The monomial part, the non-monomial part and the companion ideal

Let $(\mathcal{I},E,\mu )$ be a marked ideal on the smooth $K$-scheme $X$, $\mathcal{I}$ generically nonzero on every irreducible component and $\mu \ge 1$ (Marked ideals and their support). Factoring out the monomial part, write $\mathcal{I}=M(\mathcal{I})\cdot N(\mathcal{I})$, where $M(\mathcal{I})$ is the monomial part of $\mathcal{I}$ with respect to $E$, i.e. the product of powers of the invertible ideal sheaves of the components of the members of $E$ that divide $\mathcal{I}$, and $N(\mathcal{I})$ is the complementary part, divisible by no component of a member of $E$; for $E=\mathrm{\varnothing }$ one has $M(\mathcal{I})={\mathcal{O}}_X$ and $N(\mathcal{I})=\mathcal{I}$. If $\mathrm{supp}(\mathcal{I},E,\mu )=\mathrm{\varnothing }$, the marked ideal is already resolved in Step 2, and no companion ideal is assigned. Otherwise put ${\mathrm{ord}}_N(\mathcal{I}):=\max \{{\mathrm{ord}}_x(N(\mathcal{I})):x\in \mathrm{supp}(\mathcal{I},E,\mu )\}\mathrm{.}$ If ${\mathrm{ord}}_N(\mathcal{I})=0$, then $N(\mathcal{I})$ is a unit in a neighborhood of the support, so $(\mathcal{I},\mu )$ is locally the monomial marked ideal $(M(\mathcal{I}),\mu )$ there. Route this case directly to Step 2b; it has no non-monomial companion. When $(\mathcal{I},\mu )$ is of maximal order, $(M(\mathcal{I}),\mu )$ is an equivalent maximal-order input for Step 1; for arbitrary inputs this is the monomial branch, without a claim that it is a maximal-order companion.

Write $r={\mathrm{ord}}_N(\mathcal{I})$. It is finite in the characteristic-zero setting of this page: the closed order-superlevel sets on the Noetherian $X$ stabilize, and their intersection is empty because $N(\mathcal{I})$ is generically nonzero on each smooth component (Order functions and normal-crossings strata are upper semicontinuous, Order of an ideal sheaf at a point). Work on the open neighbourhood $X\setminus \{x:{\mathrm{ord}}_{x}N(\mathcal{I})>r\}$ of the support. Here ${\mathrm{ord}}_{x}N(\mathcal{I})\le r$ everywhere; outside this neighbourhood there are no support points to resolve. This restriction is needed when the maximum was taken only over the support.

If $r>0$, define the companion ideal $O(\mathcal{I},\mu )$ by $O(\mathcal{I},\mu ):=(N(\mathcal{I}),{\mathrm{ord}}_N(\mathcal{I}))+(M(\mathcal{I}),\mu -{\mathrm{ord}}_N(\mathcal{I}))\text{when }{\mathrm{ord}}_N(\mathcal{I})<\mu ,$ and to $O(\mathcal{I},\mu ):=(N(\mathcal{I}),{\mathrm{ord}}_N(\mathcal{I}))$ when ${\mathrm{ord}}_N(\mathcal{I})\ge \mu$; the sum is that of Addition and multiplication of marked ideals. In the sum case, $0<{\mathrm{ord}}_N(\mathcal{I})<\mu$, so both marks are positive; the sum clause uses AC (The Axiom of Choice). On this neighbourhood the companion is of maximal order: in the sum case its $N^{\mu -r}$ summand has order at most $r(\mu -r)$, which is the sum marking, and in the other case $N$ has order at most $r$. Its support is $\mathrm{supp}(O(\mathcal{I},\mu ))=\mathrm{supp}(\mathcal{I},E,\mu )\cap \{x:{\mathrm{ord}}_x(N(\mathcal{I}))={\mathrm{ord}}_N(\mathcal{I})\}\mathrm{.}$ The support formula follows from the marked-sum intersection rule and exact additivity $\mathrm{ord}(MN)=\mathrm{ord}(M)+\mathrm{ord}(N)$ in the regular-local associated graded domain (associated graded ring of a regular local ring). A companion selects the maximal residual-order part of the support and need not be equivalent to the original input, even if that input is of maximal order. For example, on ${\mathbb{A}}^2⨿{\mathbb{A}}^2$ take $I=(x^2)$ on the first component and $I=(u,v{)}^2$ on the second, mark $2$, and boundary $V(x)$ on the first component. The original maximal-order support is $V(x)⨿\{(0,0)\}$; its companion has support only $\{(0,0)\}$ on the second component. Resolving $O(\mathcal{I},\mu )$ lowers the maximal order of the non-monomial part (Step 2a of the algorithm below).