Transversal Intersection of Monomial Ideals

TL;DR

This paper establishes conditions under which monomial ideals intersect transversally and provides a simplicial characterization of this intersection property.

Contribution

It introduces new criteria for transversal intersection of monomial ideals and offers a simplicial framework to understand this phenomenon.

Findings

Derived conditions for transversal intersection

Provided a simplicial characterization

Enhanced understanding of monomial ideal interactions

Abstract

In this paper we prove conditions for transversal intersection of monomial ideals and derive a simplicial characterization of this phenomenon.

Equations

$$\mathrm{supp}(T)=\{i\mid x_i\mathrm{divides}m\text{for some}m\in T\}.$$

$$S=\{\mathrm{lcm}(f,g)\mid f\in G(I),g\in G(J)\mathrm{and}\text{mbox}\mathrm{supp}(f)\cap \text{mbox}\mathrm{supp}(g)\ne \varnothing \}.$$

$$=\sum _{1\le k\le i}(-1{)}^{k-1}\frac{\text{mbox}\mathrm{lcm}(m_{j_1},\cdots ,m_{j_i})}{\text{mbox}\mathrm{lcm}(m_{j_1},\cdots ,\widehat{m_{j_k}},\cdots ,m_{j_i})}(e_{j_1}\wedge \cdots \wedge \widehat{e_{j_k}}\wedge \cdots \wedge e_{j_i}).$$

$$(\mathbb{T}(I{)}_{\cdot }\otimes \mathbb{T}(J{)}_{\cdot }{)}_r$$

$$\text{mbox}\mathrm{mdeg}\left((e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime })\right)$$

$$=\text{mbox}\mathrm{mdeg}\left(\text{mbox}\mathrm{lcm}(m_{k_1},\dots ,m_{k_i})\cdot \text{mbox}\mathrm{lcm}(m_{k_1^{\prime }}^{\prime },\dots ,m_{k_j^{\prime }}^{\prime })\right).$$

$$(e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime })\mapsto (e_{k_1}\wedge \cdots \wedge e_{k_i}\wedge e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime }).$$

$$\text{mbox}\mathrm{mdeg}(e_{k_1}\wedge \cdots \wedge e_{k_i}\wedge e_{k_1^{\prime }}\wedge \cdots \wedge e_{k_j^{\prime }})$$

$$=$$

$$(e_{k_1}\wedge e_{k_2}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}\wedge e_{k_2^{\prime }}\wedge \cdots \wedge e_{k_j^{\prime }}),\mathrm{with}i+j=r.$$

$$({\psi }_{r-1}\circ d_r)\left((e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime })\right)$$

$$=$$

$$V_t=\frac{\text{mbox}\mathrm{lcm}(m_{k_1},\dots ,m_{k_i})\cdot \text{mbox}\mathrm{lcm}(m_{k_1^{\prime }}^{\prime },\dots ,m_{k_j^{\prime }}^{\prime })}{\text{mbox}\mathrm{lcm}(m_{k_1},\dots ,\widehat{m_{k_t}},\dots ,m_{k_i})\cdot \text{mbox}\mathrm{lcm}(m_{k_1^{\prime }}^{\prime },\dots ,m_{k_j^{\prime }}^{\prime })}=Q_t\forall 1\le t\le i.$$

$$\left(\delta (I{)}_i(e_{k_1}\wedge \cdots \wedge e_{k_i})\right)\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime })$$

$$=$$

$${\psi }_{r-1}\left(\delta (I{)}_i(e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime })\right)$$

$$=$$

$${\psi }_{r-1}\left((e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (\delta (J{)}_j(e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime }))\right)$$

$$=$$

$$W_l=\frac{\text{mbox}\mathrm{lcm}(m_{k_1},\dots ,m_{k_i},m_{k_1^{\prime }}^{\prime },\dots ,m_{k_j^{\prime }}^{\prime })}{\text{mbox}\mathrm{lcm}(m_{k_1},\dots ,m_{k_i},m_{k_1^{\prime }}^{\prime },\dots ,\widehat{m_{k_l^{\prime }}^{\prime }},\dots ,m_{k_j^{\prime }}^{\prime })},1\le l\le j.$$

$$(\ast )$$

$$=$$

$$+$$

$$\sum _{k=1}^{i+j}(-1{)}^k\frac{\text{mbox}\mathrm{lcm}(w_1,\dots ,w_{i+j})}{\text{mbox}\mathrm{lcm}(w_1,\dots ,\widehat{w_k},\dots ,w_{i+j})}(E_1\wedge \cdots \wedge \widehat{E_k}\cdots \wedge E_{i+j}).$$

$$({\delta }_r\circ {\psi }_r)((e_{k_1}\wedge \cdots \wedge e_{k_i})\otimes (e_{k_1^{\prime }}^{\prime }\wedge \cdots \wedge e_{k_j^{\prime }}^{\prime }))$$

$$=$$

$$\Gamma ({\Delta }_1\ast {\Delta }_2)=\{{\sigma }_1\cup {\sigma }_2\mid {\sigma }_1\in \Gamma ({\Delta }_1),{\sigma }_2\in \Gamma ({\Delta }_2)\}.$$

$$0⟶K^{{\Gamma }_{m-1}(\Delta )}\stackrel{{\partial }_{m-1}}{⟶}\cdots K^{{\Gamma }_i(\Delta )}\stackrel{{\partial }_i}{⟶}{K}^{{\Gamma }_{i-1}(\Delta )}\stackrel{{\partial }_{i-1}}{⟶}\cdots \stackrel{{\partial }_0}{⟶}{K}^{{\Gamma }_{-1}(\Delta )}⟶0.$$

$${\partial }_i(e_{\sigma })=\sum _{j\in \sigma }sign(j,\sigma )e_{\sigma -j},0\le i\le m-1;$$

$$\partial (\overline{v_j})=\sum _{1\le s\le q}{\alpha }_{sj}\overline{u_s}$$

$${\mathbb{G}}_i={\oplus }_{1\le i\le p}R\left(\mathrm{mdeg}({\mathrm{v}}_{\mathrm{j}})\right),\Delta (v_j)={\Sigma }_{1\le s\le q}{\alpha }_{sj}\frac{\mathrm{mdeg}({\mathrm{v}}_{\mathrm{j}})}{\mathrm{mdeg}({\mathrm{u}}_{\mathrm{s}})}{u}_s.$$

$$\mathbb{M}(I{)}_{\cdot }\otimes \mathbb{M}(J{)}_{\cdot }\cong \mathbb{T}(I{)}_{\cdot }\otimes \mathbb{T}(J{)}_{\cdot }\cong \mathbb{T}(I+J{)}_{\cdot }$$

$$\mathcal{C}({\Delta }_I{)}_{\cdot }\otimes \mathcal{C}({\Delta }_J{)}_{\cdot }\cong \mathcal{C}({\Delta }_I\ast {\Delta }_J{)}_{\cdot }$$

Full text

The first author is a post-doctoral research fellow under the research project EMR/2015/000776 sponsored by the SERB, Government of India.

The second author is the corresponding author, who is supported by the research project EMR/2015/000776 sponsored by the SERB, Government of India.

1. Introduction

The transversal intersection phenomenon for monomial ideals in the polynomial ring happens to be extremely interesting and it turns out that it is equivalent to having disjoint supports for their minimal generating sets; see Theorem 2.2. The Taylor complex which resolves (possibly non-minimally) a monomial ideal has been understood completely for ideals of the form $I+J$, when $I$ and $J$ are monomial ideals intersecting transversally; see Theorem 3.2. As an application of this theorem we prove in Corollary 3.3 that for two ideals $I$ and $J$ in $R$ intersecting transversally, the ideal $I+J$ is resolved minimally by the complex $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$, if $\mathbb{M}(I)_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ denote minimal free resolutions of $I$ and $J$ respectively. Minimal free resolutions for ideals of the form $I+J$ have an interesting structure when $I$ and $J$ intersect transversally and are supported simplicially; see 4.3.

2. Monomial Ideals

Let $R=K[x_{1},x_{2},\cdots,x_{n}]$, where $x_{i}$’s are indeterminates over the field $K$. Let ${\rm Mon}(R)$ denote the set of all monomials in $R$. Every nonzero polynomial $f\in R$ is a unique $K$-linear combination of monomials given by $f=\sum_{v\in{\rm Mon}(R)}a_{v}v$. Let $m(f):=\{v\in{\rm Mon}(R)\mid a_{v}\neq 0\}$. An ideal $I$ in $R$ is said to be a monomial ideal if it is generated by monomials of $R$. We list down some standard facts on monomial ideals.

Proposition 2.1**.**

1. (1)

$I$* is a monomial ideal if and only if $f\in I\Rightarrow m(f)\subset I$.* 2. (2)

Let $\{u_{1},\cdots,u_{m}\}$ be a monomial generating set of an ideal $I$, where $u_{i}$’s are monomials. A monomial $v\in I$ if and only if $v=u_{i}w$, for some $1\leq i\leq m$. 3. (3)

Each monomial ideal $I$ has a unique minimal monomial set of generators $G(I)$.

Proof.

See [1].∎

Definition 1**.**

Let $\emptyset\neq T\subset{\rm Mon}(R)$. We define

[TABLE]

If $T=\{m\}$, we simply write ${\rm supp}(m)$ instead of ${\rm supp}(\{m\})$. If $S$ and $T$ are two nonempty subsets of ${\rm Mon}(R)$, then, ${\rm supp}(S)\cap{\rm supp}(T)=\emptyset$ if and only if ${\rm supp}(f)\cap{\rm supp}(g)=\emptyset$ for every $f\in S$ and $g\in T$.

Definition 2**.**

We say that the ideals $I$ and $J$ of $R$ intersect transversally if $I\cap J=IJ$.

Theorem 2.2**.**

Let $I$ and $J$ be two monomial ideals of $R$. Then, $I\cap J=IJ$ if and only if $\mbox{\rm supp}(G(I))\cap\mbox{\rm supp}(G(J))=\emptyset$.

Proof.

Let $I\cap J=IJ$. Consider the set

[TABLE]

If $S=\emptyset$, then for every $f\in G(I)$ and $g\in G(J)$ we have $\mbox{\rm supp}(f)\cap\mbox{\rm supp}(g)=\emptyset$, which proves that $\mbox{\rm supp}(G(I))\cap\mbox{\rm supp}(G(J))=\emptyset$.

Suppose that $S\neq\emptyset$. We define a partial order $\leq$ on $S$ in the following way: Given $s,t\in S$, we define $s\leq t$ if and only $s\mid t$. Let $\mathbf{m}$ be a minimal element (by Zorn’s lemma) of $S$; then there exists $f,g$ such that $f\in G(I)$ and $g\in G(J)$ such that ${\rm\mbox{lcm}}(f,g)=\mathbf{m}$. Since ${\rm\mbox{lcm}}(f,g)\in I\cap J$ and $I\cap J=IJ$, we have ${\rm\mbox{lcm}}(f,g)=\mathbf{m}\in IJ$. By Proposition 2.1, there exist $h_{1}\in G(I)$ and $h_{2}\in G(J)$, such that $h_{1}h_{2}\mid\mathbf{m}$, since the generating set of $IJ$ is the set $G(I)G(J)=\{uv\mid u\in G(I),v\in G(J)\}$.

We prove that $\mbox{\rm supp}(h_{1})\cap\mbox{\rm supp}(h_{2})=\emptyset$. If this is not the case, then ${\rm\mbox{lcm}}(h_{1},h_{2})\neq h_{1}h_{2}$, in fact ${\rm\mbox{lcm}}(h_{1},h_{2})<h_{1}h_{2}$ and ${\rm\mbox{lcm}}(h_{1},h_{2})\mid\mathbf{m}$; contradicting minimality of $\mathbf{m}$ in $S$.

We now prove that $\mbox{\rm supp}(f)\cap\mbox{\rm supp}(h_{2})\neq\emptyset$. We know that $h_{1}h_{2}\mid{\rm\mbox{lcm}}(f,g)=\mathbf{m}$. Therefore, if $\mbox{\rm supp}(f)\cap\mbox{\rm supp}(h_{2})=\emptyset$, then $h_{2}\mid g$; which contradicts minimality of the generating set $G(J)$. Similarly we can prove that $\mbox{\rm supp}(g)\cap\mbox{\rm supp}(h_{1})\neq\emptyset$. Now, $h_{1}h_{2}\mid\mathbf{m}={\rm\mbox{lcm}}(f,g)$ implies that $h_{2}\mid\mathbf{m}={\rm\mbox{lcm}}(f,g)$. Moreover, $f\mid{\rm\mbox{lcm}}(f,g)$. Therefore, ${\rm\mbox{lcm}}(f,h_{2})\mid{\rm\mbox{lcm}}(f,g)$. Similarly, ${\rm\mbox{lcm}}(g,h_{1})\mid{\rm\mbox{lcm}}(f,g)$. Now if ${\rm\mbox{lcm}}(f,h_{2})<\mathbf{m}$ or ${\rm\mbox{lcm}}(g,h_{1})<{\rm\mbox{lcm}}(f,g)=\mathbf{m}$, then we have a contradiction, since ${\rm\mbox{lcm}}(f,g)$ and ${\rm\mbox{lcm}}(g,h_{1})$ both are in the set $S$ and $\mathbf{m}$ is a minimal element in $S$. Therefore, we have must have ${\rm\mbox{lcm}}(f,h_{2})={\rm\mbox{lcm}}(g,h_{1})={\rm\mbox{lcm}}(f,g)=\mathbf{m}=h_{1}h_{2}w$ (say) and $\dfrac{fh_{2}}{\rm{gcd}(f,h_{2})}=\dfrac{gh_{1}}{\gcd(g,h_{1})}=h_{1}h_{2}w$, therefore $\dfrac{f}{\gcd(f,h_{2})}=h_{1}w$. Hence, $h_{1}\mid f$ and that contradicts minimality of the generating set $G(I)$. Hence, $S=\emptyset$ and we are done.

Conversely, let us assume that $\mbox{\rm supp}(G(I))\cap\mbox{\rm supp}(G(J))=\emptyset$. Without loss of generality, we can assume that $\mbox{\rm supp}(G(I))=\{1,2,\ldots,k\}$ and $\mbox{\rm supp}(G(J))=\{k+1,k+2,...,n\}$. Let $f\in I\cap J$, such that $f=\sum_{v\in Mon(R)}a_{v}v$. We have $v\in I\cap J$ for all $v\in m(f)$. It is therefore enough to show that if $m$ is a monomial and $m\in I\cap J$, then $m\in IJ$. Let $m\in I\cap J$; there exist $m_{I}\in G(I)$ and $m_{J}\in G(J)$ such that $m_{I}\mid m$ and $m_{J}\mid m$. Let $m_{I}m_{1}=m$; then $m_{J}\mid m=m_{I}m_{1}$. We know that $\mbox{\rm supp}(m_{I})\cap\mbox{\rm supp}(m_{J})=\emptyset$, therefore $m_{J}\mid m_{1}$ and it follows that $m_{1}=m_{J}m_{2}$ and $m=m_{I}m_{J}m_{2}$. Therefore, $m\in IJ$ since $m_{I}m_{J}\in IJ$. ∎

Theorem 2.3**.**

Let $I$ and $J$ be two ideals of $R$ such that w.r.t. some monomial order $\preceq$, $\mbox{\rm supp}(\rm{Lt}(I))\cap\mbox{\rm supp}(\rm{Lt}(J))=\emptyset$. Then $I$ and $J$ intersect transversally.

Proof.

Let $\mathcal{G}_{I}$ and $\mathcal{G}_{J}$ be Gröbner bases with respect to the monomial order $\preceq$ of the ideals $I$ and $J$ respectively. Again we assume that $f\in I\cap J$, then there exist polynomials $p\in\mathcal{G}_{I}$ and $q\in\mathcal{G}_{J}$ such that $\rm{Lt}(p)\mid\rm{Lt}(f)$ and $\rm{Lt}(q)\mid\rm{Lt}(f)$. Since $\mbox{\rm supp}(\rm{Lt}(I))\cap\mbox{\rm supp}(\rm{Lt}(J))=\emptyset$, we have $\rm{Lt}(p)\cdot\rm{Lt}(q)\mid\rm{Lt}(f)$. After division we write $f=pq+r$, then $r\in I\cap J$ and $\rm{Lt}(r)\preceq\rm{Lt}(f)$. we can apply same process on $r$, and after finite stage we get $f\in IJ$. ∎

3. The Taylor complex

We define the multidegree of a monomial $x^{\mathbf{a}}=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$ to be the $n$-tuple $(a_{1},\ldots,a_{n})$, denoted by $\text{mdeg}(\mathbf{x}^{\mathbf{a}})$. Hence $R=\oplus_{m\in{\rm Mon}(R)}Km$ has a direct sum decomposition as $K$-vector spaces $Km$, where $Km=\{cm\mid c\in K\}$.

Definition 3** (The Taylor complex).**

Let $M$ be a monomial ideal of $R$ minimally generated by the monomials $m_{1},m_{2},\cdots,m_{p}$. The Taylor complex for $M$ is given by $\mathbb{T}(M)_{\centerdot}$ as follows: $\mathbb{T}(M)_{i}$ is the free $R$-module generated by $e_{j_{1}}\wedge e_{j_{2}}\wedge\cdots\wedge e_{j_{i}}$, for all $1\leq j_{1}<\cdots<j_{i}\leq p$, where $\{e_{1},\ldots,e_{p}\}$ is the standard basis for the free $R$-module $R^{p}$. The differential $\delta(M)_{\cdot}$ is given by the following:

$\delta(M)_{i}(e_{j_{1}}\wedge\cdots\wedge e_{j_{i}})$

[TABLE]

We define ${\rm mdeg}(e_{j_{1}}\wedge\cdots\wedge e_{j_{i}})={\rm\mbox{mdeg}}({\rm\mbox{lcm}}(m_{j_{1}}\cdots m_{j_{i}}))$. The Taylor complex $(\mathbb{T}(M)_{\centerdot},\delta(M)_{\centerdot})$ defined above is in deed a chain complex of free $R$ modules and gives a free resolution for the $R$-module $R/M$; see section 26 in [2].

Theorem 3.1**.**

The Taylor complex $(\mathbb{T}(M)_{\centerdot},\delta(M)_{\centerdot})$ defined above is a free resolution (though not minimal) of the monomial ideal M; called the Taylor resolution of $M$.

Proof.

See Theorem 26.7 in [2].∎

Theorem 3.2**.**

Let $I$ and $J$ be two monomial ideals of $R$ intersecting transversally. Let $\mathbb{T}(I)_{\centerdot}$ and $\mathbb{T}(J)_{\centerdot}$ be the Taylor resolutions of $I$ and $J$ respectively. Then, the Taylor’s resolution of $I+J$ is isomorphic to the complex $\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}$ and hence $\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}$ is acyclic.

Proof.

We know that $\mathbb{T}(I)_{i}=R^{\binom{p}{i}}$ and $\mathbb{T}(J)_{j}=R^{\binom{q}{j}}$, where $p$ and $q$ denote the minimal number of generators of $I$ and $J$ respectively. Then,

[TABLE]

A basis of $(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}$ is given by $(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes(e^{\prime}_{k^{\prime}_{1}}\wedge\cdots\wedge e^{\prime}_{k^{\prime}_{j}})$, such that $i+j=r$ and $1\leq k_{1}<k_{2}<\cdots<k_{i}\leq p$ and $1\leq k^{\prime}_{1}<k^{\prime}_{2}<\cdots<k^{\prime}_{j}\leq q$. Now for each $r$, the free module $(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}$ can be graded with the help of mdeg as follows. We define

[TABLE]

This defines a multi-graded structure for the complex $\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}$.

Let $G(I)=\{m_{1},m_{2}\cdots,m_{p}\}$ and $G(J)=\{m^{\prime}_{1},m^{\prime}_{2}\cdots,m^{\prime}_{q}\}$ denote the minimal set of generators for $I$ and $J$ respectively. Then $I+J$ is minimally generated by $G(I)\cup G(J)$, by Theorem 2.2, since $G(I)$ and $G(J)$ are of disjoint support. We will now show that the tensor product complex $\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}$is isomorphic to the Taylor resolution $\mathbb{T}(I+J)_{\centerdot}$. We define $\psi_{r}:(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}\longrightarrow\left(\mathbb{T}(I+J)_{\centerdot}\right)_{r}$ as

[TABLE]

Moreover,

[TABLE]

since $G(I)$ and $G(J)$ are with disjoint supports. Hence, the map is a graded isomorphism between the free modules $(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}$ and $\left(\mathbb{T}(I+J)_{\centerdot}\right)_{r}$. We therefore have the following diagram of complexes:

\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{r}}$$\scriptstyle{\psi_{r}}$$\textstyle{(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{r-1}}$$\textstyle{\cdots}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\left(\mathbb{T}(I+J)_{\centerdot}\right)_{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta(I+J)_{r}}$$\textstyle{\left(\mathbb{T}(I+J)_{\centerdot}\right)_{r-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots}

We now show that $\delta_{r}\circ\psi_{r}=\psi_{r-1}\circ d_{r}$. Let us take an arbitrary basis element of $(\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot})_{r}$, which is of the form

[TABLE]

Then,

[TABLE]

Let us write

$Q_{t}=\dfrac{{\rm\mbox{lcm}}(m_{k_{1}},\ldots,m_{k_{i}})}{{\rm\mbox{lcm}}(m_{k_{1}},\ldots,\widehat{m_{k_{t}}},\ldots,m_{k_{i}})}$,   $1\leq t\leq i$;

$V_{t}=\dfrac{{\rm\mbox{lcm}}(m_{k_{1}},\ldots,m_{k_{i}},m^{\prime}_{k^{\prime}_{1}},\ldots,m^{\prime}_{k^{\prime}_{j}})}{{\rm\mbox{lcm}}(m_{k_{1}},\ldots,\widehat{m_{k_{t}}},\ldots,m_{k_{i}},m^{\prime}_{k^{\prime}_{1}},\ldots,m^{\prime}_{k^{\prime}_{j}})}$,   $1\leq t\leq i$;

We know that $G(I)$ and $G(J)$ have disjoint supports, therefore,

[TABLE]

Then

[TABLE]

and

[TABLE]

The last equality in the above expression follows from the fact that $G(I)$ and $G(J)$ have disjoint supports. Similarly, it can be proved that

[TABLE]

where

[TABLE]

Hence,

[TABLE]

We now introduce $i+j$ new symbols $w_{s}$, such that $w_{s}=m_{k_{s}}$ for $s\leq i$ and $w_{s}=m^{\prime}_{k^{\prime}_{s-i}}$ for $s>i$. We also introduce another set of $i+j$ symbols $E_{s}$, such $E_{s}=e_{k_{s}}$ for $s\leq i$ and $E_{s}=e^{\prime}_{k^{\prime}_{s-i}}$ for $s>i$. Hence, the expression after the last equality in (*) can be written in a compact form as

[TABLE]

Now

[TABLE]

Hence, the diagram is commutative.∎

Corollary 3.3**.**

Let $I$ and $J$ be ideals in $R$ such that $IJ=I\cap J$. Let $\mathbb{M}(I)_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ denote minimal free resolutions of $I$ and $J$ respectively. Then, $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$ is a minimal free resolution of the ideal $I+J$.

Proof.

The minimal free resolutions $\mathbb{M}(I)_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ are direct summands of the Taylor resolutions $\mathbb{T}(I)_{\centerdot}$ and $\mathbb{T}(J)_{\centerdot}$ respectively. It follows that $(\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot})$ is a direct summand of $\mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}=\mathbb{T}(I+J)_{\centerdot}$. Hence, $(\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot})$ is a free resolution of $I+J$ and it is minimal since $\mathbb{M}(I)_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ are both minimal.∎

4. A Simplicial characterization of transversal intersection of monomial ideals

We first introduce the basic definitions of a simplicial complex; see [1].

Definition 4**.**

A simplicial complex $\Delta$ on the vertex set $\{1,\ldots,m\}$ is a collection of subsets called faces or simplices, satisfying the following condition: $\sigma\in\Delta$ and $\tau\subset\sigma$ implies that $\tau\in\Delta$. A simplex $\sigma\in\Delta$ of cardinality $|\sigma|=r+1$ has dimension $r$ and it is called an $r$* face* of $\Delta$.

Definition 5**.**

A facet is a maximal simplex of a simplicial complex $\Delta$. Let $\Gamma(\Delta)$ denote the set of all facets of $\Delta$. Then, the vertex set of $\Delta$ and the set $\Gamma(\Delta)$ determine $\Delta$ completely.

Definition 6**.**

A standard simplicial complex of dimension $m-1$ on the vertex set $\{1,\ldots,m\}$ is the simplicial complex whose $\Gamma(\Delta)=\{\{1,\ldots,m\}\}$.

Definition 7**.**

Let $\Delta_{1}$ and $\Delta_{2}$ be simplicial complexes on disjoint vertex sets. Let the vertext set of $\Delta_{1}$ be $\{1,\ldots,m\}$ and that of $\Delta_{2}$ be $\{m+1,\ldots,m+p\}$. The join of simplicial complexes $\Delta_{1}$ and $\Delta_{2}$ is the simplicial complex $\Delta_{1}*\Delta_{2}$, whose vertex set is $\{1,\ldots,m+p\}$ and

[TABLE]

We now define a frame as a complex of $K$-vector spaces with a fixed basis, which encodes the minimal free resolution of a monomial ideal; see [2].

Definition 8**.**

An $r-$frame $\mathbb{U}_{\centerdot}$ is a complex of finite $K$ vector spaces with differential $\partial$ and a fixed basis satisfying the following:

1. (i)

$\mathbb{U}_{i}=0$ for $i<0$; 2. (ii)

$\mathbb{U}_{0}=K$; 3. (iii)

$\mathbb{U}_{1}=K^{r}$, with basis $\{w_{1},\ldots,w_{r}\}$; 4. (iv)

$\partial(w_{j})=1$, for all $j=1,\ldots,r$.

Construction of a frame of a simplicial complex. Let $\Delta$ be a simplicial complex on the vertex set $\{1,\ldots,m\}$. For each integer $i$, let $\Gamma_{i}(\Delta)$ be the set of all $i$ dimensional faces of $\Delta$ and let $K^{\Gamma_{i}(\Delta)}$ denote the $K$-vector space generated by the basis elements $e_{\sigma}$, for $\sigma\in\Gamma_{i}(\Delta)$. The chain complex of $\Delta$ over $K$ is the complex $\mathcal{C}(\Delta)_{\centerdot}$,

[TABLE]

The boundary maps $\partial_{i}$ are defined as

[TABLE]

such that $sign(j,\sigma)=(-1)^{t-1}$, where $j$ is the $t$-th element of the set $\sigma\subset\{1,\ldots,m\}$ written in increasing order. If $i<-1$ or $i>m-1$, then $K^{\Gamma_{i}(\Delta)}=0$ and $\partial_{i}=0$. We see that the chain complex of a simplicial complex is an $m$-frame.

If $\Delta$ is a standard simplicial complex of dimension $m-1$ on the vertex set $\{1,\ldots,m\}$, then $\mathcal{C}(\Delta)_{\centerdot}$ is acyclic and $K^{\Gamma_{i}(\Delta)}=K^{\binom{m}{i}}$. Therefore, the $m$-frame of a standard simplicial complex of dimension $m-1$ is nothing but the Koszul complex.

Definition 9**.**

Let $M=\{m_{1},m_{2},\cdots,m_{r}\}$ be a set of monomials in $R$. An $M-$complex $\mathbb{G}_{\centerdot}$ is a multigraded complex of finitely generated free multigraded $R$ modules with differential $\Delta$ and a fixed multihomogeneous basis with multidegrees that satisfy

1. (i)

$\mathbb{G}_{i}=0$   for   $i<0$; 2. (ii)

$\mathbb{G}_{0}=R$; 3. (iii)

$\mathbb{G}_{1}=R(m_{1})\oplus\cdots\oplus R(m_{r})$; 4. (iv)

$\Delta(w_{j})=m_{j}$ for each basis vector $w_{j}\in\mathbb{G}_{1}$.

Construction. Let $\mathbb{U}_{\centerdot}$ be an $r-$frame and $M=\{m_{1},m_{2},\cdots,m_{r}\}$ be a set of monomials in $R$. The $M-$homogenization of $\mathbb{U}_{\centerdot}$ is defined to be the complex $\mathbb{G}_{\centerdot}$ with $\mathbb{G}_{0}=R$ and $\mathbb{G}_{1}=R(m_{1})\oplus\cdots\oplus R(m_{r})$. Let $\bar{v_{1}},\cdots,\bar{v_{p}}$ and $\bar{u_{1}},\cdots,\bar{u_{q}}$ be the given bases of $\mathbb{U}_{i}$ and $\mathbb{U}_{i-1}$ respectively. Let $u_{1},u_{2},\cdots,u_{q}$ be the basis of $\mathbb{G}_{i-1}$ chosen in the previous step of induction. We take $v_{1},\cdots,v_{p}$ to be a basis of $\mathbb{G}_{i}$. If

[TABLE]

with coefficients $\alpha_{sj}\in k$, then set $\rm{mdeg}(v_{j})={\rm\mbox{lcm}}\left(mdeg(u_{s})|\alpha_{sj}\neq 0\right)$, where ${\rm\mbox{lcm}}(\emptyset)=1$. We define

[TABLE]

Definition 10**.**

Let $I\subset R$ be a monomial ideal with $G(I)=\{m_{1},\ldots,m_{r}\}$. We say that a minimal free resolution of $I$ is supported by a simplicial complex $\Delta$ on the vertex set $\{1,\ldots,r\}$ if it is isomorphic to the $G(I)$ homogenization of the chain complex $\mathcal{C}(\Delta)_{\centerdot}$ of $\Delta$.

Lemma 4.1**.**

Let $I$ and $J$ be two monomial ideals such that $I\cap J=IJ$, $|G(I)|=r$ and $|G(J)|=s$, and whose minimal free resolutions are supported by standard simplicial complexes $\Delta_{I}$ and $\Delta_{J}$ on the vertex sets $\{1,\ldots,r\}$ and $\{r+1,\ldots,r+s\}$ respectively. Then minimal free resolution of $I+J$ is supported by the standard simplicial complex $\Delta_{I}*\Delta_{J}$.

Proof.

We have seen that if $\Delta$ is a standard simplicial complex of dimension $m-1$ on the vertex set $\{1,\ldots,m\}$, then the $m$-frame of $\Delta$ is nothing but the Koszul complex. Therefore, it follows that if a minimal free resolution $\mathbb{M}(I)_{\centerdot}$ of $I$ is supported by the standard simplicial complex $\Delta_{I}$, then the it is actually isomorphic to the Taylor complex $\mathbb{T}(I)_{\centerdot}$. Since $I\cap J=IJ$, by the Theorem 2.2 we have $G(I)\cap G(J)=\emptyset$, then $|G(I)\cup G(J)|=r+s$. Again $\Delta_{I}*\Delta_{J}$ is also standard simplicial complex of dimension $r+s-1$. Therefore $G(I)\cup G(J)$-homogenization of $\mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$ is isomorphic to the Taylor complex $\mathbb{T}(I+J)_{\centerdot}$. Thus by Theorem 3.2, we have

[TABLE]

Again by the Corollary 3.3, $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$ is minimal free resolution of $I+J$. Therefore minimal free resolution of $I+J$ is supported by the standard simplicial complex $\Delta_{I}*\Delta_{J}$∎

Lemma 4.2**.**

Let $\Delta_{I}$ be a simplicial complex on vertex $\{1,\ldots,r\}$ and $\Delta_{J}$ be another simplicial complex on vertex $\{r+1,\ldots,r+s\}$,i.e., the vertex sets of $\Delta_{I}$ and $\Delta_{J}$ are disjoint. Then

[TABLE]

Proof.

Let $\Gamma(\Delta_{I})=\{\gamma_{1},\ldots,\gamma_{l_{1}}\}$ and $\Gamma(\Delta_{J})=\{\sigma_{1},\ldots,\sigma_{l_{2}}\}$. We have $\Gamma(\Delta_{I}*\Delta_{J})=\{\gamma_{i}\cup\sigma_{j}\mid 1\leq i\leq l_{1},1\leq j\leq l_{2}\}$. Now assuming $\binom{s}{t}=0$ for $s<t$, we have $\mid\Gamma_{p}(\Delta_{I})\mid=\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\mid}{p}$, for all $1\leq p\leq l_{1}$ and $\mid\Gamma_{q}(\Delta_{J})\mid=\sum_{j=1}^{l_{2}}\binom{\mid\sigma_{j}\mid}{q}$ for all $1\leq q\leq l_{2}$. On the other hand, $\mid\Gamma_{t}(\Delta_{I}*\Delta_{J})\mid=\sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\cup\sigma_{j}\mid}{t}=\sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\mid+\mid\sigma_{j}\mid}{t}=\sum_{p+q=t}\sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\mid}{p}\binom{\mid\sigma_{j}\mid}{q}$. Therefore the map

[TABLE]

defined by

[TABLE]

is an isomorphism, where $\gamma\in\Delta_{I},\mid\gamma\mid=p$ and $\sigma\in\Delta_{J},\mid\gamma\mid=q$ and $p+q=t$. We therefore have to show the following diagram commutes

{(\mathcal{C}(\Delta_{I})_{\cdot}\otimes\mathcal{C}(\Delta_{J})_{\cdot})_{t}}$${(\mathcal{C}(\Delta_{I})_{\cdot}\otimes\mathcal{C}(\Delta_{J})_{\cdot})_{t-1}}$${\mathcal{C}(\Delta_{I}*\Delta_{J})_{t}}$${\mathcal{C}(\Delta_{I}*\Delta_{J})_{t-1}}$$\scriptstyle{(\partial(\Delta_{I})\otimes\partial(\Delta_{J}))_{t}}$$\scriptstyle{\theta_{t}}$$\scriptstyle{\theta_{t-1}}$$\scriptstyle{(\partial(\Delta_{I}*\Delta_{J}))_{t}}

Let $\gamma\in\Delta_{I},\mid\gamma\mid=p$ and $\sigma\in\Delta_{J},\mid\gamma\mid=q$ such that $p+q=t$. Then,

[TABLE]

Theorem 4.3**.**

Let $I$ and $J$ be two monomial ideals such that $I\cap J=IJ$, $|G(I)|=r$ and $|G(J)|=s$, and their minimal free resolutions are supported by the simplicial complexes $\Delta_{I}$ and $\Delta_{J}$ on the vertex sets $\{1,\ldots,r\}$ and $\{r+1,\ldots,r+s\}$ respectively. Then minimal free resolution of $I+J$ is supported by the simplicial complex $\Delta_{I}*\Delta_{J}$.

Proof.

We have $I\cap J=IJ$, therefore $\mbox{\rm supp}(G(I))\cap\mbox{\rm supp}(G(J))=\emptyset$, by Theorem 2.2. Again, by Lemma 4.2,

[TABLE]

Since $\mathbb{M}(I)_{\centerdot}$ is the $G(I)$-homogenization of the complex $\mathcal{C}(\Delta_{I})_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ is the $G(J)$-homogenization of the complex $\mathcal{C}(\Delta_{J})_{\centerdot}$, we can proceed in the same way as lemma 4.2 to prove that $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$ is the $G(I)\cup G(J)$–homogenization of the complex $\mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$. Again by the Corollary 3.3, $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$ is minimal free resolution of $I+J$. Therefore, minimal free resolution of $I+J$ is supported by the simplicial complex $\Delta_{I}*\Delta_{J}$. ∎

Funding Sources

This research is supported by the research project EMR/2015/000776 sponsored by the SERB, Government of India.