1 Radical supports

Let n be a positive integer and \(m=(m_1,\dots , m_n)\in {{\mathbb {N}}}_{>0}^n\). Let

$$\begin{aligned} S(m)=K[x_{ij}\mid 1\le j\le n,\ 1\le i\le m_j] \end{aligned}$$

be a polynomial ring over a field K endowed with the standard \({{\mathbb {Z}}}^n\)-grading induced by setting \(\deg (x_{ij})=e_j\), where \(e_j\in {{\mathbb {Z}}}^n\) is the j-th standard basis vector. We will denote S(m) only by S if there is no danger of confusion.

Given \(A\subseteq [n]=\{1,2,\dots , n\}\) we will denote by \(S_A\) the \({{\mathbb {Z}}}^n\)-graded homogeneous component of degree \(\sum _{i\in A} e_i\) of S. An element \(f\in S_A\) will be said to have multidegree A.

Remark 1.1

Let \(\emptyset \ne A\subseteq [n]\) and let \(f\in S_A\). If \(f\ne 0\) then it cannot have multiple factors in its factorization into irreducible factors. Hence (f) is radical and the same obviously holds if \(f=0\).

More generally we have:

Lemma 1.2

Consider a collection \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) of non-empty subsets of [n] such that \(A_i\cap A_j=\emptyset\) for all \(i\ne j\). Then for every \(m\in {{\mathbb {N}}}_{>0}^n\), for every field K and for every choice of \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) the ideal \(I=(f_1,\dots , f_s)\) of S is radical.

Proof

We introduce a term order \(\sigma\) in S. Then the leading term of the nonzero \(f_i\)’s are pairwise coprime and squarefree. Hence \(f_1,\dots , f_s\) form a Gröbner basis with \({{\,\mathrm{in}}}(I)\) radical and I is radical as well. \(\square\)

For generalities on Gröbner bases and the transfer of properties under Gröbner deformation we refer the reader to [4] or, for a complete picture, to the forthcoming [5].

The simple assertion of Lemma 1.2 suggests the following definition.

Definition 1.3

A collection (repetitions are allowed) \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) of non-empty subsets of [n] is a radical support with respect to a field K if for every \(m=(m_1,\dots , m_n)\in {{\mathbb {N}}}_{>0}^n\) and for every choice of \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) the ideal \(I=(f_1,\dots , f_s)\) of S is radical. Furthermore we say that \({{\mathcal {A}}}\) is a radical support if it is a radical support for every field K.

Our goal is to provide a combinatorial characterization of radical supports. We stress that in Definition 1.3 the required property should hold for every \(m\in {{\mathbb {N}}}_{>0}^n\). For example if \(m=(1,1,\dots ,1)\) then \({{\mathbb {Z}}}^n\)-graded ideals of S are indeed monomial ideals and hence I is radical if - and only if - its minimal generators have squarefree degrees (independently of the field).

As we will see, the characterization of radical supports is related to the notion of Cartwright-Sturmfels ideals. This notion has its roots in the work of Bernstein, Boocher, Brion, Cartwright, Conca, Sturmfels, Villarreal and Zelevisky [1,2,3, 6, 7, 15, 16] among others. The theory of Cartwright-Sturmfels ideals has been developed in a series of papers by the authors of the present manuscript [8,9,10,11,12] and studied further by Conca and Welker in [13]. Here we quickly revise the definition and the main features of Cartwright-Sturmfels ideals referring the readers to [8,9,10,11,12] for more general statements and for the proofs.

2 Cartwright–Sturmfels ideals

We follow the notation of the previous section. The group \(G={{\,\mathrm{GL}\,}}_{m_1}(K)\times \cdots \times {{\,\mathrm{GL}\,}}_{m_n}(K)\) acts on S as the group of multigraded K-algebra automorphisms. Let \(B=B_{m_1}(K)\times \cdots \times B_{m_n}(K)\) be the Borel subgroup of G, consisting of the upper triangular matrices in G. Let I be a multigraded ideal of S and let \(\sigma\) be a term order on S with \(x_{ij}>x_{kj}\) for all j and all \(i<k\).

If K is infinite we may define, as in the standard \({{\mathbb {Z}}}\)-graded situation, the (multigraded) generic initial ideal \({{\,\mathrm{gin}}}(I)\) of I with respect to \(\sigma\) as \({{\,\mathrm{in}}}_\sigma (g(I))\) with g a “generic" element in G. It turns out that \({{\,\mathrm{gin}}}(I)\) is Borel fixed, i.e., fixed by the action of B.

Radical Borel-fixed ideals play an important role in the definition of Cartwright–Sturmfels ideals.

Definition 2.1

We let \({{\,\mathrm{B-rad}}}(S)\) be the set of radical monomial ideals J of S such that for every monomial \(f\in S\) and i, j such that \(x_{ij}f\) is a generator of J one has \(x_{kj}f\in J\) for all \(k<i\).

The ideals in \({{\,\mathrm{B-rad}}}(S)\) are Borel-fixed. Indeed, at least if K is infinite, \({{\,\mathrm{B-rad}}}(S)\) is the set of Borel-fixed radical ideals, hence the name.

Let M be a finitely generated \({{\mathbb {Z}}}^n\)-graded S-module and assume for simplicity that \(M_a=0\) if \(a\not \in {{\mathbb {N}}}^n\).

The multigraded Hilbert series \(\mathrm{HS}_M(z_1,\dots ,z_n)\) of M has a rational expression

$$\begin{aligned} {{\,\mathrm{HS}}}_M(z_1,\dots ,z_n)=\frac{{{\mathbb {K}}}_M(z_1,\dots ,z_n)}{\prod _{i=1}^n(1-z_i)^{m_i}}. \end{aligned}$$

Here \({{\mathbb {K}}}_M(z_1,\dots ,z_n)\) is a polynomial with coefficients in \({{\mathbb {Z}}}\) known as the K-polynomial of M. The dual K-polynomial \({{\mathbb {K}}}^*_M(z_1,\dots ,z_n)\) of M is defined as

$$\begin{aligned} {{\mathbb {K}}}^*_M(z_1,\dots ,z_n)={{\mathbb {K}}}_M(1-z_1,\dots ,1-z_n). \end{aligned}$$

If n is clear from the context we will use \({{\,\mathrm{HS}}}_M(z)\) for \({{\,\mathrm{HS}}}_M(z_1,\dots ,z_n)\) and similarly will use \({{\mathbb {K}}}_M(z)\) and \({{\mathbb {K}}}^*_M(z)\).

Definition 2.2

Let I be a multigraded ideal of S. Then I is Cartwright-Sturmfels if there exists \(J\in {{\,\mathrm{B-rad}}}(S)\) such that \({{\,\mathrm{HS}}}_{S/I}(z)={{\,\mathrm{HS}}}_{S/J}(z)\).

We denote by \({{\,\mathrm{CS}}}(S)\) the family of Cartwright-Sturmfels ideals of S. It turns out that the ideal J that appears in Definition 2.2 is uniquely determined by (the multigraded Hilbert series of) I, see [8, Theorem 3.5]. This leads to the following characterization.

Proposition 2.3

Assume that K is infinite. Then I is Cartwright-Sturmfels if and only if \({{\,\mathrm{gin}}}(I) \in {{\,\mathrm{B-rad}}}(S)\).

For every \(j\in [n]\) we rename \(y_j\) the variable \(x_{1j}\) and let \(T=K[y_1,\dots , y_n] \subseteq S\) with induced (fine) \({{\mathbb {Z}}}^n\)-graded structure, i.e. \(\deg y_{j}=e_j\in {{\mathbb {Z}}}^n\). Let \({{\mathcal {M}}}(T,m)\) be the set of the monomial ideals of T generated by monomials in the y’s, whose exponent vector is bounded from above by \(m=(m_1,\dots ,m_n)\).

The two sets \({{\,\mathrm{B-rad}}}(S)\) and \({{\mathcal {M}}}(T,m)\) of monomial ideals are in one-to-one correspondence via the map

  1. (1)

    \(\psi : {{\mathcal {M}}}(T,m) \rightarrow {{\,\mathrm{B-rad}}}(S)\) that sends \(E \in {{\mathcal {M}}}(T,m)\) to \(\psi (E)=J={{\,\mathrm{pol}}}(E)^*\). Here \({{\,\mathrm{pol}}}(E)\) is the polarization of E and the star \(^*\) denotes the Alexander dual.

When K is infinite, the inverse of \(\psi\) can be defined as follows.

  1. (2)

    \(\phi :{{\,\mathrm{B-rad}}}(S) \rightarrow {{\mathcal {M}}}(T,m)\) sends \(J\in {{\,\mathrm{B-rad}}}(S)\) to \(\phi (J)=E\), where \(E \in {{\mathcal {M}}}(T,m)\) is uniquely determined by the property \({{\,\mathrm{gin}}}(J^*)=ES\). Here \(J^*\) is the Alexander dual of J and ES is the extension of E to S.

When K is finite, one can take an infinite extension F of K. The map \(\phi\) is defined on F and \(\phi (J)=E\) is an ideal generated by monic monomials. The same monomials generate the ideal \(\psi ^{-1}(J)\in {{\mathcal {M}}}(T,m)\).

Proposition 2.4

Let K be an arbitrary field. Given \(J\in {{\,\mathrm{B-rad}}}(S)\) and \(E\in {{\mathcal {M}}}(T,m)\) one has that J corresponds to E in the bijective correspondence above if and only if \({{\mathbb {K}}}^*_{S/J}(z)={{\mathbb {K}}}_E(z)\).

Furthermore we have

Lemma 2.5

Let \(m=(m_1,\dots ,m_n), q=(q_1,\dots ,q_n)\in {{\mathbb {N}}}_{>0}^n\). Let K and L be fields, let \(I\subseteq S=K[x_{ij} : j\in [n] \text{ and } i\le m_j]\) and \(J\subseteq R=L[x_{ij} : j\in [n] \text{ and } i\le q_j]\) be multigraded ideals such that \({{\mathbb {K}}}_{S/I}(z)={{\mathbb {K}}}_{R/J}(z)\). Then \(I\in {{\mathrm{CS}}}(S)\) if and only if \(J\in {{\mathrm{CS}}}(R){.}\)

Proof

Assume \(I\in {{\mathrm{CS}}}(S)\). Then there exists \(H\in {{\mathrm{B-rad}}}(S)\) such that I and H have the same Hilbert series, in particular \({{\mathbb {K}}}_{S/I}(z)={{\mathbb {K}}}_{S/H}(z)\). Then set \(r=(r_1,\dots , r_n)\) with \(r_j= \max \{m_j,q_j\}\) for all \(j\in [n]\) and \(T=L[x_{ij} : j\in [n] \text{ and } i\le r_j]\). Then we have the inclusion \(R\subseteq T\). We may consider the ideal \(H'\) of T generated by the monomial generators of H. Since \(H\in {{\mathrm{B-rad}}}(S)\) we have \(H'\in {{\mathrm{B-rad}}}(T)\). We can consider the extensions JT of J. Obviously the K-polynomial does not change under a polynomial extension and under the passage from H to \(H'\). Hence

$$\begin{aligned} {{\mathbb {K}}}_{T/JT}(z)={{\mathbb {K}}}_{R/J}(z)={{\mathbb {K}}}_{S/I}(z)={{\mathbb {K}}}_{S/H}(z)={{\mathbb {K}}}_{T/H'}(z). \end{aligned}$$

Hence JT has the same Hilbert series of \(H'\), that is, \(JT \in {{\mathrm{CS}}}(T)\). It remains to prove that \(JT \in {{\mathrm{CS}}}(T)\) implies \(J\in {{\mathrm{CS}}}(R)\). First assume that L is infinite. The computation of generic initial ideals commute with polynomial extensions, i.e. \({{\,\mathrm{gin}}}(JT)={{\mathrm{gin}}}(J)T\). Since \({{\,\mathrm{gin}\,}}(JT)\in {{\mathrm{B-rad}}}(T)\) we have that \({{\,\mathrm{gin}}}(J)\in {{\mathrm{B-rad}}}(R)\) and hence \(J\in {{\mathrm{CS}}}(R)\). If L is finite then we can consider an infinite extension of L, compute the gin of J in with coefficients in the extension and consider the outcome in the original polynomial ring R. In this way, repeating the argument above, we obtain an ideal in \({{\,\mathrm{B-rad}}}(R)\) with the Hilbert series of J. Hence \(J\in {{\mathrm{CS}}}(R)\) in this case as well. \(\square\)

Cartwright-Sturmfels ideals remains Cartwright-Sturmfels under arbitrary \({{\mathbb {Z}}}^n\)-graded linear section. This is proved in [9] under the assumption that the base field is infinite but the result holds in general as the reader can easily check.

Proposition 2.6

Let L be an ideal of S that is generated by \({{\mathbb {Z}}}^n\)-graded linear forms. We identify \(R=S/(L)\) with a polynomial ring with the induced \({{\mathbb {Z}}}^n\)-graded structure and \(J=I+(L)/(L)\) with an ideal of R. If \(I\in {{\mathrm{CS}}}(S)\), then \(J\in {{\mathrm{CS}}}(R)\).

We conclude with a definition:

Definition 2.7

A collection (repetitions are allowed) \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) of non-empty subsets of [n] is a Cartwright-Sturmfels support if for every \(m=(m_1,\dots , m_n)\in {{\mathbb {N}}}_{>0}^n\) and for every choice of \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) the ideal \(I=(f_1,\dots , f_s)\) of S is Cartwright-Sturmfels.

3 Support for regular sequences and the graph associated to the support

For later applications, we establish some simple facts.

Lemma 3.1

Let \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) be a collection of non-empty subsets of [n]. The following are equivalent:

  1. (1)

    There exist \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) such that \(f_1,\dots , f_s\) is a regular sequence and \((f_1,\dots , f_s)\) is radical.

  2. (2)

    There exist \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) such that \(f_1,\dots , f_s\) is a regular sequence.

  3. (3)

    For every \(j\in [n]\) we have \(|\{ v : j\in A_v\}|\le m_j\).

Proof

That condition (1) implies condition (2) is obvious. For (2) implies (3) suppose there exist \(f_1\in S_{A_1}, \dots , f_s\in S_{A_s}\) such \(f_1,\dots , f_s\) is a regular sequence. For a given \(j\in [n]\) the ideal \(( f_v : j\in A_v)\) is generated by regular sequence and contained in the ideal generated by \(S_{e_j}\). Hence we have that \(|\{ v : j\in A_v\}|\le m_j\). Finally for (3) implies (1) for every \(v\in [s]\) and every \(j\in A_v\) we set

$$\begin{aligned} k_{j,v}=|\{ w: w\le v \text{ and } j\in A_w\}|. \end{aligned}$$

By assumption \(|\{ v : j\in A_v\}|\le m_j\) and hence \(k_{j,v}\le m_j\). Therefore we may set

$$\begin{aligned} f_v=\prod _{j \in A_v} x_{k_{j,v},j} \in S_{A_v} \end{aligned}$$

for every \(v\in [s]\). Notice that \(k_{j,v}\ge 1\), since \(j\in A_v\). Moreover, \(x_{k_{j,v},j}=x_{k_{i,w},i}\) implies \(i=j\) and if \(v>w\), then \(k_{j,v}>k_{j,w}\). Hence \(f_1,\dots , f_s\) is a set of pairwise coprime and squarefree monomials. In other words, \(f_1,\dots , f_s\) is a (monomial) regular sequence that generates a radical ideal. \(\square\)

Lemma 3.2

Suppose \(I=(f_1,\dots , f_c)\) is a non-radical ideal of a ring R. Then for every \(1\le t\le c\) the ideal \(J=(f_1,\dots , f_{t-1}, xf_{t}, xf_{t+1}, \dots , xf_c)\) of the polynomial extension R[x] is not radical.

Proof

Consider on R[x] the graded structure associated to the assignment \(\deg x=1\) and \(\deg a=0\) for all \(a\in R\). Then J is homogeneous and its homogeneous components \(J_i\) with \(i>0\) are all equal to I. Let \(a\in R\) be such that \(a\not \in I\) and \(a^2\in I\). Then \(ax\not \in J\) and \((ax)^2 \in J\) so that J is not radical. \(\square\)

Lemma 3.3

If \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) is a radical support with respect to a field K then every subcollection of \({{\mathcal {A}}}\) is a radical support with respect to K.

Proof

It suffices to observe that in Definition 1.3 some of the polynomials \(f_i\in S_{A_i}\) can be taken equal to 0. \(\square\)

Given a collection \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) of subsets of [n] we define a non-oriented graph \(G({{\mathcal {A}}})\) with possibly multiple edges and no loops as follows. The vertices of \(G({{\mathcal {A}}})\) are labelled by elements in \([s]=\{1,2,\dots ,s\}\) and we put an edge labelled with \(j\in [n]\) between vertex v and vertex w if \(j\in A_v\cap A_w\). This will be denoted by

figure a

A cycle in \(G({{\mathcal {A}}})\) is a sequence of vertices and edges of \(G({{\mathcal {A}}})\)

figure b

with \(v_1,\dots , v_k\) distinct.

A k-cycle is a cycle which involves k edges and k distinct vertices.

We are interested in cycles with non-constant edge labels, i.e., cycles with edge labels \(j_1,\dots , j_k\) such that \(j_a\ne j_b\) for at least a pair a, b in [k]. We observe the following.

Lemma 3.4

If \(G({{\mathcal {A}}})\) has a cycle with non-constant edge labels, then it has a cycle where all the edge labels are distinct.

Proof

By assumption there is a cycle, say a k-cycle, with non-constant edge labels, call it C. If \(k=2\) then C has distinct edge labels and we are done. So we may assume \(k>2\). We may also assume that C has some repeated edge labels (otherwise, we are done). Say the edge label 1 appears more than once in C. Since the edge labels are not constant, the edge label 1 must be adjacent at least once to a different edge label, say 2. Up to a “rotation" we may hence assume that C looks like this:

figure c

We know that there is at least another label equal to 1, say from \(v_p\) to \(v_{p+1}\), so that

figure d

with \(3\le p\le k\) and \(p+1=1\) if \(p=k\). Then \(1\in A_{v_2} \cap A_{v_p}\), hence the edge

figure e

is in \(G({{\mathcal {A}}})\). Therefore we have a shorter cycle in \(G({{\mathcal {A}}})\)

figure f

with non-constant edge labels. Iterating the process, we produce a cycle where all the edge labels are distinct. \(\square\)

Lemma 3.5

Let \(p\in {{\mathbb {N}}}\) with \(p\ge 2\) and K be any field. Set \(S=K[x_1,\dots ,x_p, y_1,\dots , y_p]\) with \({{\mathbb {Z}}}^p\)-graded structure given by \(\deg x_i=\deg y_i=e_i\) for every \(i=1,\dots , p\). Let

$$\begin{aligned} I=(x_{i+1}y_i-x_iy_{i+1} : i=1,2,\dots , p-1)+(y_1y_p). \end{aligned}$$

Then I is not radical. In particular,

$$\begin{aligned} {{\mathcal {A}}}=\{ \{1,2\}, \{2,3\}, \{3,4\},\dots , \{p-1,p\}, \{1,p\} \} \end{aligned}$$

is not a radical support with respect to K.

Proof

A straightforward application of Buchberger’s Algorithm shows that the reduced Gröbner basis of I with respect to the degree reverse lexicographic term order induced by the total order on the variables

$$\begin{aligned} x_1>\dots>x_p>y_1>\dots >y_p \end{aligned}$$

is obtained by adding to the given generators the monomials

$$\begin{aligned} (x_1y_p) y_2, (x_1y_p) x_2y_3, (x_1y_p) x_2x_3y_4, \dots , (x_1y_p) x_2x_3\cdots x_{p-1}y_p. \end{aligned}$$

Therefore \((x_1y_p) x_2x_3\cdots x_{p-1}y_p\in I\) and \(x_1 x_2x_3\cdots x_{p-1}y_p \not \in I\). This shows that I is not radical, hence \({{\mathcal {A}}}\) is not a radical support with respect to K. \(\square\)

4 The main result

We are ready to formulate and prove our main result.

Theorem 4.1

Given a collection \({{\mathcal {A}}}=\{ A_1,\dots , A_s\}\) of non-empty subsets of [n] we have that the following conditions are equivalent.

  1. (1)

    \({{\mathcal {A}}}\) is a radical support.

  2. (2)

    \({{\mathcal {A}}}\) is a radical support for at least one field K.

  3. (3)

    The graph \(G({{\mathcal {A}}})\) has no cycles with distinct edge labels.

  4. (4)

    The graph \(G({{\mathcal {A}}})\) has no cycles with non-constant edge labels.

  5. (5)

    There exist a field K, \(m=(m_1,\dots ,m_n)\) and a regular sequence \(f_1,\dots , f_s\) of degrees \({{\mathcal {A}}}\) in S such that the ideal \((f_1,\dots , f_s)\) is Cartwright-Sturmfels.

  6. (6)

    For every field K, every \(m=(m_1,\dots ,m_n)\) such that \(m_j\ge |\{v\in [s] : j\in A_v\}|\) for all j, and every \(f_1,\dots , f_s\) regular sequence of degrees \({{\mathcal {A}}}\) in S the ideal \((f_1,\dots , f_s)\) is Cartwright-Sturmfels.

  7. (7)

    \({{\mathcal {A}}}\) is a Cartwright-Sturmfels support.

Proof

It is clear that (1) implies (2). We prove that (2) implies (3) by contradiction. That is, we assume that (2) holds and that \(G({{\mathcal {A}}})\) has a cycle with distinct edge labels and derive a contradiction. Renaming vertices and edges if needed, we may hence assume that \(G({{\mathcal {A}}})\) contains the p-cycle

figure g

with \(p>1\). By 3.3 the subcollection \(\{A_1,\dots , A_p\}\) is a radical support for K. Notice that by construction

$$\begin{aligned} A_1\supseteq \{1,2\}, \ A_2\supseteq \{2,3\}, \ \dots , A_{p-1} \supseteq \{p-1,p\} \text{ and } A_p\supseteq \{1,p\}. \end{aligned}$$
(4.1)

If in (4.1) we have all equalities then by  3.5 we have polynomials \(f_1,f_2,\dots , f_p\) of degree \(\{1,2\}, \{2,3\}, \dots , \{p-1,p\}, \{1,p\}\) in a multigraded polynomial ring over K that generate a non-radical ideal and hence we have a contradiction. If instead some inclusions in (4.1) are strict we still consider the polynomials \(f_1,\dots , f_p\) above and set

$$\begin{aligned} B= (\cup _{i=1}^p A_i )\setminus \{1,\dots , p\}. \end{aligned}$$

For every \(u\in B\) we pick a new variable \(x_u\) of degree \(\{u\}\) and set

$$\begin{aligned} f_i'=f_i (\prod _{u\in B\cap A_i} x_u). \end{aligned}$$

Now by the iterated application of  3.2 we know that the polynomials \(f_1',\dots , f_p'\) generate a non-radical ideal and have degrees \(A_1,\dots , A_p\) respectively. A contradiction.

It follows from 3.4 that (3) implies (4).

Equivalence of (5) and (6) follows from 2.5, since the K-polynomial of a regular sequence generated by elements of degree \({{\mathcal {A}}}\) is

$$\begin{aligned} {{\mathbb {K}}}_{{\mathcal {A}}}(z)=\prod _{v=1}^s (1-\prod _{j\in A_v} z_j) \in {{\mathbb {Z}}}[z_1,\dots ,z_n], \end{aligned}$$
(4.2)

hence it depends only on \({{\mathcal {A}}}\).

Now we prove that (4) implies (5). For \(j\in [n]\), let \(m_j=|\{v\in [s] : j\in A_v\}|\). Let K be any field. By 3.1 there exists a regular sequence \(f_1,\dots , f_s\) in S(m) of degrees \({{\mathcal {A}}}\). Set \(I=(f_1,\dots ,f_s)\). Then the K-polynomial of S/I is as in 4.2. In order to prove I is a Cartwright-Sturmfels ideal, by 2.4, it is suffices to exhibit a monomial ideal E in the polynomial ring \(T=K[y_{1},y_{2},\dots , y_{n}]\) equipped with the (fine) \({{\mathbb {Z}}}^n\)-grading \(\deg y_i=e_i \in {{\mathbb {Z}}}^n\) such that:

  1. (i)

    the K-polynomial of E is \({{\mathbb {K}}}^*_{{\mathcal {A}}}(z)\)

  2. (ii)

    for every j, the largest exponent of \(y_j\) in the generators of E is bounded from above by \(m_j\).

We have

$$\begin{aligned} {{\mathbb {K}}}^*_{{\mathcal {A}}}(z)={{\mathbb {K}}}_{{\mathcal {A}}}(1-z_1,\dots , 1-z_n)= \prod _{v=1}^s G_v(z_1,\dots , z_n) \end{aligned}$$

with

$$\begin{aligned} G_v(z_1,\dots , z_n)=(1-\prod _{j\in A_v} (1-z_j) )=\sum _{\emptyset \ne B\subseteq A_v } (-1)^{|B|+1} \prod _{j\in B} z_j. \end{aligned}$$

Notice that the ideal \(E_v=( y_j : j\in A_v)\) is resolved by the truncated Koszul complex \({{\mathcal {K}}}^*(E_v)\) associated to the variables \(y_j\) with \(j\in A_v\) and hence its \({{\mathbb {Z}}}^n\)-graded Hilbert series is

$$\begin{aligned} \frac{G_v(z_1,\dots , z_n)}{\prod _{i=1}^n (1-z_i)}. \end{aligned}$$

Indeed the full Koszul complex \({{\mathcal {K}}}(E_v)\) resolves \(T/E_v\) and we take the truncated Koszul complex obtained from \({{\mathcal {K}}}(E_v)\) by removing the component in homological position 0 and shifting the remaining homological positions by 1 to get a T-resolution of the ideal \(E_v\).

We claim that, under the assumption on \(G({{\mathcal {A}}})\) from (4), the ideal

$$\begin{aligned} E=\prod _{v=1}^s E_v \end{aligned}$$

has the properties (i) and (ii). Indeed, (ii) is satisfied by construction. We now prove (i).

The minimal free resolution of the product of any collection of ideals \(I_1,\dots , I_v\) generated by linear forms is described in [14]. It is proved that such a resolution is obtained as a subcomplex, supported on a specific polymatroid, of the tensor product of the truncated Koszul complexes \({{\mathcal {K}}}^*(I_j)\). We notice that assumption (4) on \(G({{\mathcal {A}}})\) is equivalent to the assumption that the ideal E has \(\prod _{v=1}^s |A_v|\) generators. Then the results in [14] show that the minimal free resolution of E is the tensor product \({{\mathcal {K}}}^*(E_1)\otimes {{\mathcal {K}}}^*(E_2) \otimes \cdots \otimes {{\mathcal {K}}}^*(E_v)\). This implies (i) and concludes the proof that (4) implies (5).

Since every Cartwright–Sturmfels ideal is radical, then (7) implies (1).

It remains to prove that (5) implies (7). Let K be any field, \(S=K[x_{ij}\mid 1\le j\le n,\ 1\le i\le m_j]\) be a \({{\mathbb {Z}}}^n\)-graded polynomial ring over K and I an ideal generated by elements \(f_1,\dots , f_s\) of degrees \(A_1,\dots , A_s\). For \(i=1,\dots , s\) we introduce new variables \(t_{ij}\) for every pair \(i\in [s]\) and \(j\in [n]\) such that \(j\in A_i\), with \(\deg t_{ij}=e_j\in {{\mathbb {Z}}}^n\). Then in the \({{\mathbb {Z}}}^n\)-graded polynomial ring

$$\begin{aligned} R=S[t_{ij} : i\in [s] \ \ j\in A_i] \end{aligned}$$

we consider the polynomials

$$\begin{aligned} g_i=f_i+\prod _{j\in A_i} t_{ij} \end{aligned}$$

of degree \(A_i\) and observe that the \(g_1,\dots , g_s\) form a regular sequence. Indeed, the leading term of \(g_i\) with respect to the lexicographic order with the t’s larger than the x’s is \(\prod _{j\in A_i} t_{ij}\). Since the leading terms of the \(g_1,\dots , g_s\) are pairwise coprime, we have that they form a Gröbner basis and \(g_1,\dots , g_s\) form a regular sequence of elements of degree \({{\mathcal {A}}}\) is the \({{\mathbb {Z}}}^n\)-graded polynomial ring R. Hence \((g_1,\dots , g_s)\in {{\mathrm{CS}}}(R)\) by assumption. The \(g_i\)’s specialize to to the \(f_i\)’s modulo the multigraded ideal of linear forms \((t_{ij} )\). Hence \(I\in {{\mathrm{CS}}}(S)\) by 2.6. \(\square\)