Lemma 54.15.3. In the situation above let $X' \to X$ be the blowing up of $X$ in $p$. Let $Y', Z' \subset X'$ be the strict transforms of $Y, Z$. If $\mathcal{O}_{Y, p}$ is regular, then

1. $Y' \to Y$ is an isomorphism,

2. $Y'$ meets the exceptional fibre $E \subset X'$ in one point $q$ and $m_ q(Y \cap E) = 1$,

3. if $q \in Z'$ too, then $m_ q(Y \cap Z') < m_ p(Y \cap Z)$.

Proof. Since $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, p}$ is surjective and $\mathcal{O}_{Y, p}$ is a discrete valuation ring, we can pick an element $x_1 \in \mathfrak m_ p$ mapping to a uniformizer in $\mathcal{O}_{Y, p}$. Choose an affine open $U = \mathop{\mathrm{Spec}}(A)$ containing $p$ such that $x_1 \in A$. Let $\mathfrak m \subset A$ be the maximal ideal corresponding to $p$. Let $I, J \subset A$ be the ideals defining $Y, Z$ in $\mathop{\mathrm{Spec}}(A)$. After shrinking $U$ we may assume that $\mathfrak m = I + (x_1)$, in other words, that $V(x_1) \cap U \cap Y = \{ p\}$ scheme theoretically. We conclude that $p$ is an effective Cartier divisor on $Y$ and since $Y'$ is the blowing up of $Y$ in $p$ (Divisors, Lemma 31.33.2) we see that $Y' \to Y$ is an isomorphism by Divisors, Lemma 31.32.7. The relationship $\mathfrak m = I + (x_1)$ implies that $\mathfrak m^ n \subset I + (x_1^ n)$ hence we can define a map

$\psi : A[\textstyle {\frac{\mathfrak m}{x_1}}] \longrightarrow A/I$

by sending $y/x_1^ n \in A[\frac{\mathfrak m}{x_1}]$ to the class of $a$ in $A/I$ where $a$ is chosen such that $y \equiv ax_1^ n \bmod I$. Then $\psi$ corresponds to the morphism of $Y \cap U$ into $X'$ over $U$ given by $Y' \cong Y$. Since the image of $x_1$ in $A[\frac{\mathfrak m}{x_1}]$ cuts out the exceptional divisor we conclude that $m_ q(Y', E) = 1$. Finally, since $J \subset \mathfrak m$ implies that the ideal $J' \subset A[\frac{\mathfrak m}{x_1}]$ certainly contains the elements $f/x_1$ for $f \in J$. Thus if we choose $f \in J$ whose image $\overline{f}$ in $A/I$ has minimal valuation equal to $m_ p(Y \cap Z)$, then we see that $\psi (f/x_1) = \overline{f}/x_1$ in $A/I$ has valuation one less proving the last part of the lemma. $\square$

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