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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BI7. Beware of the difference between the letter 'O' and the digit '0'.