The Stacks project

Proof. Say $T = \{ x_1, \ldots , x_ r\}$. Denote $I_ i$ the stalk of $\mathcal{I}$ at $x_ i$. Set

This is finite as $\mathcal{O}_ X/\mathcal{I}$ is supported on $T$ and hence $\mathcal{O}_{X, x_ i}/I_ i$ has support equal to $\{ \mathfrak m_{x_ i}\}$ (see Algebra, Lemma 10.62.3). We are going to use induction on $\sum n_ i$. If $n_ i = 0$ for all $i$, then $\mathcal{I} = \mathcal{O}_ X$ and we are done.

Suppose $n_ i > 0$. Let $X' \to X$ be the blowing up of $X$ in $x_ i$ (see discussion above the lemma). Since $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x_ i}) \to X$ is flat we see that $X' \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x_ i})$ is the blowup of the ring $\mathcal{O}_{X, x_ i}$ in the maximal ideal, see Divisors, Lemma 31.32.3. Hence the square in the commutative diagram

is cartesian. Let $E \subset X'$ and $E' \subset \text{Proj}(\bigoplus \nolimits _{d \geq 0} \mathfrak m_{x_ i}^ d)$ be the exceptional divisors. Let $d \geq 1$ be the integer found in Lemma 54.3.5 for the ideal $\mathcal{I}_ i \subset \mathcal{O}_{X, x_ i}$. Since the horizontal arrows in the diagram are flat, since $E' \to E$ is surjective, and since $E'$ is the pullback of $E$, we see that

(some details omitted). Set $\mathcal{I}' = \mathcal{I}\mathcal{O}_{X'}(dE) \subset \mathcal{O}_{X'}$. Then we see that $\mathcal{O}_{X'}/\mathcal{I}'$ is supported in finitely many closed points $T' \subset |X'|$ because this holds over $X \setminus \{ x_ i\}$ and for the pullback to $\text{Proj}(\bigoplus \nolimits _{d \geq 0} \mathfrak m_{x_ i}^ d)$. The final assertion of Lemma 54.3.5 tells us that the sum of the lengths of the stalks $\mathcal{O}_{X', x'}/\mathcal{I}'\mathcal{O}_{X', x'}$ for $x'$ lying over $x_ i$ is $< n_ i$. Hence the sum of the lengths has decreased.

By induction hypothesis, there exists a sequence

of blowups at closed points lying over $T'$ such that $\mathcal{I}'\mathcal{O}_{X'_ n}$ is invertible. Since $\mathcal{I}'\mathcal{O}_{X'}(-dE) = \mathcal{I}\mathcal{O}_{X'}$, we see that $\mathcal{I}\mathcal{O}_{X'_ n} = \mathcal{I}'\mathcal{O}_{X'_ n}(-d(f')^{-1}E)$ where $f' : X'_ n \to X'$ is the composition. Note that $(f')^{-1}E$ is an effective Cartier divisor by Divisors, Lemma 31.32.11. Thus we are done by Divisors, Lemma 31.13.7. $\square$