Proof Assume that $R = [0, r]$ is relevant to $\mathcal {F}$ and choose $a \in \mathbb {Z}_+^q$ such that $a_i^{} \ge r_i^{}$ . Equation (2) can be rewritten as

$$\begin{align*}\mathcal{F}(a) = \sum_{0 \le b_1 \le a_1} \!\!\! \cdots \!\!\! \sum_{0 \le b_i \le r_i} \!\!\! \cdots \!\!\! \sum_{0 \le b_q \le a_q} \!\!\! \operatorname{\Delta} \mathcal{F}(b) + \sum_{0 \le b_1 \le a_1} \!\!\! \cdots \!\!\! \sum_{r_i < b_i \le a_i} \!\!\! \cdots \!\!\! \sum_{0 \le b_q \le a_q} \!\!\! \operatorname{\Delta} \mathcal{F}(b). \end{align*}$$

The first summation equals $\mathcal {F}(a_1^{}, \dots , r_i^{}, \dots , a_q^{})$ , again by virtue of Equation (2). The second summation vanishes because $\operatorname {\Delta } \mathcal {F}(b) = 0$ if b lies outside R.

Conversely, assume that for every index $i \in \{1, \dots , q \}$ and for every $a \in \mathbb {Z}^q$ such that $a_i^{} \ge r_i^{}$ , we have the equation $\mathcal {F}(a) = \mathcal {F}(a_1^{}, \dots , r_i^{}, \dots , a_q^{})$ . This is equivalent to saying that for every index $i \in \{1, \dots , q \}$ and for every $a \in \mathbb {Z}^q$ such that $a_i^{}> r_i^{}$ , we have the equation $\mathcal {F}(a) = \mathcal {F}(a - e_i^{})$ . Choose $a \in \mathbb {Z}_+^q \setminus R$ . There is an index i such that $a_i^{}> r_i^{}$ . Equation (1) can be rewritten in the form

$$ \begin{align*} \operatorname{\Delta} \mathcal{F}(a) &= \big(\mathcal{F}(a) - \mathcal{F}(a - e_i^{}) \big) \\ & + \sum_{1 \le p \le q - 1} (-1)^p \!\!\! \sum_{\substack{1 \le j_1 < \dots < j_p \le q \\ j_1, \dots, j_p \neq i}} \big(\mathcal{F}(a - e_{j_1}^{} - \dots - e_{j_p}^{}) - \mathcal{F}(a - e_i^{} - e_{j_1}^{} - \dots - e_{j_p}^{}) \big). \end{align*} $$

All terms in parentheses vanish, hence, $\operatorname {\Delta } \mathcal {F}(a) = 0$ . Thus, R is relevant to $\mathcal {F}$ .

Proof Applying Lemma 2.2 repeatedly, we obtain the equations

$$\begin{align*}\mathcal{F}(a) = \mathcal{F}(r_1^{}, a_2^{}, \dots, a_q^{}) = \mathcal{F}(r_1{}, r_2^{}, a_3^{}, \dots, a_q^{}) = \dots = \mathcal{F}(r).\\[-34pt] \end{align*}$$

Proof Let us write $R = [0, r]$ and $r = (r_1^{}, \dots , r_q^{})$ . Given $b \in B_R^{}$ , there is an integer $p = p_b^{} \in \{ 1, \dots , q \}$ and there are indices $1 \le i_1^{} < \dots < i_p^{} \le q$ such that $b_i^{} < r_i^{}$ for all $i \in \{ 1, \dots , q \} \setminus \{ i_1^{}, \dots , i_p^{} \}$ and $b_{i_1}^{} = r_{i_1}^{}$ , $\dots $ , $b_{i_p}^{} = r_{i_p}^{}$ . We consider the set

$$\begin{align*}A(b) = \{ a \in \mathbb{Z}_+^q \mid a_{i_1}^{} \ge r_{i_1}^{}, \dots, a_{i_p}^{} \ge r_{i_p}^{}, \, a_i^{} = b_i^{} \ \text{for} \ i \in \{ 1, \dots, q \} \setminus \{ i_1^{}, \dots, i_p^{} \} \}. \end{align*}$$

According to Lemma 2.2, $\mathcal {F}(a) = \mathcal {F}(b)$ for all $a \in A(b)$ . For $d \ge | b |,$ consider the set

$$\begin{align*}A_d^{}(b) = \{ a \in A(b) \mid | a | = d \}. \end{align*}$$

We now recall the fact that, for a fixed nonnegative integer s, the number of integer solutions to the equation $c_1^{} + \dots + c_p^{} = s$ , with unknowns $c_i^{} \ge 0$ , is ${s + p - 1 \choose p - 1}$ . For $a \in A_d^{}(b),$ we have the equation $(a_{i_1}^{} - r_{i_1}^{}) + \dots + (a_{i_p}^{} - r_{i_p}^{}) = d - | b |$ , hence

$$\begin{align*}| A_d^{}(b) | = {d - | b | + p - 1 \choose p - 1}. \end{align*}$$

Assume now that $d \ge | r |$ . The decomposition $\{ a \in \mathbb {Z}_+^q \mid | a | = d \} = \bigsqcup _{b \in B_R} A_d^{}(b)$ leads us to the following expression for $\overline {\mathcal {F}}(d)$ :

$$\begin{align*}\sum_{\substack{a \in \mathbb{Z}_+^q \\ | a | = d}} \mathcal{F}(a) = \sum_{b \in B_R} \sum_{a \in A_d(b)} \!\! \mathcal{F}(a) = \sum_{b \in B_R} \mathcal{F}(b) | A_d^{}(b) | = \sum_{b \in B_R} \mathcal{F}(b) {d - | b | + p_b^{} - 1 \choose p_b^{} - 1}. \end{align*}$$

The r.h.s. is a polynomial function in d with rational coefficients. The assertion about the dominant term follows from the fact that $p_r^{} = q$ while $p_b^{} < q$ for $b \in B_R^{} \setminus \{ r \}$ .

Proof We adopt the notation from the proof of Lemma 2.5. For $d \ge \tau ,$ we denote

$$\begin{align*}C_d^{} = \{ a \in \mathbb{Z}_+^q \mid | a | = d \} \setminus \bigcup_{b \in B_R \setminus T} A_d^{}(b). \end{align*}$$

By hypothesis, $\mathcal {F}(a) = \mathcal {F}(r)$ if $a \in T$ . According to Lemma 2.2, $\mathcal {F}(a) = \mathcal {F}(r)$ if a lies in $A(b)$ for some $b \in B_R^{} \cap T$ . Thus, $\mathcal {F}(a) = \mathcal {F}(r)$ if $a \in C_d^{}$ . For $d \ge \tau ,$ we calculate

$$ \begin{align*} \overline{\mathcal{F}}(d) & = \sum_{a \in \mathbb{Z}_+^q, | a | = d} \mathcal{F}(a) \\ & = \sum_{a \in C_d} \mathcal{F}(a) + \sum_{b \in B_R \setminus T} \sum_{a \in A_d(b)} \mathcal{F}(a) \\ & = \mathcal{F}(r) | C_d^{} | + \sum_{b \in B_R \setminus T} \mathcal{F}(b) | A_d^{}(b) | \\ & = \mathcal{F}(r) \left[ {d + q - 1 \choose q - 1} - \sum_{b \in B_R \setminus T} {d - | b | + p_b^{} - 1 \choose p_b^{} - 1} \right] \\ & \quad + \sum_{b \in B_R \setminus T} \mathcal{F}(b) {d - | b | + p_b^{} - 1 \choose p_b^{} - 1}. \end{align*} $$

This is a polynomial expression in the variable d, which, in view of Lemma 2.5, must coincide with $\mathcal {P}(d)$ .

Proof It is well-known that $\operatorname {H}^1(\mathcal {I}_X^{}(a))$ vanishes if $a_1^{}, \dots , a_q^{}$ are sufficiently large (see [Reference Hartshorne19, Theorem III.5.2]). Consequently, in view of Equation (3), $\operatorname {\mathcal {H}}_X^{}(a) = \operatorname {length}(X)$ if $a_1^{}, \dots , a_q^{}$ are sufficiently large. We saw in Lemma 2.3 that $\operatorname {\mathcal {H}}_X^{}$ is constant on $[r, \infty )$ . We conclude that $\operatorname {\mathcal {H}}_X^{}$ takes the value $\operatorname {length}(X)$ on $[r, \infty )$ .

Proof We denote by $\{ \epsilon _1^{}, \dots , \epsilon _q^{} \}$ the standard basis of the $\mathbb {K}$ -vector space $E = \mathbb {K}^q$ . For each $k \in \{ 1, \dots , q \},$ we endow $\wedge ^k E \otimes _{\mathbb {K}}^{} M$ with a $\mathbb {Z}^q$ -grading as follows: if $h \in M$ is $\mathbb {Z}^q$ -homogeneous and $1 \le i_1^{} < \dots < i_k^{} \le q$ , then

$$\begin{align*}\deg(\epsilon_{i_1}^{} \wedge \dots \wedge \epsilon_{i_k}^{} \otimes h) = \deg(h) + \!\!\!\!\! \sum_{i \in \{ 1, \dots, q \} \setminus \{ i_1^{}, \dots, i_k^{} \}} \!\!\!\!\! e_i^{}. \end{align*}$$

The Koszul complex $K(u_1^{}, \dots , u_q^{}) \otimes M$ introduced at Notation 3.4 becomes a complex of $\mathbb {Z}^q$ -graded $\mathbb {S}$ -modules. According to [Reference Eisenbud5, Corollary 17.5], $K(u_1^{}, \dots , u_q^{}) \otimes M$ is exact, by virtue of the fact that $\{ u_1^{}, \dots , u_q^{} \}$ is M-regular. We have an isomorphism $M \simeq \wedge ^q E \otimes M$ of $\mathbb {Z}^q$ -graded $\mathbb {S}$ -modules given by $h \mapsto \epsilon _1^{} \wedge \dots \wedge \epsilon _q^{} \otimes h$ . The cokernel of the last map in the Koszul complex is thus isomorphic to $M / (u_1^{}, \dots , u_q^{}) M$ . The lemma follows from the additivity of the Hilbert function on short exact sequences.

Proof Assume that X is a complete intersection, so $I(X)$ is generated by a regular sequence $\{ f_1^{}, \dots , f_n^{} \}$ , where $n = n_1^{} + \dots + n_q^{}$ . Write $d_k^{} = (d_k^1, \dots , d_k^q) = \deg (f_k^{})$ . Arguing as in the proof of Lemma 3.5, we can show that $K(f_1^{}, \dots , f_n^{}) \otimes \mathbb {S}$ is a resolution of $\mathbb {S} / I(X)$ , hence, by the additivity of the Hilbert function on short exact sequences,

$$\begin{align*}\operatorname{\mathcal{H}}_X^{}(a) = \operatorname{\mathcal{H}}_{\mathbb{S}}^{}(a) + \sum_{1 \le p \le n} (- 1)^p \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le n} \operatorname{\mathcal{H}}_{\mathbb{S}}^{}(a - d_{k_1}^{} - \dots - d_{k_p}^{}). \end{align*}$$

Assume now that $a = (a_1^{}, 0, \dots , 0)$ . We have the equation

(4) $$ \begin{align} \operatorname{\mathcal{H}}_{Z_1}^{}(a_1^{}) = \operatorname{\mathcal{H}}_X^{}(a) = \operatorname{\mathcal{H}}_{\mathbb{S}_1}^{}(a_1^{}) + \sum_{1 \le p \le n} (- 1)^p \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le n} \operatorname{\mathcal{H}}_{\mathbb{S}}^{}(a - d_{k_1}^{} - \dots - d_{k_p}^{}). \end{align} $$

Put $\{ g_1^{}, \dots , g_m^{} \} = \{ f_1^{}, \dots , f_n^{} \} \cap \mathbb {S}_1^{}$ and write $e_k^{} = \deg (g_k^{}) \in \mathbb {Z}$ for $1 \le k \le m$ . Notice that $\{ g_1^{}, \dots , g_m^{} \}$ is $\mathbb {S}_1^{}$ -regular, hence, $m \le n_1^{}$ . Let $Z \subset \mathbb {P}^{n_1}$ be the subscheme defined by the ideal $(g_1^{}, \dots , g_m^{})$ . If for some indices $1 \le \mu \le p$ and $2 \le i \le q,$ we have $d_{k_\mu }^i> 0$ , then $\operatorname {\mathcal {H}}_{\mathbb {S}}^{}(a - d_{k_1}^{} - \dots - d_{k_p}^{}) = 0$ . Discarding the superfluous terms, we can rewrite Equation (4) in the form

$$\begin{align*}\operatorname{\mathcal{H}}_{Z_1}^{}(a_1^{}) = \operatorname{\mathcal{H}}_{\mathbb{S}_1}^{}(a_1^{}) + \sum_{1 \le p \le m} (- 1)^p \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le m} \operatorname{\mathcal{H}}_{\mathbb{S}_1}^{}(a_1^{} - e_{k_1}^{} - \dots - e_{k_p}^{}). \end{align*}$$

The r.h.s. equals $\operatorname {\mathcal {H}}_Z^{}(a_1^{})$ because $K(g_1^{}, \dots , g_m^{}) \otimes \mathbb {S}_1^{}$ is exact. Thus, $\operatorname {\mathcal {H}}_{Z_1}^{} = \operatorname {\mathcal {H}}_Z^{}$ . By construction, $Z_1^{}$ is a subscheme of Z, hence, $Z = Z_1^{}$ , and hence, $m = n_1^{}$ . Moreover, $Z_1^{}$ is a complete intersection. The same argument works for all $Z_i^{}$ , so each of them is a complete intersection. We have proved that $I(X) = I(Z_1^{}) + \dots + I(Z_q^{})$ , hence $X = Z_1^{} \times \dots \times Z_q^{}$ .

Proof Write $r = \operatorname {rin}(Z)$ . To prove statement (i), we apply Proposition 3.2 and Lemma 3.1. In light of Equation (3), we have $\dim _{\mathbb {K}}^{} \operatorname {H}^1(\mathcal {I}_Z^{}(a)) = s - \operatorname {\mathcal {H}}_Z^{}(a) = 0$ for $a \ge r$ . This proves (ii). To prove statement (iii), we employ the exact sequence

$$\begin{align*}0 \longrightarrow \mathcal{I}_Z^{} \longrightarrow \mathcal{O}_{\mathbb{P}^n}^{} \longrightarrow \mathcal{O}_Z^{} \longrightarrow 0. \end{align*}$$

Applying the long exact cohomology sequence, we obtain the exact sequence

$$\begin{align*}\operatorname{H}^{m - 1}(\mathcal{O}_Z^{}) \longrightarrow \operatorname{H}^m(\mathcal{I}_Z^{}(a)) \longrightarrow \operatorname{H}^m(\mathcal{O}_{\mathbb{P}^n}^{}(a)). \end{align*}$$

The group on the left vanishes because $\mathcal {O}_Z^{}$ has support of dimension zero. The group on the right vanishes for $a \ge -n$ . Thus, the group in the middle also vanishes.

(iv) According to Proposition 3.2, $\operatorname {\mathcal {H}}_Z^{}$ increases on the interval $[0, r]$ . By definition, $\operatorname {\mathcal {H}}_Z^{}(0) = 1$ , hence, $\operatorname {\mathcal {H}}_Z^{}(r) \ge r + 1$ , and hence, in view of statement (i), $s \ge r + 1$ .

(v) For $0 \le a \le s - 1,$ we have $\operatorname {\mathcal {H}}_Z^{}(a) = a + 1$ because there are no forms of degree a vanishing on a subscheme $Z \subset \mathbb {P}^1$ of length s. Thus, $\operatorname {\mathcal {H}}_Z^{}$ increases on the interval $[0, s - 1]$ forcing the inequality $s - 1 \le r$ . The reverse inequality was obtained at (iv).

(vi) From statements (ii) and (iii), we see that $\operatorname {H}^m(\mathcal {I}_Z^{}(r + 1 - m)) = \{ 0 \}$ if $m \ge 1$ . From the definition of r and from Equation (3), it follows that $\operatorname {H}^1(\mathcal {I}_Z^{}(r - 1)) \neq \{ 0 \}$ .

(vii) In the case when $n = 1,$ we have $r = s - 1 = \delta (Z)$ . Assume that $n \ge 2$ . Write $d_i^{} = \deg (f_i^{})$ . The hypothesis that $\{ f_1^{}, \dots , f_n^{} \}$ be a regular sequence implies that $K(f_1^{}, \dots , f_n^{}) \otimes \mathbb {S}$ is a resolution of $\mathbb {S} / I(Z)$ . Consult the proof of Lemma 3.5. Thus,

$$\begin{align*}\operatorname{\mathcal{H}}_Z^{}(a) = \operatorname{\mathcal{H}}_{\mathbb{S}}^{}(a) + \sum_{1 \le p \le n} (- 1)^p \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le n} \operatorname{\mathcal{H}}_{\mathbb{S}}^{}(a - d_{k_1}^{} - \dots - d_{k_p}^{}). \end{align*}$$

Differentiating both sides, we obtain the equation

$$\begin{align*}\operatorname{\Delta} \operatorname{\mathcal{H}}_Z^{}(a) = {a + n - 1 \choose n - 1} + \sum_{1 \le p \le n} (- 1)^p \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le n} {a - d_{k_1}^{} - \dots - d_{k_p}^{} + n - 1 \choose n - 1}. \end{align*}$$

We need to prove that $\operatorname {\Delta } \operatorname {\mathcal {H}}_Z^{}(a) = 0$ for $a \ge \delta (Z) + 1$ , that is, we need to prove the combinatorial equation

(5) $$ \begin{align} {a + n - 1 \choose n - 1} = \sum_{1 \le p \le n} (- 1)^{p + 1} \!\!\!\!\! \sum_{1 \le k_1 < \dots < k_p \le n} {a - d_{k_1}^{} - \dots - d_{k_p}^{} + n - 1 \choose n - 1} \end{align} $$

for $a \ge \delta (Z) + 1$ . Put $A = \{ 1, 2, \dots , a + 1 \}$ . For $1 \le k \le n$ , consider mutually disjoint subsets $D_k^{} \subset A$ with $d_k^{}$ elements. Such subsets exist because, by hypothesis, $a + 1 \ge d_1^{} + \dots + d_n^{}$ . Let $\mathcal {C}$ be the set of repeat combinations of $n - 1$ elements chosen in A. Let $\mathcal {C}_k^{}$ be the set of repeat combinations of $n - 1$ elements chosen in $A \setminus D_k^{}$ . The l.h.s. of Equation (5) equals $\left | \mathcal {C} \right |$ and

$$\begin{align*}{a - d_{k_1}^{} - \dots - d_{k_p}^{} + n - 1 \choose n - 1} = \left| \mathcal{C}_{k_1}^{} \cap \dots \cap \mathcal{C}_{k_p}^{} \right|. \end{align*}$$

Applying the inclusion–exclusion principle, we deduce that the r.h.s. of Equation (5) equals $\left | \mathcal {C}_1^{} \cup \dots \cup \mathcal {C}_n^{} \right |$ . Equation (5) reduces to proving that $\mathcal {C} = \mathcal {C}_1^{} \cup \dots \cup \mathcal {C}_n^{}$ . Take $(c_1^{}, \dots , c_{n - 1}^{}) \in \mathcal {C}$ . By construction, the sets $D_1^{}, \dots , D_n^{}$ are mutually disjoint, hence, there is k such that $c_i^{} \notin D_k^{}$ for all indices $1 \le i \le n - 1$ . Thus, $(c_1^{}, \dots , c_{n - 1}^{}) \in \mathcal {C}_k^{}$ .

Proof By symmetry, we may assume that $i = 1$ . By virtue of the Künneth formula,

$$\begin{align*}\operatorname{H}^m(\mathcal{I}_{W_1}^{}(a)) \simeq \bigoplus_{m_1 + \dots + m_q = m} \operatorname{H}^{m_1}(\mathcal{I}_{Z_1}^{}(a_1^{})) \otimes \operatorname{H}^{m_2}(\mathcal{O}_{\mathbb{P}^{n_2}}^{}(a_2^{})) \otimes \cdots \otimes \operatorname{H}^{m_q}(\mathcal{O}_{\mathbb{P}^{n_q}}^{}(a_q^{})). \end{align*}$$

By hypothesis, $a_1^{} \ge r_1^{}$ , hence, in view of Proposition 3.8, $\operatorname {H}^{m_1}(\mathcal {I}_{Z_1}^{}(a_1^{})) = \{ 0 \}$ for $m_1^{} \ge 1$ . By hypothesis, $a_2^{} \ge 0, \dots , a_q^{} \ge 0$ , hence, the higher cohomology groups of $\mathcal {O}_{\mathbb {P}^{n_2}}^{}(a_2^{}), \dots , \mathcal {O}_{\mathbb {P}^{n_q}}^{}(a_q^{})$ also vanish. We deduce that $\operatorname {H}^m(\mathcal {I}_{W_1}^{}(a)) = \{ 0 \}$ for $m \ge 1$ . From the short exact sequence

$$\begin{align*}0 \longrightarrow \mathcal{I}_{W_1}^{} \longrightarrow \mathcal{I}_X^{} \longrightarrow \mathcal{I}_{X, W_1}^{} \longrightarrow 0 \end{align*}$$

of sheaves on $\mathbb {V,}$ we obtain the long exact sequence

$$\begin{align*}\{ 0 \} = \operatorname{H}^1(\mathcal{I}_{W_1}^{}(a)) \longrightarrow \operatorname{H}^1(\mathcal{I}_X^{}(a)) \longrightarrow \operatorname{H}^1(\mathcal{I}_{X, W_1}^{}(a)) \longrightarrow \operatorname{H}^2(\mathcal{I}_{W_1}^{}(a)) = \{ 0 \}. \end{align*}$$

The middle arrow becomes an isomorphism. The line bundle $\mathcal {O}_{\mathbb {P}^{n_1}}^{}(a_1^{})$ is trivial on $Z_1^{}$ because the latter is supported on finitely many points. We obtain the isomorphism $\mathcal {I}_{X, W_1}^{}(a) \simeq \mathcal {I}_{X, W_1}^{}(0, a_2^{}, \dots , a_q^{})$ , which leads us to the desired isomorphism

$$\begin{align*}\operatorname{H}^1(\mathcal{I}_X^{}(a)) \simeq \operatorname{H}^1(\mathcal{I}_{X, W_1}^{}(0, a_2^{}, \dots, a_q^{})).\\[-34pt] \end{align*}$$

Proof From the decomposition $W_i^{} = \mathbb {W}_1^{} \sqcup \dots \sqcup \mathbb {W}_m^{}$ , we obtain the decomposition

$$\begin{align*}\operatorname{H}^1(\mathcal{I}_{X, W_i}^{}(a - a_i^{} e_i^{})) \simeq \bigoplus_{1 \le k \le m} \operatorname{H}^1(\mathcal{I}_{Y_k, \mathbb{W}_k}^{}(a_1^{}, \dots, \widehat{a}_i^{}, \dots, a_q^{})). \end{align*}$$

Applying Equation (3) and Lemma 4.1, we calculate

$$ \begin{align*} \operatorname{\mathcal{H}}_X^{}(a) & = \operatorname{length}(X) - \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_X^{}(a)) \\ & = \operatorname{length}(X) - \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_{X, W_i}^{} (a - a_i^{} e_i^{})) \\ & = \sum_{1 \le k \le m} \operatorname{length}(Y_k^{}) - \sum_{1 \le k \le m} \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_{Y_k, \mathbb{W}_k}^{}(a_1^{}, \dots, \widehat{a}_i^{}, \dots, a_q^{})) \\ & = \sum_{1 \le k \le m} \operatorname{\mathcal{H}}_{Y_k}^{}(a_1^{}, \dots, \widehat{a}_i^{}, \dots, a_q^{}).\\[-40pt] \end{align*} $$

Proof Write $\operatorname {rem}(X) = (r_1^{}, \dots , r_q^{})$ . Consider $a \in \mathbb {Z}_+^q$ satisfying the condition $a_i^{} \ge r_i^{}$ for some index $i \in \{ 1, \dots , q \}$ . According to Lemma 4.1, the expression $\dim _{\mathbb {K}}^{} \operatorname {H}^1(\mathcal {I}_X^{}(b))$ remains constant as $b_i^{}$ varies in the interval $[r_i^{}, \infty )$ and $b_j^{}$ are nonnegative fixed integers for all indices $j \in \{ 1, \dots , q \} \setminus \{ i \}$ . Thus,

$$\begin{align*}\dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_X^{}(a)) = \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_X^{}(a_1^{}, \dots, r_i^{}, \dots, a_q^{})). \end{align*}$$

Applying Equation (3), we calculate

$$ \begin{align*} \operatorname{\mathcal{H}}_X^{}(a) & = \operatorname{length}(X) - \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_X^{}(a)) \\ & = \operatorname{length}(X) - \dim_{\mathbb{K}}^{} \operatorname{H}^1(\mathcal{I}_X^{}(a_1^{}, \dots, r_i^{}, \dots, a_q^{})) \\ & = \operatorname{\mathcal{H}}_X^{}(a_1^{}, \dots, r_i^{}, \dots, a_q^{}). \end{align*} $$

From Lemma 2.2, we deduce that $R(X)$ is relevant to $\operatorname {\mathcal {H}}_X^{}$ . We cannot shrink $R(X)$ to a smaller rectangular relevant domain because, as seen at Proposition 3.2, the function $\operatorname {\mathcal {H}}_X^{}(a_i^{} e_i^{}) = \operatorname {\mathcal {H}}_{Z_i}^{}(a_i^{})$ increases on the interval $[0, r_i^{}]$ .

Proof Statement (i) follows from Theorem 4.4 and Lemma 3.1. To prove (ii) and (iii), we argue precisely as in the proof of Proposition 3.8(ii and iii). Statements (iv) and (v) follow from their counterparts at Proposition 3.8. Statement (vi) follows from Propositions 3.6 and 3.8(vii).

Proof Clearly, the Poincarè polynomial associated to $\operatorname {\mathcal {H}}_X^{}$ , introduced at Definition 2.6, coincides with $\mathcal {P}_X^{}$ . We apply Lemma 2.5 to $\operatorname {\mathcal {H}}_X^{}$ and to its relevant domain $[0, \operatorname {rem}(X)]$ . We deduce that $\mathcal {P}_X^{} = \overline {\operatorname {\mathcal {H}}}_X^{}$ on $[| \operatorname {rem}(X) |, \, \infty )$ , and that $\mathcal {P}_X^{}$ has dominant term $\operatorname {\mathcal {H}}_X^{}(\operatorname {rem}(X)) d^{q - 1} / (q - 1)!$ It follows that $\operatorname {rin}(X) \le | \operatorname {rem}(X) |$ . In accordance with Proposition 4.5(i), $\operatorname {\mathcal {H}}_X^{}(\operatorname {rem}(X)) = \operatorname {length}(X)$ . This proves the second assertion.

Proof We apply Lemma 2.7 to the function $\mathcal {F} = \operatorname {\mathcal {H}}_X^{}$ . The hypothesis of Lemma 2.7 is satisfied because $\operatorname {\mathcal {H}}_X^{}$ takes the value s on the region $\{ a \in R(X) \mid | a | \ge \tau \}$ . We deduce that $\overline {\operatorname {\mathcal {H}}}_X^{}(d) = \mathcal {P}_X^{}(d)$ for $d \ge \tau $ , forcing the inequality $\operatorname {rin}(X) \le \tau $ .

Proof According to Corollary 4.7, $\deg (\overline {X}) = s$ , hence, the two inequalities above are equivalent. Assume that $| \operatorname {rem}(X) | \le s - 1$ . Applying Corollary 4.7, we get the inequalities $\operatorname {rin}(X) \le | \operatorname {rem}(X) | \le s - 1$ . Assume now that $s - 1 < | \operatorname {rem}(X) |$ . According to [Reference Maican20, Corollary 4.5], $\operatorname {\mathcal {H}}_X^{}(a) = s$ if $a \in R(X)$ and $| a | \ge s - 1$ . Applying Lemma 4.8 with $\tau = s - 1,$ we obtain the inequality $\operatorname {rin}(X) \le s - 1$ .

Proof Put $k = \dim \mathbb {K}[\mathbf {x}] / \mathfrak {p} \cap \mathbb {K}[\mathbf {x}]$ and $l = \dim \mathbb {K}[\mathbf {y}] / \mathfrak {p} \cap \mathbb {K}[\mathbf {y}]$ . The assertion is equivalent to the inequality $\dim \mathbb {K}[\mathbf {x}, \mathbf {y}] / \mathfrak {p} \le k + l$ (see [Reference Eisenbud5, Corollary 13.4]). Applying the Noether normalization theorem [Reference Eisenbud5, Theorem 13.3] we deduce that there are linearly independent one-forms $u_1^{}, \dots , u_k^{} \in \mathbb {K}[\mathbf {x}]$ , respectively, $v_1^{}, \dots , v_l^{} \in \mathbb {K}[\mathbf {y}]$ such that $\mathbb {K}[\mathbf {x}] / \mathfrak {p} \cap \mathbb {K}[\mathbf {x}]$ is integral over $\mathbb {K}[\mathbf {u}]$ and $\mathbb {K}[\mathbf {y}] / \mathfrak {p} \cap \mathbb {K}[\mathbf {y}]$ is integral over $\mathbb {K}[\mathbf {v}]$ . Here $\mathbf {u} = (u_1^{}, \dots , u_k^{})$ and $\mathbf {v} = (v_1^{}, \dots , v_l^{})$ . The extension of algebras $\mathbb {K}[\mathbf {u}, \mathbf {v}] / \mathfrak {p} \cap \mathbb {K}[\mathbf {u}, \mathbf {v}] \subset \mathbb {K}[\mathbf {x}, \mathbf {y}] / \mathfrak {p}$ is integral, hence, the two algebras have the same dimension (see [Reference Eisenbud5, Theorem A, p. 286]). Thus, $\dim \mathbb {K}[\mathbf {x}, \mathbf {y}] / \mathfrak {p} \le \dim \mathbb {K}[\mathbf {u}, \mathbf {v}] = k + l$ .

Proof Recall that $I(X) \cap \mathbb {K}[x_{ij}^{} \mid 0 \le j \le n_i^{}]$ defines a zero-dimensional subscheme $Z_i^{}$ in $\mathbb {P}^{n_i}$ . By hypothesis, $u_i^{}$ does not vanish at any point of $\operatorname {red}(Z_i^{})$ , hence, $(u_i^{}) + I(Z_i^{})$ defines the empty subscheme in $\mathbb {P}^{n_i}$ . Thus, $(U_i^{}) \subset \operatorname {rad}((u_i^{}) + I(Z_i^{})) \subset \operatorname {rad}((u_i^{}) + I(X))$ .

Proof Set $\mathbf {x}_i^{} = (x_{i0}^{}, \dots , x_{i n_i})$ . Performing induction on k, we will construct a regular sequence $\{ u_1^{}, \dots , u_k^{} \}$ for $\mathbb {S} / I(X)$ with $u_i^{} \in U_i^0$ . To start the induction, choose $u_1^{}$ in $U_1^0$ and recall Remark 5.2. For the induction step, assume that $k \in \{ 1, \dots , q - 1 \}$ and that $\{ u_1^{}, \dots , u_k^{} \}$ has already been constructed. Write $J = (u_1^{}, \ldots, u_k^{}) + I(X)$ . Choose a prime ideal $\mathfrak {p}$ that is associated to J. By hypothesis, $\mathbb {S} / I(X)$ is Cohen–Macaulay of dimension q, hence, $\mathbb {S} / J$ is Cohen–Macaulay of dimension $q - k$ . According to [Reference Eisenbud5, Corollary 18.14], J is unmixed, hence, $\dim (\mathfrak {p}) = q - k$ and $\operatorname {ht}(\mathfrak {p}) = n_1^{} + \dots + n_q^{} + k$ . According to Lemma 5.5, $(U_1^{}, \dots , U_k^{})$ lies in $\operatorname {rad}(J)$ , so it is contained in $\mathfrak {p}$ . We claim that $\operatorname {ht}(\mathfrak {p} \cap \mathbb {K}[\mathbf {x}_i^{}]) = n_i^{}$ for each index $i \in \{ k + 1, \dots , q \}$ . Indeed, the inequalities $\operatorname {ht}(\mathfrak {p} \cap \mathbb {K}[\mathbf {x}_i^{}]) \ge n_i^{}$ follow from the fact that $\mathfrak {p} \cap \mathbb {K}[\mathbf {x}_i^{}]$ contains the ideal of the projection of X onto $\mathbb {P}^{n_i}$ . According to Lemma 5.4, we have the inequality

$$\begin{align*}\operatorname{ht}(\mathfrak{p}) \ge \sum_{1 \le i \le q} \operatorname{ht}(\mathfrak{p} \cap \mathbb{K}[\mathbf{x}_i^{}]) \quad \text{or, equivalently}, \ \sum_{k + 1 \le i \le q} n_i^{} \ge \sum_{k + 1 \le i \le q} \operatorname{ht}(\mathfrak{p} \cap \mathbb{K}[\mathbf{x}_i^{}]). \end{align*}$$

This proves the claim. The claim implies that $U_{k + 1}^{}$ is not contained in $\mathfrak {p}$ . The same is true for all associated primes of J. Since $\mathbb {K}$ is infinite, we can choose $u_{k + 1}^{} \in U_{k + 1}^0$ such that $u_{k + 1}^{}$ does not lie in any associated prime of J. Thus, $u_{k + 1}$ is a non-zerodivisor for $\mathbb {S} / J$ , hence, $\{ u_1^{}, \dots , u_{k + 1}^{} \}$ is regular relative to $\mathbb {S} / I(X)$ .

Proof In view of Remark 5.3, we have $\mathfrak {p}_k^{} = (U_1^{}, \dots , U_p^{}) + (U_{k, p+1}^{}, \dots , U_{k, q}^{})$ . This is clearly a prime ideal. Some of these ideals may coincide, however, if $\mathfrak {p}_k^{} \neq \mathfrak {p}_l^{}$ , then $\mathfrak {p}_k^{} \nsubseteq \mathfrak {p}_l^{}$ and $\mathfrak {p}_l^{} \nsubseteq \mathfrak {p}_k^{}$ . The lemma reduces to proving that

$$\begin{align*}\bigcap_{1 \le k \le m} \mathfrak{p}_k^{} = \operatorname{rad}((u_1^{}, \dots, u_p^{}) + I(X)). \end{align*}$$

The inclusion “ $\supset $ ” is obvious, so we focus on proving the reverse inclusion. We denote by $\mathfrak {r}$ the ideal on the r.h.s. Take $f \in \bigcap _{1 \le k \le m} \mathfrak {p}_k^{}$ and write $f = g + h$ , where g lies in the ideal $(U_1^{}, \dots , U_p^{})$ and h is a polynomial in the variables $x_{ij}^{}$ , $p + 1 \le i \le q$ , $0 \le j \le n_i^{}$ . According to Lemma 5.5, $(U_1^{}, \dots , U_p^{}) \subset \mathfrak {r}$ , hence, $g \in \mathfrak {r}$ . By construction,

$$\begin{align*}h \in \bigcap_{1 \le k \le m} (U_{ki}^{} \mid p + 1 \le i \le q) \subset \bigcap_{1 \le k \le m} I(P_k^{}) = \operatorname{rad}(I(X)) \subset \mathfrak{r}. \end{align*}$$

We conclude that f lies in $\mathfrak {r}$ .

Proof If G is lower-triangular, then the lemma follows from the definition of an M-regular sequence. According to [Reference Eisenbud5, Corollary 17.5 and Theorem 17.6], a sequence $\{ w_1^{}, \dots , w_p^{} \} \subset S$ of homogeneous elements is M-regular if and only if the Koszul complex $K(w_1^{}, \dots , w_p^{}) \otimes M$ , introduced at Notation 3.4, is exact. Permutations of $\{ w_1^{}, \dots , w_p^{} \}$ yield isomorphic Koszul complexes. Thus, every permutation of $\{ v_1^{}, \dots , v_p^{} \}$ remains an M-regular sequence. The lemma follows from the fact that the lower triangular matrices and the row permutations generate $\operatorname {GL}_p^{}(\mathbb {K})$ .

Proof Write $\operatorname {red}(X) = \{ P_1^{}, \dots , P_m^{} \}$ . According to Lemma 5.7, $\mathfrak {p}_k^{} = (U_1^{}) + I(P_k^{})$ for $1 \le k \le m$ are the minimal prime ideals containing $(u_1^{}) + I(X)$ . We denote $U' = \operatorname {span} \{ u_2^{}, \dots , u_q^{} \}$ . Performing induction on $l \in \{ 2, \dots , q - 1 \}$ , we will construct a regular sequence $\{ u_1^{}, v_2^{}, \dots , v_l^{} \}$ for $\mathbb {S} / I(X)$ with $v_i^{} \in U'$ . To start the induction, we consider the set $\mathfrak {A}_1^{}$ of associated primes to $(u_1^{}) + I(X)$ . We claim that $U'$ is not contained in any $\mathfrak {p} \in \mathfrak {A}_1^{}$ . To prove this, we argue by contradiction. Assume that $U' \subset \mathfrak {p}$ and that $\mathfrak {p} \in \mathfrak {A}_1^{}$ . This ideal must contain one of the minimal associated primes to $(u_1^{}) + I(X)$ , say $\mathfrak {p}_k^{} \subset \mathfrak {p}$ . Thus, $U_1^{} \subset \mathfrak {p}$ and $U_{ki}^{} \subset \mathfrak {p}$ for $2 \le i \le q$ . It follows that

$$ \begin{align*} U & = U_1^{} + U_2^{} + \dots + U_q^{} \\ & = U_1^{} + \operatorname{span} \{ u_2^{}, U_{k2}^{} \} + \dots + \operatorname{span} \{ u_q^{}, U_{kq}^{} \} \\ & = U_1^{} + U_{k2}^{} + \dots + U_{kq}^{} + U' \subset \mathfrak{p}, \end{align*} $$

hence, $\mathfrak {m} = (U) \subset \mathfrak {p}$ , so every element of $\mathfrak {m}$ is a zerodivisor for $\mathbb {S} / ((u_1^{}) + I(X))$ . On the other hand, by Remark 5.2, $u_1^{}$ is a non-zerodivisor for $\mathbb {S} / I(X)$ , hence,

$$\begin{align*}\operatorname{depth}(\mathbb{S} / ((u_1^{}) + I(X))) = \operatorname{depth}(\mathbb{S} / I(X)) - 1 = q - 2 \ge 1. \end{align*}$$

We have reached a contradiction, which proves the claim. We obtain a regular sequence $\{ u_1^{}, v_2^{} \}$ relative to $\mathbb {S} / I(X)$ by choosing $v_2^{} \in U' \setminus \bigcup _{\mathfrak {p} \in \mathfrak {A}_1} (\mathfrak {p} \cap U')$ .

We now perform the induction step. Assume that $l \in \{ 2, \dots , q - 2 \}$ and that $\{ u_1^{}, v_2^{}, \dots , v_l^{} \}$ has already been constructed. We denote by $\mathfrak {A}_l^{}$ the set of associated primes to the ideal $(u_1^{}, v_2^{}, \dots , v_l^{}) + I(X)$ . Arguing as above, we can prove that $U'$ is not contained in any $\mathfrak {p}$ from $\mathfrak {A}_l^{}$ . Indeed, $\mathfrak {p}_k^{} \subset \mathfrak {p}$ for some k, so, if $U' \subset \mathfrak {p}$ , then $U \subset \mathfrak {p}$ . It would follow that every element of $\mathfrak {m}$ is a zerodivisor for $\mathbb {S} / ((u_1^{}, v_2^{}, \dots , v_l^{}) + I(X))$ . On the other hand, this ring has depth $q - 1 - l \ge 1$ . This would yield a contradiction. Choosing $v_{l + 1}^{} \in U' \setminus \bigcup _{\mathfrak {p} \in \mathfrak {A}_l} (\mathfrak {p} \cap U'),$ we obtain a regular sequence $\{ u_1^{}, v_2^{}, \dots , v_{l + 1}^{} \}$ relative to $\mathbb {S} / I(X)$ . This completes the induction step.

Thus far, we have constructed an $\mathbb {S} / I(X)$ -regular sequence $\{ u_1^{}, v_2^{}, \dots , v_{q - 1}^{} \}$ such that $v_2^{}, \dots , v_{q - 1}^{}$ are linearly independent vectors in $U'$ . Write $v_l^{} = \sum _{2 \le i \le q} \lambda _{li}^{} u_i^{}$ . The matrix $\Lambda = (\lambda _{li}^{})_{2 \le l \le q - 1, \, 2 \le i \le q}^{}$ has maximal rank. To simplify notation, we assume that the minor obtained by deleting the last column of $\Lambda $ is non-zero. We now apply Lemma 5.8 to the $\mathbb {Z}$ -graded ring $\mathbb {S}$ , to the $\mathbb {Z}$ -graded module $M = \mathbb {S} / ((u_1^{}) + I(X))$ and to the M-regular sequence $\{ v_2^{}, \dots , v_{q - 1}^{} \}$ . We take G to be the inverse of $(\lambda _{li}^{})_{2 \le l, \, i \le q - 1}^{}$ . We obtain an M-regular sequence of the form $\{ u_i^{} + \kappa _i^{} u_q^{} \mid 2 \le i \le q - 1 \}$ . In general, if the minor obtained by deleting column p of $\Lambda $ is non-zero, then we obtain a regular sequence as in the proposition.

Proof Performing induction on i, we will show that $\{ u_1^{}, \dots , u_i^{} \}$ is $\mathbb {S} / I(X)$ -regular. By Remark 5.2, $u_1^{}$ is a non-zerodivisor for $\mathbb {S} / I(X)$ . Assume that $i \in \{1, \dots , q -1 \}$ and that $\{ u_1^{}, \dots , u_i^{} \}$ is $\mathbb {S} / I(X)$ -regular. By hypothesis, $\mathbb {S} / I(X)$ is Cohen–Macaulay, hence, $\mathbb {S} / ((u_1^{}, \dots , u_i^{}) + I(X))$ is also Cohen–Macaulay. Per [Reference Eisenbud5, Corollary 18.14], this ring is unmixed. Thus, the associated primes of $(u_1^{}, \dots , u_i^{}) + I(X)$ are precisely the minimal primes. According to Lemma 5.7, they are of the form $\mathfrak {p}_k^{} = (U_1^{}, \dots , U_i^{}) + I(P_k^{})$ . By construction, $u_{i + 1}^{}$ lies outside all ideals $\mathfrak {p}_k^{}$ , hence, $u_{i + 1}^{}$ is a non-zerodivisor of $\mathbb {S} / ((u_1^{}, \dots , u_i^{}) + I(X))$ , and hence, $\{ u_1^{}, \dots , u_{i + 1}^{} \}$ is regular relative to $\mathbb {S} / I(X)$ .

Proof The inclusion “ $\subset $ ” is obvious, so we concentrate on proving the reverse inclusion. We perform induction on p. Assume that $p = 1$ . Take $f \in (u_1^{}) \cap I$ and write $f = u_1^{} g$ . In $S / I,$ we have the relations $u_1^{} \hat {g} = \hat {u}_1^{} \hat {g} = \hat {f} = 0$ . By hypothesis, $u_1^{}$ is a non-zerodivisor for $S / I$ , hence, $\hat {g} = 0$ , that is, $g \in I$ , and hence, $f \in (u_1^{}) I$ . Assume that $p> 1$ and that the lemma is true for the $S / I$ -regular sequence $\{ u_1^{}, \dots , u_{p - 1}^{} \}$ . Take $f \in (u_1^{}, \dots , u_p^{}) \cap I$ and write $f = u_1^{} g_1^{} + \dots + u_p^{} g_p^{}$ . In $S / ((u_1^{} + \dots + u_{p - 1}^{}) + I),$ we have the relations $ u_p^{} \hat {g}_p^{} = \hat {u}_p^{} \hat {g}_p^{} = \hat {f} - \hat {u}_1^{} \hat {g}_1^{} - \dots - \hat {u}_{p - 1}^{} \hat {g}_{p - 1}^{} = 0. $ By hypothesis, $u_p^{}$ is a non-zerodivisor for $S / ((u_1^{} + \dots + u_{p - 1}^{}) + I)$ , hence, $\hat {g}_p^{} = 0$ . Write $g_p^{} = u_1^{} h_1^{} + \dots + u_{p - 1}^{} h_{p - 1}^{} + h_p^{}$ , where $h_p^{} \in I$ . From the relation

$$\begin{align*}f - u_p^{} h_p^{} = u_1^{} (g_1^{} + u_p^{} h_1^{}) + \dots + u_{p - 1}^{} (g_{p - 1}^{} + u_p^{} h_{p - 1}^{}), \end{align*}$$

we see that $f - u_p^{} h_p^{} \in (u_1^{}, \dots , u_{p - 1}^{}) \cap I$ . By the induction hypothesis, this ideal coincides with $(u_1^{}, \dots , u_{p - 1}) I$ . We conclude that $f \in (u_1^{}, \dots , u_p^{}) I$ .

Proof By hypothesis, $u_1^{}$ is a non-zerodivisor in S, hence, $u_1^{}$ is a non-zerodivisor for I. Take $i \in \{ 1, \dots , p \}$ . We apply Lemma 5.11 to the $S / I$ -regular sequence $\{ u_1^{}, \dots , u_i^{} \}$ . We deduce that $I / (u_1^{}, \dots , u_i^{})I$ is isomorphic, as an S-module, to an ideal of $S / (u_1^{}, \dots , u_i^{})$ . By hypothesis, $u_{i + 1}^{}$ is a non-zerodivisor for $S / (u_1^{}, \dots , u_i^{})$ , hence, $u_{i + 1}^{}$ is a non-zerodivisor for $I / (u_1^{}, \dots , u_i^{})I$ .

Proof We assume that $q = 2$ . According to Remark 5.2, $\{ u_1^{} \}$ is regular for $\mathbb {S} / I(X)$ . Clearly, $\{ u_1^{}, u_2^{} \}$ is $\mathbb {S}$ -regular. From Lemma 5.12, we deduce that $\{ u_1^{}, u_2^{} \}$ is $I(X)$ -regular. We assume that $q \ge 3$ . As in Proposition 5.9, we consider an $\mathbb {S} / I(X)$ -regular sequence

$$\begin{align*}\{ w_1^{}, \dots, w_{q - 1}^{} \} = \{ u_i^{} + \kappa_i^{} u_p^{} \mid i \in \{ 2, \dots, q \} \setminus \{ p \} \}. \end{align*}$$

Here, $\kappa _1^{} = 0$ . We set $w_q^{} = u_p^{}$ . Clearly, $\{ w_1^{}, \dots , w_q^{} \}$ is $\mathbb {S}$ -regular. From Lemma 5.12, we deduce that $\{ w_1^{}, \dots , w_q^{} \}$ is also $I(X)$ -regular. Let us now consider the column vectors $\mathbf {u} = (u_1^{}, \dots , u_q^{})^{\small \mathrm {T}}$ and $\mathbf {w} = (w_1^{}, \dots , w_q^{})^{\small \mathrm {T}}$ . By construction, we have $\mathbf {w} = \Lambda \mathbf {u}$ for some $\Lambda \in \operatorname {GL}_q^{}(\mathbb {K})$ . We apply Lemma 5.8 to the $I(X)$ -regular sequence $\{ w_1^{}, \dots , w_q^{} \}$ . We take $G = \Lambda ^{-1}$ . We conclude that $\{ u_1^{}, \dots , u_q^{} \}$ is regular relative to $I(X)$ .

Proof According to Proposition 5.10, $\{ u_1^{}, \dots , u_q^{} \}$ is regular relative to $\mathbb {S} / I(X)$ . Clearly, $\{ u_1^{}, \dots , u_q^{}, v_i^{} \}$ is $\mathbb {S}$ -regular. The proposition follows from Lemma 5.12.

Proof Proposition 5.6 provides an $\mathbb {S} / I(X)$ -regular sequence $\{ u_1^{}, \dots , u_q^{} \}$ . Clearly, $\{ u_1^{}, \dots , u_q^{}, v_i^{} \}$ is $\mathbb {S}$ -regular. The proposition follows from Lemma 5.12.

Proof According to Remark 5.2, $u_i^{}$ is regular for $\mathbb {S} / I(X)$ . Clearly, $\{ u_i^{}, v_j^{} \}$ is $\mathbb {S}$ -regular. The proposition follows from Lemma 5.12.

Proof By hypothesis, $\operatorname {red}(X) = \{ P \}$ for a closed point $P \in \mathbb {V}$ . By the multiprojective version of Hilbert’s Nullstellensatz, $I(P) = \operatorname {rad}(I(X))$ . As the radical of a monomial ideal, $I(P)$ itself is monomial. But $I(P)$ is also a prime ideal, hence, $I(P)$ is generated by a subset of the set of variables. Say $I(P) = (x_{ij}^{} \mid 1 \le i \le q,\ 1 \le j \le n_i^{})$ . If $x_{i0}^{} \zeta \in I(X)$ for a monomial $\zeta $ , then, since $x_{i0}^{}$ does not vanish at P, $\zeta $ must lie in $I(X)$ . This shows that the minimal generators of $I(X)$ are monomials in the same variables that generate $I(P)$ . It has now become clear that $\{ x_{i0}^{} \mid 1 \le i \le q \}$ is a regular sequence for $\mathbb {S} / I(X)$ , hence, $\operatorname {depth}(\mathbb {S} / I(X)) \ge q$ , and hence, $\operatorname {depth}(\mathbb {S} / I(X)) = q$ .

Proof Assume that X is ACM. According to Proposition 5.6, there exists a regular sequence $\{ u_1^{}, \dots , u_q^{} \}$ relative to $\mathbb {S} / I(X)$ with $u_i^{} \in U_i^{}$ . In view of Lemma 3.5, $\operatorname {\Delta } \operatorname {\mathcal {H}}_X^{}$ is the Hilbert function of $A = \mathbb {S} / ((u_1^{}, \dots , u_q^{}) + I(X))$ . Performing a linear change of coordinates on each $\mathbb {P}^{n_i}$ , we may assume that $u_i^{} = x_{i0}^{}$ , therefore, A can be regarded as a $\mathbb {Z}^q$ -graded quotient of $\mathbb {S}_0^{}$ . According to Theorem 4.4, $\operatorname {\mathcal {H}}_A^{}$ vanishes outside a rectangular domain, hence, $\dim _{\mathbb {K}}^{} A$ is finite, and hence, A is an artinian $\mathbb {K}$ -algebra.

Conversely, we assume that we are given an artinian $\mathbb {Z}^q$ -graded algebra $A = \mathbb {S}_0^{} / I_0^{}$ . We recall Notation 6.4. We choose a monomial well-ordering on $\mathbb {S}_0^{}$ and we apply Theorem 6.5 to the $\mathbb {Z}^q$ -homogeneous ideal $I_0^{}$ . We find a monomial ideal $J_0^{} = \operatorname {lead}(I_0^{})$ such that $\operatorname {\mathcal {H}}_A^{} = \operatorname {\mathcal {H}}_{\mathbb {S}_0 / J_0}$ . In particular, $\mathbb {S}_0^{} / J_0^{}$ is an artinian algebra, hence, $\operatorname {rad}(J_0^{}) = (x_{ij}^{} \mid 1 \le i \le q, \, 1 \le j \le n_i^{})$ . Let $J \subset \mathbb {S}$ be the ideal generated by $J_0^{}$ . Since J is generated by monomials that do not involve the variables $x_{i0}$ , $1 \le i \le q$ , it is obvious that J is saturated. Thus, $J = I(X)$ for a zero-dimensional subscheme $X \subset \mathbb {V,}$ which is concentrated on the point given by the ideal $(x_{ij}^{} \mid 1 \le i \le q, \, 1 \le j \le n_i^{})$ . In view of Lemma 6.3, X is ACM. Since J is generated by monomials that do not involve the variables $x_{i0}$ , $1 \le i \le q$ , it is obvious that $\{ x_{10}^{}, \dots , x_{q0}^{} \}$ is $\mathbb {S} / J$ -regular. In view of Lemma 3.5, $\operatorname {\Delta } \operatorname {\mathcal {H}}_X^{}$ must be the Hilbert function of $\mathbb {S} / ((x_{10}^{}, \dots , x_{q0}^{}) + J) = \mathbb {S}_0^{} / J_0^{}$ .

Proof Note that $\mathbb {S}_0^{} = \mathbb {K} [x_{i1}^{} \mid 1 \le i \le q]$ , where $\deg (x_{i1}^{}) = e_i^{}$ for all i. An ideal of $\mathbb {S}_0^{}$ is $\mathbb {Z}^q$ -homogeneous if and only if it is monomial. If $A = \mathbb {S}_0^{} / I_0^{}$ is an artinian quotient by a monomial ideal, then $I_0^{}$ must contain minimal generators of the form $x_{11}^{s_1}, \dots , x_{q1}^{s_q}$ . Let $c_1^{}, \dots , c_m^{}$ be the degrees of the remaining minimal generators of $I_0^{}$ , if any. Write

$$\begin{align*}Q = [0, s_1^{} - 1] \times \dots \times [0, s_q^{} - 1] \setminus \bigcup_{1 \le k \le m} [c_k^{}, (s_1^{} - 1, \dots, s_q^{} - 1)]. \end{align*}$$

We have $\operatorname {\mathcal {H}}_A^{} = \mathcal {X}_Q^{}$ Conversely, for any quasi-rectangular domain $Q \subset \mathbb {Z}^q$ , we can find an artinian $\mathbb {Z}^q$ -graded quotient $A = \mathbb {S}_0^{} / I_0^{}$ such that $\operatorname {\mathcal {H}}_A^{} = \mathcal {X}_Q^{}$ .

Proof Recall Notation 5.1. As per Proposition 5.13, we can construct $I(X)$ -regular sequences $\{ u_1^{}, \dots , u_q^{} \}$ with generic $u_i^{} \in U_i^{}$ . We write $N = I(X) / (u_1^{}, \dots , u_q^{}) I(X)$ . By applying Lemma 3.5, we deduce that $\operatorname {\Delta } \operatorname {\mathcal {H}}_{I(X)}^{} = \operatorname {\mathcal {H}}_N^{}$ . This function takes, of course, only nonnegative values. We have proved statement (i). The same argument applies to statement (v), except that this time we consider the $I(X)$ -regular sequence $\{ u_{i_1}^{}, \dots , u_{i_p}^{} \}$ .

If $I(X)_a^{} \neq \{ 0 \}$ , then we can choose $u_i^{}$ such that the subvariety given by the ideal $(u_1^{}, \dots , u_q^{})$ is not contained in the zero-set of $I(X)_a$ . Thus, $N_a^{} \neq \{ 0 \}$ , hence, $\operatorname {\Delta } \operatorname {\mathcal {H}}_{I(X)}^{}(a)> 0$ . This proves statement (ii). The same argument applies to statement (vi), except that this time we consider the ideal $(u_{i_1}^{}, \dots , u_{i_p}^{})$ . Statement (iii) follows from the equation $\operatorname {\Delta } \operatorname {\mathcal {H}}_X^{} = \operatorname {\Delta } \operatorname {\mathcal {H}}_{\mathbb {S}}^{} - \operatorname {\Delta } \operatorname {\mathcal {H}}_{I(X)}^{}$ and from (i).

Assume that $\operatorname {\Delta } \operatorname {\mathcal {H}}_X^{}(a) = \operatorname {\Delta } \operatorname {\mathcal {H}}_{\mathbb {S}}^{}(a)$ for some $a \in \mathbb {Z}_+^q$ . Thus, $\operatorname {\Delta } \operatorname {\mathcal {H}}_{I(X)}^{}(a) = 0$ . From part (ii), we deduce that $I(X)_a^{} = \{ 0 \}$ . A fortiori, $I(X)_b^{} = \{ 0 \}$ for $b \in [0, a]$ , hence, $\operatorname {\mathcal {H}}_{I(X)}^{} = 0$ on $[0, a]$ , and hence, $\operatorname {\mathcal {H}}_X^{} = \operatorname {\mathcal {H}}_{\mathbb {S}}^{}$ on $[0, a]$ . This proves statement (iv).

Proof Consider the $I(X)$ -regular sequence $\{ u_1^{}, \dots , u_q^{}, v_i^{} \}$ from Proposition 5.15. Write $N = I(X) / (u_1^{}, \dots , u_q^{}, v_i^{}) I(X)$ . By analogy with Lemma 3.5, we can prove that $\operatorname {\Delta } \operatorname {\Delta }_i^{} \operatorname {\mathcal {H}}_{I(X)}^{} = \operatorname {\mathcal {H}}_N^{}$ . This function takes only nonnegative values, proving statement (i). Assume that $I(X)_a^{} \neq \{ 0 \}$ and $n_i^{} \ge 2$ . According to Proposition 5.14, $u_1^{}, \dots , u_q^{}$ and $v_i^{}$ can be chosen generically. We choose them in such a way that the subvariety given by the ideal $(u_1^{}, \dots , u_q^{}, v_i^{})$ is not contained in the zero-set of $I(X)_a^{}$ . Thus, $N_a^{} \neq \{ 0 \}$ , hence, $\operatorname {\mathcal {H}}_N^{}(a)> 0$ . This proves statement (ii). Statement (iii) follows from the equation $\operatorname {\Delta }_i^{} \operatorname {\mathcal {H}}_A^{} = \operatorname {\Delta } \operatorname {\Delta }_i^{} \operatorname {\mathcal {H}}_{\mathbb {S}}^{} - \operatorname {\Delta } \operatorname {\Delta }_i^{} \operatorname {\mathcal {H}}_{I(X)}^{}$ and from (i). To prove the remaining statements, we can argue as in the proof of Proposition 7.1. Note also that (v) follows from the formula

$$\begin{align*}\operatorname{\Delta} \operatorname{\mathcal{H}}_{I(X)}^{} = \sum_{0 \le k \le a_i} \operatorname{\Delta} \operatorname{\Delta}_i^{} \operatorname{\mathcal{H}}_{I(X)}^{}(a - k e_i^{}) \end{align*}$$

and from (i). In the case when $n_i^{} \ge 2$ , statement (vi) follows from the above formula and from (i) and (ii). Statement (ix) follows from Equation (7) and from (v). Statement (x) follows from Equation (7) and from (v) and (vi).

Proof Substituting $n_i^{} = 1$ into Equation (6), we obtain $\operatorname {\Delta } \operatorname {\mathcal {H}}_{\mathbb {S}}^{} (a) = 1$ for $a \in \mathbb {Z}_+^q$ . Substituting this expression into Propositions 7.1(iii) and 7.2(vii) yields statement (i). Substituting this expression into Propositions 7.1(iv) and 7.2(viii) yields (ii).

Proof We use the $I(X)$ -regular sequence $\{ u_i^{}, v_j^{} \}$ provided by Proposition 5.16 and we repeat the arguments from the proof of Proposition 7.1. Statement (vii) follows from the fact that $\{ u_i^{} \}$ is regular for $\mathbb {S} / I(X)$ (see Remark 5.2).

Proof In view of Lemma 2.2, we must show that $\operatorname {\mathcal {H}}_X^{}(a) = \operatorname {\mathcal {H}}_X^{}(a - (a_i^{} - s_i^{} + 1) e_i^{})$ if $a_i^{} \ge s_i^{} - 1$ . Equivalently, we must show that $\operatorname {\Delta }_i^{} \operatorname {\mathcal {H}}_X^{}(a) = 0$ if $a_i^{} \ge s_i^{}$ . By symmetry, it is enough to prove that $\operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(a) = 0$ if $a \in \mathbb {Z}_+^q$ and $a_1^{} \ge s_1^{}$ . Given $a \in \mathbb {Z}_+^q$ , put $\sigma (a) = a_2^{} + \dots + a_q^{}$ . We perform induction on $\sigma (a)$ . To begin the induction, assume that $\sigma (a) = 0$ . Write $Z_1^{} = \mathrm {pr}_1^{}(X)$ . According to Proposition 3.8(i and v), we have the equation $\operatorname {\mathcal {H}}_X^{} (b_1^{}, 0 , \dots , 0) = \operatorname {\mathcal {H}}_{Z_1}^{}(b_1^{}) = s_1^{}$ for $b_1^{} \ge s_1^{} - 1$ . This leads to the desired outcome $\operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(a) = 0$ . We now perform the induction step. Assume that $\sigma (a)> 0$ . To simplify notation, assume that $a_2^{}, \dots , a_p^{}$ are positive and $a_{p + 1}^{}, \dots , a_q^{}$ are zero for some $p \in \{ 2, \dots , q \}$ . From the definition of the partial difference functions, we get

$$\begin{align*}\operatorname{\Delta}_1^{} \operatorname{\mathcal{H}}_X^{}(a) = \operatorname{\Delta}_{1, \dots, p}^{} \operatorname{\mathcal{H}}_X^{}(a) + \sum_{1 \le k \le p - 1} \operatorname{\Delta}_{1, \dots, k}^{} \operatorname{\mathcal{H}}_X^{}(a - e_{k + 1}^{}). \end{align*}$$

Since $\sigma (a - e_{k + 1}^{}) < \sigma (a)$ , $\operatorname {\Delta }_{1, \dots , k}^{} \operatorname {\mathcal {H}}_X^{}(a - e_{k + 1}^{})$ is a finite sum of expressions of the form $\pm \operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(b)$ , with $\sigma (b) < \sigma (a)$ and with $b_1^{} = a_1^{}$ . By the induction hypothesis, these expressions vanish. We obtain the equations

$$ \begin{align*} \operatorname{\Delta}_1^{} \operatorname{\mathcal{H}}_X^{}(a) & = \operatorname{\Delta}_{1, \dots, p}^{} \operatorname{\mathcal{H}}_X^{}(a) \\ & = (a_{p + 1}^{} + 1) \cdots (a_q^{} + 1) - \operatorname{\Delta}_{1, \dots, p}^{} \operatorname{\mathcal{H}}_{I(X)}^{}(a) \\ & = 1 - \operatorname{\Delta}_{1, \dots, p}^{} \operatorname{\mathcal{H}}_{I(X)}^{}(a). \end{align*} $$

We know that $I(Z_1^{})_{s_1}^{} \neq \{ 0 \}$ . It follows that $I(X)_a^{} \neq \{ 0 \}$ . Since we are assuming that X is ACM or sub-ACM, we may apply Propositions 7.1(vi) and 7.2(x) in order to obtain the inequality $\operatorname {\Delta }_{1, \dots , p}^{} \operatorname {\mathcal {H}}_{I(X)}^{}(a)> 0$ . A fortiori, $\operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(a) \le 0$ . According to Proposition 7.4(vii), the reverse inequality $\operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(a) \ge 0$ is also satisfied. We obtain the desired outcome $\operatorname {\Delta }_1^{} \operatorname {\mathcal {H}}_X^{}(a) = 0$ . This concludes the induction step.