Controlled transforms preserve maximal order on nonempty transformed schemes

If $\mu =0$, maximal order forces $\mathcal{I}={\mathcal{O}}_X$, so the controlled transform is ${\mathcal{O}}_{X^{\mathrm{\prime }}}$, of order zero at every point. On a nonempty $X^{\mathrm{\prime }}$ it is nonzero; on the empty scheme the asserted order bound is vacuous. Now assume $\mu \ge 1$. Off the exceptional divisor the blow-up is an isomorphism, so the controlled transform has order at most $\mu$ there. It remains to check points over $C$.

Fix $c\in C$ and write the regular center locally as $J={\mathcal{I}}_C=(x_1,\dots ,x_q)$ in regular coordinates transverse to $C$, with additional coordinates along $C$. Since every point of $C$ has ${\mathrm{ord}}_c(\mathcal{I})=\mu$, $\mathcal{I}\subseteq J^{\mu }$ near $C$ by Controlled transforms are well defined. The normal associated-graded ring is a polynomial ring over ${\mathcal{O}}_C$, by Associated graded algebra of an ideal generated by a regular sequence. At each $c$, equality of the order gives some $f\in \mathcal{I}$ whose initial transverse form $F\in {\mathrm{Sym}}^{\mu }(J\mathrm{/}{J}^2)\otimes \kappa (c)$ is nonzero. In any blow-up chart over $c$ with exceptional equation $y=x_j$, the restriction of $y^{-\mu }{\sigma }^{\ast }(f)$ to the exceptional fiber is the nonzero dehomogenization of $F$, a polynomial of degree at most $\mu$. Its order at any point of that fiber is at most $\mu$. Indeed, for a nonzero polynomial $P$ of total degree $d\le \mu$, choose a nonzero top-degree monomial $c{Z}^{\alpha }$. The Hasse derivative of index $\alpha$ is the nonzero constant $c$. Hasse derivatives extend to the local polynomial ring by substituting $Z\mapsto Z+T$ and inverting denominators as formal series. Their higher Leibniz rule sends ${\mathfrak{m}}^N$ into ${\mathfrak{m}}^{N-\mathrm{\mid }\alpha \mathrm{\mid }}$, because at most $\mathrm{\mid }\alpha \mathrm{\mid }$ factors of a product can receive positive derivative index. Thus $P\in {\mathfrak{m}}^{d+1}$ would force the unit $c$ into $\mathfrak{m}$, a contradiction. This establishes the degree bound at every prime and in every characteristic; the order of the full local section is no greater than that of its restriction. Thus the controlled transform has order at most $\mu$ at every point over $C$, and is nonzero when $X^{\mathrm{\prime }}\mathrm{\ne }\mathrm{\varnothing }$, since the zero ideal at any point has infinite order. This proves the stated maximal-order conclusion with its empty-scheme qualification.