Strict-transform equation by removing the maximal exceptional power

Write the homogeneous decomposition of $f$ about the origin as $f=f_m+f_{m+1}+\cdots$, with $f_m\mathrm{\ne }0$ the leading form by [F1]. Substituting $y=xs$ gives $f(x,xs)={\sum }_{d\ge m}{x}^d{f}_d(1,s)=x^m(f_m(1,s)+x{f}_{m+1}(1,s)+x^2{f}_{m+2}(1,s)+\cdots )=x^{m}g(x,s)$, where $g(x,s):={\sum }_{d\ge m}{x}^{d-m}{f}_d(1,s)\in k[x,s]$.

The constant term in $x$ of $g$ is $f_m(1,s)$, which is nonzero: the distinct degree-$m$ monomials $x^{m-j}{y}^j$ become the distinct monomials $s^j$, so their nonzero coefficient vector cannot vanish. Consequently $g(0,s)=f_m(1,s)\mathrm{\ne }0$, and the exact power of $x$ dividing $f(x,xs)$ is $m$; since $E=V(x)$ on this chart by [F3], the equation of the total transform on the chart is $x^{m}g$ with $g$ not divisible by $x$.

By [F4] the total transform is ${\pi }^{\ast }C=C^{\mathrm{\prime }}+mE$, and in the chart its local equation is the product of a local equation of $C^{\mathrm{\prime }}$ with the $m$-th power $x^m$ of the exceptional equation; by step 2.1 the local equation of the total transform is $x^{m}g$ with $x\nmid g$, so the strict transform is cut out by $g=0$ in this chart, as claimed. The same computation with the roles of $x$ and $y$ interchanged, using the second chart with $x=yu$, gives the symmetric description $f(yu,y)=y^m\tilde{g}(u,y)$ with $\tilde{g}(u,0)=f_m(u,1)$ and strict transform $\tilde{g}=0$; the two chart equations glue to the strict transform by [F4] and Strict transform of a closed subscheme, since they are the saturations of the total transform by the exceptional equation on each chart.

A closed point $q$ of $E$ in the first chart corresponds to an irreducible polynomial $p(s)$, and its ambient maximal ideal is $(x,p(s))$. Let $e$ be the exponent of $p$ in the nonzero polynomial $f_m(1,s)$. Its image in $k[s{]}_{(p)}$ lies in $(p{)}^e\setminus (p{)}^{e+1}$. If $g$ belonged to $(x,p{)}^{e+1}$ in the local chart ring, reduction modulo $x$ would put that polynomial in $(p{)}^{e+1}$, a contradiction. Thus the order of $g$ is at most $e\le \mathrm{deg}{f}_m(1,s)\le m$. Points of $E$ outside the strict transform have unit equation and order zero. The second chart gives the identical bound, covering also the point at infinity. This proves the bound for every closed point, with arbitrary residue field; at the generic point of $E$, $g$ is a unit as well.

Steps 3.1 and 3.2 prove the assertions: the strict transform equation in each chart is obtained from the total transform by dividing by the largest power of the exceptional equation, which is exactly $x^m$ in the first chart and $y^m$ in the second, with the leading form evaluated at $(1,s)$ (respectively $(u,1)$) as the value along $E$, and the strict transform has multiplicity at most $m$ at every point of $E$.