1. Introduction
Let 
$(A,\mathfrak{m} )$ be a non-regular Cohen–Macaulay local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. If N is an A-module of finite length then 
$\ell(N)$ denotes its length. Let M be a maximal Cohen–Macaulay (= MCM) A-module. The function 
$t_I(M, n) = \ell(\operatorname{Tor}^A_1(M, A/I^{n+1}))$ is of polynomial type, see [Reference Theodorescu12, Corollary 4] (also see [Reference Puthenpurakal8, Proposition 17]). Let 
$t_I^M(z) \in \mathbb{Q}[z]$ be such that 
$t_I^M(n) = t_I(M, n)$ for all 
$n \gg 0$. It is easily shown that 
$\deg t_I^M(z) \leq d - 1$. In [Reference Puthenpurakal8, Theorem 18], we proved that 
$\deg t_\mathfrak{m} ^M(z) = d-1$ for any non-free MCM A-module. It was also shown that if I is a parameter ideal then 
$t_I(M, n) = 0$ for all 
$n \geq 0$, see [Reference Puthenpurakal8, Remark 20]. In general, it is a difficult question to determine the degree of 
$t_I^M(z)$ and the answer is known only for a few classes of ideals and modules, see [Reference Katz and Theodorescu6, 3.5] for some examples. The fact that 
$\deg t_\mathfrak{m} ^M(z) = d-1$ for non-free MCM has an important consequence in the study of associated graded modules (with respect to 
$\mathfrak{m} $) of MCM A-modules, see [Reference Puthenpurakal9].
In this paper, we prove few surprising results. Recall A is said to be a hypersurface ring if its completion 
$\widehat {A} = Q/(f)$ where 
$(Q, \mathfrak{n} )$ is a regular local ring and 
$f \in \mathfrak{n} ^2$ is non-zero. We set the degree of the zero polynomial to be −1. We show
Theorem 1.1. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$r_I \geq -1$ (depending only on I) such that if M is any non-free MCM A-module then 
$\deg t_I^M(z) = r_I$.
1.1.
For the Ext functors, we prove an analogous result. It is known that if M is a finitely generated A-module the function 
$n \rightarrow \ell(\operatorname{Ext}_A^1(M, A/I^{n+1}))$ is of polynomial type say of degree 
$s_I^M$, see [Reference Theodorescu12, Corollary 4]. We prove
Theorem 1.2. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$s_I \geq -1$ (depending only on I) such that if M is any non-free MCM A-module then 
$s_I^M = s_I$.
It is also known that if M is a finitely generated A-module the function 
$n \rightarrow \ell(\operatorname{Ext}_A^{d+1}(A/I^{n+1}, M))$ is of polynomial type say of degree 
$e_I^M$, see [Reference Theodorescu12, Theorem 5]. Let 
$\operatorname{Spec}^0(A) = \operatorname{Spec}(A) \setminus \{\mathfrak{m} \}$. We prove
Theorem 1.3. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$e_I \geq -1$ (depending only on I) such that if M is any non-free MCM A-module free on 
$\operatorname{Spec}^0(A)$ then 
$e_I^M = e_I$.
See Remark 4.1 on why in Theorem 1.3 we need to restrict to the case of MCM modules free on 
$\operatorname{Spec}^0(A)$ while in Theorems 1.1 and 1.2 we do not have such restriction.
Finally we show
1.2.
Let 
$G_I(A) = \bigoplus_{n \geq 0}I^n/I^{n+1}$ be the associated graded ring of I. Let M be an A-module. Let 
$G_I(M) = \bigoplus_{n \geq 0}I^nM/I^{n+1}M$ be the associated graded module of M with respect to I, considered as a module over 
$G_I(A)$. We studied the function 
$t_I(M, n)$ as it is useful in the study of associated graded modules. The following result shows that the reverse direction is very strong at least for hypersurfaces; i.e. some properties of 
$G_I(M)$ for some MCM M has a strong impact on rI.
Theorem 1.5. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Suppose there exists a non-free MCM A-module M with 
$G_I(M)$ unmixed and equi-dimensional. Then 
$r_I = d -1$ or 
$r_I = -1$. Furthermore if 
$r_I = -1$ then 
$\operatorname{projdim} A/I^n \lt \infty$ for all 
$n \geq 1$.
Technique used to prove the result: We first note that the function 
$t_I(M,n)$ is a function on 
$\operatorname{\underline{CM}}(A)$ the stable category of MCM A-modules. We also note that 
$\operatorname{\underline{CM}}(A)$ is a triangulated category [Reference Buchweitz3, 4.4.1]. Let 
$\operatorname{\underline{CM}}_0(A)$ be the thick subcategory of MCM A-modules which are free on the punctured spectrum 
$\operatorname{Spec}^0(A)$ of A. The crucial ingredient in our proofs is that 
$\operatorname{\underline{CM}}_0(A)$ has no proper thick subcategories, see [Reference Takahashi11, 6.6]. We first prove Theorem 1.1 for non-free MCM modules in 
$\operatorname{\underline{CM}}_0(A)$ and then prove for all non-free MCM A-modules by using an induction on 
$\dim \operatorname{\underline{Hom}}_A(M, M)$. The techniques to prove Theorems 1.2 and 1.3 are similar.
Here is an overview of the contents of this paper. In § 2, we discuss a few preliminaries that we need. In § 3, we prove Theorems 1.1–1.3 when M is free on the punctured spectrum of A. In § 4, we prove Theorems 1.1 and 1.2. In § 5, we prove Theorem 1.4. Finally in § 6, we prove Theorem 1.5.
2. Preliminaries
In this section, we discuss a few preliminary results that we need. We use [Reference Neeman7] for notation on triangulated categories. However, we will assume that if 
$\mathcal{C}$ is a triangulated category then 
$\operatorname{Hom}_\mathcal{C}(X, Y)$ is a set for any objects 
$X, Y$ of 
$\mathcal{C}$.
2.1.
Let 
$\mathcal{C} $ be a skeletally small triangulated category with shift operator ∑ and let 
$\mathbb{I} (\mathcal{C} )$ be the set of isomorphism classes of objects in 
$\mathcal{C} $. By a weak triangle function on 
$\mathcal{C} $, we mean a function 
$\xi \colon \mathbb{I} (\mathcal{C} ) \rightarrow \mathbb{Z} $ such that
(1)
$\xi(X) \geq 0$ for all
$X \in \mathcal{C} $.(2)
$\xi(0) = 0$.(3)
$\xi(X \oplus Y) = \xi(X) + \xi(Y)$ for all
$X, Y \in \mathcal{C} $.(4)
$\xi(\Sigma X ) = \xi(X)$ for all
$X \in \mathcal{C} $.(5) If
$X \rightarrow Y \rightarrow Z \rightarrow \Sigma X $ is a triangle in
$\mathcal{C} $ then
$\xi(Z) \leq \xi(X) + \xi(Y)$.
2.2.
Set
\begin{equation*}\ker \xi = \{X \mid \xi(X) = 0 \}.\end{equation*}The following result (essentially an observation) is a crucial ingredient in our proof of Theorem 1.1.
Lemma 2.1. (see [Reference Puthenpurakal10], 2.3) (with hypotheses as above)
$\ker \xi $ is a thick subcategory of 
$\mathcal{C} $.
2.3.
Let 
$(A,\mathfrak{m} )$ be a hypersurface ring and let I be an 
$\mathfrak{m} $-primary ideal in A. Let M be an MCM A-module. Set for 
$n \geq 0$
\begin{align*}
t_I(M,n) &= \ell(\operatorname{Tor}^A_1(M, A/I^{n+1})) \\
s_I(M,n) &= \ell(\operatorname{Ext}^1_A(M, A/I^{n+1})) \\
e_I(M, n) &= \ell(\operatorname{Ext}^{d+1}_A( A/I^{n+1}, M)).
\end{align*} Let 
$\Omega^i_A(M)$ denote the 
$i\mathrm{th}$-syzygy of M. We prove
Lemma 2.2. (with hypotheses as above)
(1) For all
$n \geq 0$, the functions
$t_I(-,n), s_I(-,n)$ and
$e_I(-,n)$ are functions on
$\operatorname{\underline{CM}}(A)$.(2) For all
$n \geq 0$, we have
$t_I(M,n) = t_I(\Omega^1_A(M),n), s_I(M,n) = s_I(\Omega^1_A(M),n)$ and
$e_I(M,n) = e_I(\Omega^1_A(M),n)$.
Proof. (1) Let 
$E = M \oplus F = N \oplus G$ where 
$F, G$ are free A-modules. Then by definition 
$t_I(E, n) = t_I(M,n) = t_I(N,n)$. Thus 
$t_I(-, n)$ is a function on 
$\operatorname{\underline{CM}}(A)$.
The proof for assertions on 
$s_I(-, n)$ and 
$e_I(-, n)$ are similar.
(2) We may assume that M has no free summands. Set 
$N = \Omega_A^1(M)$. Let 
$ 0 \rightarrow N \rightarrow F \rightarrow M \rightarrow 0$ be the minimal presentation of M with 
$F = A^r$. Then note as A is a hypersurface ring and M is MCM without free summands we get that a minimal presentation of N is as follows 
$0 \rightarrow M \rightarrow G \rightarrow N \rightarrow 0$ where 
$G = A^r$, see [Reference Eisenbud4, 6.1(ii)]. By using the first exact sequence, we get
\begin{equation*}
0 \rightarrow \operatorname{Tor}^A_1(M, A/I^{n+1}) \rightarrow N/I^{n+1}N \rightarrow F/I^{n+1}F \rightarrow M/I^{n+1} \rightarrow 0.
\end{equation*}So we have
\begin{equation*}
t_I(M,n) = \ell(N/I^{n+1}N) + \ell(M/I^{n+1}M) - r\ell(A/I^{n+1}A).
\end{equation*} Using the second exact sequence, we find that 
$t_I(M,n) = t_I(N,n)$. The result follows.
The proof for assertions on 
$s_I(-, n)$ and 
$e_I(-, n)$ is similar.
2.4.
The only fact we used in the above lemma was that 
$\Omega^2_A(M) \cong M$ if M is an MCM A-module with no free summands and that A is Gorenstein. Thus, the assertions of the lemma hold if A is a local complete intersection and M is a 2-periodic MCM A-module.
3.
$\operatorname{\underline{CM}}_0(A)$
In this section, we give proofs of Theorems 1.1–1.3 when M is free on 
$\operatorname{Spec}^0(A)$.
Theorem 3.1. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$r_I \geq -1$ (depending only on I) such that if 
$M \in \operatorname{\underline{CM}}_0(A)$ is non-zero then 
$\deg t_I^M(z) = r_I$.
Proof. We first note that for any MCM M we have 
$\deg t_I^M(z) \leq d - 1$, see [Reference Theodorescu12, Corollary 4]. Recall the degree of the zero polynomial to be −1. Set
\begin{equation*} r = \max \{\deg t_I^M(z) \mid M \in \operatorname{\underline{CM}}_0(A) \}. \end{equation*} If 
$r = -1$ then we have nothing to prove. So assume 
$r \geq 0$. For 
$M \in \operatorname{\underline{CM}}_0(A)$ define
\begin{equation*}
\xi_I(M) = \lim_{n \rightarrow \infty}\frac{r!}{n^r} t_I(M,n).
\end{equation*} We note that 
$\xi_I(M) \geq 0$ and is zero precisely when 
$\deg t_I(M, z) \lt r$.
Claim: 
$\xi_I(-)$ is a weak triangle function on 
$\operatorname{\underline{CM}}_0(A)$, see § 2.1.
Assume the claim for the time being. Then 
$\ker \xi$ is a thick subcategory of 
$\operatorname{\underline{CM}}_0(A)$. Also if 
$\deg t_I^L(z) = r$ then 
$L \notin \ker \xi$. So 
$\ker \xi \neq \operatorname{\underline{CM}}_0(A)$. As 
$\operatorname{\underline{CM}}_0(A)$ has no proper thick subcategories, see [Reference Takahashi11, 6.6], it follows that 
$\ker \xi = 0$. Therefore 
$\deg t_I^M(z) = r$ for all M ≠ 0 in 
$\operatorname{\underline{CM}}_0(A)$.
It remains to show ξI is a weak triangle function on 
$\operatorname{\underline{CM}}_0(A)$. The first three conditions are trivial to satisfy. By Lemma 2.2(2), it follows that 
$\xi_I(\Omega^{-1}_A(M)) = \xi_I(M)$. Let 
$ L \rightarrow M \rightarrow N \rightarrow \Omega^{-1}(L)$ is a triangle in 
$\operatorname{\underline{CM}}_0(A)$ then note that we have a short exact sequence of A-modules
\begin{equation*} 0 \rightarrow M \rightarrow N \oplus F \rightarrow \Omega^{-1}(L) \rightarrow 0, \quad \text{where}\ F\ \text{is free.}\end{equation*}Therefore, we have an inequality
\begin{equation*} t_I(N,n) \leq t_I(M, n) + t_I( \Omega^{-1}(L), n).\end{equation*}The result follows.
The following two results can be proved similarly as in Theorem 3.1. We have to use that 
$\deg s_I^M(z) \leq d -1$ (see [Reference Theodorescu12, Corollary 4]) and that 
$\deg e_I^M(z) \leq d$ (see [Reference Theodorescu12, Corollary 7]).
Theorem 3.2. Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$s_I \geq -1$ (depending only on I) such that if 
$M \in \operatorname{\underline{CM}}_0(A)$ is non-zero then 
$\deg s_I^M(z) = s_I$.
Theorem 3.3. (= Theorem 1.3) Let 
$(A,\mathfrak{m} )$ be a hypersurface local ring of dimension 
$d \geq 1$ and let I be an 
$\mathfrak{m} $-primary ideal. Then there is an integer 
$e_I \geq -1$ (depending only on I) such that if 
$M \in \operatorname{\underline{CM}}_0(A)$ is non-zero then 
$\deg e_I^M(z) = e_I$.
4. Proofs of Theorems 1.1 and 1.2
In this section, we give proofs of Theorems 1.1 and 1.2. We need a few preliminaries.
4.1.
Let M be any finitely generated A-module. Set
$L_i(M) = \bigoplus_{n \geq 0}\operatorname{Tor}^A_i(M, A/I^{n+1})$ for 
$i \geq 0$. Let 
$\mathcal{R} = A[It]$ be the Rees algebra of I. We have an exact sequence of 
$\mathcal{R} $-modules
\begin{equation*} 0 \rightarrow \mathcal{R} (1) \rightarrow A[t](1) \rightarrow L_0(A) \rightarrow 0.\end{equation*} Tensoring with M yields an inclusion 
$0 \rightarrow L_1(M) \subseteq \mathcal{R} (1)\otimes M$ and isomorphisms 
$L_i(M) \cong \operatorname{Tor}^A_{i-1}(\mathcal{R} (1), M)$ for 
$i \geq 2$. It follows that 
$L_i(M)$ are finitely generated 
$\mathcal{R} $-module for all 
$i \geq 1$. We note that if 
$\Omega^A_2(M) \cong M$ then we have 
$L_i(M) \cong L_{i+2}(M)$ for all 
$i \geq 1$.
4.2.
We also need the following notion. Let 
$M \in \operatorname{\underline{CM}}(A)$. Let
\begin{equation*}
\operatorname{\underline{Supp}}(M) = \{P \mid M_P \ \text{is not free} \ A_P-\text{module} \}.
\end{equation*} If I is an ideal in A then set 
$V(I) = \{P \in \operatorname{Spec}(A) \mid P \supseteq I \}$. It is readily verified that 
$\operatorname{\underline{Supp}}(M) = V(\operatorname{ann} \operatorname{\underline{Hom}}(M, M))$.
Proof of Theorem 1.1.
By Theorem 3.1, we have that there exists rI such that for any non-free MCM module 
$E \in \operatorname{\underline{CM}}_0(A)$ we have 
$\deg t_I^E(z) = r_I$.
Claim: For any non-free MCM A-module M, we have 
$\deg t_I^M(z) = r_I$.
We prove this assertion by induction on 
$\dim \operatorname{\underline{Supp}}(M)$. If 
$\dim \operatorname{\underline{Supp}}(M) = 0$ then M is free on 
$\operatorname{Spec}^0(A)$. In this case, we have nothing to show.
Now assume 
$\dim \operatorname{\underline{Supp}}(M) \gt 0$. As 
$L_1(M)_n, L_2(M)_n$ have finite length for all n and as 
$L_1(M), L_2(M)$ are finitely generated 
$\mathcal{R} $-modules it follows that there exists l such that 
$\mathfrak{m} ^l L_i(M)_n = 0$ for all n and for 
$i = 1, 2$. As M has period two it follows that 
$\mathfrak{m} ^l L_i(M)_n = 0$ for all 
$i \geq 1$ and all 
$n \geq 0$.
Let
\begin{equation*}x \in \mathfrak{m} ^l \setminus \bigcup_{\stackrel{P \supseteq \operatorname{ann} \operatorname{\underline{Hom}}(M, M)}{P \ \text{minimal}}} P.\end{equation*} Let 
$M \xrightarrow{x} M \rightarrow N \rightarrow \Omega^{-1}(M)$ be a triangle in 
$\operatorname{\underline{CM}}(A)$. It is readily verified that support of 
$\operatorname{\underline{Hom}}(N, N)$ is contained in the intersection of support of 
$\operatorname{\underline{Hom}}(M,M)$ and 
$M/x M$. So 
$\dim \operatorname{\underline{Supp}}(N) \leq \dim \operatorname{\underline{Supp}}(M) -1$. It is also not difficult to prove that N is not free A-module. By induction hypotheses 
$\deg t_I^N(z) = r_I$. By the structure of triangles in 
$\operatorname{\underline{CM}}(A)$, see [Reference Buchweitz3, 4.4.1], we have an exact sequence 
$0 \rightarrow G \rightarrow N \rightarrow M/xM \rightarrow 0$ with G-free. It follows that 
$L_3(N) = L_3(M/xM)$. We also have an exact sequence 
$0 \rightarrow M \xrightarrow{x} M \rightarrow M/xM \rightarrow 0$. As 
$x \in \operatorname{ann} L_i(M)$, it follows that we have an exact sequence
\begin{equation*} 0 \rightarrow L_3(M) \rightarrow L_3(M/xM) \rightarrow L_2(M) \rightarrow 0.\end{equation*} As the Hilbert function of 
$L_3(M)$ and 
$L_2(M)$ is identical, see Lemma 2.2(2), we get that 
$2t_I^M(z) = t_I^N(z)$. It follows that 
$\deg t_I^M(z) = r_I$. By induction, the result follows.
4.3.
To prove Theorem 1.2, we need a few preliminaries. Let M be a finitely generated Cohen–Macaulay A-module of dimension r. Let
$E^i(M) = \bigoplus_{n \geq 0}\operatorname{Ext}^i_A(M, A/I^{n+1})$. The exact sequence of 
$\mathcal{R} $-modules
\begin{equation*} 0 \rightarrow \mathcal{R} (1) \rightarrow A[t](1) \rightarrow L_0(A) \rightarrow 0,\end{equation*} induces an isomorphism 
$E^i(M) \cong \operatorname{Ext}^{i+1}_A(M, \mathcal{R} (1))$ for all 
$i \gt d -r$. In particular, 
$E^i(M)$ are finitely generated 
$\mathcal{R} $-modules for all 
$i \gt d -r$. We note that if 
$\Omega^A_2(M) \cong M$ then we have 
$E^i(M) \cong E^{i+2}(M)$ for all 
$i \geq 1$. The proof of Theorem 1.2 is mostly similar to the proof of Theorem 1.1. So we mostly sketch the proof.
Sketch of a proof of Theorem 1.2.
By Theorem 3.2, we have that there exists sI such that for any non-free MCM module 
$L \in \operatorname{\underline{CM}}_0(A)$ we have 
$\deg s_I^L(z) = s_I$.
Claim: For any non-free MCM A-module M, we have 
$\deg s_I^M(z) = s_I$.
We prove this assertion by induction on 
$\dim \operatorname{\underline{Supp}}(M)$. If 
$\dim \operatorname{\underline{Supp}}(M) = 0$ then M is free on 
$\operatorname{Spec}^0(A)$. In this case, we have nothing to show.
Now assume 
$\dim \operatorname{\underline{Supp}}(M) \gt 0$. As 
$E^1(M)_n, E^2(M)_n$ have finite length for all n and as 
$E^1(M), E^2(M)$ are finitely generated 
$\mathcal{R} $-modules it follows that there exists l such that 
$\mathfrak{m} ^l E^i(M)_n = 0$ for all n and for 
$i = 1, 2$. As M has period two it follows that 
$\mathfrak{m} ^l E^i(M)_n = 0$ for all 
$i \geq 1$ and all 
$n \geq 0$. Let
\begin{equation*}x \in \mathfrak{m} ^l \setminus \bigcup_{\stackrel{P \supseteq \operatorname{ann} \operatorname{\underline{Hom}}(M, M)}{P \ \text{minimal}}} P.\end{equation*} Let 
$M \xrightarrow{x} M \rightarrow N \rightarrow \Omega^{-1}(M)$ be a triangle in 
$\operatorname{\underline{CM}}(A)$. As before we have 
$\dim \operatorname{\underline{Supp}}(N) \leq \dim \operatorname{\underline{Supp}}(M) -1$ and N is not free. By induction hypotheses 
$\deg s_I^N(z) = s_I$. By the structure of triangles in 
$\operatorname{\underline{CM}}(A)$, see [Reference Buchweitz3, 4.4.1], we have an exact sequence 
$0 \rightarrow G \rightarrow N \rightarrow M/xM \rightarrow 0$ with G-free. It follows that 
$E^3(N) = E^3(M/xM)$. We also have an exact sequence 
$0 \rightarrow M \xrightarrow{x} M \rightarrow M/xM \rightarrow 0$. As 
$x \in \operatorname{ann} E^i(M)$ it follows that we have an exact sequence
\begin{equation*} 0 \rightarrow E^2(M) \rightarrow E^3(M/xM) \rightarrow E^3(M) \rightarrow 0.\end{equation*} As the Hilbert function of 
$E^3(M)$ and 
$E^2(M)$ is identical, see Lemma 2.2(2), we get that 
$2s_I^M(z) = s_I^N(z)$. It follows that 
$\deg s_I^M(z) = s_I$. By induction, the result follows.
Remark 4.1. Consider 
$U^i(M) = \bigoplus_{n \geq 0}\operatorname{Ext}^i_{A}(A/I^{n+1}, M)$. Then for 
$i \geq d+1$ it is possible to give a natural 
$\mathcal{R} $-module structure on 
$U^i(M)$. However, with this structure, 
$U^i(M)$ is NOT finitely generated (note if 
$xt \in \mathcal{R} _1$ then 
$x_1t U^i(M)_n \subseteq U^i(M)_{n-1}$). Thus, it is not possible to extend the result in Theorem 3.3 to all MCM modules.
5. Proof of Theorem 1.4
In this section, we give a proof of Theorem 1.4. To prove the result note that we may assume the residue field k of A is infinite (otherwise take the extension 
$A[t]_{\mathfrak{m} A[t]}$ of A).
5.1.
Recall 
$x \in I$ is M-superficial with respect to I if there exists c such that 
$(I^{n+1}M \colon x) \cap I^cM = I^nM$ for all 
$n \gg 0$. If 
$\operatorname{depth} M \gt 0$ then (as I is 
$\mathfrak{m} $-primary) it follows that x is M-regular and 
$(I^{n+1}M \colon x) = I^nM$ for 
$n \gg 0$. If k is infinite then I-superficial elements with respect to M exists. In fact in this case, there exists a non-empty open set UM in the Zariski topology of 
$I/\mathfrak{m} I$ such that if the image of x is in UM then x is I-superficial with respect to M.
5.2.
Let 
$E = \bigoplus_{n \geq 0}E_n$ be a finitely generated graded module over the Rees algebra 
$\mathcal{R} = A[It]$. Assume En has finite length for all n. There exists 
$xt \in \mathcal{R} _1$ such that xt is E-filter regular, i.e. 
$(0 \colon_E xt)_n = 0$ for 
$n \gg 0$. In fact in this case, there exists a non-empty open set UE in the Zariski topology of 
$I/\mathfrak{m} I$ such that if the image of x is in UE then xt is E-filter regular.
5.3.
Assume 
$\dim A \geq 2$ and assume M is an MCM A-module. Also assume A is Gorenstein. Let 
$x \in I$ be such that it is 
$M \oplus A$-superficial and xt is 
$L_1(M)$, 
$E^i(M)$-filter regular for 
$i = 1, 2$. We note that 
$(I^{n+1} \colon x) = I^n$ and 
$(I^{n+1}M \colon x) = I^nM$ for 
$n \gg 0$. Set 
$B = A/(x)$ and 
$J = I/(x)$. We have an exact sequence for 
$n \geq 1$
\begin{equation*}
0 \rightarrow \ker \alpha_n \rightarrow A/I^n \xrightarrow{\alpha_n} A/I^{n+1} \rightarrow B/J^{n+1} \rightarrow 0,
\end{equation*} where 
$\alpha_n(a + I^n) = xa + I^{n+1}$. We note that 
$\ker \alpha_n = (I^{n+1} \colon x)/I^n = 0$ for 
$n \gg 0$. Set 
$N = M/xM$. Thus for 
$n \gg 0$ we have an exact sequence
\begin{align*}
\operatorname{Tor}^A_1(M, A/I^n) &\xrightarrow{\alpha^1_n} \operatorname{Tor}^A_1(M, A/I^{n+1}) \rightarrow \operatorname{Tor}^B_1(N, B/J^{n+1}) \rightarrow \\
M/I^nM &\xrightarrow{\alpha^0_n} M/I^{n+1}M \rightarrow N/J^{n+1}N \rightarrow 0.
\end{align*} We note that 
$\ker \alpha^0_n = (I^{n+1}M \colon x)/I^nM = 0$ for 
$n \gg 0$. Furthermore, the map 
$\alpha^1_n$ is the 
$n\mathrm{th}$-component of the multiplication map by 
$xt \in \mathcal{R} _1$ on 
$L_1(M)$. As xt is 
$L_1(M)$-filter regular, it follows that 
$\ker \alpha^1_n = 0$ for 
$n \gg 0$. Thus for 
$n \gg 0$, we have an exact sequence.
5.4.
\begin{equation*}0 \rightarrow \operatorname{Tor}^A_1(M, A/I^n) \xrightarrow{\alpha^1_n} \operatorname{Tor}^A_1(M, A/I^{n+1}) \rightarrow \operatorname{Tor}^B_1(N, B/J^{n+1}) \rightarrow 0.\end{equation*} Similarly, as xt is also 
$E_1(M)$ and 
$E_2(M)$-filter regular, we have an exact sequence.
5.5.
\begin{equation*}0 \rightarrow \operatorname{Ext}^1_A(M, A/I^n) \xrightarrow{\beta^1_n} \operatorname{Ext}_A^1(M, A/I^{n+1}) \rightarrow \operatorname{Ext}_B^1(N, B/J^{n+1}) \rightarrow 0.\end{equation*}5.6.
We note that if A is a hypersurface and 
$I \subseteq \mathfrak{m} ^2$ is an 
$\mathfrak{m} $-primary ideal then if we go mod an I-superficial element x then the resulting ring 
$B = A/(x)$ is no longer a hypersurface. It is, however, a complete intersection. Thus we prove the following result:
Theorem 5.1. Let 
$(A,\mathfrak{m} )$ be a complete intersection of dimension 
$d \geq 1$. Let I be an 
$\mathfrak{m} $-primary ideal. Let M be a non-free MCM A-module with bounded Betti numbers. Then 
$\deg t^M_I(z) = \deg s^M_I(z)$.
Proof. We prove the result by induction on dimension d of A. We may assume M has no free summands. As M has bounded Betti numbers, it follows that M is 2-periodic, i.e. 
$\Omega^2_A(M) \cong M$, (this is essentially contained in [Reference Avramov1, 9.2.1]; use 
$M \cong \Omega^1(\Omega^{-1}(M))$).
We first consider the case when d = 1. We note that in this case 
$\deg t^M_I(z) = 0 $ or 
$ = -1$. Similarly, we have 
$\deg s^M_I(z) = 0 $ or 
$ = -1$. We show that 
$\deg t^M_I(z) = -1 $ if and only if 
$\deg s^M_I(z) = -1 $. Suppose 
$\deg t^M_I(z) = -1 $. So 
$t_I^M(z) = 0$ and therefore 
$t_I(M, n) = 0$ for 
$n \gg 0$ (say from 
$n \geq m$). As 
$t_I(M, n) = t_I(\Omega^1_A(M), n)$ for all n and M is 2-periodic, we have 
$\operatorname{Tor}^A_i(M, A/I^{n +1}) = 0$ for all 
$i \geq 1$ and for 
$n \geq m$. It follows from [Reference Avramov and Buchweitz2, Theorem III] that 
$\operatorname{Ext}^i(M, A/I^{n +1}) = 0$ for all 
$i \gg 0$ and for all 
$n \geq m$. As 
$s_I(M, n) = s_I(\Omega^1_A(M), n)$ for all n and M is 2-periodic, we have 
$s_I(M, n) = 0$ for all 
$n \geq m$. It follows that 
$s_I^M(z) = 0$. Thus 
$\deg s^M_I(z) = -1 $. The proof of when 
$\deg s^M_I(z) = -1 $ then 
$\deg t^M_I(z) = -1 $ is similar and left to the reader.
We now assume that 
$d \geq 2$ and the result is proved when 
$\dim A = d -1$. A proof similar to above shows that 
$\deg t^M_I(z) = -1 $ if and only if 
$\deg s^M_I(z) = -1 $. So assume 
$\deg t^M_I(z) \geq 0$ (and so 
$\deg s^M_I(z) \geq 0 $). We may assume that the residue field of A is infinite. Let 
$x \in I$ be 
$M \oplus A$-superficial and also assume that 
$xt \in \mathcal{R} _1$ is 
$L_1(M) \oplus E_1(M) \oplus E_2(M)$-filter regular. We note that 
$B = A/(x)$ is a complete intersection and 
$N = M/xM$ is an MCM B-module with bounded Betti numbers. Set 
$J = I/(x)$. So it follows from our induction hypotheses that 
$\deg t^N_J(z) = \deg s^N_J(z)$. As 
$\deg t^M_I(z) \geq 0$, it follows from § 5.4 that 
$\deg t^N_J(z) = \deg t^M_I(z) -1 $. Similarly, it follows from § 5.5 that 
$\deg s^N_J(z) = \deg s^M_I(z) -1 $. The result follows.
We now give
6. Proof of Theorem 1.5
In this section, we give
Proof of Theorem 1.5.
Let 
$\mathcal{R} = A[It]$. Recall 
$L_0(M) = \bigoplus_{n \geq 0}M/I^{n+1}M$. By [Reference Puthenpurakal9, 5.6], we have
\begin{equation*}\operatorname{Ass}_\mathcal{R} G_I(M) = \operatorname{Ass}_\mathcal{R} L_0(M).\end{equation*} As A is a hypersurface ring, in particular, it is a Gorenstein ring. So there exists an MCM A-module N and an exact sequence 
$0 \rightarrow M \rightarrow F \rightarrow N \rightarrow 0$ with F free. Notice 
$L_1(N)$ is a 
$\mathcal{R} $-submodule of 
$L_0(M)$. In particular, if 
$L_1(N) \neq 0$ then it is necessarily of dimension d by our assumption on M. So 
$r_I = d - 1$ if 
$L_1(N) \neq 0$. If 
$L_1(N) = 0$ then notice for all 
$n \geq 0$ we have 
$t_I(N,n) = t_I(\Omega^A_1(N), n) = 0$. So for all 
$n \geq 0$ we have 
$\operatorname{Tor}^A_i(N, A/I^{n+1}) = 0$ for all 
$i \geq 1$. As N does not have finite projective dimension it follows from [Reference Huneke and Wiegand5, 1.9] that 
$\operatorname{projdim} A/I^{n+1} $ is finite for all 
$n \geq 0$.
Remark 6.1. The assumption that 
$G_I(M)$ is equi-dimensional is a mild one. It is not difficult to show that if A is also universally catenary then 
$G_I(M)$ is equi-dimensional for any MCM A-module M (so is automatic in our case as A is a hypersurface and so universally catenary). The assumption that 
$G_I(M)$ is un-mixed is not mild. There are plenty of examples when 
$\operatorname{depth} G_I(M) = 0$.
Acknowledgements
I thank the referee for a careful reading and many pertinent comments.
