Total transform equals strict transform plus multiplicity times the exceptional divisor

At $p$ put $A={\mathcal{O}}_{S,p}$, $\mathfrak{m}=(x,y)$ and $\kappa =\kappa (p)$. The equation $f\in {\mathfrak{m}}^m\setminus {\mathfrak{m}}^{m+1}$ has nonzero leading form $f_m(X,Y)$ in $\kappa [X,Y]$. Write the finite ideal-power expression $f={\sum }_{i+j=m}{c}_{ij}{x}^i{y}^j$ with $c_{ij}\in A$. By [F9], $A$ is a domain and $x,y$ form a regular sequence. The chart is $B_x=A[T]\mathrm{/}(xT-y)=A[y\mathrm{/}x]$: if $xh=(xT-y)q$, reduction modulo $x$ and regularity of $y$ modulo $x$ give $q=x{q}_1$, and cancellation gives $h=(xT-y)q_1$; hence the incidence quotient has no $x$-power torsion and [F4, F7] identify it with the chart. In this chart, the ideal-power expression gives $f=x^{m}g$, where $g=\sum c_{ij}{T}^j$ and $g<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>x=f_m(1,T)\mathrm{\ne }0$. The ring $B_x$ is a domain and $B_x\mathrm{/}x{B}_x=\kappa [T]$, so $x$ is prime and does not divide $g$.

If $xh=gq$ in $B_x$, primality of $x$ forces $q=x{q}_1$, and cancellation gives $h=g{q}_1$. Therefore $(g):x=(g)$, and iteration gives $(x^{m}g):x^{\mathrm{\infty }}=(g)$. The strict-transform chart is thus $V(g)$. Since $g$ is a nonzero element of a domain, it is a Cartier equation, and the factorization $f=x^{m}g$ gives the total-transform identity there. In the second chart $B_y=A[U]\mathrm{/}(yU-x)$ the identical argument gives $f=y^{m}h$, with $h<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>y=f_m(U,1)\mathrm{\ne }0$. On the overlap $y=xT$, so cancellation of $x^m$ gives $g=T^{m}h$; the equations differ by a unit and glue. Off the exceptional curve the blowup is the identity, as follows by inverting the chart denominators. Hence globally $C^{\mathrm{\prime }}$ is effective Cartier and ${\pi }^{\ast }C=C^{\mathrm{\prime }}+mE$.

On $E\cong {\mathbb{P}}_{\kappa }^1$, these equations cut the homogeneous divisor of the nonzero degree-$m$ form $f_m$. To include the entire projective line, set $d=\mathrm{deg}{f}_m(1,T)\le m$. Its zeros in this affine chart have total degree $d$, since $\kappa [T]\mathrm{/}(f_m(1,T))$ has dimension $d$ (zero if $d=0$); this counts local lengths times residue degrees. At the omitted point, $U=0$, the identity $f_m(U,1)=U^m{f}_m(1,1\mathrm{/}U)$ gives order $m-d$. Thus the complete zero-cycle degree over $\kappa$ is $d+(m-d)=m$. When $[\kappa :k]$ is finite, each residue degree over $k$ is that degree times its degree over $\kappa$, giving $m[\kappa :k]$. No splitting or separability assumption is used.