The Stacks project

Proof. Let $X' \to \mathop{\mathrm{Spec}}(A)$ be the blowup, in other words

is the Proj of the Rees algebra. This in particular shows that $X'$ is integral and that $X' \to \mathop{\mathrm{Spec}}(A)$ is a projective modification. Let $X$ be the normalization of $X'$. Since $A$ is Nagata, we see that $\nu : X \to X'$ is finite (Morphisms, Lemma 29.54.11). Let $E' \subset X'$ be the exceptional divisor and let $E \subset X$ be the inverse image. Let $\mathcal{I}' \subset \mathcal{O}_{X'}$ and $\mathcal{I} \subset \mathcal{O}_ X$ be their ideal sheaves. Recall that $\mathcal{I}' = \mathcal{O}_{X'}(1)$ (Divisors, Lemma 31.32.13). Observe that $\mathcal{I} = \nu ^*\mathcal{I}'$ and that $E$ is an effective Cartier divisor (Divisors, Lemma 31.13.13). We are trying to show that $\nu $ is an isomorphism. As $\nu $ is finite, it suffices to show that $\mathcal{O}_{X'} \to \nu _*\mathcal{O}_ X$ is an isomorphism. If not, then we can find an $n \geq 0$ such that

for example because we can recover quasi-coherent $\mathcal{O}_{X'}$-modules from their associated graded modules, see Properties, Lemma 28.28.3. By the projection formula we have

the last equality by Lemma 54.9.3. On the other hand, there is clearly an injection $\mathfrak m^ n \to H^0(X', (\mathcal{I}')^ n)$. Since $H^0(X', (\mathcal{I}')^ n)$ is torsion free we conclude equality holds for all $n$, hence $X = X'$. $\square$