}:C_n[r]\to D_n/rD_n.$$

We claim that there are exact sequences $$0\to D_n[r]\to \ker(\psi_n^r)\to \ker(\theta_n^r)\to 0, \tag{5.1}$$ $$0\to \operatorname{coker}(\theta_n^r)\to \operatorname{coker}(\psi_n^r)\to C_n/rC_n\to 0. \tag{5.2}$$

To prove (5.1), note from (3.5) that $$\ker\beta_n=D_n[r], \qquad \operatorname{im}\beta_n=\ker q_n.$$ Thus $$\ker\psi_n^r$$

\beta_n^{-1}(C_n[r]). Its image in $P_n$ is $$\operatorname{im}\beta_n\cap C_n[r]$$

\ker q_n\cap C_n[r]

\ker(\theta_n^r). Since the kernel of $\beta_n$ is $D_n[r]$, we obtain (5.1).

For (5.2), observe that $$\operatorname{coker}\psi_n^r$$

\big(\overline{I^n}/r\overline{I^n}\big)\big/\gamma_n(\operatorname{im}\beta_n). There is an exact sequence $$0\to \gamma_n(P_n)/\gamma_n(\operatorname{im}\beta_n)\to \operatorname{coker}\psi_n^r\to \operatorname{coker}\gamma_n\to 0.$$ Since $$\ker\gamma_n=C_n[r]$$ by (3.6), one has $$\gamma_n(P_n)/\gamma_n(\operatorname{im}\beta_n) \cong P_n/(C_n[r]+\operatorname{im}\beta_n) \cong \operatorname{coker}(\theta_n^r),$$ while $\operatorname{coker}\gamma_n=C_n/rC_n$. Hence (5.2) follows.

Finally, since $\alpha_n$ is surjective with kernel $\mathcal E_n$ by (3.4), we have $$0\to \mathcal E_n\to \ker\phi_n^r\to \ker\psi_n^r\to 0, \tag{5.3}$$ $$\operatorname{coker}\phi_n^r\cong \operatorname{coker}\psi_n^r. \tag{5.4}$$

Step 2: proof of the index formula

By (5.3) and (5.4), it is enough to prove $$\lambda(\ker\psi_n^r)=\lambda(\operatorname{coker}\psi_n^r). \tag{5.5}$$

From (5.1) and (5.2), $$\lambda(\ker\psi_n^r)$$

\lambda(D_n[r])+\lambda(\ker\theta_n^r), \tag{5.6} $$\lambda(\operatorname{coker}\psi_n^r)$$

\lambda(\operatorname{coker}\theta_n^r)+\lambda(C_n/rC_n). \tag{5.7} Also, $$\lambda(\ker\theta_n^r)-\lambda(\operatorname{coker}\theta_n^r)$$

\lambda(C_n[r])-\lambda(D_n/rD_n). \tag{5.8} Using (2.6), $$\lambda(C_n[r])=\lambda(C_n/rC_n), \qquad \lambda(D_n[r])=\lambda(D_n/rD_n). \tag{5.9}$$ Substituting (5.8) and (5.9) into (5.6) and (5.7) gives (5.5). Hence $$\lambda(\ker\phi_n^r)-\lambda(\operatorname{coker}\phi_n^r)$$

\lambda(\mathcal E_n),λ(kerϕnr​)−λ(cokerϕnr​)=λ(En​), which proves (4.1).

Step 3: the exact decomposition of $\nu_r(n)$

Since $$B_n/rB_n \cong \frac{\overline{I^n}/I^n}{r(\overline{I^n}/I^n)} \cong \frac{\overline{I^n}}{I^n+r\overline{I^n}},$$ we have $$\lambda(B_n/rB_n)=\nu_r(n). \tag{5.10}$$

Now apply multiplication by $r$ to $$0\to D_n\to B_n\to C_n\to 0.$$ Because all three modules have finite length, we get the long exact sequence $$0\to D_n[r]\to B_n[r]\to C_n[r]\xrightarrow{\partial_n^r} D_n/rD_n\to B_n/rB_n\to C_n/rC_n\to 0. \tag{5.11}$$ Break it into short exact pieces: $$0\to D_n[r]\to B_n[r]\to \ker\partial_n^r\to 0, \tag{5.12}$$ $$0\to \ker\partial_n^r\to C_n[r]\to \operatorname{im}\partial_n^r\to 0, \tag{5.13}$$ $$0\to \operatorname{im}\partial_n^r\to D_n/rD_n\to B_n/rB_n\to C_n/rC_n\to 0. \tag{5.14}$$ Taking lengths and using (2.6) yields $$\lambda(B_n/rB_n)$$

\lambda(D_n[r])+\lambda(C_n[r])-\lambda(\operatorname{im}\partial_n^r). By definition, $\tau_r(n)=\lambda(\operatorname{im}\partial_n^r)$, so $$\nu_r(n)=d_r(n)+\kappa_r(n)-\tau_r(n),$$ which proves (4.3).

The cokernel description (4.2) follows directly from (5.10) and (5.4).

Step 4: the intersection formula for $\tau_r(n)$

Take $x+U_n\in C_n[r]$. Then $x\in \overline{I^n}$ and $rx\in U_n$. Under the connecting map, $$x+U_n\longmapsto rx+(I^n+rU_n)\in U_n/(I^n+rU_n)\cong D_n/rD_n.$$ Therefore the image consists exactly of residue classes represented by elements of $$U_n\cap r\overline{I^n}$$ modulo $I^n+rU_n$. Since $I^n\subseteq U_n$, we have $$I^n+(U_n\cap r\overline{I^n})=U_n\cap (I^n+r\overline{I^n}),$$ hence $$\operatorname{im}\partial_n^r \cong \frac{U_n\cap (I^n+r\overline{I^n})}{I^n+rU_n}.$$ Taking lengths proves (4.4).

Step 5: asymptotic detection of $R_1$

First suppose $\mathcal C=0$. Then $\kappa_r(n)=0$ for all $n$ and $\tau_r(n)=0$ for all $n$. Since $\dim \mathcal D\le d-1$, one has $$d_r(n)=O(n^{d-2}),$$ hence by (4.3), $$\nu_r(n)=O(n^{d-2}).$$ So $\deg \nu_r(n)\le d-2$.

Now suppose $\mathcal C\neq 0$. Then $\dim \mathcal C=d$. We claim $$\dim(0:$${\mathcal C}r)=d. \tag{5.15}dim(0:C​r)=d.(5.15) Let $P$ be a minimal prime of $\mathcal C$. Every graded piece $C_n$ has finite length, so $$P\cap A=\mathfrak m.$$ Since $r\in \mathfrak m$, we have $r\in P$. Because minimal primes of a finite module are associated primes, there exists $z\in \mathcal C$ with $$\operatorname{Ann}(z)=P.$$ Then $rz=0$, so $z\in 0:${\mathcal C}r. Thus $P\in \operatorname{Supp}(0:${\mathcal C}r)P∈Supp(0:C​r) and $$\dim(0:$${\mathcal C}r)\ge \dim \mathcal U(I)/P=d. The reverse inequality is automatic, so (5.15) holds.

Therefore $\kappa_r(n)=\lambda(C_n[r])$ is eventually a polynomial of degree $d-1$. On the other hand, $$d_r(n)=O(n^{d-2}),\qquad \tau_r(n)=O(n^{d-2}),$$ because $\tau_r(n)$ is the length of a subquotient of $D_n/rD_n$ and $\dim \mathcal D\le d-1$. Hence from (4.3), $$\nu_r(n)=\kappa_r(n)+O(n^{d-2}),$$ so $\deg \nu_r(n)=d-1$.

Since $\mathcal R(I)$ is $R_1$ if and only if $\mathcal C=0$, this proves (4.5) and (4.6).

Step 6: the case $\mathcal D=0$

If $\mathcal D=0$, then every $D_n=0$. Hence $d_r(n)=0$ and the boundary map $$\partial_n^r:C_n[r]\to D_n/rD_n$$ is zero, so $\tau_r(n)=0$. Thus (4.3) becomes $$\nu_r(n)=\kappa_r(n),$$ proving (4.7).

This completes the proof of Theorem A.

Proof of Theorem B

We now assume $$J_{\mathrm{fib}}=0.$$

Step 1: a degreewise exact sequence through the first-coefficient algebra

The composite $$\operatorname{Sym}_A^n(I)\twoheadrightarrow I^n\hookrightarrow U_n$$ has kernel $\mathcal E_n$ and cokernel $D_n$, so there is an exact sequence $$0\to \mathcal E_n\to \operatorname{Sym}_A^n(I)\to U_n\to D_n\to 0. \tag{5.16}$$

Tensoring (5.16) with $k$ gives an exact segment $$\operatorname{Tor}_1^A(k,D_n)\xrightarrow{\alpha_n} \mathcal E_n/\mathfrak m\mathcal E_n \xrightarrow{\beta_n} \operatorname{Sym}_k^n(V) \xrightarrow{\sigma_n} U_n/\mathfrak m U_n \to D_n/\mathfrak m D_n\to 0. \tag{5.17}$$

Tensoring $$0\to I^n\to U_n\to D_n\to 0$$ with $k$ gives $$\operatorname{Tor}_1^A(k,D_n)\xrightarrow{\gamma_n} I^n/\mathfrak m I^n \xrightarrow{\iota_n} U_n/\mathfrak m U_n \to D_n/\mathfrak m D_n\to 0. \tag{5.18}$$

Since $J_{\mathrm{fib}}=0$, the natural map $$\pi_n:\operatorname{Sym}_k^n(V)\to I^n/\mathfrak m I^n$$ is an isomorphism for every $n$. Therefore $$\sigma_n=\iota_n\circ \pi_n.$$ Hence $$\ker\sigma_n\cong \ker\iota_n=\operatorname{im}\gamma_n. \tag{5.19}$$ So $\ker\sigma_n$ is a quotient of $\operatorname{Tor}_1^A(k,D_n)$.

From (5.17), the cokernel of $\alpha_n$ is exactly $\ker\sigma_n$. Therefore $$0\to \operatorname{im}\alpha_n\to \mathcal E_n/\mathfrak m\mathcal E_n\to \ker\sigma_n\to 0. \tag{5.20}$$ Both $\operatorname{im}\alpha_n$ and $\ker\sigma_n$ are quotients of $\operatorname{Tor}_1^A(k,D_n)$. Consequently $$\dim_k(\mathcal E_n/\mathfrak m\mathcal E_n)\le 2,\dim_k\operatorname{Tor}_1^A(k,D_n). \tag{5.21}$$

Step 2: uniform control of $\operatorname{Tor}_1^A(k,D_n)$

Set $$b_1(A):=\dim_k\operatorname{Tor}_1^A(k,k).$$ If $M$ is a finite-length $A$-module, then $$\dim_k\operatorname{Tor}_1^A(k,M)\le b_1(A),\lambda(M). \tag{5.22}$$ Indeed, choose a composition series for $M$ with simple quotients $k$ and induct on $\lambda(M)$.

Applying (5.22) to $M=D_n$ and combining with (5.21) gives $$\mu_A(\mathcal E_n):=\dim_k(\mathcal E_n/\mathfrak m\mathcal E_n) \le 2b_1(A),\lambda(D_n). \tag{5.23}$$

If $\mathcal E=0$, then there is nothing more to prove. Assume $\mathcal E\neq 0$. Since $\mathcal E$ is annihilated by $\mathfrak c(I)$ and $I$ is $\mathfrak m$-primary, the conductor is $\mathfrak m$-primary in this nontrivial case. Choose $s\ge 1$ such that $$\mathfrak m^s\subseteq \mathfrak c(I). \tag{5.24}$$ Then $$\mathfrak m^s\mathcal E_n=0$$ for every $n$. Hence a minimal generating set of $\mathcal E_n$ gives a surjection $$(A/\mathfrak m^s)^{\mu_A(\mathcal E_n)}\twoheadrightarrow \mathcal E_n,$$ so $$\lambda(\mathcal E_n)\le \lambda(A/\mathfrak m^s),\mu_A(\mathcal E_n). \tag{5.25}$$ Combining (5.23) and (5.25), we get $$\lambda(\mathcal E_n)\le C,\lambda(D_n), \tag{5.26}$$ where $$C:=2,\lambda(A/\mathfrak m^s),b_1(A).$$ This proves (4.8).