2.1 Rees algebras and the analytic spread

In this subsection, we include the definition and some of the properties of Rees algebras of modules. We also define the analytic spread of modules. For more details see [Reference Eisenbud, Huneke and Ulrich17] and [Reference Saia40].

Henceforth, we denote by ${\operatorname {Sym}}_R(M),$ the symmetric algebra of the R-module M, or simply ${\operatorname {Sym}}(M)$ when the base ring is clear. We also denote by $\tau _R(M)$ the R-torsion of M, that is, $\tau _R(M)=\{x\in M\mid (0:_R x) \text { contains nonzero divisors of } R\}$ .

Definition 2.2 If M has a rank, the Rees algebra of M is defined as

$$ \begin{align*} {\mathcal R}(M):={\operatorname{Sym}}(M)/\tau_R({\operatorname{Sym}}(M)). \end{align*} $$

The above definition coincides with the usual one for ideals, that is, ${\mathcal R}(I)=R[It]=\oplus _{n\in {\mathbb {N}}} I^nt^n$ , although we note that the latter does not require the rank assumption.

Remark 2.3. Assume M has a rank, then the natural map ${\mathcal R}(M)\to {\mathcal R}(M/\tau _R(M))$ is an isomorphism, that is,

$$ \begin{align*} {\operatorname{Sym}}(M)/\tau_R({\operatorname{Sym}}(M))\cong {\operatorname{Sym}}(M/\tau_R(M))/\tau_R\big({\operatorname{Sym}}(M/\tau_R(M))\big). \end{align*} $$

To see this, we note that since ${\operatorname {Sym}}(M/\tau _R(M))/\tau _R\big ({\operatorname {Sym}}(M/\tau _R(M))\big )$ is torsion-free, the kernel of the natural map $\varphi : {\operatorname {Sym}}(M)\to {\operatorname {Sym}}(M/\tau _R(M))/\tau _R\big ({\operatorname {Sym}}(M/\tau _R(M))\big )$ contains $\tau _R({\operatorname {Sym}}(M))$ . On the other hand, since M has a rank, $M\otimes _R Q(R)$ is free and then $\varphi \otimes _R Q(R)$ is an isomorphism. Thus, $\ker (\varphi )$ has rank zero which is equivalent to being contained in $\tau _R({\operatorname {Sym}}(M))$ .

We also note that $M/\tau _R(M)$ is a torsion-free module with a rank, then it is contained in a free R-module. The latter implies that when dealing with the Rees algebra of a module with a rank, one may always assume it is contained in a free module.

In the following proposition, we recall some facts about the dimension and associated primes of Rees algebras. Following the notation from Remark 2.4, let ${\mathcal T} := R[t_{1},\ldots , t_r]$ . For any $I \in {\operatorname {Spec}} R,$ we denote by $I'$ the ${\mathcal R}$ -ideal $I {\mathcal T}\cap {\mathcal R}(M)$ .

We are now ready to define the analytic spread.

The following proposition will be needed in several of our arguments.

Proposition 2.7 ([Reference Saia40, 2.3]). Let M be an R-module having a rank. Then

$$ \begin{align*} {\mathrm{rank}} (M) {\leqslant} \ell(M) {\leqslant} {\operatorname{dim}} R + {\mathrm{rank}} (M) -1. \end{align*} $$

2.2 Integral closure of modules

In this subsection, we include the definition of integral closure of modules and some basic properties of it. We restrict ourselves to the case of torsion-free modules with a rank. For more details see [Reference Huneke and Swanson27, Chapter 16] and [Reference Vasconcelos45, Chapter 8].

The integral closure of modules admits several characterizations. The following theorem relates the integral closure of modules with the integral closure of ideals. As far as the authors are aware, this result had not appeared in the literature in this generality (see [Reference Gaffney18, 1.7], [Reference Huneke and Swanson27, 16.3.2], [Reference Rees38, 1.2], [Reference Vasconcelos45, 8.66] for related statements).

For the proof of the theorem, we need the following lemma whose proof is essentially the same as [Reference Gaffney18, 1.6]. We include here the details for completeness.

We are now ready to prove the theorem.

Proof of Theorem 2.10

We begin with (1) $\Rightarrow $ (2). Let ${\mathfrak {p}}$ be a minimal prime of R and V a DVR or a field between $R/{\mathfrak {p}}$ and $R_{\mathfrak {p}}/{\mathfrak {p}} R_{\mathfrak {p}}$ . Since $hV\in MV$ , for every $i\,{\geqslant }\, 1$ we have

$$ \begin{align*} {\mathbf I}_i(M+Rh)V ={\mathbf I}_i(MV+R(hV))\subseteq {\mathbf I}_i(MV)= {\mathbf I}_i(M)V. \end{align*} $$

Thus ${\mathbf I}_i(M+Rh)\subseteq \overline {{\mathbf I}_i(M)}$ and (2) follows.

Since (2) $\Rightarrow $ (3) is clear, it suffices to show (3) $\Rightarrow $ (1). Let ${\mathfrak {p}}$ be a minimal prime of R and for a submodule $N\subseteq R^q$ let $N(R/{\mathfrak {p}})$ its image in $(R/{\mathfrak {p}})^q$ . By assumption, we have that $M(R/{\mathfrak {p}})$ and $(M+Rh)(R/{\mathfrak {p}})$ both have rank r. In particular, ${\mathbf I}_r(M)(R/{\mathfrak {p}})={\mathbf I}_r(M(R/{\mathfrak {p}}))\neq 0$ , and likewise ${\mathbf I}_r(M+Rh)(R/{\mathfrak {p}})\neq 0$ . Let V a DVR or a field between $R/{\mathfrak {p}}$ and $R_{\mathfrak {p}}/{\mathfrak {p}} R_{\mathfrak {p}}$ . Then by the assumption and Lemma 2.11, applied to $R/{\mathfrak {p}}$ , we have

$$ \begin{align*} ({\mathbf I}_r(M)V)hV=({\mathbf I}_r(M)h)V\subseteq ({\mathbf I}_r(M+Rh)M)V=({\mathbf I}_r(M+Rh)V)MV=({\mathbf I}_r(M)V)MV. \end{align*} $$

Thus $hV\in MV$ . We conclude $h\in \overline {M}$ , as desired.

As an immediate consequence of Theorem 2.10, we have the following result.

Corollary 2.12. Let R be a Noetherian ring and let $M\subseteq R^p$ be a submodule having a rank. Let $r={\mathrm {rank}}(M)$ . Then

$$ \begin{align*} \overline M=\big\{ h\in R^p: {\mathrm{rank}}(M + R h)=r \quad\text{and}\quad{\mathbf I}_r(M+R h)\subseteq \overline{{\mathbf I}_r(M)}\big\}. \end{align*} $$

Assume R is local of dimension d and let $\lambda (-)$ denote the length function of R-modules. If $\lambda (R^p/M)<\infty $ , we say M has finite colength and in this case the limit

$$ \begin{align*} e(M)=(d+p-1)!\lim_{n\to \infty}\frac{\lambda([{\operatorname{Sym}}(R^p)]_n/[{\mathcal R}(M)]_n)}{n^{d+p-1}} \end{align*} $$

is called the Buchsbaum–Rim multiplicity of M. It is known that if R is Cohen Macaulay and M is generated by $d+p-1$ elements, then $e(M)=\lambda (R^p/M)=\lambda (R/{\mathbf I}_p(M))$ (see for instance [Reference Gaffney19, p. 214]).

We recall the following numerical characterization of integral closures due to Rees [Reference Matsumura37] in the case of ideals and Katz [Reference Katz30] for modules.

Remark 2.14. Let $M\subseteq R^p$ be a submodule. In general, we have

(2.1) $$ \begin{align} \overline M\subseteq \overline{M_1\oplus \cdots \oplus M_p}=\overline{M_1}\oplus \cdots \oplus \overline{M_p}. \end{align} $$

However, the first inclusion in (2.1) might be strict. For instance, consider the submodule of ${\mathcal O}_2^2$ generated by the columns of the matrix

$$ \begin{align*} \left[\begin{matrix} x+y & x^3 & y^3 \\ x & y & x \end{matrix}\right]. \end{align*} $$

It is clear that $x^3\in \overline {M_1}$ , $x\in \overline {M_2}$ . Let $h=[\begin {matrix} x^3 & x \end {matrix}]^T$ , we can see that $h\notin \overline M$ . By Theorem 2.10, we have that

$$ \begin{align*} h\in\overline M \Longleftrightarrow {\mathbf I}_2(M+ {\mathcal O}_2 h)\subseteq \overline {{\mathbf I}_2(M)} \Longleftrightarrow e({\mathbf I}_2(M))=e({\mathbf I}_2(M+{\mathcal O}_2 h)), \end{align*} $$

where the last equivalence follows from Theorem 2.13. However $e({\mathbf I}_2(M))=8$ and $e({\mathbf I}_2(M+ {\mathcal O}_2 h))=6$ , as can be computed using Singular [Reference De Jong14]. Hence $h\notin \overline M$ .

Another argument leading to the conclusion that $h\notin \overline M$ is the following. We have that $e(M)=7$ and $e(M+ {\mathcal O}_2 h)=5$ , computed again using Singular. Since these multiplicities are different, it follows that $h\notin M$ , by Theorem 2.13. Moreover, by using Macaulay2 [Reference Grayson and Stillman21] (see Remark 4.25), it is possible to prove that $\overline M$ is generated by the columns of the matrix

$$ \begin{align*} \left[\begin{matrix} x+y & x^3 & y^3&x^3y^2 \\ x & y & x&x+y \end{matrix}\right]. \end{align*} $$

That is, $\overline {M}=M+{\mathcal O}_2[\begin {matrix} x^3y^2& x+y \end {matrix}]^T. $

Given an analytic map $\varphi :({\mathbb C}^m,0)\to ({\mathbb C}^n,0)$ , we denote by $\varphi ^*$ the morphism ${\mathcal O}_n\to {\mathcal O}_m$ given by $\varphi ^*(h)=h\circ \varphi $ , for all $h\in {\mathcal O}_n$ . For submodules of ${\mathcal O}_n^p$ , we have the following alternative definition of integral closure.

Example 2.16. It is also possible to check that $h\notin \overline M$ in the example from Remark 2.14 by considering the arc $\varphi :({\mathbb C},0)\to ({\mathbb C}^2,0)$ given by $\varphi (t)=(-t+t^3, t)$ , for all $t\in {\mathbb C}$ . We have that $\varphi ^*(h)=[\begin {matrix} (-t+t^3)^3 & -t+t^3 \end {matrix}]^T $ and that $\varphi ^*(M)$ is generated by the columns of the matrix

$$ \begin{align*} \left[\begin{matrix} t^3 & (-t+t^3)^3 & t^3 \\ -t+t^3 & t & -t+t^3 \end{matrix}\right]. \end{align*} $$

We note that the first and third columns of the previous matrix coincide. If $\varphi ^*(h)\in \varphi ^*(M)$ , then we would have

(2.2) $$ \begin{align} {\mathbf I}_2 \left[\begin{matrix} t^3 & (-t+t^3)^3 \\ -t+t^3 & t \end{matrix}\right]= {\mathbf I}_2 \left[\begin{matrix} t^3 & (-t+t^3)^3 & (-t+t^3)^3 \\ -t+t^3 & t & -t+t^3 \end{matrix}\right]. \end{align} $$

The ideal on the left of (2.2) is equal to $( t^4)$ and the ideal on the right of (2.2) is equal to $( t^6)$ . Hence $\varphi ^*(h)\notin \varphi ^*(M)$ and by Theorem 2.15 it follows that $h\not \in \overline {M}$ .

We finish this section with the following relation between integral closures and projections.

3.1 Multigraded algebras and multiprojective spectrum

In this subsection, we recall several facts about multigraded algebras and their multihomogeneous spectrum, we refer the reader to [Reference Hyry28] for more information. We start by setting up some notation.

Let $p\in {\mathbb {Z}}_{>0}.$ We denote by ${\mathbf {n}}$ the vector $(n_1,\ldots , n_p)\in {\mathbb {N}}^p$ . For convenience, we also set $\mathbf {0}=(0,\ldots , 0)$ and $\mathbf {1}=(1,\ldots , 1)$ where each of these vectors belongs to ${\mathbb {N}}^p$ . We call the sum $n_1+\cdots +n_p$ the total degree of ${\mathbf {n}}$ and denote it by $|{\mathbf {n}}|$ .

Let R be a Noetherian ring and $A=\oplus _{{\mathbf {n}}\in {\mathbb {N}}^p}A_{{\mathbf {n}}}$ a Noetherian ${\mathbb {N}}^p$ -graded algebra with $A_{\mathbf {0}}=R$ and generated by the elements of total degree one (standard graded). We denote by $A^{\Delta }$ the diagonal subalgebra of A, that is, $A^{\Delta }=\oplus _{n\in {\mathbb {N}}}A_{n\mathbf {1}}$ . For every $1{\leqslant } i{\leqslant } p,$ we write $A^{(i)}=\oplus _{n_i=0}A_{{\mathbf {n}}}$ . We also consider the following ${\mathbb {N}}^p$ -homogeneous A-ideals $A^{+}_i=\oplus _{n_i>0}A_{{\mathbf {n}}}$ for $1{\leqslant } i{\leqslant } p$ and $ A^{+}=\oplus _{n_1,\ldots , n_p>0}A_{{\mathbf {n}}}.$ We write ${\operatorname {Proj}}^p A = \{P\in {\operatorname {Spec}} A\mid P$ is ${\mathbb {N}}^p$ -homogeneous, and $A^+\not \subset P\}.$ The dimension of ${\operatorname {Proj}}^p A$ is one minus the maximal length of an increasing chain of elements of ${\operatorname {Proj}}^p A$ , $P_0\subsetneq P_1\subsetneq \cdots \subsetneq P_d$ . The relation between the dimensions of ${\operatorname {Proj}}^p A$ and A is explained in the following lemma.

It is possible to give ${\operatorname {Proj}}^p A$ a structure of scheme and to show that it is isomorphic to ${\operatorname {Proj}}^1 A^{\Delta }$ (see [Reference Hartshorne22, Part II, Exercise 5.11] and also [Reference Hayasaka24, Lemma 3.2] and [Reference Kirby and Rees33, Lemma 7.1]). For the reader’s convenience, we provide a proof of the following particular result which suffices for our applications.

We end this subsection with the following lemma that will be used in the proofs of our main results.

3.2 Multigraded Rees algebras

In this subsection, we describe a standard multigraded structure for the Rees algebras of direct sums of modules.

Definition 3.4. Let $M_1,\ldots ,M_p$ be R-modules having a rank. We define a natural standard ${\mathbb {N}}^p$ -graded structure on ${\mathcal R}(M_1\oplus \cdots \oplus M_p)$ . By [Reference Eisenbud16, A2.2.c], we have

$$ \begin{align*} {\operatorname{Sym}}(M_1\oplus\cdots \oplus M_p)\cong \bigotimes_{i=1}^p{\operatorname{Sym}}(M_i), \end{align*} $$

and since each of the algebras ${\operatorname {Sym}}(M_i)$ has a standard ${\mathbb {N}}$ -grading, we can combine these to an ${\mathbb {N}}^p$ -grading of ${\mathcal R}(M_1\oplus \cdots \oplus M_p)$ by setting $[\bigotimes _{i=1}^p{\operatorname {Sym}}(M_i)]_{\mathbf {n}}=\bigotimes _{i=1}^p {\operatorname {Sym}}(M_i)_{n_i}.$

Proposition 3.5. Let $M_1,\ldots , M_p$ be R-modules having a rank and set ${\mathcal R}'={\mathcal R} (M_1\oplus \cdots \oplus M_{p-1})$ . Then there is a natural graded ${\mathcal R}'$ -isomorphism

$$ \begin{align*} {\mathcal R}(M_1\oplus\cdots \oplus M_p)\cong {\mathcal R}(M_p\otimes_R {\mathcal R}'). \end{align*} $$

Proof We claim that for any R-module M with a rank we have $\tau _{{\mathcal R}'}(M\otimes _R {\mathcal R}')$ is equal to the image T of $\tau _R(M)\otimes _R {\mathcal R}'$ in $M\otimes _{R} {\mathcal R}'$ . First observe that by Proposition 2.5(1), $M\otimes _R {\mathcal R}'$ has a rank as ${\mathcal R}'$ -module and it is equal to ${\mathrm {rank}} (M)$ . Now, consider a short exact sequence

$$ \begin{align*} 0\to \tau_R(M)\to M\to F\cong R^r\to 0. \end{align*} $$

By tensoring with ${\mathcal R}'$ it follows that T contains $\tau _{{\mathcal R}'}(M\otimes _R {\mathcal R}')$ . On the other hand, T has rank zero as ${\mathcal R}'$ -module (since ${\mathrm {rank}} (\tau _R(M))=0$ ), then it must be ${\mathcal R}'$ -torsion. The claim follows. We obtain the following natural maps

$$ \begin{align*} {\mathcal S} :=\bigotimes_{i=1}^p{\operatorname{Sym}}_R(M_i)&\xrightarrow{\text{[16, A2.2.c]}} {\mathcal R}'\otimes_R {\operatorname{Sym}}_R(M_p)\\ &\stackrel{\text{[16, A2.2.c]}}{\cong}{\operatorname{Sym}}_{{\mathcal R}'}(M_p\otimes_{R} {\mathcal R}' )\\ &\,\,\,\,\,\,\,\,\,\,\xrightarrow{\text{onto}} \,\,\,\,\,\,\,\, {\operatorname{Sym}}_{{\mathcal R}'}(M_p\otimes_{R} {\mathcal R}' )/\tau_R( {\operatorname{Sym}}_{{\mathcal R}'}(M_p\otimes_{R} {\mathcal R}' ))\\ &\,\,\,\,\,\,\,\,\,\stackrel{\text{claim}}{=} \,\,\,\,\,\,\,\,{\mathcal R}(M_p\otimes_{R} {\mathcal R}'). \end{align*} $$

Clearly the kernel of the composition of these maps contains $\tau _R({\mathcal S})$ and, since tensoring by $Q(R)$ leads to an monomorphism, this kernel must be equal to $\tau _R({\mathcal S})$ . The result follows.

3.3 Main results about the analytic spread of modules

This subsection contains the main results of this section. We assume $(R,{\mathfrak {m}},k)$ is a Noetherian local ring.

The following is the main theorem of this section. This result, in particular, allows us to recover, and extend, the results in [Reference Hyry29, Lemma 4.7], [Reference Lejeune and Teissier35, 5.5], and [Reference Saia40, 2.3].

Theorem 3.6. Let M and N be R-modules having a rank. Then

$$ \begin{align*} \max\{\ell(M)+{\mathrm{rank}}(N), \ell(N)+ {\mathrm{rank}}(M)\} {\leqslant} \ell(M\oplus N){\leqslant} \ell(M)+\ell(N). \end{align*} $$

Proof We may assume M and N are torsion-free and hence contained in free R-modules (Remark 2.3). If either M or N has rank zero, then it has to be the zero module. Then we may assume they both have positive rank. Consider the following natural surjective maps

$$ \begin{align*} {\mathcal R}(M)\otimes_R {\mathcal R}(N)\xrightarrow{\alpha} {\mathcal R}(M)\otimes_R {\mathcal R}(N)/\tau_R({\mathcal R}(M)\otimes_R {\mathcal R}(N))\xleftarrow{\beta} {\operatorname{Sym}}(M)\otimes_R {\operatorname{Sym}}(N).\end{align*} $$

Since $Q(R)\otimes _R \beta $ is an isomorphism and the image of $\beta $ is torsion-free, it follows that $\ker \beta \subseteq \tau _R({\operatorname {Sym}}(M)\otimes _R {\operatorname {Sym}}(N))\subseteq \ker \beta .$ Then we obtain a surjective map

$$ \begin{align*} {\mathcal R}(M)\otimes_R {\mathcal R}(N)&\,\,\,\,\,\,\,\,\,\,\xrightarrow{ \text{onto}}\,\,\,\,\,\,\,\,\,\,{\operatorname{Sym}}(M)\otimes_R {\operatorname{Sym}}(N)/\tau_R({\operatorname{Sym}}(M)\otimes_R {\operatorname{Sym}}(N))\\ & \stackrel{\text{[16, A2.2.c]}}{\cong} {\operatorname{Sym}}(M\oplus N)/\tau_R({\operatorname{Sym}}(M\oplus N))\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,\,\,{\mathcal R}(M\oplus N). \end{align*} $$

By tensoring this map by $k,$ we observe that ${\mathcal F}(M\oplus N)$ is a quotient of ${\mathcal F}(M)\otimes _k {\mathcal F}(N)$ , and since the latter is a tensor product of affine algebras, it has dimension ${\operatorname {dim}} {\mathcal F}(M)+{\operatorname {dim}} {\mathcal F}(N)=\ell (M)+\ell (N)$ . The right-hand inequality follows.

We now show the left-hand inequality. Set ${\mathcal R} = {\mathcal R}(M\oplus N) $ . Following the multigrading in Definition 3.4, we have ${\mathcal R}^{(1)}={\mathcal R}(M)$ . We also observe that ${\mathcal R} \cong {\mathcal R}(N')$ , where $N'=N\otimes _R {\mathcal R}^{(1)}$ (Proposition 3.5). Fix ${\mathfrak {p}}\in {\operatorname {Proj}}^1 {\mathcal R}^{(1)}$ such that ${\mathfrak {p}}\cap R={\mathfrak {m}}$ and ${\operatorname {dim}} {\mathcal R}^{(1)}/{\mathfrak {p}}= \ell (M)$ , which exists by Lemma 3.3 and the fact that $\ell (M)\,{\geqslant }\, 1$ (Proposition 2.7). By Proposition 2.5(1), $(N')_{\mathfrak {p}}$ is an ${\mathcal R}^{(1)}_{\mathfrak {p}}$ -module with the same rank as N; let e be this rank. Then,

(3.1) $$ \begin{align} {\operatorname{dim}} Q({\mathcal R}^{(1)}/{\mathfrak{p}}) \otimes_{{\mathcal R}^{(1)}} {\mathcal R} = \ell((N^{\prime})_{\mathfrak{p}}){\geqslant} e \quad (\text{Proposition 2.7}). \end{align} $$

Therefore, by Lemma 3.3, there exist $P_0\subsetneq \cdots \subsetneq P_{e-1}$ in ${\operatorname {Proj}}^2 {\mathcal R}$ with $P_i\cap {\mathcal R}^{(1)} = {\mathfrak {p}}$ for every $0{\leqslant } i{\leqslant } e-1$ . We have an inclusion of domains $A={\mathcal R}^{(1)}/{\mathfrak {p}} \hookrightarrow B:={\mathcal R}/P_0$ and Lemma 3.3 implies ${\operatorname {dim}} Q(A )\otimes _A B\,{\geqslant }\, e$ . Hence, ${\operatorname {dim}} Q(A /{\mathfrak {p}}')\otimes _A B\,{\geqslant }\, e$ for every ${\mathfrak {p}}'\in {\operatorname {Proj}}^1 A$ ([Reference Eisenbud16, 14.8(b)]). Choose a ${\mathfrak {p}}'$ that avoids a general element of A (cf. [Reference Eisenbud16, 14.5]) and that ${\operatorname {dim}} A/{\mathfrak {p}}' = 1$ ; such ${\mathfrak {p}}'$ exists by Hilbert’s Nullstellensatz. Additionally, choose $P_{e-1}'\in {\operatorname {Proj}}^2 B$ such that $P_{e-1}'\cap A={\mathfrak {p}}'$ and its image in $ Q(A /{\mathfrak {p}}')\otimes _A B$ has height ${\geqslant }\, e-1$ (Lemma 3.3). Then, from [Reference Eisenbud16, 14.5], we obtain

(3.2) $$ \begin{align} {\operatorname{dim}} {\operatorname{Proj}}^1 A ={\operatorname{dim}} A_{{\mathfrak{p}}'} ={\operatorname{dim}} B_{P_{e-1}'} - {\operatorname{dim}} B_{P_{e-1}'}/{\mathfrak{p}}' B_{P_{e-1}'} {\leqslant} {\operatorname{dim}} {\operatorname{Proj}}^2 B -e+1. \end{align} $$

Thus,

$$ \begin{align*} \ell(M\oplus N)={\operatorname{dim}} {\mathcal R}\otimes_R k {\geqslant} {\operatorname{dim}} B{\geqslant} {\operatorname{dim}} {\operatorname{Proj}}^2 B +2{\geqslant} {\operatorname{dim}} {\operatorname{Proj}}^1 A+e+1 = \ell(M)+e.\end{align*} $$

Where the second inequality follows from Lemma 3.1(1). Likewise, $\ell (M\oplus N)\,{\geqslant }\, \ell (N)+{\mathrm {rank}} (M)$ , finishing the proof.

In the following corollary, we observe that if a module satisfies equality in one of the inequalities in Proposition 2.7, then we obtain a closed formula for the analytic spread of its direct sum with any other module.

In the following corollary, we relate the analytic spread of direct sums and products of ideals and modules. We remark that the estimates for the analytic spread in [Reference Branco Correia and Zarzuela5, 6.5 6.8] and [Reference Branco Correia and Zarzuela6, 5.9] follow from our next result.

Corollary 3.10. Let $I_1,\ldots , I_{p-1}$ be R-ideals for some $p\,{\geqslant }\, 1$ and let M be an R-module, all of positive rank. Then

$$ \begin{align*} \ell(I_1\ldots I_{p-1} M)+p-1&=\ell(I_1\oplus \cdots \oplus I_{p-1}\oplus M)\\ &{\geqslant} \max_{1{\leqslant} i{\leqslant} p-1}\{\ell(I_i)+{\mathrm{rank}} (M)-1,\ell(M)\}+p-1. \end{align*} $$

Proof As the inequality follows directly from Theorem 3.6, it suffices to show the equality.

We may assume M is torsion-free (Remark 2.3). We proceed by induction on $p\,{\geqslant }\, 1$ , the case $p=1$ being clear. Now, assume $p\,{\geqslant }\, 2$ and set $A={\mathcal F}(I_1\oplus \cdots \oplus I_{p-1}\oplus M)$ . Notice that A has a natural ${\mathbb {N}}^{p}$ -graded structure (Definition 3.4). Moreover, $A^{\Delta }={\mathcal F}(I_1\ldots I_{p-1}M)$ , $A^{(i)}={\mathcal F}(I_1\oplus \cdots \oplus I_{i-1}\oplus I_{i+1} \oplus \cdots \oplus I_{p-1}\oplus M)$ for every $1{\leqslant } i{\leqslant } p-1$ , and $A^{(p)}={\mathcal F}(I_1\oplus \cdots \oplus I_{p-1})$ . Hence, ${\operatorname {dim}} A^{(i)} <{\operatorname {dim}} A$ for every $1{\leqslant } i{\leqslant } p$ (Theorem 3.6). Therefore, by Lemmas 3.1(2) and 3.2 we have

$$ \begin{align*} {\operatorname{dim}} A = {\operatorname{dim}} {\operatorname{Proj}}^{p} A + p ={\operatorname{dim}} {\operatorname{Proj}}^{1} A^{\Delta} + p = {\operatorname{dim}} A^{\Delta} +p-1, \end{align*} $$

and the result follows.

The following example extends [Reference Vasconcelos45, 8.6]. Here we are able to provide a formula for the analytic spread of a certain class of modules.

Example 3.11. Let $A_1,\ldots , A_p$ be standard graded k-algebras and for each $i=1,\ldots ,p $ let $I_i$ be an $R_i$ -ideal of positive rank and generated by elements of degree $\delta _i$ . Consider $A=A_1\otimes _k \cdots \otimes _k A_p$ and identify each $I_i$ with its image in A. Then $I_1\cdots I_p$ is generated in degree $\delta _1+\cdots +\delta _p$ and its minimal number of generators is the dimension of the k-vector space $[I_1]_{\delta _1}\otimes _k\cdots \otimes _k [I_p]_{\delta _p}$ , that is, $\prod _{i=1}^p {\operatorname {dim}}_k [I_i]_{\delta _i}=\prod _{i=1}^p\mu (I_i)$ , where $\mu (-)$ denotes minimal number of generators. Likewise, for every $n\in {\mathbb {N}}$ , we have $\mu ((I_1\cdots I_p)^n)=\prod _{i=1}^p\mu (I_i^n)$ . Hence, $\ell (I_1\cdots I_p)-1=(\ell (I_1)-1)+\cdots +(\ell (I_p)-1)$ ([Reference Bruns and Koch9, 4.1.3]). From Corollary 3.10, we conclude that

$$ \begin{align*} \ell(I_1\oplus\cdots \oplus I_p)=\ell(I_1)+\cdots+\ell(I_p). \end{align*} $$

In the following corollary we recover, and slightly extend, the results in [Reference Lejeune and Teissier35, 5.5] (see also [Reference Huneke and Swanson27, 8.4.4] and [Reference Hyry29, Lemma 4.7]). We recall that the analytic spread of an ideal is defined as $\ell (I)={\operatorname {dim}} {\mathcal R}(I)\otimes _R k$ regardless of any rank assumption.

Proof For (1), assume I is not nilpotent. If J is nilpotent, that is, $\ell (J)=0$ , the inequality clearly holds. Otherwise, for any ${\mathfrak {p}}$ minimal prime of R that does not contain $IJ,$ we have

$$ \begin{align*} \ell(I)+\ell(J){\geqslant} \ell(I(R/{\mathfrak{p}}))+\ell(J(R/{\mathfrak{p}}))>\ell(IJ(R/{\mathfrak{p}})) \end{align*} $$

where the first inequality follows from [Reference Huneke and Swanson27, 5.1.7] and the second one from Theorem 3.6 and Corollary 3.10. The result then follows from [Reference Huneke and Swanson27, 5.1.7]. Similarly, for (2), let ${\mathfrak {p}}$ be a minimal prime of R such that $\ell (I)=\ell (I(R/{\mathfrak {p}}))$ ([Reference Huneke and Swanson27, 5.1.7]), then

$$ \begin{align*} \ell(IJ){\geqslant} \ell(IJ(R/{\mathfrak{p}})){\geqslant} \ell(I(R/{\mathfrak{p}}))=\ell(I), \end{align*} $$

where the second inequality follows from Corollary 3.10. Likewise, $\ell (IJ)\,{\geqslant }\, \ell (J)$ , and the result follows.

Our results allow us to build a minimal reduction of a direct sum of multiple copies of an ideal I as we show in the next corollary. This result extends [Reference Vasconcelos45, 8.67] to arbitrary ideals. Moreover, the computation of integral closure in [Reference Kodiyalam34, 3.5] follows from this result.

Given elements $a_1,\ldots , a_s\in R$ and an integer $p\,{\geqslant }\, 1$ , we define the matrix

$$ \begin{align*} A^p(a_1,\ldots, a_s):=\begin{pmatrix} a_1&a_2&a_3&\cdots &0&0&0\\ 0&a_1&a_2&\cdots &0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&a_{s-1}&a_s&0\\ 0&0&0&\cdots&a_{s-2}&a_{s-1}&a_s \end{pmatrix}. \end{align*} $$

Corollary 3.13. Let I be an R-ideal of positive rank and let s be its analytic spread. Fix $p\in {\mathbb {Z}}_{>0}$ and consider the R-module

$$ \begin{align*} M=\underbrace{I\oplus\cdots\oplus I}_{p \text{ times}}. \end{align*} $$

Then, $\ell (M)=s+p-1$ and given any (minimal) reduction $(a_1,\ldots , a_s)\subseteq I$ , the R-submodule of $R^p$ generated by the columns of the matrix $A^p(a_1,\ldots , a_s)$ is a minimal reduction of M.

Proof Let U be the module generated by the columns of this matrix and notice that $U\subseteq M$ . We first show that U is a reduction of M. For this, note that by [Reference Bruns and Vetter10, p. 15], ${\mathbf I}_p(U)=I^p$ , and the latter is clearly also equal to ${\mathbf I}_p(M)$ . By Theorem 2.10, it follows that U is a reduction of M.

It remains to show $\ell (M)=s+p-1$ , but this follows from Corollary 3.10 since

$$ \begin{align*} \ell(M)=\ell(I^p)+p-1=s+p-1, \end{align*} $$

finishing the proof.

Example 3.14. Let I be a monomial ideal of ${\mathcal O}_2$ . Let $\Gamma _+(I)$ denote the Newton polyhedron of I (see the definition of this notion before Example 4.14) and let $\{(a_1,b_1),\, (a_2,b_2),\,\ldots ,\, (a_n,b_n)\}\subset {\mathbb {N}}^2$ be the set of vertices of $\Gamma _+(I)$ , with $n\,{\geqslant }\, 2$ and $a_1<a_2<\cdots <a_n$ and $b_1>b_2>\cdots >b_n$ . Consider the polynomials of ${\mathbb C}[x,y]$ given by

$$ \begin{align*} g_1=\displaystyle\sum_{i \text{ is odd}}x^{a_i}y^{b_i} \quad\text{and}\quad g_2=\displaystyle\sum_{i \text{ is even}}x^{a_i}y^{b_i}.\end{align*} $$

By [Reference Bivià-Ausina, Fukui and Saia4] (see also [Reference Crispin Quiñonez13, 3.6] or [Reference Chan and Liu12, 3.7]), the ideal $(g_1,g_2)$ is a reduction of I. Thus, by Corollary 3.13, the module generated by the columns of $A^p(g_1,g_2)$ is a minimal reduction of the module $M=I\oplus \cdots \oplus I\subset {\mathcal O}_2^p$ .

4.1 Integrally decomposable modules

Let M be a submodule of $R^p$ and let $r={\mathrm {rank}}(M)$ . We identify M with any matrix of generators and denote by $\Lambda _M$ the set of vectors $(i_1,\ldots , i_r)\in {\mathbb Z}^r_{>0}$ such that $1{\leqslant } i_1<\cdots <i_r{\leqslant } p$ and there exists some nonzero minor of M formed from rows $i_1,\ldots , i_r$ .

We remark that, under the conditions of the above definition, if ${\texttt L}\in \Lambda _M$ and we write ${\texttt L}=(i_1,\ldots , i_r)$ , where $1{\leqslant } i_1<\cdots <i_r{\leqslant } p$ , then $\overline {M_{\texttt L}}$ is decomposable if and only if $\overline {M_{\texttt L}}=(\overline {M_{\texttt L}})_{i_1} \oplus \cdots \oplus (\overline {M_{\texttt L}})_{i_r}$ . In particular, we observe that Definition 4.1 constitutes a void condition when ${\mathrm {rank}}(M)=1$ .

Proposition 4.3. Let M be submodule of $R^p$ and let $r={\mathrm {rank}}(M)$ . Then M is integrally decomposable if and only if

(4.1) $$ \begin{align} \overline {M_{\texttt L}}=\overline{M_{i_1}}\oplus \cdots \oplus \overline{M_{i_r}}. \end{align} $$

for all ${\texttt L}=(i_1,\ldots ,i_r)\in \Lambda _M$ , where $1{\leqslant } i_1<\cdots <i_r{\leqslant } p$ .

Proof Since $M_{\texttt L}$ is a submodule of $R^r$ of rank r, for all ${\texttt L}\in \Lambda _M,$ it suffices to show the result in the case $r=p$ . So let us assume that ${\mathrm {rank}}(M)=p$ . In general we have the following inclusions:

$$ \begin{align*} \overline M\subseteq (\overline M)_1\oplus \cdots \oplus (\overline M)_p\subseteq \overline{(\overline M)_1}\oplus \cdots \oplus \overline{(\overline M)_p}=\overline{M_1}\oplus \cdots \oplus \overline{M_p} \end{align*} $$

where the last equality is an application of Lemma 4.2. This shows that if relation (4.1) holds, then $\overline M$ is decomposable.

Conversely, if $\overline M$ is decomposable, then $\overline M=(\overline M)_1\oplus \cdots \oplus (\overline M)_p$ . Taking integral closures in this equality it follows that

$$ \begin{align*} \overline M=\overline{\overline M}=\overline{(\overline M)_1\oplus \cdots \oplus (\overline M)_p}= \overline{(\overline M)_1}\oplus \cdots \oplus \overline{(\overline M)_p}=\overline{M_1}\oplus \cdots \oplus \overline{M_p} \end{align*} $$

again by Lemma 4.2, and thus equality (4.1) follows.

In the following proposition we characterize integrally decomposable modules in terms of their ideals of minors.

Let R be a Noetherian local ring of dimension d and let $I_1,\ldots , I_d$ be a family of ideals of R of finite colength. We denote by $e(I_1,\ldots , I_d)$ the mixed multiplicity of the family of ideals $I_1,\ldots , I_d$ (see [Reference Huneke and Swanson27, p. 339]). We recall that when the ideals $I_1,\ldots , I_d$ coincide with a given ideal I of finite colength, then $e(I_1,\ldots , I_d)=e(I)$ , where $e(I)$ is the multiplicity of I, in the usual sense.

Let $(i_1,\ldots , i_p)\in {\mathbb Z}_{\,{\geqslant }\, 0}^p$ , for some $p{\leqslant } d$ , such that $i_1+\cdots +i_p=d$ . We denote by $e_{i_1,\ldots , i_p}(I_1,\ldots , I_p)$ the mixed multiplicity $e(I_1,\ldots , I_1,\ldots , I_p,\ldots , I_p)$ where $I_j$ is repeated $i_j$ times, for all $j=1,\ldots , p$ .

Let M be a submodule of $R^p$ of finite colength. Following [Reference Bivià-Ausina2, p. 418], we define

$$ \begin{align*} \delta(M)=\sum_{\substack{i_1+\cdots +i_p=d\\ i_1,\ldots, i_p{\geqslant} 0 }}e_{i_1,\ldots, i_p}(M_1,\ldots, M_p). \end{align*} $$

We remark that the condition that M has finite colength in $R^p$ implies that $M_i$ has finite colength in R, for all $i=1,\ldots , p$ .

By a result of Kirby and Rees in [Reference Kirby and Rees32, p. 444] (see also [Reference Bivià-Ausina2, p. 417]), we have that $e(I_1\oplus \cdots \oplus I_p) =\delta (I_1\oplus \cdots \oplus I_p)$ , for any family of ideals $I_1,\ldots ,I_p$ of R of finite colength. Therefore $\delta (M)=e(M_1\oplus \cdots \oplus M_p)$ .

Proof Let us fix any ${\texttt L}=(i_1,\ldots , i_r)\in \Lambda _M$ . By Proposition 4.3, the submodule $\overline {M_{\texttt L}}\subseteq R^r$ is decomposable if and only if $\overline {M_{\texttt L}}=\overline {M_{i_1}}\oplus \cdots \oplus \overline {M_{i_r}}$ . Let us recall that

$$ \begin{align*} \overline{M_{i_1}}\oplus \cdots \oplus \overline{M_{i_r}}= \overline{\overline{M_{i_1}}\oplus \cdots \oplus \overline{M_{i_r}}}. \end{align*} $$

Thus $M_{\texttt L}$ is integrally decomposable if and only if $M_{\texttt L}$ is a reduction of $\overline {M_{i_1}}\oplus \cdots \oplus \overline {M_{i_r}}$ , which is to say that $e(M_{\texttt L})=e(M_{i_1}\oplus \cdots \oplus M_{i_r})$ , by Theorem 2.13. But $e(M_{i_1}\oplus \cdots \oplus M_{i_r})=\delta (M_{\texttt L})$ , thus the result follows.

For a submodule of $R^p$ , we introduce the following objects.

Definition 4.6. Let $M\subseteq R^p$ be a submodule of rank r. We define the ideal

$$ \begin{align*} J_M=\sum_{(i_1,\ldots, i_r)\in\Lambda_M}M_{i_1}\ldots M_{i_r} \end{align*} $$

and the following modules

$$ \begin{align*} Z(M)&=\big\{ h\in R^p: {\mathrm{rank}}(M)={\mathrm{rank}}(M + R h) \big\}\\ C(M)&=Z(M)\cap \big(\overline{M_1}\oplus \cdots\oplus \overline{M_p}\big). \end{align*} $$

From Remarks 2.9 and 2.14, it follows that $\overline M$ is always contained in $C(M)$ but this containment can be strict. We ask the following question.

The following is the main theorem of this section, here we provide a partial answer to Question 4.8 by showing that integrally decomposable modules satisfy this equality.

We remark that $\text {(3)}\nRightarrow \text {(2)}$ . In particular $\text {(3)}\nRightarrow \text {(1)}$ in general. This is shown in Example 4.26. In a wide variety of examples of modules $M\subseteq R^3$ with ${\mathrm {rank}}(M)=2,$ we have verified that M is integrally decomposable when $\overline {{\mathbf I}_r(M)}=\overline {J_M}$ . However, we have not yet found a proof or a counterexample of the implication $\text {(2)}\Rightarrow \text {(1)}$ ; we conjecture that this implication holds in general.

We present the proof of Theorem 4.9 after the following remark and lemma.

Remark 4.10. We observe that $\overline {{\mathbf I}_r(M)}\subseteq \overline {J_M}$ . In general, this inclusion might be strict. For instance, consider the submodule $M\subseteq {\mathcal O}_2^3$ generated by the columns of the following matrix

$$ \begin{align*} \left[\begin{matrix} x^2 & xy & x^3 \\ y^2 & y^2 & y^2 \\ x+y & 2y & x^2+y \end{matrix}\right]. \end{align*} $$

Notice that $M_1=( x^2, xy)$ , $M_2=( y^2),$ and $M_3=( x,y)$ . We see that ${\mathrm {rank}}(M)=2$ and

$$ \begin{align*} \overline{J_M}=\overline{M_1M_2+M_1M_3+M_2M_3}=\overline{( x^3, y^3)}. \end{align*} $$

However, ${\mathbf I}_2(M)=(x^2y, xy^2, y^3)$ . Therefore $\overline {{\mathbf I}_r(M)}$ is strictly contained in $\overline {J_M}$ .

We need one more lemma prior presenting the proof of the theorem.

We are now ready to present the proof of Theorem 4.9.

Proof of Theorem 4.9

We begin with $\text {(1)}\Rightarrow \text {(2)}$ . From ${\mathbf I}_r(M)=\sum _{{\texttt L}\in \Lambda _M}{\mathbf I}_r(M_{\texttt L})$ and Proposition 4.4, we obtain

$$ \begin{align*} \overline{{\mathbf I}_r(M)}=\overline{\sum_{{\texttt L}\in\Lambda_M} {\mathbf I}_r(M_{\texttt L})}=\overline{\sum_{{\texttt L}\in\Lambda_M}\overline{{\mathbf I}_r(M_{\texttt L})}}=\overline{\sum_{{\texttt L}\in\Lambda_M}\overline{\prod_{i\in {\texttt L}}M_i}}= \overline{J_M}. \end{align*} $$

We continue with $\text {(2)}\Rightarrow \text {(3)}$ . The inclusion $\overline M\subseteq C(M)$ follows immediately from Remarks 2.9 and 2.14, then we need to show the reverse inclusion. Let $h\in C(M)$ , we claim that ${\mathbf I}_r(M+R h)\subseteq \overline {{\mathbf I}_r(M)}$ . We note that if the claim holds then h is integral over M, by Theorem 2.10, finishing the proof.

Now we prove the claim. Identify M with a matrix of generators and let g be a nonzero minor of size r of the matrix $[M|h]$ with row set ${\texttt L}=\{i_1,\ldots , i_r\}$ . By Lemma 4.11, we have ${\mathrm {rank}}(M_{\texttt L})={\mathrm {rank}}(M_{\texttt L}|h_{\texttt L})$ . In particular, the matrix $M_{\texttt L}$ has some nonzero minor of order r. This implies that ${\texttt L}\in \Lambda _M$ . Since $h\in \overline {M_1}\oplus \cdots \oplus \overline {M_p}$ , we have $g\in \prod _{i\in {\texttt L}}\overline {M_i}\subseteq \overline {\prod _{i\in {\texttt L}}M_i} \subseteq \overline {J_M}$ ([Reference Huneke and Swanson27, 1.3.2]). Therefore, ${\mathbf I}_r(M+R h)\subseteq \overline {J_M}=\overline {{\mathbf I}_r(M)}$ , and the claim follows.

Let us suppose that $r=p$ . In this case, $C(M)=\overline {M_1}\oplus \cdots \overline {M_p}$ and therefore the equivalence of the conditions follows as a direct consequence of Propositions 4.3.

The following result shows a procedure to compute the module $Z(M)$ with the aid of Singular [Reference De Jong14] or other computational algebra programs. If N is a submodule of $R^p$ , then we denote by $N^T$ the transpose of any matrix whose columns generate N.

Lemma 4.12. Let R be an integral domain and let M be an $p\times m$ matrix with entries in R. Then

$$ \begin{align*} \{h\in R^p: {\mathrm{rank}}(M)={\mathrm{rank}}([M\mid h])\}=\ker\big((\ker(M^T))^T\big). \end{align*} $$

Proof Let $Q(R)$ be the field of fractions of R and let $\ker _{Q(R)}(-)$ the kernel of matrices computed over $Q(R)$ .

Clearly the rank of a matrix over R is equal to the rank as a matrix over $Q(R)$ . Let $h\in R^p$ , then by the dimension theorem for matrices, we have

$$ \begin{align*}{\mathrm{rank}}(M)&={\mathrm{rank}}(M^T)=p-{\operatorname{dim}} \ker_{Q(R)}(M^T),\quad \text{and}\quad\\ {\mathrm{rank}}([M\mid h])&=p-{\operatorname{dim}} \ker_{Q(R)}([M\mid h]^T). \end{align*} $$

Since we always have $\ker _{Q(R)}([M\mid h]^T) \subseteq \ker _{Q(R)}(M^T)$ , it follows that

$$ \begin{align*} {\mathrm{rank}}(M)={\mathrm{rank}}([M\mid h]) & \Longleftrightarrow \ker_{Q(R)}([M\mid h]^T) = \ker_{Q(R)}(M^T)\\&\Longleftrightarrow h^Tv=0\quad \text{for every } v\in \ker_{Q(R)}(M^T)\\& \Longleftrightarrow h^Tv=0\quad \text{for every } v\in \ker(M^T)\\& \Longleftrightarrow h\in \ker\big((\ker(M^T))^T\big). \end{align*} $$

This finishes the proof.

In the next example we show an application of Theorem 4.9 in order to compute the integral closure of a module. First, we introduce some concepts.

Let us fix coordinates $x_1,\ldots , x_n$ for ${\mathbb C}^n$ . If $n=2$ , we simply write $x,y$ instead of $x_1, x_2$ . If $\mathbf {k}=(k_1,\ldots , k_n)\in {\mathbb N}^{n}$ , then we denote the monomial $x_{1}^{k_{1}}\cdots x_{n}^{k_{n}}$ by $x^{\mathbf {k}}$ . If $f\in {\mathcal O}_n$ and $f=\sum _{\mathbf k} a_{\mathbf k} x^{\mathbf k}$ is the Taylor expansion of f around the origin, then the support of f, denoted by ${\operatorname {supp}}(f)$ , is the set $\{\mathbf k\in {\mathbb N}^n: a_{\mathbf k}\neq 0\}$ . The support of a nonzero ideal I of ${\mathcal O}_n$ is the union of the supports of the elements of I. We denote this set by ${\operatorname {supp}}(I)$ .

Given a subset $A\subseteq \mathbb R^n_{\,{\geqslant }\, 0}$ , the Newton polyhedron determined by A, denoted by $\Gamma _+(A)$ , is the convex hull of the set $\{\mathbf k+\mathbf v: \mathbf k\in A, \mathbf v\in \mathbb R^n_{\,{\geqslant }\, 0}\}$ . The Newton polyhedron of f is defined as $\Gamma _+(f)=\Gamma _+({\operatorname {supp}}(f))$ . For an ideal I of ${\mathcal O}_n$ , the Newton polyhedron of I is defined as $\Gamma _+(I)=\Gamma _+({\operatorname {supp}}(I))$ . It is well-known that $\Gamma _+(I)=\Gamma _+(\overline I)$ (see for instance [Reference Bivià-Ausina, Fukui and Saia4, p. 58]).

Let ${\mathbf w}\in {\mathbb Z}^n_{\,{\geqslant }\, 0}$ and let $f\in {\mathcal O}_n$ , $f\neq 0$ . We define $d_{\mathbf w}(f)=\min \{\langle {\mathbf w},\mathbf k\rangle : \mathbf k\in {\operatorname {supp}}(f)\}$ , where $\langle {\mathbf w},\mathbf k\rangle $ denotes the usual scalar product. If $f=0$ then we set $d_{\mathbf w}(f)=+\infty $ . We say that a nonzero $f\in {\mathcal O}_n$ is weighted homogeneous with respect to $\; {\mathbf w}$ when $\langle {\mathbf w},\mathbf k\rangle =d_{\mathbf w}(f)$ , for all ${\mathbf k}\in {\operatorname {supp}}(f)$ .

Example 4.14. Let us consider the submodule M of ${\mathcal O}_2^3$ generated by the columns of the following matrix:

$$ \begin{align*} \left[\begin{matrix} x^2y & xy^3 & x^2+y^5 \\ xy^3 & x^2+y^5 & x^2y \\ x^2y-xy^3 & xy^3-x^2-y^5 & x^2+y^5-x^2y \end{matrix}\right]. \end{align*} $$

We observe that ${\mathrm {rank}}(M)=2$ and $\Lambda _M=\{(1,2),(1,3),(2,3)\}$ . Using Singular [Reference De Jong14] we verified that $M_{12}$ , $M_{13}$ and $M_{23}$ have finite colength and $e(M_{12})=e(M_{13})=e(M_{23})=33$ .

Let $I=M_1=M_2=(x^2y, xy^3, x^2+y^5)$ . We have $e(I)=11=e(M_3)$ . Since $M_3\subseteq I$ , it follows that $\overline I=\overline {M_3}$ . Hence $e(M_1, M_2)=e(M_1, M_3)=e(M_2, M_3)=e(I)=11$ . This fact shows that $\delta (M_{12})=e(M_1)+e(M_1,M_2)+e(M_2)=3e(I)=33=\delta (M_{13})=\delta (M_{23})$ . Therefore M is integrally decomposable, by Proposition 4.5.

By Theorem 4.9, the integral closure of M is expressed as

$$ \begin{align*} \overline M=\big\{h \in \overline I\oplus \overline I \oplus\overline I: {\mathrm{rank}}(M+{\mathcal O}_2h)=2\big\}. \end{align*} $$

Let $L=(x^2+y^5, xy^3, x^2y, x^3, y^6)$ . We observe that $IL=L^2$ , therefore I is a reduction of L. Hence $L\subseteq \overline I$ . Let us see that equality holds.

Let $f=x^2+y^5$ . We observe that f is weighted homogeneous with respect to ${\mathbf w}=(5,2)$ . Let N denote the ideal of ${\mathcal O}_2$ generated by all monomials $x^{k_1}y^{k_2}$ , where $k_1, k_2\in {\mathbb Z}_{\,{\geqslant }\, 0}$ , such that $d_{\mathbf w}(x^{k_1}y^{k_2})=5k_1+2k_2\,{\geqslant }\, 11$ . Then $L=(f)+N$ .

Let $g\in \overline I$ . In particular $\Gamma _+(g)\subseteq \Gamma _+(\overline I)=\Gamma _+(I)=\Gamma _+(x^2, y^5)$ . Let $g_1$ denote the part of lowest degree with respect to w in the Taylor expansion of g, and let $g_2=g-g_1$ . Then $d_{\mathbf w}(g_1)\,{\geqslant }\, 10$ and $d_{\mathbf w}(g_2)\,{\geqslant }\, 11$ . In particular $g_2\in N\subseteq L$ . Then $g\in L$ if and only if $g_1\in L$ .

We may assume that ${\operatorname {supp}}(g_1)\subseteq \{(2,0), (0,5)\}$ , as otherwise $g\in L$ . If ${\operatorname {supp}}(g_1)$ is equal to $\{(2,0)\}$ or to $\{(0,5)\}$ , then the ideal $(f, g_1)$ has finite colength and $e(f, g_1)=10$ , which is a contradiction, since $(f,g_1)\subseteq \overline I$ and $e(I)=11$ . Therefore $g_1=\alpha x^2+\beta y^5$ , for some $\alpha , \beta \in {\mathbb C}\setminus \{0\}$ . If $\alpha \neq \beta $ , we would have that $(f,g_1)$ is an ideal of finite colength and $e(f, g_1)=10$ . Therefore $\alpha =\beta $ , which means that $g_1\in (f)\subseteq L$ . Therefore $\overline I\subseteq L$ .

By Theorem 4.9, we have that $\overline M=Z(M)\cap (\overline I\oplus \overline I\oplus \overline I)$ . The module $Z(M)$ can be computed by means of Lemma 4.11. Thus we obtain that $Z(M)$ is generated by the columns of the matrix

$$ \begin{align*} \left[\begin{array}{rr} 1 & 0 \\ 0 & -1 \\ 1 & 1 \end{array}\right]. \end{align*} $$

We have seen before that $\overline I=L$ . Let us remark that $\{x^2+y^5, xy^3, y^6\}$ is a minimal system of generators of L. Then, by intersecting the modules $Z(M)$ and $L\oplus L\oplus L$ , we finally obtain that $\overline M$ is generated by the columns of the following matrix:

$$ \begin{align*} \left[\begin{array}{cccccc} x^2+y^5 & xy^3 & y^6 & x^2+y^5 & xy^3 & y^6 \\ x^2+y^5 & xy^3 & y^6 & 0 & 0 & 0 \\ 0 & 0 & 0 & x^2+y^5 & xy^3 & y^6 \end{array}\right]. \end{align*} $$

In the next subsection, we will introduce an important class of modules that are integrally decomposable.

4.2 Newton nondegenerate modules

Let us fix coordinates $x_1,\ldots , x_n$ for ${\mathbb C}^n$ . Let M be a submodule of ${\mathcal O}_n^p$ and let us identify M with any matrix of generators of M. We recall that $M_i$ is the ideal of ${\mathcal O}_n$ generated by the elements of ith row of M. We define the Newton polyhedron of M as

$$ \begin{align*} \Gamma_+(M)=\Gamma_+\left(\prod_{i=1}^pM_i\right)=\Gamma_+(M_1)+\cdots+\Gamma_+(M_p)=\{{\mathbf k}_1+\cdots+{\mathbf k}_p: {\mathbf k}_i\in \Gamma_+(M_i),\,\,\text{for all }i\}. \end{align*} $$

We denote by $\mathscr F_c(\Gamma _+(M))$ the set of compact faces of $\Gamma _+(M)$ (see [Reference Bivià-Ausina2, p. 408] or [Reference Bivià-Ausina3, p. 397] for details).

Let I be an ideal of ${\mathcal O}_n$ . We denote by $I^0$ the ideal by all monomials $x^{\mathbf k}$ such that $\mathbf k\in \Gamma _+(I)$ . We refer to this ideal as the term ideal of I. If I is the zero ideal, then we set $\Gamma _+(I)=\emptyset $ and $I^0=0$ . Recall that an ideal is said to be monomial if it admits a generating system formed by monomials. It is known that if I is a monomial ideal, then $\overline I=I^0$ (see [Reference Eisenbud16, p. 141], [Reference Huneke and Swanson27, p. 11], or [Reference Teissier43, p. 219]). The ideals I for which $\overline I$ is generated by monomials are characterized in [Reference Rees39] and are called Newton nondegenerate ideals (see also [Reference Bivià-Ausina3], [Reference Bivià-Ausina, Fukui and Saia4], or [Reference Teissier43, p. 242]).

In [Reference Bivià-Ausina2], the Carles Bivià-Ausina introduced and studied the notion of Newton nondegenerate modules of maximal rank. Here we extend this concept to modules of submaximal rank.

Let $f\in {\mathcal O}_n$ and let $f=\sum _{\mathbf k} a_{\mathbf k} x^{\mathbf k}$ be the Taylor expansion of f around the origin. If $\Delta $ is any compact subset of $\mathbb R^n_{\,{\geqslant }\, 0}$ , then we denote by $f_{\Delta }$ the polynomial resulting as the sum of all terms $a_{\mathbf k} x^{\mathbf k}$ such that ${\mathbf k}\in \Delta $ . If no such ${\mathbf k}$ exist, then we set $f_{\Delta }=0$ .

In particular, if I is an ideal of ${\mathcal O}_n$ and $g_1,\ldots , g_s$ denotes a generating system of I, then I is Newton nondegenerate if and only if $\{x\in {\mathbb C}^n: (g_1)_{\Delta }(x)=\cdots =(g_s)_{\Delta }(x)=0\}\subseteq \{x\in {\mathbb C}^n: x_1\cdots x_n=0\}$ , for any $\Delta \in \mathscr F_c(\Gamma _+(I))$ .

The following result follows from [Reference Bivià-Ausina2, 3.7, 3.8] and it characterizes the Newton nondegeneracy of submodules of ${\mathcal O}_n^p$ of maximal rank.

As an immediate consequence of Theorem 4.16, the following result follows.

Therefore, we see from the previous result that if M is Newton nondegenerate, then it is integrally decomposable. The converse does not hold in general, as Example 4.14 shows.

From the results of the previous section we obtain the following combinatorial interpretation for the analytic spread of Newton nondegenerate modules of maximal rank.

Corollary 4.18. Let $M\subseteq {\mathcal O}_n^p$ be a Newton nondegenerate module of rank p, then

$$ \begin{align*} \ell(M)=\max\big\{{\operatorname{dim}}(\Delta): \Delta\in \mathscr F_c(\Gamma_+(M))\big\}+p. \end{align*} $$

Proof We may assume R has infinite residue field and then $\ell (M)=\ell (\overline M)$ (see Remark 3.7). Moreover $\overline M=M_1^0\oplus \cdots \oplus M_p^0$ , since M is Newton nondegenerate. Therefore $\ell (M)=\ell (M_1^0\oplus \cdots \oplus M_p^0)=\ell (M_1^0\cdots M_p^0)+p-1$ , where the last equality is an application of Corollary 3.10. By [Reference Bivià-Ausina1, Theorem 2.3] we have

$$ \begin{align*} \ell(M_1^0\cdots M_p^0)=\max\big\{{\operatorname{dim}}(\Delta): \Delta\in \mathscr F_c\big(\Gamma_+(M_1^0\cdots M_p^0)\big)\big\}+1. \end{align*} $$

Since $\Gamma _+(M)=\Gamma _+(M_1^0\cdots M_p^0)$ the result follows.

Example 4.19. Let M be the submodule of ${\mathcal O}_2^2$ generated by the columns of the following matrix

$$ \begin{align*} M=\left[\begin{matrix} x^3 & xy & y^3 & y^3 \\ x^5 & x^2y & xy^2 & x^5+x^2y \end{matrix}\right]. \end{align*} $$

We observe that ${\mathrm {rank}}(M)=2$ and ${\mathbf I}_2(M)$ is a Newton nondegenerate ideal. Moreover $\overline {{\mathbf I}_2(M)}=\overline {(xy^5, x^2y^3, x^3y^2, x^5y, x^8)}=\overline {M_1M_2}.$ Therefore $\overline M=M_1^0\oplus M_2^0$ , by Corollary 4.17 and $\ell (M)=\ell (\overline {M})=3$ , by Corollary 4.18.

Analogously to Definition 4.6, for a submodule of ${\mathcal O}_n^p$ we introduce the following objects.

Definition 4.20. Let $M\subseteq {\mathcal O}^n_p$ and let $r={\mathrm {rank}}(M)$ . We define

$$ \begin{align*} H_M=\sum_{(i_1,\ldots, i_r)\in\Lambda_M}M_{i_1}^0\ldots M_{i_r}^0, \end{align*} $$

and

$$ \begin{align*} C^0(M)=Z(M)\cap \big(M_1^0\oplus \cdots\oplus M_p^0\big), \end{align*} $$

where we recall that $Z(M)=\big \{ h\in R^p: {\mathrm {rank}}(M)={\mathrm {rank}}(M + R h) \big \}$ .

We remark that $H_M$ is a monomial ideal and $\Gamma _+(H_M)=\Gamma _+(J_M)$ . Therefore $\overline {H_M}=J_M^0$ , where $J_M^0$ is the ideal of ${\mathcal O}_n$ generated by the monomials $x^k$ such that $k\in \Gamma _+(J_M)$ . We also remark that $C(M)\subseteq C^0(M)$ . The following result follows from Theorem 4.9 and Corollary 4.17.

In the following example we use Corollary 4.21 to compute the integral closure of a family of modules.

Example 4.23. Let us consider the submodule $M\subseteq {\mathcal O}_2^3$ generated by the columns of the following matrix:

$$ \begin{align*} \left[\begin{matrix} x^a & xy & y^a \\ y^a & x^a & xy \\ x^a+y^a & xy+x^a & y^a+xy \end{matrix}\right], \end{align*} $$

where $a\in {\mathbb Z}_{\,{\geqslant }\, 2}$ . We remark that ${\mathrm {rank}}(M)=2$ . Let $J=( x^a, xy, y^a)$ . The ideal J is integrally closed and $M_1^0=M_2^0=M_3^0=J$ . An elementary computation shows that ${\mathbf I}_2(M)=(xy^{a+1}-x^{2a}, x^{a+1}y-y^{2a}, x^2y^2)$ and that ${\mathbf I}_2(M)$ is Newton nondegenerate. Moreover $J_M^0=\overline {( x^{2a}, x^2y^2, y^{2a})}$ and then $\overline {{\mathbf I}_2(M)}=J_M^0$ , since $\Gamma _+({\mathbf I}_2(M))=\Gamma _+(J_M^0)$ . Therefore, by Corollary 4.21, we conclude that $\overline M=C^0(M)=C(M)$ . Given any element $h=(h_1,h_2,h_3)\in {\mathcal O}_2^3$ , we have that ${\mathrm {rank}}(M)={\mathrm {rank}}(M+{\mathcal O}_2 h)$ if and only if $h_3=h_1+h_2$ (Lemma 4.12). Therefore,

$$ \begin{align*}\overline M&=\big\{ h\in M_1^0\oplus M_2^0 \oplus M_3^0: {\mathrm{rank}}(M)={\mathrm{rank}}(M+{\mathcal O}_2 h) \big\}\\&=\big\{ [h_1\hspace{0.4cm}h_2\hspace{0.4cm}h_1+h_2]^T\in {\mathcal O}_2^3: h_1,h_2\in J \big\}. \end{align*} $$

Therefore, a minimal generating system of $\overline M$ is given by the columns of the folowing matrix

$$ \begin{align*} \left[\begin{matrix} x^a & xy & y^a & 0 & 0 & 0 \\ 0 & 0 & 0 & x^a & xy & y^a\\ x^a & xy & y^a & x^a & xy & y^a \end{matrix}\right]. \end{align*} $$

Example 4.24. Let M be the submodule of ${\mathcal O}_2^2$ generated by the columns of the following matrix

$$ \begin{align*} \left[\begin{matrix} x^3 & x^2y \\ x(x+y) & y(x+y) \end{matrix}\right]. \end{align*} $$

We observe that ${\mathrm {rank}}(M)=1$ . The ideal ${\mathbf I}_1(M)$ is given by

$$ \begin{align*} {\mathbf I}_1(M)=( x^3, x^2y, x(x+y), y(x+y))=M_1+M_2=J_M. \end{align*} $$

We have $\Gamma _+({\mathbf I}_1(M))=\Gamma _+(x^2, y^2)$ . Let $\Delta $ denote the unique compact face of dimension $1$ of $\Gamma _+(x^2, y^2)$ . Hence $(x^3)_{\Delta }=0$ , $(x^2y)_{\Delta }=0$ , $(x(x+y))_{\Delta }=x(x+y)$ and $(y(x+y))_{\Delta }=y(x+y)$ . Since the line of equation $y=-x$ is contained in the set of solutions of the system $x(x+y)=y(x+y)=0$ , we conclude that $I_1(M)$ is Newton degenerate. Therefore $\overline {{\mathbf I}_1(M)}\neq J^0_M$ (otherwise $I_1(M)$ would be a reduction of the monomial ideal $J_M^0$ and hence $I_1(M)$ would be Newton nondegenerate, which is not the case). Let us observe that $\overline {{\mathbf I}_1(M)}=(x(x+y), y(x+y))+{\mathbf m}_2^3$ . Therefore, by Corollary 2.12 and applying Singular [Reference De Jong14] (see Remark 4.13), we deduce that $\overline M=M$ .

By computing explicitly a generating system of $C^0(M)=Z(M)\cap (M_1^0\oplus M_2^0)$ , we also obtain that $C^0(M)=M$ and hence $C^0(M)=\overline M$ . Then $\text {(3)}\nRightarrow \text {(2)}$ in Corollary 4.21.

The following two examples are motivated by Example 5.8 of [Reference Kodiyalam34].

Example 4.26. Let us consider the submodule $M\subseteq {\mathcal O}_2^3$ generated by the columns of the following matrix

$$ \begin{align*} \left[\begin{matrix} x^2 & y & 0 \\ 0 & x & y^2 \\ x^2 & x+y & y^2 \\ \end{matrix}\right]. \end{align*} $$

The rank of M is $2$ and ${\mathbf I}_2(M)=(x^3, x^2y^2, y^3)$ . Thus $\overline {{\mathbf I}_2(M)}={\mathbf m}_2^3$ . By Corollary 2.12, we have $\overline M=Z(M)\cap A(M)$ , where

(4.2) $$ \begin{align} A(M)=\big\{h=[h_1\quad h_2\quad h_3]^T\in{\mathcal O}_2^3: {\mathbf I}_2(M,h)\subseteq {\mathbf m}_2^{3} \big\}. \end{align} $$

In general, the submodule $Z(M)$ can be computed by using Singular [Reference De Jong14], as explained in Remark 4.13. In this case it is immediate to see that

$$ \begin{align*} Z(M)=\left[\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ \end{matrix}\right]. \end{align*} $$

In (4.2) the minors of size $2$ of the matrix $(M,h)$ are $x^2h_2, yh_2-xh_1, y^2h_1, x^2(h_3-h_1), yh_3-(x+y)h_1, xh_3-(x+y)h_2$ and $y^2(h_3-h_2)$ . Then $A(M)$ is equal to the intersection of the following submodules of ${\mathcal O}_2^3$ :

$$ \begin{align*} N_1&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: x^2h_2 \in {\mathbf m}_2^3 \big\}\\ N_2&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: yh_2-xh_1 \in {\mathbf m}_2^3 \big\}\\ N_3&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: y^2h_1 \in {\mathbf m}_2^3 \big\}\\ N_4&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: x^2(h_3-h_1) \in {\mathbf m}_2^3 \big\}\\ N_5&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: yh_3-(x+y)h_1 \in {\mathbf m}_2^3 \big\}\\ N_6&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: xh_3-(x+y)h_2 \in {\mathbf m}_2^3 \big\}\\ N_7&=\big\{h=[h_1\hspace{0.3cm}h_2\hspace{0.3cm}h_3]^T\in{\mathcal O}_2^3: y^2(h_3-h_2) \in {\mathbf m}_2^3 \big\}. \end{align*} $$

Each of the above submodules can be computed with Singular. For instance, to obtain a generating system of $N_5$ we can use the following procedure. Let S denote the quotient ring ${\mathcal O}_2/{\mathbf m}_2^3$ and let us consider the submodule of $S^3$ given by ${\operatorname {syz}}_S(-x-y, 0, y)=\{(g_1,g_2, g_3)\in S^3: (-x-y)g_1+yg_3=0\}$ . Once we have obtained a matrix of generators of ${\operatorname {syz}}_S(-x-y, 0, y)$ with Singular, if B is any submodule of ${\mathcal O}_2^3$ whose image in $S^3$ generates ${\operatorname {syz}}_S(-x-y, 0, y)$ , then $N_5=B+ ({\mathbf m}_2^3\oplus {\mathbf m}_2^3\oplus {\mathbf m}_2^3)$ . Therefore it follows that $N_5$ is generated by the columns of the matrix

$$ \begin{align*} \left[\begin{matrix} y^2 & y^2 & xy-y^2 & x^2-xy+y^2 & y & 0 \\ 0 & 0 & y^2 & 0 & 0 & 1 \\ 0 & y^2 & -y^2 & y^2 & x+y & 0 \end{matrix}\right]. \end{align*} $$

By computing a minimal generating system of $Z(M)\cap N_1\cap \cdots \cap N_7$ , it follows that

$$ \begin{align*} \left[\begin{matrix} x^2 & xy & y^2 & y & 0 \\ 0 & 0 & 0 & x & y^2 \\ x^2 & xy & y^2 & x+y & y^2 \\ \end{matrix}\right]. \end{align*} $$

We remark that $\overline M_1=M_1$ , $\overline M_2=M_2$ and $\overline M_3=(x+y)+{\mathbf m}_2^2$ . Therefore, a computation with Singular shows that the module $C(M)$ , which is defined as $Z(M)\cap (\overline {M_1}\oplus \overline {M_2}\oplus \overline {M_3})$ , is equal to $\overline M$ .

However we have the strict inclusion $\overline {{\mathbf I}_2(M)}\subseteq \overline {J_M}$ in this case, since $J_M={\mathbf m}_2^2$ . Hence we have $\text {(3)}\nRightarrow \text {(2)}$ in Theorem 4.9.

The inequality $\overline {{\mathbf I}_2(M)}\neq \overline {J_M}$ implies that M is not integrally decomposable, by Theorem 4.9. Actually, none of the submodules $\overline {M_{\{1,2\}}}$ , $\overline {M_{\{1,3\}}}$ and $\overline {M_{\{2,3\}}}$ are integrally decomposable, by Proposition 4.5, since they $\delta (M_{1,2})=\delta (M_{1,3})=\delta (M_{2,3})=5$ and $e(M_{1,2})=e(M_{1,3})=e(M_{2,3})=8$ .

Example 4.27. Let us consider the submodule $M\subseteq {\mathcal O}_2^2$ generated by the columns of the following matrix:

$$ \begin{align*} \left[\begin{matrix} x^a & y^b & 0 \\ 0 & x^c & y^d \\ \end{matrix}\right], \end{align*} $$

where $a,b,c,d\in {\mathbb Z}_{\,{\geqslant }\, 1}$ . Let $I={\mathbf I}_2(M)=(x^{a+c}, x^ay^d, y^{b+d})$ . Since the ideals $M_1$ and $M_2$ are generated by monomials we have, from Theorem 4.16, that

(4.3) $$ \begin{align} \overline M \text{ is decomposable} &\Longleftrightarrow \overline M=M_1^0\oplus M_2^0\nonumber\\ &\Longleftrightarrow \overline M \text{ is Newton nondegenerate}\nonumber\\ &\Longleftrightarrow I \text{ is Newton nondegenerate and } \Gamma_+(I)=\Gamma_+(M_1M_2). \end{align} $$

Therefore $\overline M$ is not decomposable if and only if $\Gamma _+(I)$ is strictly contained in $\Gamma _+(M_1M_2)$ . We see that $\Gamma _+(M_1M_2)=\Gamma _+(x^{a+c}, x^ay^d, y^{b+d},x^cy^b)$ . Let us observe that $\Gamma _+(I)=\Gamma _+(x^{a+c}, y^{b+d})$ if and only if $ad\,{\geqslant }\, bc$ .

Let us suppose first that $ad\,{\geqslant }\, bc$ . Then $\Gamma _+(I)$ is strictly contained in $\Gamma _+(M_1M_2)$ if and only if $(c,b)$ lies below the line determined by the two vertices of $\Gamma _+(I)$ , which is to say that $ad>bc$ .

If $ad<bc$ , then the Newton boundary of $\Gamma _+(I)$ is equal to the union of two segments and $(c,b)$ belongs to the interior of $\Gamma _+(I)$ . Hence $\Gamma _+(I)=\Gamma _+(M_1M_2)$ and this implies that $\overline M$ is decomposable by (4.3).

Thus we have shown that $\overline M$ is not decomposable if and only if $ad>bc$ . In this case, we have $\Gamma _+(I)= \Gamma _+(x^{a+c}, y^{b+d})$ . Let ${\mathbf w}=(b+d, a+c)$ . By Corollary 2.12, we obtain that

(4.4) $$ \begin{align} \overline M=\big\{h=[h_1\hspace{0.4cm}h_2]^T\in{\mathcal O}_2^2:\hspace{0.2cm} &d_{\mathbf w}(x^ah_2){\geqslant} (a+c)(b+d),\\ & d_{\mathbf w}(y^bh_2-x^ch_1){\geqslant} (a+c)(b+d) \hspace{0.5cm}\text{and}\nonumber\\ &d_{\mathbf w}(y^dh_1){\geqslant} (a+c)(b+d)\big\}.\nonumber \end{align} $$

Once positive integer values are assigned to $a,b,c,d$ , it is possible to obtain a generating system of $\overline M$ with Singular [Reference De Jong14] by following an analogous procedure as in Example 4.26.

In the following example, we show an example of a nondecomposable integrally closed submodule of ${\mathcal O}_2^2$ of rank $2$ whose ideal of maximal minors is not simple (i.e., it is factorized as the product of two proper integrally closed ideals).

Example 4.28. Let M be the submodule of ${\mathcal O}_2^2$ generated by the columns of the following matrix:

$$ \begin{align*} \left[\begin{matrix} x^5 & xy & y^5 \\ y^2 & x+y & y^2 \\ \end{matrix}\right]. \end{align*} $$

We observe that $e(M_1)=10$ , $e(M_2)=2$ and $e(M_1, M_2)=2$ . Therefore $\delta (M)=14$ . However $e(M)=22$ . Then $\overline M$ is not decomposable by Proposition 4.5. In particular, since $\overline M$ is a submodule of ${\mathcal O}_2^2$ , $\overline M$ does not split as the direct sum of two proper ideals of ${\mathcal O}_2$ . Let $I={\mathbf I}_2(M)=( -xy^3+xy^5+y^6, -x^5y^2+y^7, -xy^3+x^6+x^5y)$ . By Corollary 2.12 it follows that

$$ \begin{align*} \overline M=\big\{h\in{\mathcal O}_2^2: {\mathbf I}_2(M+{\mathcal O}_2 h)\subseteq \overline{{\mathbf I}_2(M)}\big\}. \end{align*} $$

An easy computation shows that I is Newton nondegenerate and $\Gamma _+(I)=\Gamma _+(x^6, xy^3, y^6)$ . The ideal generated by all monomials $x^{k_1}y^{k_2}$ such that $(k_1,k_2)\in \Gamma _+(I)$ is $J=(x^6, x^5y, x^3y^2, xy^3, y^6)$ . Hence $\overline I=J$ and this implies that

$$ \begin{align*} \overline M=\big\{h=[h_1\hspace{0.4cm}h_2]^T\in{\mathcal O}_2^2: x^5h_2-y^2h_1,\, y^5h_2-y^2h_1,\,\, xyh_2-(x+y)h_1\in J\big\}. \end{align*} $$

Hence $\overline M=N_1\cap N_2 \cap N_3$ , where

$$ \begin{align*} N_1&=\big\{h=[h_1\hspace{0.3cm}h_2]^T\in{\mathcal O}_2^3: x^5h_2-y^2h_1 \in J \big\}\\ N_2&=\big\{h=[h_1\hspace{0.3cm}h_2]^T\in{\mathcal O}_2^3: y^5h_2-y^2h_1 \in J \big\}\\ N_3&=\big\{h=[h_1\hspace{0.3cm}h_2]^T\in{\mathcal O}_2^3: xyh_2-(x+y)h_1 \in J \big\}. \end{align*} $$

As in Example 4.26, using Singular [Reference De Jong14], we obtain that

$$ \begin{align*} N_1&=\left[\begin{matrix} y^4 & x^3 & xy & 0 & 0 \\ 0 & 0 & 0 & y & x \end{matrix}\right] \hspace{2cm} N_2 =\left[\begin{matrix} y^4 & x^3 & xy & y^3 \\ 0 & 0 & 0 & 1 \end{matrix}\right] \\ N_3&= \left[\begin{matrix} y^5 & x^4y-x^3y^2+x^2y^3-xy^4 & x^5-x^4y+x^3y^2-x^2y^3+xy^4 &x^2y^2-xy^3 & 0 & xy \\ 0 & 0 & 0 & 0 & y^2 & x+y \\ \end{matrix}\right]. \end{align*} $$

Using Singular again, we have $N_3\subseteq N_1\cap N_2$ . Therefore $\overline M=N_3$ . As we have discussed before, $\overline M$ is not decomposable and obviously it is integrally closed. However, we have that

$$ \begin{align*} {\mathbf I}_2(\overline M)=\overline{{\mathbf I}_2(M)}=(x^6, x^5y, x^3y^2, xy^3, y^6)=(x, y^3)(x^5, x^4y, x^2y^2, y^3). \end{align*} $$

That is, the ideal ${\mathbf I}_2(\overline M)$ is not simple. We also refer to [Reference Hayasaka23] for another type and wide class of examples of integrally closed and nondecomposable submodules $N\subseteq {\mathcal O}_2^2$ of rank $2$ for which the ideal ${\mathbf I}_2(N)$ is not simple (these examples are motivated by a question raised by Kodiyalam in [Reference Kodiyalam34, p. 3572] about the converse of Theorem 5.7 of [Reference Kodiyalam34]).