A point blowup lowers pairwise contact order by one and separates transverse branches

The regular curve germ $Y$ has prime ideal $P$ with $A\mathrm{/}P$ regular of dimension one. Its cotangent space has dimension one, so the kernel of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to \mathfrak{m}\mathrm{/}(P+{\mathfrak{m}}^2)$ is nonzero. Choose $y\in P$ with nonzero cotangent class and extend it to a parameter system $x,y$. The ring $A\mathrm{/}(y)$ is a one-dimensional regular local domain, hence a DVR. The prime $P\mathrm{/}(y)$ must be zero, since its quotient has dimension one, whereas the only nonzero prime in a DVR is maximal and has zero-dimensional quotient. Thus $P=(y)$. The same argument gives a principal equation $z$ of $Z$. In the DVR $A\mathrm{/}(y)$, $\bar{x}$ is a uniformizer and the contact order is $n={\mathrm{ord}}_{\bar{x}}\bar{z}$.

Write ${\mathrm{\ell }}_Y$ and ${\mathrm{\ell }}_Z$ for the leading forms of $y$ and $z$ in the symmetric algebra of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, so ${\mathrm{\ell }}_Y=y$ and ${\mathrm{\ell }}_Z=\alpha x+\beta y$ with $(\alpha ,\beta )\mathrm{\ne }(0,0)$ by the regularity of $Z$ at $p$ in [F5]. Since $\bar{z}=z(x,0)=\alpha x+O(x^2)$ in the DVR $A\mathrm{/}(y)$ with uniformizer $x$, the order is $n=1$ exactly when $\alpha \mathrm{\ne }0$: the tangent directions of $Y$ and $Z$ at $p$, cut out by ${\mathrm{\ell }}_Y$ and ${\mathrm{\ell }}_Z$, agree exactly when ${\mathrm{\ell }}_Z$ is a nonzero multiple of $y$, that is exactly when $\alpha =0$ and $\beta \mathrm{\ne }0$; hence $n=1$ if and only if the tangent directions differ, and in that case $\beta$ may be zero or not, while for $n>1$ the two curves have the common tangent direction cut out by $y$.

The first chart is $A[T]\mathrm{/}(xT-y)$: if $xg=(xT-y)h$, reduction modulo $x$ and regularity of $y$ modulo $x$ give $h=x{h}_1$, then cancellation gives $g=(xT-y)h_1$. Thus no $x$-power torsion remains in the incidence quotient. Work in the first chart $\mathrm{Spec}A[T]\mathrm{/}(xT-y)$, $T=y\mathrm{/}x$, so $y=xT$ and $E=V(x)$; by [F2] and [F3] this chart contains the point of $E$ corresponding to the tangent direction cut out by $y$, namely $T=0$, and the other chart covers the remaining points, so the two charts together see all of $E$. By [F4] applied to $Y$ and $Z$, whose local equations have multiplicity one at $p$, one has ${\pi }^{\ast }Y=Y^{\mathrm{\prime }}+E$ and ${\pi }^{\ast }Z=Z^{\mathrm{\prime }}+E$ as identities of effective Cartier divisors, well defined by [F6]; and $Y^{\mathrm{\prime }}$ meets $E$ in the reduced point cut out by ${\mathrm{\ell }}_Y$, $Z^{\mathrm{\prime }}$ in the reduced point cut out by ${\mathrm{\ell }}_Z$; explicitly in this chart the total transform of $Z$ is $V(z)$ with $z=xh$, $h=z\mathrm{/}x\in A[y\mathrm{/}x]$, and $Z^{\mathrm{\prime }}=V(h)$.

If $n=1$, then $\alpha \mathrm{\ne }0$ by step 2.1, so $h(0,0)=\alpha \mathrm{\ne }0$: the strict transform $Z^{\mathrm{\prime }}$ does not pass through the point $Y^{\mathrm{\prime }}\cap E=\{T=0\}$, and its intersection with $E$ is cut out by ${\mathrm{\ell }}_Z$ at the point of $E$ corresponding to the tangent direction of $Z$, which differs from that of $Y$ by step 2.1. On the open complement of $Z^{\mathrm{\prime }}\cap Y^{\mathrm{\prime }}$ (a closed subset missing $E$) the curves $Y^{\mathrm{\prime }}$ and $Z^{\mathrm{\prime }}$ are disjoint: they meet $E$ at distinct points and are therefore disjoint near $E$.

If $n>1$, then $\alpha =0$ and $\beta \mathrm{\ne }0$, so ${\mathrm{\ell }}_Z=\beta y$: both $Y^{\mathrm{\prime }}$ and $Z^{\mathrm{\prime }}$ meet $E$ at the single point $q$ corresponding to the common tangent direction cut out by $y$, namely $T=0$; and $h(x,0)=\bar{z}\mathrm{/}x=x^{n-1}u(x)$ for a unit $u$ of the DVR $A\mathrm{/}(y)$, because ${\mathrm{ord}}_x\bar{z}=n$ by step 1.1. The ambient local ring at $q$ is regular: the domain chart $B=A[T]\mathrm{/}(xT-y)$ has $B\mathrm{/}xB=\kappa (p)[T]$, and ${\mathcal{O}}_{S^{\mathrm{\prime }},q}=B_{(x,T)}$ has maximal ideal generated by $x,T$. The strict prime chain $(0)⊊(x)⊊(x,T)$ gives dimension at least two, while [F8] bounds it above by its embedding dimension at most two. Hence it is regular, with $x,T$ a cotangent basis. Since $h<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>x=\beta T$ is a nonzero linear form, $Z^{\mathrm{\prime }}$ is regular at $q$ by [F5], and $Y^{\mathrm{\prime }}=V(T)$ is regular there; in the DVR ${\mathcal{O}}_{Y^{\mathrm{\prime }},q}\cong (A\mathrm{/}(y))$ with uniformizer $x$, the ideal of $Z^{\mathrm{\prime }}$ is generated by $h(x,0)=x^{n-1}u(x)$, so the contact order of $Y^{\mathrm{\prime }}$ and $Z^{\mathrm{\prime }}$ at $q$ is $n-1$. Finally each of $Y^{\mathrm{\prime }},Z^{\mathrm{\prime }}$ meets $E$ in a $0$-cycle of degree one by [F4], so every intersection of a strict transform with $E$ has order one.

Steps 4.1 and 4.2 prove the two assertions: for transverse branches ($n=1$) the strict transforms meet $E$ at distinct points and are disjoint near $E$, while for $n>1$ they meet at the common tangent direction with contact order $n-1$, every intersection with $E$ having order one.