Lemma 54.3.5. Let $(A, \mathfrak m)$ be a regular local ring of dimension $2$. Let $f : X \to S = \mathop{\mathrm{Spec}}(A)$ be the blowing up of $A$ in $\mathfrak m$. Let $\mathfrak m^ n \subset I \subset \mathfrak m$ be an ideal. Let $d \geq 0$ be the largest integer such that

$I \mathcal{O}_ X \subset \mathcal{O}_ X(-dE)$

where $E$ is the exceptional divisor. Set $\mathcal{I}' = I\mathcal{O}_ X(dE) \subset \mathcal{O}_ X$. Then $d > 0$, the sheaf $\mathcal{O}_ X/\mathcal{I}'$ is supported in finitely many closed points $x_1, \ldots , x_ r$ of $X$, and

\begin{align*} \text{length}_ A(A/I) & > \text{length}_ A \Gamma (X, \mathcal{O}_ X/\mathcal{I}') \\ & \geq \sum \nolimits _{i = 1, \ldots , r} \text{length}_{\mathcal{O}_{X, x_ i}} (\mathcal{O}_{X, x_ i}/\mathcal{I}'_{x_ i}) \end{align*}

Proof. Since $I \subset \mathfrak m$ we see that every element of $I$ vanishes on $E$. Thus we see that $d \geq 1$. On the other hand, since $\mathfrak m^ n \subset I$ we see that $d \leq n$. Consider the short exact sequence

$0 \to I\mathcal{O}_ X \to \mathcal{O}_ X \to \mathcal{O}_ X/I\mathcal{O}_ X \to 0$

Since $I\mathcal{O}_ X$ is globally generated, we see that $H^1(X, I\mathcal{O}_ X) = 0$ by Lemma 54.3.4. Hence we obtain a surjection $A/I \to \Gamma (X, \mathcal{O}_ X/I\mathcal{O}_ X)$. Consider the short exact sequence

$0 \to \mathcal{O}_ X(-dE)/I\mathcal{O}_ X \to \mathcal{O}_ X/I\mathcal{O}_ X \to \mathcal{O}_ X/\mathcal{O}_ X(-dE) \to 0$

By Divisors, Lemma 31.15.8 we see that $\mathcal{O}_ X(-dE)/I\mathcal{O}_ X$ is supported in finitely many closed points of $X$. In particular, this coherent sheaf has vanishing higher cohomology groups (detail omitted). Thus in the following diagram

$\xymatrix{ & & A/I \ar[d] \\ 0 \ar[r] & \Gamma (X, \mathcal{O}_ X(-dE)/I\mathcal{O}_ X) \ar[r] & \Gamma (X, \mathcal{O}_ X/I\mathcal{O}_ X) \ar[r] & \Gamma (X, \mathcal{O}_ X/\mathcal{O}_ X(-dE)) \ar[r] & 0 }$

the bottom row is exact and the vertical arrow surjective. We have

$\text{length}_ A \Gamma (X, \mathcal{O}_ X(-dE)/I\mathcal{O}_ X) < \text{length}_ A(A/I)$

since $\Gamma (X, \mathcal{O}_ X/\mathcal{O}_ X(-dE))$ is nonzero. Namely, the image of $1 \in \Gamma (X, \mathcal{O}_ X)$ is nonzero as $d > 0$.

To finish the proof we translate the results above into the statements of the lemma. Since $\mathcal{O}_ X(dE)$ is invertible we have

$\mathcal{O}_ X/\mathcal{I}' = \mathcal{O}_ X(-dE)/I\mathcal{O}_ X \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(dE).$

Thus $\mathcal{O}_ X/\mathcal{I}'$ and $\mathcal{O}_ X(-dE)/I\mathcal{O}_ X$ are supported in the same set of finitely many closed points, say $x_1, \ldots , x_ r \in E \subset X$. Moreover we obtain

$\Gamma (X, \mathcal{O}_ X(-dE)/I\mathcal{O}_ X) = \bigoplus \mathcal{O}_ X(-dE)_{x_ i}/I\mathcal{O}_{X, x_ i} \cong \bigoplus \mathcal{O}_{X, x_ i}/\mathcal{I}'_{x_ i} = \Gamma (X, \mathcal{O}_ X/\mathcal{I}')$

because an invertible module over a local ring is trivial. Thus we obtain the strict inequality. We also get the second because

$\text{length}_ A(\mathcal{O}_{X, x_ i}/\mathcal{I}'_{x_ i}) \geq \text{length}_{\mathcal{O}_{X, x_ i}}(\mathcal{O}_{X, x_ i}/\mathcal{I}'_{x_ i})$

as is immediate from the definition of length. $\square$

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