Lemma 54.9.4. In Situation 54.9.1 the blowup of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$ is normal.

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

$X' = \text{Proj}(A \oplus \mathfrak m \oplus \mathfrak m^2 \oplus \ldots ).$

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 29.54.10). 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

$H^0(X', (\mathcal{I}')^ n) \not= H^0(X', (\nu _*\mathcal{O}_ X) \otimes (\mathcal{I}')^ n)$

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

$H^0(X', (\nu _*\mathcal{O}_ X) \otimes (\mathcal{I}')^ n) = H^0(X, \nu ^*(\mathcal{I}')^ n) = H^0(X, \mathcal{I}^ n) = \mathfrak m^ n$

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 $X = X'$. $\square$

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