Proof of Lemma 1.5

Let $v = v_{\mathfrak {p}}(M)$ and $w = \operatorname {\mathrm {indeg}}\big ( \operatorname {\mathrm {ann}}_M(\mathfrak {p})/\operatorname {\mathrm {ann}}_M(\mathfrak {p})\cap \Gamma _V(M) \big )$ . Then, by definition of $v_{\mathfrak {p}}(M)$ , there exists a nonzero element $x \in M_v$ such that $\mathfrak {p} = (0 :_R x)$ . Hence, $\mathfrak {p} x=0$ , i.e., $x \in \operatorname {\mathrm {ann}}_M(\mathfrak {p})$ . We prove that $x\notin \operatorname {\mathrm {ann}}_M(\mathfrak {p})\cap \Gamma _V(M)$ , equivalently, $x\notin \Gamma _V(M)$ . If $V=R$ , then $\Gamma _V(M)=0$ , and there is nothing to prove. We may assume that $V\not =R$ , i.e., $X_{\mathfrak {p}}\not =\emptyset $ . If possible, assume that $x\in \Gamma _V(M)$ . Then $V^mx=0$ for some integer $m\geqslant 1$ . Thus, $V^m \subseteq (0 :_R x) = \mathfrak {p}$ , which implies that $V\subseteq \mathfrak {p}$ . Hence, since $V = \prod _{\mathfrak {q}\in X_{\mathfrak {p}}}\mathfrak {q}$ , it follows that $\mathfrak {q}\subseteq \mathfrak {p}$ for some $\mathfrak {q}\in X_{\mathfrak {p}}$ . This is a contradiction because $\mathfrak {p} \subsetneq \mathfrak {q}$ (by the definition of $X_{\mathfrak {p}}$ ). So $x\notin \operatorname {\mathrm {ann}}_M(\mathfrak {p})\cap \Gamma _V(M)$ . Thus, the image of x in $\operatorname {\mathrm {ann}}_M(\mathfrak {p})/\operatorname {\mathrm {ann}}_M(\mathfrak {p})\cap \Gamma _V(M)$ is a nonzero homogeneous element of degree v. Therefore, $v\geqslant w$ . It remains to prove the other inequality, i.e., $v\leqslant w$ .

Suppose $y\in M_w$ induces a nonzero element of $\operatorname {\mathrm {ann}}_M(\mathfrak {p})/\operatorname {\mathrm {ann}}_M(\mathfrak {p})\cap \Gamma _V(M)$ . Then $y\in \operatorname {\mathrm {ann}}_M(\mathfrak {p})\smallsetminus \Gamma _V(M)$ . In particular, $\mathfrak {p} y=0$ , i.e., $\mathfrak {p}\subseteq (0:_R y)$ . If the equality holds, i.e., $\mathfrak {p} = (0:_R y)$ , then $v\leqslant w$ . So it is enough to prove that $\mathfrak {p} = (0:_R y)$ . If possible, assume that $\mathfrak {p}\subsetneq (0:_R y)$ . Note that $(0:_R y)$ is a proper ideal, i.e., $\operatorname {\mathrm {Ass}}_R(R/(0:_R y))\not =\emptyset $ . Considering the map $R\to M$ given by $r\mapsto ry$ , there is an injective R-module homomorphism $R/(0:_R y) \hookrightarrow M$ . So $\operatorname {\mathrm {Ass}}_R(R/(0:_R y)) \subseteq \operatorname {\mathrm {Ass}}_R(M)$ . Thus, for each $\mathfrak {q}\in \operatorname {\mathrm {Ass}}_R(R/(0:_R y))$ , one has that $\mathfrak {p}\subsetneq (0:_R y) \subseteq \mathfrak {q} \in \operatorname {\mathrm {Ass}}_R(R/(0:_R y)) \subseteq \operatorname {\mathrm {Ass}}_R(M)$ , which yields that $\mathfrak {q}\in X_{\mathfrak {p}}$ . Therefore, $\operatorname {\mathrm {Ass}}_R(R/(0:_R y)) \subseteq X_{\mathfrak {p}}$ . In particular, $X_{\mathfrak {p}}\not =\emptyset $ . Set $V_1 := \prod _{\mathfrak {q}\in \operatorname {\mathrm {Ass}}_R(R/(0:_R y))}\mathfrak {q}$ . Hence, from a primary decomposition of $(0:_R y)$ , one deduces that there is an integer $m\geqslant 1$ such that $V_1^m \subseteq (0 :_R y)$ , i.e., $V_1^m y = 0$ . Since $\operatorname {\mathrm {Ass}}_R(R/(0:_R y)) \subseteq X_{\mathfrak {p}}$ , it follows that $V\subseteq V_1$ . So $V^m y \subseteq V_1^m y = 0$ . Thus, $y\in \Gamma _V(M)$ , which is a contradiction. Therefore, $\mathfrak {p} = (0:_R y)$ .

2.1. With Setup 1.1, further assume that R is standard graded. Set $R_+:=\bigoplus _{n\geqslant 1}R_n$ . The Castelnuovo–Mumford regularity of M is given by

$$\begin{align*}\operatorname{\mathrm{reg}}(M) := \max\{ \operatorname{\mathrm{end}}\big( H_{R_+}^i(M)\big) + i : 0\leqslant i \leqslant \dim(R) \}, \end{align*}$$

where $H_{R_+}^i(M)$ denotes the ith local cohomology module of M with respect to the ideal $R_+$ . Note that $H_{R_+}^0(M) = \Gamma _{R_+}(M)$ . It follows that $\operatorname {\mathrm {end}}\big ( \Gamma _{R_+}(M) \big ) \leqslant \operatorname {\mathrm {reg}}(M)$ . Interested readers can look at [Reference Bruns, Conca, Raicu and Varbaro4, Chapter 8] and [Reference Bruns, Conca and Varbaro5] for more details on this topic.

Proof. Let $\mathfrak {p}\in \operatorname {\mathrm {Ass}}_R(M)$ and $v = v_{\mathfrak {p}}(M)$ . Then there exists $x \in M_v$ such that $\mathfrak {p} = (0 :_R x)$ . Since $R_+\subseteq \mathfrak {p}$ , it follows that $R_+x=0$ . Thus, $ x \in \Gamma _{R_+}(M) \cap M_v = \big (\Gamma _{R_+}(M)\big )_v $ , which implies that $ v \leqslant \operatorname {\mathrm {end}}\big ( \Gamma _{R_+}(M) \big ) \leqslant \operatorname {\mathrm {reg}}(M) $ . This proves the first part. If $\operatorname {\mathrm {depth}}(R_+,M)=0$ , then $R_+ \subseteq \bigcup _{\mathfrak {q}\in \operatorname {\mathrm {Ass}}_R(M)}\mathfrak {q}$ , which yields that $R_+\subseteq \mathfrak {r}$ for some $\mathfrak {r}\in \operatorname {\mathrm {Ass}}_R(M)$ (by prime avoidance), and hence $v(M)\leqslant v_{\mathfrak {r}}(M)\leqslant \operatorname {\mathrm {reg}}(M)$ .

Proof. Let $\mathfrak {p}\in \operatorname {\mathrm {Ass}}_R(L)$ , and $v=v_{\mathfrak {p}}(L)$ . Then there exists $x\in L_v$ such that $\mathfrak {p} = (0:_Rx)$ . Since $x\in L_v \subseteq M_v$ , one obtains that $\mathfrak {p}\in \operatorname {\mathrm {Ass}}_R(M)$ and $v_{\mathfrak {p}}(M)\leqslant v = v_{\mathfrak {p}}(L)$ . It remains to prove the second part. Since $v(0)=\infty $ , we may assume that $L\neq 0$ , equivalently, $\operatorname {\mathrm {Ass}}_R(L)$ is nonempty. Let $w=v(L)$ . Then $w=v_{\mathfrak {q}}(L)$ for some $\mathfrak {q}\in \operatorname {\mathrm {Ass}}_R(L)$ . Since $\operatorname {\mathrm {Ass}}_R(L)\subseteq \operatorname {\mathrm {Ass}}_R(M)$ , it follows that $\mathfrak {q}\in \operatorname {\mathrm {Ass}}_R(M)$ . By the first part, $v_{\mathfrak {q}}(M)\leqslant v_{\mathfrak {q}}(L)=w$ . Hence, $v(M) \leqslant v_{\mathfrak {q}}(M) \leqslant w=v(L)$ .

2.6. In [Reference McAdam and Eakin18, Prop. 2], McAdam-Eakin proved that if $\mathcal {R}$ is a Noetherian standard graded algebra over R, then $\operatorname {\mathrm {Ass}}_R(\mathcal {R}_n)$ is eventually constant. This proof can be easily modified to give the result for a finitely generated graded $\mathcal {R}$ -module $\mathcal {M}$ , see [Reference West22, Thm. 3.4]. Note that the Rees module $\mathscr {R}(I,M)$ is finitely generated graded over the Rees algebra $\mathscr {R}(I)$ . Since N is a submodule of M, the Rees module $\mathscr {R}(I,N)$ is a graded submodule of $\mathscr {R}(I,M)$ . So the quotient $\mathscr {H} := \bigoplus _{n\geqslant 0}I^nM/I^nN$ is a finitely generated graded module over $\mathscr {R}(I)$ . Thus, by [Reference West22, Thm. 3.4], it follows that the set $\operatorname {\mathrm {Ass}}_R(I^nM/I^nN)$ is constant for all $n\gg 0$ .

2.7. With Setup 1.1, let $\deg (x_i)=f_i$ for $1\leqslant i\leqslant d$ . Let $\mathscr {R}(J) = \bigoplus _{n\in \mathbb {N}}J^n$ be the Rees algebra of J. We consider it as an $\mathbb {N}^2$ -graded ring by setting the $(n,l)$ th graded component of $\mathscr {R}(J)$ as the lth graded component of $J^n$ for each $(n,l)\in \mathbb {N}^2$ . Then $\mathscr {R}(J)$ can be written as $ \mathscr {R}(J) = R_0[x_1,\ldots ,x_d,y_1,\ldots ,y_c] $ , where $ \deg (x_i) = (0,f_i) $ for $ 1 \leqslant i \leqslant d $ and $ \deg (y_j) = (1,d_j) $ for $ 1 \leqslant j \leqslant c $ . Consider the Rees module $\mathscr {R}(I,M) = \bigoplus _{n\in \mathbb {N}}I^nM $ . By convention, $I^nM = 0$ whenever $n<0$ . Setting $\mathscr {R}(I,M)_{(n,l)} :=(I^nM)_l$ for $(n,l)\in \mathbb {Z}^2$ , we make $\mathscr {R}(I,M)$ a $\mathbb {Z}^2$ -graded module over $\mathscr {R}(J)$ . Since J is a reduction ideal of I, the module $\mathscr {R}(I,M)$ is finitely generated over $\mathscr {R}(J)$ . With similar gradation, $\mathscr {R}(I,N)$ is a $\mathbb {Z}^2$ -graded $\mathscr {R}(J)$ -submodule of $\mathscr {R}(I,M)$ . So the quotient $\mathscr {H} := \mathscr {R}(I,M)/\mathscr {R}(I,N)$ is a finitely generated $\mathbb {Z}^2$ -graded module over $\mathscr {R}(J)$ .

Proof. Note that $x_i\mathscr {L}_{(n,*)}\subseteq \mathscr {L}_{(n,*)}$ for each $n \in \mathbb {Z}$ and $ 1 \leqslant i \leqslant d $ . Thus, restricting the scalars from T to R, the set $ \mathscr {L}_{(n,*)} $ forms a $\mathbb {Z}$ -graded R-module. From the construction of $ \mathscr {L}_{(n,*)} $ , the bigraded module $ \mathscr {L} $ can be written as $ \mathscr {L} = \bigoplus _{n\in \mathbb {Z}} \mathscr {L}_{(n,*)} $ . Since $y_j \mathscr {L}_{(n,*)} \subseteq \mathscr {L}_{(n+1,*)}$ , we may consider $ \mathscr {L} = \bigoplus _{n\in \mathbb {Z}} \mathscr {L}_{(n,*)} $ as a $\mathbb {Z}$ -graded module over an $\mathbb {N}$ -graded ring $T=R[y_1,\ldots ,y_c]$ , where $ \deg (y_j) = 1 $ for $ 1 \leqslant j \leqslant c $ , and R is the $0$ th graded component of T in this gradation. Since we are only changing the grading (from bigraded to graded), $ \mathscr {L} = \bigoplus _{n\in \mathbb {Z}} \mathscr {L}_{(n,*)} $ is also finitely generated as a graded module over $T=R[y_1,\ldots ,y_c]$ . Consequently, by [Reference West22, Thm. 3.4], there exists $n_0$ such that $\operatorname {\mathrm {Ass}}_R\left ( \mathscr {L}_{(n,*)} \right ) = \operatorname {\mathrm {Ass}}_R\left ( \mathscr {L}_{(n_0,*)} \right )$ for all $n\geqslant n_0$ . Denote $\mathcal {A}_{\mathscr {L}} := \operatorname {\mathrm {Ass}}_R\left ( \mathscr {L}_{(n_0,*)} \right )$ . Clearly, if $\mathcal {A}_{\mathscr {L}}$ is an empty-set, then $\mathscr {L}_{(n,*)} = 0$ for all $n\geqslant n_0$ . In the other case, $\mathcal {A}_{\mathscr {L}}\neq \emptyset $ , and we have that $\mathscr {L}_{(n,*)} \neq 0$ for all $n\geqslant n_0$ . Since $\mathscr {L}_{(n,*)}\not =0$ for $n\gg 0$ , we must have that $(y_1,\ldots ,y_c)\not \subseteq \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {L})}$ , and hence $\delta $ is a finite number.

(1) Consider an $\mathbb {N}^2$ -graded polynomial ring $ S = R_0 [X_1,\ldots ,X_d,Y_1,\ldots ,Y_c] $ over $R_0$ , where $ \deg (X_i) = (0,f_i) $ for $ 1 \leqslant i \leqslant d $ and $ \deg (Y_j) = (1,d_j) $ for $ 1 \leqslant j \leqslant c $ . There is a natural graded ring homomorphism $S\to T$ . Via this homomorphism, $\mathscr {L}$ is a finitely generated $ \mathbb {Z}^2 $ -graded $ S $ -module as well. Hence, by [Reference Chardin, Ghosh and Nemati7, Prop. 3.1]Footnote ¹ , there exist $ a_1\in \{ d_j : 1 \leqslant j \leqslant c \} $ and $ b_1 \in \mathbb {Z} $ such that

(2.1) $$ \begin{align} \operatorname{\mathrm{indeg}}\left( \mathscr{L}_{(n,*)} \right) = \operatorname{\mathrm{indeg}}\big( \operatorname{\mathrm{Tor}}_0^R( \mathscr{L}_{(n,*)},R_0) \big) = a_1n + b_1 \mbox{ for all } n \gg 0. \end{align} $$

We show that $a_1 = d_{\delta }$ , where $\delta = \inf \left \{ j:y_j\notin \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {L})}\right \}$ . As a graded module over $T=R[y_1,\ldots ,y_c]$ , let

(2.2) $$ \begin{align} \mathscr{L} = \bigoplus_{n\in\mathbb{Z}} \mathscr{L}_{(n,*)} \text{ be generated by homogeneous elements of degree } \leqslant n_1. \end{align} $$

For $1\leqslant j < \delta $ , since $y_j \in \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {L})}$ , there exist $k_j$ such that $y_j^{k_j} \in \operatorname {\mathrm {ann}}_T(\mathscr {L})$ . Set $n_2 := (\delta -1)\cdot \max \{k_j : 1\leqslant j < \delta \}$ . So $(y_1,\dots ,y_{\delta -1})^k\mathscr {L}=0$ for all $k\geqslant n_2$ . Thus, for every $n\geqslant n_1+n_2$ , one has that

$$ \begin{align*} \mathscr{L}_{(n,*)} &= \bigoplus_{i\leqslant n_1} (y_1,\dots,y_c)^{n-i}\mathscr{L}_{(i,*)}\\ &=\bigoplus_{i\leqslant n_1} \ \bigoplus_{k \leqslant n-i} (y_1,\dots,y_{\delta-1})^k(y_\delta,\dots,y_c)^{n-i-k}\mathscr{L}_{(i,*)}\\ &=\bigoplus_{i\leqslant n_1} \ \bigoplus_{k\leqslant n_2} (y_1,\dots,y_{\delta-1})^k(y_\delta,\dots,y_c)^{n-i-k}\mathscr{L}_{(i,*)}. \end{align*} $$

Moreover, for every $n \geqslant n_1+n_2$ , note that $y_{\delta }\mathscr {L}_{(n,*)}\neq 0$ . Indeed, if $y_{\delta }\mathscr {L}_{(n,*)}= 0$ for some $n \geqslant n_1+n_2$ , then $y_{\delta }^{n-i+1}\mathscr {L}_{(i,*)} \subseteq y_{\delta }\mathscr {L}_{(n,*)} =0$ for all $i\leqslant n_1$ , and hence $y_{\delta } \in \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {L})}$ , a contradiction. Thus, one concludes that

(2.3) $$ \begin{align} \operatorname{\mathrm{indeg}}\left( \mathscr{L}_{(n+1,*)} \right) - \operatorname{\mathrm{indeg}}\left( \mathscr{L}_{(n,*)} \right) = \deg(y_{\delta}) = d_{\delta} \mbox{ for all } n \gg 0. \end{align} $$

Consequently, the statement (1) follows from (2.1) and (2.3).

(2) Let $\mathfrak {p}\in \operatorname {\mathrm {Ass}}_R\left ( \mathscr {L}_{(n,*)} \right )$ for all $n\gg 0$ , equivalently, $\mathfrak {p}\in \mathcal {A}_{\mathscr {L}}$ . Set $X_{\mathfrak {p}} := \{\mathfrak {q} \in \mathcal {A}_{\mathscr {L}} : \mathfrak {p} \subsetneq \mathfrak {q} \}$ . Let $V=R$ if $X_{\mathfrak {p}}=\emptyset $ , otherwise $V = \prod _{\mathfrak {q}\in X_{\mathfrak {p}}}\mathfrak {q}$ . Let $\mathscr {M} = \operatorname {\mathrm {ann}}_{\mathscr {L}}(\mathfrak {p})/\operatorname {\mathrm {ann}}_{\mathscr {L}}(\mathfrak {p})\cap \Gamma _V(\mathscr {L})$ . Note that $y_j \in \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {L})} \subseteq \sqrt {\operatorname {\mathrm {ann}}_T(\mathscr {M})}$ for all $1\leqslant j < \delta $ . Since $ T $ is Noetherian, and $\mathscr {L}$ is finitely generated, the (sub)quotient $\mathscr {M}$ is also a finitely generated $\mathbb {Z}^2$ -graded module over T, where the grading of $\mathscr {M}$ is induced by that of $\mathscr {L}$ . In particular, $ \mathscr {M}_{(n,*)} = \bigoplus _{l\in \mathbb {Z}} \mathscr {M}_{(n,l)}$ is same as $\operatorname {\mathrm {ann}}_{\mathscr {L}_{(n,*)}}(\mathfrak {p})/\operatorname {\mathrm {ann}}_{\mathscr {L}_{(n,*)}}(\mathfrak {p})\cap \Gamma _V({\mathscr {L}_{(n,*)}})$ . Therefore, in view of Lemma 1.5 and (1), one has that $v_{\mathfrak {p}}\left ( \mathscr {L}_{(n,*)} \right ) = \operatorname {\mathrm {indeg}}(\mathscr {M}_{(n,*)}) = a_{\mathfrak {p}}n+b_{\mathfrak {p}}$ for all $n\gg 0$ , and for some $ a_{\mathfrak {p}} \in \{ d_j : \delta \leqslant j \leqslant c \} $ and $ b_{\mathfrak {p}} \in \mathbb {Z} $ .

(3) Note that $\mathcal {A}_{\mathscr {L}}$ is a (nonempty) finite set. By the definition of v-numbers, $v\left ( \mathscr {L}_{(n,*)} \right ) = \inf \{ v_{\mathfrak {p}}\left ( \mathscr {L}_{(n,*)} \right ) : \mathfrak {p}\in \mathcal {A}_{\mathscr {L}} \}$ for all $n\geqslant n_0$ . Hence from (2) and the observation made in 2.9, the function $v\left ( \mathscr {L}_{(n,*)} \right ) $ is eventually linear, i.e., $v\left ( \mathscr {L}_{(n,*)} \right ) = a \cdot n + b_2 $ for all $ n \gg 0$ , and for some $a \in \{ d_j : \delta \leqslant j \leqslant c \} $ and $ b_2 \in \mathbb {Z} $ . Clearly, $d_{\delta } \leqslant a$ . It is enough to show that $a\leqslant d_{\delta }$ .

Since the module $\mathscr {L}$ is Noetherian, the chain of submodules

$$ \begin{align*}(0:_{\mathscr{L}} y_{\delta}) \subseteq (0:_{\mathscr{L}} y_{\delta}^2) \subseteq (0:_{\mathscr{L}} y_{\delta}^3) \subseteq \cdots,\end{align*} $$

stabilizes. So there exists $m_0\geqslant 1$ such that $(0:_{\mathscr {L}} y_{\delta }^m) = (0:_{\mathscr {L}} y_{\delta }^{m_0})$ for all $m\geqslant m_0$ . In particular, $(0:_{\mathscr {L}_{(n_1,*)}} y_{\delta }^m) = (0:_{\mathscr {L}_{(n_1,*)}} y_{\delta }^{m_0})$ for all $m\geqslant m_0$ , where $n_1$ is as in (2.2). We prove that $(0:_{\mathscr {L}_{(n_1,*)}} y_{\delta }^{m_0})$ is a proper R-submodule of $\mathscr {L}_{(n_1,*)}$ . If possible, let $(0:_{\mathscr {L}_{(n_1,*)}} y_{\delta }^{m_0})=\mathscr {L}_{(n_1,*)}$ . Then $y_{\delta }^{m_0}\mathscr {L}_{(n_1,*)}=0$ . Denote $\operatorname {\mathrm {indeg}}(\mathscr {L}):=\inf \{n:\mathscr {L}_{(n,*)}\neq 0\}$ . Setting $l:=\max \{0,-\operatorname {\mathrm {indeg}}(\mathscr {L})\}$ , for all $\operatorname {\mathrm {indeg}}(\mathscr {L})\leqslant n\leqslant n_1$ , one has that $y_{\delta }^{m_0+n_1+l}\mathscr {L}_{(n,*)}\subseteq y_{\delta }^{m_0+n+l}\mathscr {L}_{(n_1,*)} = 0$ as $n+l\geqslant 0$ . Therefore, since $\mathscr {L}$ is generated by homogeneous elements of degree $\leqslant n_1$ , one concludes that $y_{\delta }^{m_0+n_1+l}\mathscr {L}_{(n,*)}=0$ for all $n\in \mathbb {Z}$ , and hence $y_{\delta }\in \sqrt {\operatorname {\mathrm {ann}}_T \mathscr {L}}$ , a contradiction. Thus, $(0:_{\mathscr {L}_{(n_1,*)}} y_{\delta }^{m_0}) \subsetneq \mathscr {L}_{(n_1,*)}$ .

We show that $v\left ( \mathscr {L}_{(n,*)} \right )\leqslant v\Big ( \mathscr {L}_{(n_1,*)}/ \big (0:_{\mathscr {L}_{(n_1,*)}} y_\delta ^{m_0}\big )\Big )+(n-n_1)\cdot d_\delta $ for all $n\geqslant n_1+m_0$ . For every $n\geqslant m\geqslant 0$ , the natural map $\mathscr {L}_{(n-m-1,*)}(-d_{\delta }) \stackrel {y_{\delta }}{\longrightarrow } \mathscr {L}_{(n-m,*)}/\big (0:_{\mathscr {L}_{(n-m,*)}} y_\delta ^{m}\big )$ induces a graded injective R-module homomorphism

(2.4) $$ \begin{align} \dfrac{\mathscr{L}_{(n-m-1,*)}}{\big(0:_{\mathscr{L}_{(n-m-1,*)}} y_\delta^{m+1}\big)} (-d_{\delta}) \stackrel{y_{\delta}}{\longrightarrow} \dfrac{\mathscr{L}_{(n-m,*)}}{\big(0:_{\mathscr{L}_{(n-m,*)}} y_\delta^{m}\big)}. \end{align} $$

Here $M(-h)$ denotes a graded R-module with $M_{n-h}$ as its n-graded component. So, by definition of v-numbers, $v(M(-h)) = v(M) + h$ . Thus, by Proposition 2.5.(2), for every $n\geqslant m\geqslant 0$ , the map (2.4) yields that

(2.5) $$ \begin{align} v\left( \dfrac{\mathscr{L}_{(n-m,*)}}{\big(0:_{\mathscr{L}_{(n-m,*)}} y_\delta^{m}\big)} \right) \leqslant v\left( \dfrac{\mathscr{L}_{(n-m-1,*)}}{\big(0:_{\mathscr{L}_{(n-m-1,*)}} y_\delta^{m+1}\big)} \right) + d_{\delta}. \end{align} $$

Using this inequality repeatedly starting from $m=0$ , one obtains that

$$ \begin{align*} v\left( \mathscr{L}_{(n,*)} \right) & \leqslant v\left( \dfrac{\mathscr{L}_{(n-1,*)}}{\big(0:_{\mathscr{L}_{(n-1,*)}} y_\delta^{1}\big)} \right) + d_{\delta} \leqslant v\left( \dfrac{\mathscr{L}_{(n-2,*)}}{\big(0:_{\mathscr{L}_{(n-2,*)}} y_\delta^{2}\big)} \right) + 2 \cdot d_{\delta} \leqslant \cdots \\ & \leqslant v\left( \dfrac{\mathscr{L}_{(n_1,*)}}{\big(0:_{\mathscr{L}_{(n_1,*)}} y_\delta^{n-n_1}\big)} \right) + (n-n_1)\cdot d_{\delta} \\ & = n \cdot d_{\delta} + v\Big( \mathscr{L}_{(n_1,*)}/ \big(0:_{\mathscr{L}_{(n_1,*)}} y_\delta^{m_0}\big)\Big) - n_1 \cdot d_{\delta} \;\mbox{ for all } n \geqslant n_1+m_0. \end{align*} $$

It follows that $a\leqslant d_{\delta }$ , and hence $a= d_{\delta }$ . This completes the proof.

2.9. Let r be a positive integer. Let $a_i,b_i \in \mathbb {Z}$ for $i = 1,\ldots , r $ . Define $f(n) = \inf \{ a_in+b_i : 1\leqslant i\leqslant r \}$ for all $n \in \mathbb {N}$ . Then $f(n) = an+b$ for all $n\gg 0$ , where $a := \inf \{ a_1,\ldots ,a_r \}$ and $b:=\inf \{ b_i : a_i = a, 1\leqslant i\leqslant r \}$ .

2.10. With Setup 1.1, considering the (graded) module $\mathscr {H} := \mathscr {R}(I,M)/\mathscr {R}(I,N)$ over the Rees algebra $\mathscr {R}(J)$ , set $\delta :=\inf \left \{j:y_j\notin \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)}(\mathscr {H})}, 1 \leqslant j \leqslant c \right \}$ .

Proof. With the discussion made in 2.7, $\mathscr {H} := \mathscr {R}(I,M)/\mathscr {R}(I,N)$ is a finitely generated $\mathbb {Z}^2$ -graded module over $ \mathscr {R}(J) = R_0[x_1,\ldots ,x_d,y_1,\ldots ,y_c] $ , where $ \deg (x_i) = (0,f_i) $ for $ 1 \leqslant i \leqslant d $ and $ \deg (y_j) = (1,d_j) $ for $ 1 \leqslant j \leqslant c $ . Note that $ \mathscr {H}_{(n,*)} := \bigoplus _{l\in \mathbb {Z}} \mathscr {H}_{(n,l)}$ is given by $\mathscr {R}(I,M)_{(n,*)}/\mathscr {R}(I,N)_{(n,*)}$ , which is same as $I^nM/I^nN$ . Therefore, in view of Theorem 2.8, one deduces that:

1. (1) $\operatorname {\mathrm {indeg}}(I^nM/I^nN)$ is eventually linear in n with the leading coefficient $d_\delta $ ;

2. (2) for every $\mathfrak {p}\in \mathcal {B}_N^M(I)$ , the function $v_{\mathfrak {p}}(I^nM/I^nN)$ is eventually linear in n with the leading coefficient inside the set $\{d_\delta ,\dots ,d_c\}$ ;

3. (3) $v(I^nM/I^nN)$ is eventually linear in n with the leading coefficient $d_\delta $ .

This completes the proof of the theorem.

Proof. Let $\mathfrak {p}\in \mathcal {B}_N^M(I)$ . Since $I^nM/I^nN$ is a (graded) R-submodule of $M/I^nN$ for all $n\geqslant 0$ , it follows that $\mathcal {B}_N^M(I)\subseteq \mathcal {A}_N^M(I)$ . So $\mathfrak {p}\in \mathcal {A}_N^M(I)$ . In view of Theorem 2.11, there exist $a\in \{d_\delta ,\ldots ,d_c\}$ and $b\in \mathbb {Z}$ such that $v_{\mathfrak {p}}(I^nM/I^nN) = an+b$ for all $n\gg 0$ . On the other hand, by Proposition 2.5.(1), $v_{\mathfrak {p}}(M/I^nN) \leqslant v_{\mathfrak {p}}(I^nM/I^nN)$ for all $n\gg 0$ . Combining these two results, $v_{\mathfrak {p}}(M/I^nN)\leqslant an+b$ for all $n\gg 0$ .

For the second part, in view of Proposition 2.5.(2) and Theorem 2.11, there exists $n_0$ such that $v(M/I^nN)\leqslant v(I^nM/I^nN)=d_\delta n+b'$ for all $n\geqslant n_0$ , and for some $b'\in \mathbb {Z}$ . Note that $M\neq IN$ . Otherwise, if $M=IN$ , then $M=I^nN$ for all $n\geqslant 1$ , and hence $\mathcal {A}_N^M(I)$ is an empty set, a contradiction. So $M\neq IN$ . Consequently, $M\neq I^nN$ , and $v(M/I^nN)$ is finite for every $n\geqslant 1$ . Set e as the maximum value among $b'$ and $\big (v(M/I^nN)-d_\delta n\big )$ , $1\leqslant n < n_0$ . It follows that $v(M/I^nN) \leqslant d_\delta n+e$ for all $n\geqslant 1$ .

Proof. Note that $(I^{n+1}N :_M \mathfrak {u}) \subseteq (I^{n+1}M :_M \mathfrak {u}) \subseteq (I^{n+1}M :_M I) = I^nM$ for all $n\gg 0$ , where the last equality follows from [Reference Brodmann3, Lem. (4)]. So

$$ \begin{align*} (I^{n+1}N :_M \mathfrak{u}) = (I^{n+1}N :_M \mathfrak{u}) \cap I^nM = (I^{n+1}N :_{I^nM} \mathfrak{u}) \mbox{ for all } n\gg 0. \end{align*} $$

Going modulo $I^{n+1}N$ both sides, as graded submodules of $I^nM/I^{n+1}N$ ,

(2.6) $$ \begin{align} \operatorname{\mathrm{ann}}_{M/I^{n+1}N}(\mathfrak{u}) = \operatorname{\mathrm{ann}}_{I^nM/I^{n+1}N}(\mathfrak{u}) \mbox{ for all } n\gg 0. \end{align} $$

As $(I^nM/I^{n+1}N)\cap \Gamma _{\mathfrak {a}}(M/I^{n+1}N) = \Gamma _{\mathfrak {a}}(I^nM/I^{n+1}N)$ , (2.6) further induces that

(2.7) $$ \begin{align} \operatorname{\mathrm{ann}}_{M/I^{n+1}N}(\mathfrak{u})\cap\Gamma_{\mathfrak{a}}(M/I^{n+1}N) &=\operatorname{\mathrm{ann}}_{I^nM/I^{n+1}N}(\mathfrak{u})\cap\Gamma_{\mathfrak{a}}(M/I^{n+1}N)\\ &=\operatorname{\mathrm{ann}}_{I^nM/I^{n+1}N}(\mathfrak{u})\cap\Gamma_{\mathfrak{a}}(I^nM/I^{n+1}N)\nonumber \end{align} $$

for all $n\gg 0$ . Combining (2.6) and (2.7), one obtains the desired equalities.

Proof. (1) Considering $IN$ in place of N in 2.7, one has that $\mathscr {G} = \mathscr {R}(I,M)/\mathscr {R}(I,IN)$ is a finitely generated $\mathbb {Z}^2$ -graded module over $\mathscr {R}(J)$ , where $\mathscr {R}(J)$ is an $\mathbb {N}^2$ -graded ring with the same gradation as in 2.7. Using the prime ideal $\mathfrak {p}\in \mathcal {A}_N^M(I)$ , set $X_{\mathfrak {p}} := \{\mathfrak {q} \in \mathcal {A}_N^M(I) : \mathfrak {p} \subsetneq \mathfrak {q} \}$ . Let $V=R$ if $X_{\mathfrak {p}}=\emptyset $ , otherwise $V = \prod _{\mathfrak {q}\in X_{\mathfrak {p}}}\mathfrak {q}$ . Let $\mathscr {L} = \operatorname {\mathrm {ann}}_{\mathscr {G}}(\mathfrak {p})/\operatorname {\mathrm {ann}}_{\mathscr {G}}(\mathfrak {p})\cap \Gamma _V(\mathscr {G})$ . Then $\mathscr {L}$ is also a finitely generated $\mathbb {Z}^2$ -graded module over $\mathscr {R}(J)$ , where the bigrading in $\mathscr {L}$ is induced by that of $\mathscr {G}$ . So $ \mathscr {L}_{(n,*)} = \bigoplus _{l\in \mathbb {Z}} \mathscr {L}_{(n,l)}$ is same as $\operatorname {\mathrm {ann}}_{\mathscr {G}_{(n,*)}}(\mathfrak {p})/\operatorname {\mathrm {ann}}_{\mathscr {G}_{(n,*)}}(\mathfrak {p})\cap \Gamma _V({\mathscr {G}_{(n,*)}})$ , where $\mathscr {G}_{(n,*)} = I^nM/I^{n+1}N$ . Hence, since $I\subseteq \mathfrak {p}$ , in view of Lemma 2.13,

$$ \begin{align*}\mathscr{L}_{(n,*)} = \dfrac{\operatorname{\mathrm{ann}}_{M/I^{n+1}N}(\mathfrak{p})}{\operatorname{\mathrm{ann}}_{M/I^{n+1}N}(\mathfrak{p})\cap\Gamma_{V}(M/I^{n+1}N)} \mbox{ for all } n \gg 0.\end{align*} $$

Finally, by Lemma 1.5 and Theorem 2.8.(1), there exists an integer $b_1$ such that $v_{\mathfrak {p}}(M/I^{n+1}N) = \operatorname {\mathrm {indeg}}(\mathscr {L}_{(n,*)}) = d_{\delta _{\mathfrak {p}}} n + b_1 $ for all $n \gg 0$ , where

(2.8) $$ \begin{align} \delta_{\mathfrak{p}} = \inf\left\{j:y_j\notin\sqrt{\operatorname{\mathrm{ann}}_{\mathscr{R}(J)} (\mathscr{L})}, 1 \leqslant j \leqslant c \right\}. \end{align} $$

Setting $b_{\mathfrak {p}}:=b_1-d_{\delta _{\mathfrak {p}}}$ , one obtains that $v_{\mathfrak {p}}(M/I^nN) = d_{\delta _{\mathfrak {p}}} n+b_{\mathfrak {p}}$ for all $n\gg 0$ . This proves the first part of (1). For the second part of (1), further assume that $\mathcal {B}_{IN}^M(I) = \mathcal {A}_N^M(I)$ . Analyzing the proof above, one obtains that $v_{\mathfrak {p}}(I^nM/I^{n+1}N) = v_{\mathfrak {p}}(\mathscr {G}_{(n,*)}) = \operatorname {\mathrm {indeg}}(\mathscr {L}_{(n,*)})=v_{\mathfrak {p}}(M/I^{n+1}N)$ for all $n\gg 0$ , where the last two equalities follow from Lemma 1.5 and the two different expressions of $\mathscr {L}_{(n,*)}$ .

(2) Note that $\mathcal {A}_N^M(I)$ is a finite nonempty set. Since $I^{n_0}M\subseteq N$ for some $n_0$ , it can be verified that each $\mathfrak {p}\in \mathcal {A}_N^M(I)$ satisfies $I\subseteq \mathfrak {p}$ . Hence, from the proof of (1), one observes that $\mathscr {G}_{(n,*)} = I^nM/I^{n+1}N\neq 0$ for all $n\gg 0$ (otherwise, if $\mathscr {G}_{(n,*)} = 0$ , then $\mathscr {L}_{(n,*)} = 0$ , and hence $v_{\mathfrak {p}}(M/I^{n+1}N)=\infty $ for all $n\gg 0$ , a contradiction). Thus, $\mathcal {B}_{IN}^M(I)$ is also a nonempty set. This will be used later while applying Theorem 2.11. Note that $v(M/I^{n+1}N) = \inf \{ v_{\mathfrak {p}}(M/I^{n+1}N) : \mathfrak {p}\in \mathcal {A}_N^M(I) \}$ for all $n\gg 0$ . Therefore, since each $\mathfrak {p}\in \mathcal {A}_N^M(I)$ contains I, by (1) and 2.9, one concludes that $v(M/I^{n+1}N)$ is eventually linear in n with the leading coefficient $d_\tau $ , where $\tau :=\inf \{\delta _{\mathfrak {p}}: \mathfrak {p}\in \mathcal {A}_N^M(I)\}$ and $\delta _{\mathfrak {p}}$ is described in (2.8). We prove that $\tau =\gamma $ . First, notice that in the proof of (1), if $y_j\in \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)} (\mathscr {G})}$ , then $y_j\in \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)} (\mathscr {L})}$ . This yields that $\delta _{\mathfrak {p}}\geqslant \gamma $ for every $\mathfrak {p}\in \mathcal {A}_N^M(I)$ . Hence, $\tau \geqslant \gamma $ . Second, in view of Proposition 2.5.(2), $v(M/I^{n+1}N)\leqslant v(I^nM/I^{n+1}N)$ for all $n\geqslant 0$ . Here the leading coefficient of the asymptotic linear function $v(M/I^{n+1}N)$ is same as $d_{\tau }$ , while the leading coefficients of $\operatorname {\mathrm {indeg}}(I^nM/I^{n+1}N)$ and $v(I^nM/I^{n+1}N)$ are equal to $d_\gamma $ by Theorem 2.11. Thus, comparing the leading coefficients, it follows that $d_{\tau }\leqslant d_\gamma $ , which implies that $\tau \leqslant \gamma $ . So $\tau =\gamma $ . It proves the first part of (2). Since each $\mathfrak {p}\in \mathcal {A}_N^M(I)$ contains I, the second part of (2) follows from (1).

(3) Assume that $(0 :_M y_1) = 0$ and $d_1\geqslant 1$ . In view of (2), it is enough to show that $\gamma = 1$ , i.e., $y_1 \notin \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)} (\mathscr {G})}$ . If possible, let $y_1 \in \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)} (\mathscr {G})}$ . Then $y_1^s\mathscr {G}=0$ for some $s\geqslant 1$ . Therefore, since $\mathscr {G}_{(n,*)} = I^nM/I^{n+1}N$ , one obtains that $y_1^sI^nM \subseteq I^{n+s+1}N$ for all $n\geqslant 0$ . Since J is a reduction ideal of I, there exists $n_1$ such that $JI^{n_1} = I^{n_1+1}$ , and hence $J^nI^{n_1} = I^{n_1+n}$ for all $n\geqslant 1$ . Therefore, $y_1^sI^{n_1}M \subseteq I^{n_1+s+1}N=J^{s+1}I^{n_1}N$ . Since $(0:_MI) = 0$ , $I^{n_0}M\subseteq N$ and $M\neq 0$ , it follows that $I^{n_1}N\neq 0$ and $I^{n_1}M\neq 0$ . Moreover, since $J=(y_1,\ldots ,y_c)$ and $(0 :_M y_1) = 0$ , one derives that $\operatorname {\mathrm {indeg}}(y_1^sI^{n_1}M) = sd_1+\operatorname {\mathrm {indeg}}(I^{n_1}M)$ and $\operatorname {\mathrm {indeg}}(J^{s+1}I^{n_1}N) = (s+1)d_1+\operatorname {\mathrm {indeg}}(I^{n_1}N)$ . Thus

$$ \begin{align*} sd_1+\operatorname{\mathrm{indeg}}(I^{n_1}M) &= \operatorname{\mathrm{indeg}}(y_1^sI^{n_1}M) \\ &\geqslant \operatorname{\mathrm{indeg}}(J^{s+1}I^{n_1}N) \quad [\mbox{as } y_1^sI^{n_1}M \subseteq J^{s+1}I^{n_1}N] \\ &= (s+1)d_1+\operatorname{\mathrm{indeg}}(I^{n_1}N) \\ &\geqslant (s+1)d_1+\operatorname{\mathrm{indeg}}(I^{n_1}M) \quad [\mbox{as } I^{n_1}N \subseteq I^{n_1}M], \end{align*} $$

which is a contradiction as $d_1\geqslant 1$ . So $y_1 \notin \sqrt {\operatorname {\mathrm {ann}}_{\mathscr {R}(J)} (\mathscr {G})}$ , and hence $d_\gamma = d_1$ .

Proof. Let $n\geqslant 1$ . Then, $I^nM = (X^{an},XY^b)/(XY^b)$ . It follows that

$$ \begin{align*}\operatorname{\mathrm{indeg}}(I^nM/I^{n+1}M) = \operatorname{\mathrm{indeg}}(I^nM) = an.\end{align*} $$

As $\operatorname {\mathrm {Ass}}_R(I^nM) \subseteq \operatorname {\mathrm {Ass}}_R(M) = \{ \mathfrak {p},\mathfrak {q} \}$ , and $(I^nM)_{\mathfrak {p}}=0$ , one gets that $\operatorname {\mathrm {Ass}}_R(I^nM) = \{ \mathfrak {q} \}$ . Write the images of X and Y in M as x and y, respectively. The main relation of x and y that we have in M is $xy^b=0$ . So $(0:_M I) = \mathfrak {q}^b M \neq 0$ . Note that each element of $I^nM$ can be written as $x^{an}f(x,y)$ for some polynomial $f(x,y)$ over k. Therefore, $x^{an}y^{b-1} \in (I^nM)_{an+b-1}$ and $\mathfrak {q} = (0 :_R x^{an}y^{b-1})$ . Clearly, $an+b-1$ is the least possible degree of a homogeneous element of $I^nM$ whose annihilator is $\mathfrak {q}$ . It follows that $v(I^nM) = v_{\mathfrak {q}}(I^nM) = an+(b-1)$ for all $n\geqslant 1$ . This proves (1).

The quotient $M/I^nM\cong R/(X^{an},XY^b)$ for all $n\geqslant 1$ . So $\operatorname {\mathrm {Ass}}_R(M/I^nM) = \{ \mathfrak {p},\mathfrak {m} \}$ if $an\geqslant 2$ . If $a=1$ , then $M/IM\cong R/(X)$ , hence $\operatorname {\mathrm {Ass}}_R(M/IM) = \{ \mathfrak {p} \}$ and $v_{\mathfrak {p}}(M/IM) = 0$ . In the case, when $an\geqslant 2$ , one has that $\mathfrak {p} = \big ( (X^{an},XY^b) :_R Y^b \big )$ and $\mathfrak {m} = \big ( (X^{an},XY^b) :_R X^{an-1}Y^{b-1} \big )$ . These two equalities do not hold if Y and $X^{an-1}Y^{b-1}$ are replaced, respectively, by any other homogeneous element of lower degree. Thus, one obtains (2), (4), and (5). Consequently, (6) follows.

For (3), let $n\geqslant 1$ . Note that $\operatorname {\mathrm {Ass}}_R(I^nM/I^{n+1}M) \subseteq \operatorname {\mathrm {Ass}}_R(M/I^{n+1}M) = \{\mathfrak {p},\mathfrak {m}\} $ by (2). Therefore, since $(I^nM)_{\mathfrak {p}}=0$ , one concludes that $\operatorname {\mathrm {Ass}}_R(I^nM/I^{n+1}M) = \{ \mathfrak {m} \}$ . Since $I^nM = (X^{an},XY^b)/(XY^b)$ , the element $\overline {x^{an+a-1}y^{b-1}} \in I^nM/I^{n+1}M $ has the smallest possible degree such that $\mathfrak {m} = \operatorname {\mathrm {ann}}_R(\overline {x^{an+a-1}y^{b-1}})$ . Therefore, $v(I^nM/I^{n+1}M) = an + (a+b-2)$ .

Proof. Let $n\geqslant 1$ . Since $xy=0$ , it follows that $I^n=(x^{d_1n},y^{d_2n})$ . Therefore, $R/I^n$ along with the gradation can be written as

$$\begin{align*}k \oplus (kx \oplus ky) \oplus (kx^2 \oplus ky^2) \oplus \cdots \oplus (k x^{d_1n-1} \oplus k y^{d_1n-1}) \oplus k y^{d_1n} \oplus\cdots \oplus k y^{d_2n-1}. \end{align*}$$

Clearly, $\mathfrak {m} = (0 :_R x^{d_1n-1})$ , and $d_1n-1$ is the least possible degree of a homogeneous element of $R/I^n$ whose annihilator is $\mathfrak {m}$ . So $v(R/I^n) = v_{\mathfrak {m}}(R/I^n) = d_1 n - 1$ . Since $R/I^n$ has finite length, $\operatorname {\mathrm {reg}}(R/I^n) = \operatorname {\mathrm {end}}(R/I^n) = d_2n-1$ .

Proof. Since $M\hspace{-1pt}=\hspace{-1pt}R/(X^3,XY^4)$ , the element $Z^3\hspace{-0.5pt}\in\hspace{-0.5pt} I$ is M-regular, and hence $(0:_M I)\hspace{-0.5pt}=\hspace{-0.5pt}0$ . Note that $I^{n}M = \big (I^{n}+(X^3,XY^4)\big )/(X^3,XY^4)$ , $M/I^{n}M \cong R/\big (I^{n}+(X^3,XY^4)\big )$ , and $I^{n}M/I^{n+1}M \cong \big (I^{n}+(X^3,XY^4)\big )/\big (I^{n+1}+(X^3,XY^4)\big )$ for all $n\geqslant 0$ . We use $x,y,z$ for the classes of $X,Y,Z$ in M, respectively.

(1) Let $n\geqslant 1$ . Note that I and $I^n$ annihilate $I^nM/I^{n+1}M$ and $M/I^nM$ , respectively. Therefore, every associated prime ideal of each of these modules contains $I=(X,Y^2,Z^3)$ , and hence this prime ideal must be same as $\mathfrak {m}$ .

(2) A nonzero element of the least possible degree in the module $I^{n}M/I^{n+1}M$ for $0\leqslant n \leqslant 4$ is given by $1$ , x, $x^2$ , $x^2y^2$ and $x^2y^2z^3$ of degree $0$ , $1$ , $2$ , $4$ , and $7$ , respectively. For $n\geqslant 5$ , the module $I^nM$ is generated by

$$ \begin{align*}x^2y^2(z^3)^{n-3},\ x^2(z^3)^{n-2},\ xy^2(z^3)^{n-2},\ x(z^3)^{n-1} \mbox{ and } (y^2)^{j}(z^3)^{n-j} \mbox{ for } 0\leqslant j \leqslant n,\end{align*} $$

and their total degrees in M are, respectively

$$\begin{align*}3n-5, 3n-4, 3n-3,3n-2 \mbox{ and } 3n-j \mbox{ for } 0\leqslant j \leqslant n. \end{align*}$$

Among these degrees, $2n$ is the least possible value. Thus, $\overline {y^{2n}}$ is a nonzero element of the least possible degree in $I^nM/I^{n+1}M$ proving $(2)$ .

(3) Consider $n\geqslant 0$ . By (1), $v(M/I^{n+1}M)=v_{\mathfrak {m}}(M/I^{n+1}M)$ . Moreover, in view of Lemma 1.5, one has that

(3.1) $$ \begin{align} v(M/I^{n+1}M)=v_{\mathfrak{m}}(M/I^{n+1}M)=\operatorname{\mathrm{indeg}} \big((I^{n+1}M :_M \mathfrak{m})/I^{n+1}M\big). \end{align} $$

For $n\geqslant 5$ , the monomial ideal $((X^3,XY^4)+I^{n+1})$ is minimally generated by

$$ \begin{align*} X^3,XY^4, X^2Y^2(Z^3)^{n-2}, X^2(Z^3)^{n-1}, XY^2(Z^3)^{n-1}, X(Z^3)^{n}\\ \mbox{and } (Y^2)^{j}(Z^3)^{n+1-j} \mbox{ for } 0\leqslant j \leqslant n+1. \end{align*} $$

Therefore, using [Reference Ene and Herzog12, Sec. 3.2.2], one has that

$$ \begin{align*} \begin{aligned} \Big(\big((X^3,XY^4)+I^{n+1}\big):_RX\Big)=\Big( & X^2,Y^4,XY^2(Z^3)^{n-2}, X(Z^3)^{n-1}, \\ &Y^2(Z^3)^{n-1},(Z^3)^{n}\Big),\\ \Big(\big((X^3,XY^4)+I^{n+1}\big):_RY^2\Big)=\Big(&X^3,XY^2,X^2(Z^3)^{n-2}, X(Z^3)^{n-1},\\ (&Y^2)^{j}(Z^3)^{n-j},0\leqslant j\leqslant n\Big) \; \mbox{ and}\\ \Big(\big((X^3,XY^4)+I^{n+1}\big):_R Z^3\Big)=\Big(&X^3,XY^4,X^2Y^2(Z^3)^{n-3}, X^2(Z^3)^{n-2},\\ XY^2(Z^3)^{n-2},&X(Z^3)^{n-1},(Y^2)^{j}(Z^3)^{n-j},0\leqslant j\leqslant n\Big).\\ \end{aligned} \end{align*} $$

Now $\big (\big ((X^3,XY^4)+I^{n+1}\big ):_R I\big )$ is the intersection of the three ideals shown above. Moreover, the intersection of two monomial ideals is constructed by taking the lcm of pairs of generators one from each ideal. So the resulting ideal is exactly $(X^3,XY^4)+I^n$ , which is minimally generated by

$$ \begin{align*} X^3, XY^4, X^2Y^2(Z^3)^{n-3}, X^2(Z^3)^{n-2}, XY^2(Z^3)^{n-2}, X(Z^3)^{n-1} \mbox{ and } (Y^2)^{j}(Z^3)^{n-j} \end{align*} $$

where j is varying in $0\leqslant j \leqslant n$ .

The cases $n=0,\dots ,4$ would require more attention, but instead they can be verified using any mathematical software (e.g., Macaulay2 [Reference Grayson and Stillman14]). Thus, the claim is verified.

Using the above claim, since $I\subseteq \mathfrak {m}$ , it follows that

(3.2) $$ \begin{align} (I^{n+1}M:_M\mathfrak{m}) \subseteq (I^{n+1}M:_MI) = I^{n}M \mbox{ for every } n\geqslant 0. \end{align} $$

For $n=0,1,\ldots ,4$ , using Macaulay2 [Reference Grayson and Stillman14], one obtains that a nonzero homogeneous element in $(I^{n+1}M:_M\mathfrak {m})/I^{n+1}M$ of minimum possible degree is given by

$$ \begin{align*}yz^2, xyz^2, x^2yz^2, x^2y^3z^2 \text{ and } x^2y^3z^5, \text{respectively.} \end{align*} $$

Hence, in view of (3.1), for $0\hspace{-0.5pt}\leqslant\hspace{-0.5pt} n\hspace{-0.5pt}\leqslant\hspace{-0.5pt} 4$ , $v(M/I^{n+1}M)\hspace{-0.5pt}=\hspace{-0.5pt}3,4,5,7$ , and $10$ , respectively. Let $n\hspace{-0.5pt}\geqslant\hspace{-0.5pt} 5$ . Then, since $\mathfrak {m}\subseteq (I^{n+1}M :_R y^{2n+1}z^2)$ and $y^{2n+1}z^2\notin I^{n+1}M$ , one has that $\mathfrak {m}=(I^{n+1}M :_R y^{2n+1}z^2)$ . Moreover, one can check that if $g\in I^{n}M$ with $\deg (g)<2n+3$ , then $\mathfrak {m}\not =(I^{n+1}M:_R g)$ . Therefore, using (3.1) and (3.2), it follows $v_{\mathfrak {m}}(M/I^{n+1}M)=2n+3$ .

In view of Remark 2.15 and the above claim, one obtains that $v(I^nM/I^{n+1}M)=v(M/I^{n+1}M)$ for all $n\geqslant 0$ .