The Stacks project

Proof. The dualizing module $\omega _ A$ is a finite $A$-module whose stalk at the generic point is invertible. Namely, $\omega _ A \otimes _ A K$ is a dualizing module for the fraction field $K$ of $A$, hence has rank $1$. Thus there exists a blowup $b : Y \to \mathop{\mathrm{Spec}}(A)$ such that the strict transform of $\omega _ A$ with respect to $b$ is an invertible $\mathcal{O}_ Y$-module, see Divisors, Lemma 31.35.3. By Lemma 54.5.3 we can choose a sequence of normalized blowups

such that $X_ n$ dominates $Y$. By Lemma 54.9.4 and arguing by induction each $X_ i \to X_{i - 1}$ is simply a blowing up.

We claim that $\omega _{X_ n}$ is invertible. Since $\omega _{X_ n}$ is a coherent $\mathcal{O}_{X_ n}$-module, it suffices to see its stalks are invertible modules. If $x \in X_ n$ is a regular point, then this is clear from the fact that regular schemes are Gorenstein (Dualizing Complexes, Lemma 47.21.3). If $x$ is a singular point of $X_ n$, then each of the images $x_ i \in X_ i$ of $x$ is a singular point (because the blowup of a regular point is regular by Lemma 54.3.2). Consider the canonical map $f_ n^*\omega _ A \to \omega _{X_ n}$ of Lemma 54.9.6. For each $i$ the morphism $X_{i + 1} \to X_ i$ is either a blowup of $x_ i$ or an isomorphism at $x_ i$. Since $x_ i$ is always a singular point, it follows from Lemma 54.9.7 and induction that the maps $f_ i^*\omega _ A \to \omega _{X_ i}$ is always surjective on stalks at $x_ i$. Hence

is surjective. On the other hand, by our choice of $b$ the quotient of $f_ n^*\omega _ A$ by its torsion submodule is an invertible module $\mathcal{L}$. Moreover, the dualizing module is torsion free (Duality for Schemes, Lemma 48.22.3). It follows that $\mathcal{L}_ x \cong \omega _{X_ n, x}$ and the proof is complete. $\square$