2.1. Let $\mathcal{C} $ be an essentially small triangulated category with shift operator Σ and let $\operatorname{Iso}(\mathcal{C} )$ 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 \operatorname{Iso}(\mathcal{C} ) \rightarrow \mathbb{Z} $ suchthat

1. (1) $\xi(X) \geq 0$ for all $X \in \mathcal{C} $.

2. (2) $\xi(0) = 0$.

3. (3) $\xi(X \oplus Y) = \xi(X) + \xi(Y)$ for all $X, Y \in \mathcal{C} $.

4. (4) $\xi(\Sigma X ) = \xi(X)$ for all $X \in \mathcal{C} $.

5. (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 1.2.

Proof. We have

1. (1) $0 \in \ker \xi$.

2. (2) If $X \cong Y$ and $X \in \ker \xi$. Then note $\xi(Y) = \xi(X) = 0$. So $Y \in \ker \xi$.

3. (3) If $X \in \ker \xi$ then note $\xi(\Sigma X) = \xi(X) = 0$. So $\Sigma X \in \ker \xi$. Similarly $\Sigma^{-1} X \in \ker \xi$.

4. (4) If $X \rightarrow Y \rightarrow Z \rightarrow \Sigma X$ is a triangle in $\mathcal{C} $ with $X, Y \in \ker \xi$. Then note

   \begin{equation*} 0 \leq \xi(Z) \leq \xi(X) + \xi(Y) = 0 + 0 = 0. \end{equation*}

   So $Z \in \ker \xi$.

5. (5) If $X \oplus Y \in \ker \xi$ then $\xi(X) + \xi(Y) = \xi(X \oplus Y) = 0$. As $\xi(X), \xi(Y)$ are nonnegative, it follows that $\xi(X) = \xi(Y) = 0$. Thus $X, Y \in \ker \xi$.

It follows that $\ker \xi$ is a thick subcategory of $\mathcal{C} $.

2.4. Let A be a ring. Let $K^b(\operatorname{proj} A)$ be the homotopy category of bounded complexes of projective complexes. We index complexes cohomologically,

\begin{equation*}\mathbf{X}_\bullet \colon \cdots \rightarrow \mathbf{X}_\bullet^{n-1} \rightarrow \mathbf{X}_\bullet^n \rightarrow \mathbf{X}_\bullet^{n+1} \rightarrow \cdots.\end{equation*}

We note that $\mathbf{X}_\bullet = 0$ in $K^b(\operatorname{proj} A)$ if and only if $H^*(\mathbf{X}_\bullet) = 0$. If $\mathbf{X}_\bullet = 0$ in $K^b(\operatorname{proj} A)$ then note that $H^*(X\otimes N) = 0$ for any A-module N.

2.5. Let $K^b_f(\operatorname{proj} A)$ denote the homotopy category of bounded complexes of projective complexes with finite length cohomology. We note that if $\mathbf{X}_\bullet \in K^b_f(\operatorname{proj} A)$ and N is an A-module then $H^*(\mathbf{X}_\bullet \otimes N)$ also has finite length. To see this if P is a prime ideal in A with $P \neq \mathfrak{m} $ then

\begin{equation*} H^*(\mathbf{X}_\bullet \otimes_A N)_P = H^*({\mathbf{X}_\bullet}_P \otimes_{A_P} N_P) = 0 \quad \text{as ${\mathbf{X}_\bullet}_P = 0$ in $K^b(\operatorname{proj} A_P)$}. \end{equation*}

Proof. We may assume $\mathbf{X}_\bullet$ is a minimal complex. Furthermore (after a shift), we may assume that $\mathbf{X}_\bullet^0 \neq 0$ and $\mathbf{X}_\bullet^i = 0$ for $i \geq 1$. Let $H^0(\mathbf{X}_\bullet) = E \neq 0$ since $\mathbf{X}_\bullet$ is minimal. It is straight forward to check that $H^0(\mathbf{X}_\bullet \otimes N) = E\otimes N \neq 0$. The result follows.

2.7. Suppose for an A-module M and an ideal I we have $\ell(H^i(\mathbf{X}_\bullet \otimes M/I^nM))$ has finite length for all $n \geq 1$ and for all $i \in \mathbb{Z} $. Consider the function

\begin{equation*} \psi_{\mathbf{X}_\bullet}^{M, I}(n) = \sum_{i \in \mathbb{Z} }\ell(H^i(\mathbf{X}_\bullet \otimes M/I^nM)), \quad \text{for $n \geq 1$}. \end{equation*}

By [Reference Theodorescu9, Proposition 3] we know that $\psi_{\mathbf{X}_\bullet}^{M, I}(n)$ is of polynomial type say of degree $r_I^M(X)$ and

\begin{equation*} r_I(M) \leq \dim M. \end{equation*}

2.8. Let I be an $\mathfrak{m} $-primary ideal in A and let M be an A-module. An element $x \in I$ is said to be 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 $n \gg 0$. Superficial elements exist when $k = A/\mathfrak{m} $ is infinite, (see [Reference Sally8, p. 7] for the case when M = A; the same proof generalizes).

2.9. If $\operatorname{grade}(I, M) \gt 0$ and x is M-superficial with respect to I then x is M-regular. This fact is well-known. We give a proof due to lack of a suitable reference. Let $(I^{n+1}M \colon x) \cap I^cM = I^nM$ for all $n \gg 0$. Let $u \in I$ be M-regular. If xm = 0 then $xu^cm = 0$. So $u^cm \in I^n$ for all $n \gg 0$. Thus $u^cm = 0$ and so m = 0.

2.10. A sequence $\mathbf{x} = x_1, \ldots, x_r \in M$ is said to be an M-superficial sequence if x_i is $M/(x_1, \ldots, x_{i-1})M$-superficial for $i = 1, \ldots, r$. If $\operatorname{grade}(I, M) \geq r$ then it follows from 2.9 that x is an A-regular sequence.

Proof of Theorem 1.2

By 2.6, it follows that the function $\psi_{\mathbf{X}_\bullet}^{M, I}(n) \neq 0$ for all $n \geq 1$. Thus $r_I^M(\mathbf{X}_\bullet) \geq 0$ for all $\mathbf{X}_\bullet \neq 0$. Also by 2.7, $r_I^M(\mathbf{X}_\bullet) \leq \dim A$ for any $\mathbf{X}_\bullet \in K^b_f(\operatorname{proj} A)$. Let

\begin{equation*} c = \max \{r_I^M(\mathbf{X}_\bullet) \mid \mathbf{X}_\bullet \neq 0 \}. \end{equation*}

For $\mathbf{Y}_\bullet \in K^b(\operatorname{proj} A)_f$ define

\begin{equation*}\eta(\mathbf{Y}_\bullet) = \lim_{n \rightarrow \infty} \frac{c!}{n^c} \psi_{\mathbf{Y}_\bullet}^{M, I}(n). \end{equation*}

Clearly $\eta(\mathbf{Y}_\bullet) \in \mathbb{Z} _{\geq 0}$. Furthermore if $\mathbf{Y}_\bullet \cong \mathbf{Z}_\bullet$ then clearly $\eta(\mathbf{Y}_\bullet) = \eta(\mathbf{Z}_\bullet)$. Thus, we have a function $\eta \colon \operatorname{Iso}(K^b_f(\operatorname{proj} A)) \rightarrow \mathbb{Z} $ where $\operatorname{Iso}(K^b_f(\operatorname{proj} A))$ denotes the set of isomorphism classes of objects in $K^b_f(\operatorname{proj} A)$.

Claim: η is a weak triangle function on $K^b_f(\operatorname{proj} A)$.

Assume the claim for the time-being. By 2.3, $\ker \eta$ is a thick subcategory of $K^b_f(\operatorname{proj} A)$. Let $\mathbf{X}_\bullet$ be such that $r_I^M(\mathbf{X}_\bullet) = c$. Then $\eta(\mathbf{X}_\bullet) \gt 0$. So $\mathbf{X}_\bullet \notin \ker \eta$. Thus $\ker \eta \neq K^b(\operatorname{proj} A)$. By [Reference Neeman5, Lemma 1.2], it follows that $\ker \eta = 0$. Thus $r_I^M(\mathbf{Y}_\bullet) = c$ for any $\mathbf{Y}_\bullet \neq 0$ in $K^b_f(\operatorname{proj} A)$.

It remains to prove the claim. The first four properties of definition in 2.1 are trivial to verify. Let $\mathbf{X}_\bullet \xrightarrow{f} \mathbf{Y}_\bullet \rightarrow \mathbf{Z}_\bullet \rightarrow \mathbf{X}_\bullet[1]$ be a triangle in $K^b_f(\operatorname{proj} A)$. Then $\mathbf{Z}_\bullet \cong \operatorname{cone}(f)$ and we have an exact sequence in $C^b(\operatorname{proj} A)$

\begin{equation*} 0 \rightarrow \mathbf{Y}_\bullet \rightarrow \operatorname{cone}(f) \rightarrow \mathbf{X}_\bullet[1] \rightarrow 0.\end{equation*}

As $\mathbf{X}_\bullet^i$ are free A-modules we have an exact sequence for all $n \geq 1$,

\begin{equation*} 0 \rightarrow \mathbf{Y}_\bullet \otimes M/I^nM \rightarrow \operatorname{cone}(f)\otimes M/I^nM \rightarrow \mathbf{X}_\bullet[1]\otimes M/I^nM \rightarrow 0. \end{equation*}

Taking homology we have

\begin{equation*} \psi_{\mathbf{Z}_\bullet}^{M, I}(n) \leq \psi_{\mathbf{Y}_\bullet}^{M, I}(n) + \psi_{\mathbf{X}_\bullet[1]}^{M, I}(n) \end{equation*}

for all $n \geq 1$. It follows that

\begin{equation*} \eta(\mathbf{Z}_\bullet) \leq \eta(\mathbf{Y}_\bullet) + \eta(\mathbf{X}_\bullet[1]) = \eta(\mathbf{Y}_\bullet) + \eta(\mathbf{X}_\bullet). \end{equation*}

Thus, η is a weak triangle function on $K^b_f(\operatorname{proj} A)$.

Proof of Corollary 1.6

By Theorem 1.2, we have that there exists c with $r_I^M(\mathbf{X}_\bullet) = c$ for any nonzero $\mathbf{X}_\bullet \in K^b_f(\operatorname{proj} A)$. Let L be a nonzero finite length A-module with finite projective dimension. Let $\mathbf{Y}_\bullet$ be a minimal projective resolution of L. Then $\mathbf{Y}_\bullet \in K^b_f(\operatorname{proj} A)$ and is nonzero. It follows that $r_I^M(\mathbf{Y}_\bullet) = c$. Observe that $r_I^M(\mathbf{Y}_\bullet) = t^I_M(L)$. Set $\mathbf{Y}_\bullet^* = \operatorname{Hom}_A(\mathbf{Y}_\bullet, A)$. Note that $\mathbf{Y}_\bullet^* \in K^b_f(A)$ and is nonzero. Also observe

\begin{equation*} \operatorname{Ext}^*_A(L, M/I^nM) = H^*(\operatorname{Hom}_A(\mathbf{Y}_\bullet, M/I^nM) \cong H^*( \mathbf{Y}_\bullet^*\otimes_A M/I^nM). \end{equation*}

Therefore

\begin{equation*} e^I_M(L) = r_I^M(\mathbf{Y}_\bullet^*) = c. \end{equation*}

The result follows.

4.1. Base change:

(1) We first consider a flat base change $A \rightarrow B$ where $(B, \mathfrak{n} )$ is local and $\mathfrak{n} = \mathfrak{m} B$. We claim that $r_I^M(A) = r_{IB}^{M\otimes_A B}(B)$.

Proof of Theorem 1.4

By 1.7, we have $r_I^M \leq \dim M -1$. We first do the following base-changes:

1. (1) If the residue field of A is finite then we set $B = A[X]_{\mathfrak{m} A[X]}$ then $(B,\mathfrak{n} )$ is a flat extension of A with $\mathfrak{m} B = \mathfrak{n} $ and the residue field of B is k(X) is infinite. So we replace M by $M \otimes_A B$ and I by IB (see 4.1(1)).

2. (2) We then complete A (see 4.1(1)).

3. (3) By (1), (2) we assume A is complete with an infinite residue field. Let A be a quotient of a regular local ring Q. Then, we can replace A by Q (see 4.1(2)).

4. (4) By (3), we can assume A is regular local with infinite residue field. We note $a = \operatorname{grade}(\operatorname{ann} M) = \operatorname{height} \operatorname{ann} M$. Choose $y_1, \ldots, y_a \in \operatorname{ann} M $ an A-regular sequence. By 4.1(3), we can replace A with $A/(y_1,\ldots, y_a)$.

Thus, we can assume A is Cohen–Macaulay with infinite residue field and $\dim A = \dim M \gt 0$. Let $d = \dim A$ and let $\mathbf{x} = x_1, \ldots, x_d$ be a maximal $M \oplus A$-superficial sequence with respect to I. Then as x is an A-superficial sequence with respect to I it is an A-regular sequence, see 2.10. Let $\mathbf{K}_\bullet$ be the Koszul complex on x. Then $\mathbf{K}_\bullet \in K^b_f(\operatorname{proj} A)$. We also note that as x ₁ is M-superficial with respect to I there exists c and n ₀ such that $(I^{n}M \colon x_1)\cap I^cM = I^{n-1}M$ for all $n \geq n_0$.

Set

\begin{equation*} \psi_{\mathbf{K}_\bullet, A}^{M, I}(n) = \sum_{i \in \mathbb{Z} }\ell_A(H^i(\mathbf{K}_\bullet \otimes M/I^nM)), \quad \text{for $n \geq 1$} \end{equation*}

and let r be its degree. By 2.7, $r \leq d - 1$. We note that

\begin{equation*} H^d(\mathbf{K}_\bullet \otimes M/I^nM) = \frac{I^nM \colon \mathbf{x} }{I^nM} \supseteq \frac{(I^nM \colon \mathbf{x} )\cap I^cM}{I^nM} = \frac{I^{n-1}M}{I^nM} \ (\text{for } n \geq n_0). \end{equation*}

So $\psi_{\mathbf{K}_\bullet, A}^{M, I}(n) \geq \ell(I^{n-1}M/I^nM)$ for all $n \geq n_0$. So $r \geq d -1$. Thus $r = d -1$. By Theorem 1.2, it follows that $r_I^M = r = d -1$.

Proof of Theorem 1.8

By 1.7, it suffices to consider the case when $\dim H^*(\mathbf{X}_\bullet \otimes M) \leq \dim M -1$.

We first consider the case when $\dim H^*(\mathbf{X}_\bullet \otimes M) = 0$. We prove the result by inducting on $\dim H^*(\mathbf{X}_\bullet)$. If $\dim H^*(\mathbf{X}_\bullet) = 0$ then the result follows from Theorem 1.4. If $\dim H^*(\mathbf{X}_\bullet) \gt 0$ then choose x such that map $ H^*(\mathbf{X}_\bullet) \xrightarrow{x} H^*(\mathbf{X}_\bullet)$ has finite length kernel and $\dim H^*(\mathbf{X}_\bullet)/xH^*(\mathbf{X}_\bullet) = \dim H^*(\mathbf{X}_\bullet) -1$. Consider the triangle $\mathbf{X}_\bullet \xrightarrow{x} \mathbf{X}_\bullet \rightarrow \mathbf{Y}_\bullet \rightarrow \mathbf{X}_\bullet[1]$. By taking long exact sequence of homology, we get an exact sequence

\begin{equation*} 0 \rightarrow H^*(\mathbf{X}_\bullet)/xH^*(\mathbf{X}_\bullet) \rightarrow H^*(\mathbf{Y}_\bullet) \rightarrow (0 \colon_{H^*(\mathbf{X}_\bullet)} x)[1] \rightarrow 0. \end{equation*}

It follows that $\dim H^*(\mathbf{Y}_\bullet) = \dim H^*(\mathbf{X}_\bullet) - 1$. Furthermore note

$\mathbf{Z}_\bullet = \operatorname{cone}(x, \mathbf{X}_\bullet) \cong \mathbf{Y}_\bullet$. We have an exact sequence

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet \rightarrow \mathbf{Z}_\bullet \rightarrow \mathbf{X}_\bullet[1] \rightarrow 0. \end{equation*}

As $\mathbf{X}_\bullet^i$ is free for all i we have an exact sequence for all $n \geq 0$

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet\otimes M/I^nM \rightarrow \mathbf{Z}_\bullet \otimes M/I^n M \rightarrow \mathbf{X}_\bullet[1]\otimes M/I^nM \rightarrow 0, \end{equation*}

and

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet\otimes M \rightarrow \mathbf{Z}_\bullet \otimes M \rightarrow \mathbf{X}_\bullet[1]\otimes M \rightarrow 0. \end{equation*}

By considering later short exact sequence of complexes, we get by looking at long exact sequence in homology that $\dim H^*(\mathbf{Z}_\bullet\otimes M) = 0$. So by induction hypothesis $s_I^M(\mathbf{Y}_\bullet) = \dim M -1$. By considering all $n \geq 1$ and summing all i, we get

\begin{equation*} \psi_{\mathbf{Y}_\bullet}^{M, I}(n) \leq 2 \psi_{\mathbf{X}_\bullet}^{M, I}(n). \end{equation*}

It follows that $s_I^M(\mathbf{X}_\bullet) \geq s_I^M(\mathbf{Y}_\bullet) = \dim M -1$. But $s_I^M(\mathbf{X}_\bullet) \leq \dim M -1$. The result follows.

We now assume $ 0 \lt a = \dim H^*(\mathbf{X}_\bullet \otimes M) \leq \dim M -1$ and the result is proved for complexes $\mathbf{Z}_\bullet$ with $\dim H^*(\mathbf{Z}_\bullet \otimes M) = a -1$. Choose x such that map $ H^*(\mathbf{X}_\bullet \otimes M) \xrightarrow{x} H^*(\mathbf{X}_\bullet\otimes M)$ has finite length kernel and $\dim H^*(\mathbf{X}_\bullet \otimes M)/xH^*(\mathbf{X}_\bullet \otimes M) = \dim H^*(\mathbf{X}_\bullet \otimes M) -1$. Consider the triangle $\mathbf{X}_\bullet \xrightarrow{x} \mathbf{X}_\bullet \rightarrow \mathbf{Y}_\bullet \rightarrow \mathbf{X}_\bullet[1]$. Note $\mathbf{Z}_\bullet = \operatorname{cone}(x, \mathbf{X}_\bullet) \cong \mathbf{Y}_\bullet$. We have an exact sequence

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet \rightarrow \mathbf{Z}_\bullet \rightarrow \mathbf{X}_\bullet[1] \rightarrow 0. \end{equation*}

As $\mathbf{X}_\bullet^i$ is free for all i, we have an exact sequence for all $n \geq 0$

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet\otimes M/I^nM \rightarrow \mathbf{Z}_\bullet \otimes M/I^n M \rightarrow \mathbf{X}_\bullet[1]\otimes M/I^nM \rightarrow 0, \end{equation*}

and

\begin{equation*} 0 \rightarrow \mathbf{X}_\bullet\otimes M \rightarrow \mathbf{Z}_\bullet \otimes M\rightarrow \mathbf{X}_\bullet[1]\otimes M \rightarrow 0. \end{equation*}

By considering the latter short exact sequence of complexes, we get by looking at long exact sequence in homology we get an exact sequence

\begin{equation*} 0 \rightarrow H^*(\mathbf{X}_\bullet \otimes M)/x^*H^*(\mathbf{X}_\bullet \otimes M) \rightarrow H^*(\mathbf{Z}_\bullet \otimes M) \rightarrow (0 \colon_{H^*(\mathbf{X}_\bullet \otimes M)}x)[1] \rightarrow 0. \end{equation*}

Therefore $\dim H^*(\mathbf{Z}_\bullet\otimes M) = \dim H^*(\mathbf{X}_\bullet\otimes M) - 1$. So by induction hypothesis $s_I^M(\mathbf{Y}_\bullet) = \dim M -1$. By considering all $n \geq 1$ and summing all i, we get

\begin{equation*} \psi_{\mathbf{Y}_\bullet}^{M, I}(n) \leq 2 \psi_{\mathbf{X}_\bullet}^{M, I}(n) \end{equation*}

It follows that $s_I^M(\mathbf{X}_\bullet) \geq s_I^M(\mathbf{Y}_\bullet) = \dim M -1$. But $s_I^M(\mathbf{X}_\bullet) \leq \dim M -1$. The result follows.

Proof. It is clear that $\operatorname{Supp} (M\otimes L) = \operatorname{Supp} M \cap \operatorname{Supp} L$. Thus, it follows that $\operatorname{Supp} \operatorname{Ext}^*_A(L, M) \subseteq \operatorname{Supp} (M \otimes L)$. Conversely let $P \in \operatorname{Supp} M \otimes L$. We localize at P. So it suffices to prove $\operatorname{Ext}^*(L, M) \neq 0$. By taking a minimal resolution of L, it clear that if $c = \operatorname{projdim} L$ then $\operatorname{Ext}^c_A(L,M) \neq 0$. The result follows.

Proof of Corollary 1.10

Let $\mathbf{X}_\bullet$ be a minimal projective resolution of L. Then

$t^I_M(L, n) = \ell(H^*(\mathbf{X}_\bullet \otimes M/I^nM))$. By 1.8, it follows that

\begin{equation*}t^I_M(L) = \max \{\dim H^*(\mathbf{X}_\bullet \otimes M), \dim M - 1 \}.\end{equation*}

The result follows as $\dim H^*(\mathbf{X}_\bullet \otimes M) = \dim M \otimes L$.

Set $\mathbf{X}_\bullet^* = \operatorname{Hom}_A(\mathbf{X}_\bullet, A)$. Observe

\begin{equation*} \operatorname{Ext}^*_A(L, M/I^nM) = H^*(\operatorname{Hom}_A(\mathbf{X}_\bullet, M/I^nM)) \cong H^*( \mathbf{X}_\bullet^*\otimes_A M/I^nM). \end{equation*}

So

\begin{equation*}e^I_M(L) = \max \{\dim H^*(\mathbf{X}_\bullet^* \otimes M), \dim M - 1 \}.\end{equation*}

Notice $H^*(\mathbf{X}_\bullet^* \otimes M) = \operatorname{Ext}^*_A(L, M)$. The results follows from Lemma 6.1.