Proposition 3.3 — multigraded intersection bounds

Let $K$ be any nontrivially normed field for which there is an isometric field isomorphism either $K\cong\mathbb C$, or $K\cong\mathbb C_p$ for some prime natural number $p$, where $\mathbb C_p$ is the field denoted by $\mathrm{PadicComplex}(p)$. For every product $M$ of a positive number $s$ of projective spaces of dimensions $N_i\in\mathbb N$, every multihomogeneous ideal $I_0$ in its coordinate ring $R$, every $m\in\mathbb N$, every list $(P_j)_{j=1}^m$ of polynomials, and every natural block-degree vector $D$, suppose each $P_j$ is homogeneous of some block degree $e_j$ with $(e_j)_i\leq D_i$ for every $i$. Let $I=I_0+(P_1,\ldots,P_m)$. Then both of the following statements hold: $S_{\operatorname{MaxSpec}R}(\sqrt I,D)\leq S_{\operatorname{MaxSpec}R}(\sqrt{I_0},D)$; and, for every open subset $U$ of the maximal-ideal spectrum of $R$, if $I_0$ satisfies the local condition below at every point of $U$, then $S_U(I,D)\leq S_U(I_0,D)$. These are inequalities of rational numbers. In the first inequality the component ideals are formed from the radicals; in the second they are formed from the original ideals. For this space let $R=K[X_{ij}:1\leq i\leq s,\ 0\leq j\leq N_i]$. An ideal is multihomogeneous here precisely when it contains every block-homogeneous component of every one of its elements. Write $B_{\mathrm{irr}}=\bigcap_i(X_{i0},\ldots,X_{iN_i})$. For an ideal $J$, a minimal prime $\mathfrak q$ is counted on an open set $U$ of the maximal-ideal spectrum exactly when $B_{\mathrm{irr}}\nsubseteq\mathfrak q$ and some maximal ideal $\mathfrak m\in U$ contains $\mathfrak q$. Its contributing ideal is $C_{J,\mathfrak q}=(JR_{\mathfrak q})\cap R$, obtained by extending to the localization at $\mathfrak q$ and contracting. For any ideal $L$, form the rational polynomial $Q_L$ which equals the dimension of the image of the degree-$e$ homogeneous piece in $R/L$ for every coordinatewise sufficiently large $e\in\mathbb N^s$, with $Q_L=0$ if such a polynomial does not exist. Put $d_L=\deg Q_L$, taking $\deg 0=0$, and $F_L(D)=d_L!\,[Q_L]_{d_L}(D)$. The rational number $S_U(J,D)$ is the finite sum of $F_{C_{J,\mathfrak q}}(D)$ over precisely the counted minimal primes. It does not use an externally supplied component list or multiplicity; components of all dimensions are included when they pass the two tests. For an ideal $I_0\subseteq R$ and a maximal ideal $\mathfrak m$, the local condition on $I_0$ means that $B=R_{\mathfrak m}/I_0R_{\mathfrak m}$ is the zero ring, or there exists a finite list of nonunits in $B$ forming a regular sequence on $B$ whose length, as an element of $\mathbb N\cup\{\infty,-\infty\}$, equals the Krull dimension of $B$. Regularity means that multiplication by each entry is injective after quotienting by the previous entries and that the final quotient is nonzero. The condition on $U$ requires this at every maximal ideal in $U$. No nonzero assumption is imposed on the $P_j$, no common exact degree is required, and no properness assumption is imposed on $I_0$ or $I$. The case $m=0$ gives $I=I_0$ and both inequalities are equalities. For $U=\varnothing$ the local hypothesis is vacuous and both component sums are zero. An ideal with no minimal primes contributes an empty sum, hence zero; in particular this holds for the unit ideal. Zero entries of $D$ are allowed, with $0^0=1$, and the zero fallback for a Hilbert polynomial contributes zero. A zero-dimensional projective factor is allowed; a product with no factors is not.