The Stacks project

Lemma 54.14.4. Let $(A, \mathfrak m, \kappa )$ be a Noetherian complete local ring. Assume $A$ is a normal domain of dimension $2$. Then $\mathop{\mathrm{Spec}}(A)$ has a resolution of singularities.

Proof. A Noetherian complete local ring is J-2 (More on Algebra, Proposition 15.48.7), Nagata (Algebra, Proposition 10.162.16), excellent (More on Algebra, Proposition 15.52.3), and has a dualizing complex (Dualizing Complexes, Lemma 47.22.4). Moreover, the same is true for any ring essentially of finite type over $A$. If $B$ is a excellent local normal domain, then the completion $B^\wedge $ is normal (as $B \to B^\wedge $ is regular and More on Algebra, Lemma 15.42.2 applies). In other words, the local rings which we encounter in the rest of the proof will have the required “excellency” properties required of them.

Choose $A_0 \subset A$ with $A_0$ a regular complete local ring and $A_0 \to A$ finite, see Algebra, Lemma 10.160.11. This induces a finite extension of fraction fields $K/K_0$. We will argue by induction on $[K : K_0]$. The base case is when the degree is $1$ in which case $A_0 = A$ and the result is true.

Suppose there is an intermediate field $K_0 \subset L \subset K$, $K_0 \not= L \not= K$. Let $B \subset A$ be the integral closure of $A_0$ in $L$. By induction we choose a resolution of singularities $Y \to \mathop{\mathrm{Spec}}(B)$. Let $X$ be the normalization of $Y \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$. Picture:

\[ \xymatrix{ X \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A) \ar[d] \\ Y \ar[r] & \mathop{\mathrm{Spec}}(B) } \]

Since $A$ is J-2 the regular locus of $X$ is open. Since $X$ is a normal surface we conclude that $X$ has at worst finitely many singular points $x_1, \ldots , x_ n$ which are closed points with $\dim (\mathcal{O}_{X, x_ i}) = 2$. For each $i$ let $y_ i \in Y$ be the image. Since $\mathcal{O}_{Y, y_ i}^\wedge \to \mathcal{O}_{X, x_ i}^\wedge $ is finite of smaller degree than before we conclude by induction hypothesis that $\mathcal{O}_{X, x_ i}^\wedge $ has resolution of singularities. By Lemma 54.14.3 there is a sequence

\[ Z^\wedge _{i, n_ i} \to \ldots \to Z^\wedge _{i, 1} \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x_ i}^\wedge ) \]

of normalized blowups with $Z^\wedge _{i, n_ i}$ regular. By Lemma 54.11.7 there is a corresponding sequence of normalized blowing ups

\[ Z_{i, n_ i} \to \ldots \to Z_{i, 1} \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x_ i}) \]

Then $Z_{i, n_ i}$ is a regular scheme by Lemma 54.11.2. By Lemma 54.6.5 we can fit these normalized blowing ups into a corresponding sequence

\[ Z_ n \to Z_{n - 1} \to \ldots \to Z_1 \to X \]

and of course $Z_ n$ is regular too (look at the local rings). This proves the induction step.

Assume there is no intermediate field $K_0 \subset L \subset K$ with $K_0 \not= L \not= K$. Then either $K/K_0$ is separable or the characteristic to $K$ is $p$ and $[K : K_0] = p$. Then either Lemma 54.8.6 or 54.8.10 implies that reduction to rational singularities is possible. By Lemma 54.8.5 we conclude that there exists a normal modification $X \to \mathop{\mathrm{Spec}}(A)$ such that for every singular point $x$ of $X$ the local ring $\mathcal{O}_{X, x}$ defines a rational singularity. Since $A$ is J-2 we find that $X$ has finitely many singular points $x_1, \ldots , x_ n$. By Lemma 54.9.8 there exists a finite sequence of blowups in singular closed points

\[ X_{i, n_ i} \to X_{i, n_ i - 1} \to \ldots \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x_ i}) \]

such that $X_{i, n_ i}$ is Gorenstein, i.e., has an invertible dualizing module. By (the essentially trivial) Lemma 54.6.4 with $n = \sum n_ a$ these sequences correspond to a sequence of blowups

\[ X_ n \to X_{n - 1} \to \ldots \to X \]

such that $X_ n$ is normal and the local rings of $X_ n$ are Gorenstein. Again $X_ n$ has a finite number of singular points $x'_1, \ldots , x'_ s$, but this time the singularities are rational double points, more precisely, the local rings $\mathcal{O}_{X_ n, x'_ i}$ are as in Lemma 54.12.3. Arguing exactly as above we conclude that the lemma is true. $\square$

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