Addition and multiplication of marked ideals

Order calculus. Work at $x\in X$, with $R={\mathcal{O}}_{X,x}$ and maximal ideal $\mathfrak{m}$. For ideals $A,B\subseteq R$, ${\mathrm{ord}}_x(A+B)=\min ({\mathrm{ord}}_{x}A,{\mathrm{ord}}_{x}B)$: containment of both ideals in $A+B$ gives one inequality, and $A+B\subseteq {\mathfrak{m}}^n$ forces both into ${\mathfrak{m}}^n$. Always ${\mathrm{ord}}_x(AB)\ge {\mathrm{ord}}_{x}A+{\mathrm{ord}}_{x}B$ by multiplying ideal containments. For assertion (1), if $A\mathrm{\ne }0$ has finite order $a$, choose $f\in A\setminus {\mathfrak{m}}^{a+1}$; its initial class in ${\mathrm{gr}}_{\mathfrak{m}}R$ is nonzero by [F1]. By [A1], [F5], and [F6], the associated-graded ring is a domain, so the initial class of $f^k$ is nonzero in degree $ka$; hence ${\mathrm{ord}}_x(A^k)=ka$, since the reverse inequality follows from $A\subseteq {\mathfrak{m}}^a$. The zero ideal has infinite order and its positive powers are zero, so the identity also holds there.

Supports. Put $\mu ={\prod }_j{\mu }_j$ and $e_j={\prod }_{k\mathrm{\ne }j}{\mu }_k$, all positive. By step 1.1, ${\mathrm{ord}}_x({\sum }_j{\mathcal{I}}_j^{e_j})={\min }_j{e}_j{\mathrm{ord}}_x({\mathcal{I}}_j)$. This is at least $\mu$ exactly when every ${\mathrm{ord}}_x({\mathcal{I}}_j)\ge {\mu }_j$, giving the support intersection in (1). For (2), if $x$ is in both factor supports, then ${\mathrm{ord}}_x(\mathcal{I})+{\mathrm{ord}}_x(\mathcal{J})\ge {\mu }_{\mathcal{I}}+{\mu }_{\mathcal{J}}$; the product lower bound in step 1.1 puts $x$ in the product support. This proves the stated reverse-direction inclusion, including zero marks.

Transform identities. For a blow-up with exceptional equation $y$ and center in the support of the sum, step 2.1 places that center in every summand support. Writing ${\mathcal{I}}_{j,i}=y^{-{\mu }_j}{\sigma }^{\ast }{\mathcal{I}}_j$, we have $y^{-\mu }{\sigma }^{\ast }({\sum }_j{\mathcal{I}}_j^{e_j})={\sum }_j(y^{-{\mu }_j}{\sigma }^{\ast }{\mathcal{I}}_j{)}^{e_j}={\sum }_j{\mathcal{I}}_{j,i}^{e_j}$ because $e_j{\mu }_j=\mu$. For the product, at any simultaneous admissible center, $y^{-({\mu }_{\mathcal{I}}+{\mu }_{\mathcal{J}})}{\sigma }^{\ast }(\mathcal{I}\mathcal{J})=(y^{-{\mu }_{\mathcal{I}}}{\sigma }^{\ast }\mathcal{I})(y^{-{\mu }_{\mathcal{J}}}{\sigma }^{\ast }\mathcal{J})$. Thus the corresponding controlled transforms agree. These are ideal-sheaf identities and use no choice.

Test blow-ups. The support equality in step 2.1 and transform identity in step 3.1 show inductively that a sequence is a multiple test blow-up of the sum exactly when each center is simultaneously admissible for every summand; all transformed supports agree at each stage. For the product, simultaneous admissibility puts each center in the intersection of factor supports and hence, by step 2.1, in the product support; the product transform identity then gives the induction that every simultaneous test sequence is a product test sequence.

Associativity of the sum. For either bracketing of three or more summands, step 2.1 identifies the initial supports and step 4.1 identifies the multiple test blow-ups and induced supports. The two bracketings therefore satisfy the equivalence criterion [F4], although their defining ideal sheaves need not be equal. This proves the final assertion.