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## Tag 0BGB

### 50.12. Rational double points

In Section 50.9 we argued that resolution of $2$-dimensional rational singularities reduces to the Gorenstein case. A Gorenstein rational surface singularity is a rational double point. We will resolve them by explicit computations.

According to the discussion in Examples, Section 101.17 there exists a normal Noetherian local domain $A$ whose completion is isomorphic to $\mathbf{C}[[x, y, z]]/(z^2)$. In this case one could say that $A$ has a rational double point singularity, but on the other hand, $\mathop{\mathrm{Spec}}(A)$ does not have a resolution of singularities. This kind of behaviour cannot occur if $A$ is a Nagata ring, see Algebra, Lemma 10.156.13.

However, it gets worse as there exists a local normal Nagata domain $A$ whose completion is $\mathbf{C}[[x, y, z]]/(yz)$ and another whose completion is $\mathbf{C}[[x, y, z]]/(y^2 - z^3)$. This is Example 2.5 of [Nishimura-few]. This is why we need to assume the completion of our ring is normal in this section.

Situation 50.12.1. Here $(A, \mathfrak m, \kappa)$ be a Nagata local normal domain of dimension $2$ which defines a rational singularity, whose completion is normal, and which is Gorenstein. We assume $A$ is not regular.

The arguments in this section will show that repeatedly blowing up singular points resolves $\mathop{\mathrm{Spec}}(A)$ in this situation. We will need the following lemma in the course of the proof.

Lemma 50.12.2. Let $\kappa$ be a field. Let $I \subset \kappa[x, y]$ be an ideal. Let $$a + b x + c y + d x^2 + exy + f y^2 \in I^2$$ for some $a, b, c, d, e, f \in k$ not all zero. If the colength of $I$ in $\kappa[x, y]$ is $> 1$, then $a + b x + c y + d x^2 + exy + f y^2 = j(g + hx + iy)^2$ for some $j, g, h, i \in \kappa$.

Proof. Consider the partial derivatives $b + 2dx + ey$ and $c + ex + 2fy$. By the Leibniz rules these are contained in $I$. If one of these is nonzero, then after a linear change of coordinates, i.e., of the form $x \mapsto \alpha + \beta x + \gamma y$ and $y \mapsto \delta + \epsilon x + \zeta y$, we may assume that $x \in I$. Then we see that $I = (x)$ or $I = (x, F)$ with $F$ a monic polynomial of degree $\geq 2$ in $y$. In the first case the statement is clear. In the second case observe that we can write any element in $I^2$ in the form $$A(x, y) x^2 + B(y) x F + C(y) F^2$$ for some $A(x, y) \in \kappa[x, y]$ and $B, C \in \kappa[y]$. Thus $$a + b x + c y + d x^2 + exy + f y^2 = A(x, y) x^2 + B(y) x F + C(y) F^2$$ and by degree reasons we see that $B = C = 0$ and $A$ is a constant.

To finish the proof we need to deal with the case that both partial derivatives are zero. This can only happen in characteristic $2$ and then we get $$a + d x^2 + f y^2 \in I^2$$ We may assume $f$ is nonzero (if not, then switch the roles of $x$ and $y$). After dividing by $f$ we obtain the case where the characteristic of $\kappa$ is $2$ and $$a + d x^2 + y^2 \in I^2$$ If $a$ and $d$ are squares in $\kappa$, then we are done. If not, then there exists a derivation $\theta : \kappa \to \kappa$ with $\theta(a) \not = 0$ or $\theta(d) \not = 0$, see Algebra, Lemma 10.152.2. We can extend this to a derivation of $\kappa[x, y]$ by setting $\theta(x) = \theta(y) = 0$. Then we find that $$\theta(a) + \theta(d) x^2 \in I$$ The case $\theta(d) = 0$ is absurd. Thus we may assume that $\alpha + x^2 \in I$ for some $\alpha \in \kappa$. Combining with the above we find that $a + \alpha d + y^2 \in I$. Hence $$J = (\alpha + x^2, a + \alpha d + y^2) \subset I$$ with codimension at most $2$. Observe that $J/J^2$ is free over $\kappa[x, y]/J$ with basis $\alpha + x^2$ and $a + \alpha d + y^2$. Thus $a + d x^2 + y^2 = 1 \cdot (a + \alpha d + y^2) + d \cdot (\alpha + x^2) \in I^2$ implies that the inclusion $J \subset I$ is strict. Thus we find a nonzero element of the form $g + hx + iy + jxy$ in $I$. If $j = 0$, then $I$ contains a linear form and we can conclude as in the first paragraph. Thus $j \not = 0$ and $\dim_\kappa(I/J) = 1$ (otherwise we could find an element as above in $I$ with $j = 0$). We conclude that $I$ has the form $(\alpha + x^2, \beta + y^2, g + hx + iy + jxy)$ with $j \not = 0$ and has colength $3$. In this case $a + dx^2 + y^2 \in I^2$ is impossible. This can be shown by a direct computation, but we prefer to argue as follows. Namely, to prove this statement we may assume that $\kappa$ is algebraically closed. Then we can do a coordinate change $x \mapsto \sqrt{\alpha} + x$ and $y \mapsto \sqrt{\beta} + y$ and assume that $I = (x^2, y^2, g' + h'x + i'y + jxy)$ with the same $j$. Then $g' = h' = i' = 0$ otherwise the colength of $I$ is not $3$. Thus we get $I = (x^2, y^2, xy)$ and the result is clear. $\square$

Let $(A, \mathfrak m, \kappa)$ be as in Situation 50.12.1. Let $X \to \mathop{\mathrm{Spec}}(A)$ be the blowing up of $\mathfrak m$ in $\mathop{\mathrm{Spec}}(A)$. By Lemma 50.9.4 we see that $X$ is normal. All singularities of $X$ are rational singularities by Lemma 50.8.4. Since $\omega_A = A$ we see from Lemma 50.9.7 that $\omega_X \cong \mathcal{O}_X$ (see discussion in Remark 50.7.7 for conventions). Thus all singularities of $X$ are Gorenstein. Moreover, the local rings of $X$ at closed point have normal completions by Lemma 50.11.4. In other words, by blowing up $\mathop{\mathrm{Spec}}(A)$ we obtain a normal surface $X$ whose singular points are as in Situation 50.12.1. We will use this below without further mention. (Note: we will see in the course of the discussion below that there are finitely many of these singular points.)

Let $E \subset X$ be the exceptional divisor. We have $\omega_E = \mathcal{O}_E(-1)$ by Lemma 50.9.7. By Lemma 50.9.5 we have $\kappa = H^0(E, \mathcal{O}_E)$. Thus $E$ is a Gorenstein curve and by Riemann-Roch as discussed in Algebraic Curves, Section 49.5 we have $$\chi(E, \mathcal{O}_E) = 1 - g = -(1/2) \deg(\omega_E) = (1/2)\deg(\mathcal{O}_E(1))$$ where $g = \dim_\kappa H^1(E, \mathcal{O}_E) \geq 0$. Since $\deg(\mathcal{O}_E(1))$ is positive by Varieties, Lemma 32.43.15 we find that $g = 0$ and $\deg(\mathcal{O}_E(1)) = 2$. It follows that we have $$\dim_\kappa (\mathfrak m^n/\mathfrak m^{n + 1}) = 2n + 1$$ by Lemma 50.9.5 and Riemann-Roch on $E$.

Choose $x_1, x_2, x_3 \in \mathfrak m$ which map to a basis of $\mathfrak m/\mathfrak m^2$. Because $\dim_\kappa(\mathfrak m^2/\mathfrak m^3) = 5$ the images of $x_i x_j$, $i \geq j$ in this $\kappa$-vector space satisfy a relation. In other words, we can find $a_{ij} \in A$, $i \geq j$, not all contained in $\mathfrak m$, such that $$a_{11} x_1^2 + a_{12} x_1x_2 + a_{13}x_1x_3 + a_{22} x_2^2 + a_{23} x_2x_3 + a_{33} x_3^2 = \sum a_{ijk} x_ix_jx_k$$ for some $a_{ijk} \in A$ where $i \leq j \leq k$. Denote $a \mapsto \overline{a}$ the map $A \to \kappa$. The quadratic form $q = \sum \overline{a}_{ij} t_i t_j \in \kappa[t_1, t_2, t_3]$ is well defined up to multiplication by an element of $\kappa^*$ by our choices. If during the course of our arguments we find that $\overline{a}_{ij} = 0$ in $\kappa$, then we can subsume the term $a_{ij} x_i x_j$ in the right hand side and assume $a_{ij} = 0$; this operation changes the $a_{ijk}$ but not the other $a_{i'j'}$.

The blowing up is covered by $3$ affine charts corresponding to the ''variables'' $x_1, x_2, x_3$. By symmetry it suffices to study one of the charts. To do this let $$A' = A[\mathfrak m/x_1]$$ be the affine blowup algebra (as in Algebra, Section 10.69). Since $x_1, x_2, x_3$ generate $\mathfrak m$ we see that $A'$ is generated by $y_2 = x_2/x_1$ and $y_3 = x_3/x_1$ over $A$. We will occasionally use $y_1 = 1$ to simplify formulas. Moreover, looking at our relation above we find that $$a_{11} + a_{12} y_2 + a_{13} y_3 + a_{22} y_2^2 + a_{23} y_2y_3 + a_{33} y_3^2 = x_1 (\sum a_{ijk} y_iy_jy_k)$$ in $A'$. Recall that $x_1 \in A'$ defines the exceptional divisor $E$ on our affine open of $X$ which is therefore scheme theoretically given by $$\kappa[y_2, y_3]/ (\overline{a}_{11} + \overline{a}_{12} y_2 + \overline{a}_{13} y_3 + \overline{a}_{22} y_2^2 + \overline{a}_{23} y_2y_3 + \overline{a}_{33} y_3^2)$$ In other words, $E \subset \mathbf{P}^2_\kappa = \text{Proj}(\kappa[t_1, t_2, t_3])$ is the zero scheme of the quadratic form $q$ introduced above.

The quadratic form $q$ is an important invariant of the singularity defined by $A$. Let us say we are in case II if $q$ is a square of a linear form times an element of $\kappa^*$ and in case I otherwise. Observe that we are in case II exactly if, after changing our choice of $x_1, x_2, x_3$, we have $$x_3^2 = \sum a_{ijk}x_ix_jx_k$$ in the local ring $A$.

Let $\mathfrak m' \subset A'$ be a maximal ideal lying over $\mathfrak m$ with residue field $\kappa'$. In other words, $\mathfrak m'$ corresponds to a closed point $p \in E$ of the exceptional divisor. Recall that the surjection $$\kappa[y_2, y_3] \to \kappa'$$ has kernel generated by two elements $f_2, f_3 \in \kappa[y_2, y_3]$ (see for example Algebra, Example 10.26.3 or the proof of Algebra, Lemma 10.113.1). Let $z_2, z_3 \in A'$ map to $f_2, f_3$ in $\kappa[y_2, y_3]$. Then we see that $\mathfrak m' = (x_1, z_2, z_3)$ because $x_2$ and $x_3$ become divisible by $x_1$ in $A'$.

Claim. If $X$ is singular at $p$, then $\kappa' = \kappa$ or we are in case II. Namely, if $A'_{\mathfrak m'}$ is singular, then $\dim_{\kappa'} \mathfrak m'/(\mathfrak m')^2 = 3$ which implies that $\dim_{\kappa'} \overline{\mathfrak m}'/(\overline{\mathfrak m}')^2 = 2$ where $\overline{m}'$ is the maximal ideal of $\mathcal{O}_{E, p} = \mathcal{O}_{X, p}/x_1\mathcal{O}_{X, p}$. This implies that $$q(1, y_2, y_3) = \overline{a}_{11} + \overline{a}_{12} y_2 + \overline{a}_{13} y_3 + \overline{a}_{22} y_2^2 + \overline{a}_{23} y_2y_3 + \overline{a}_{33} y_3^2 \in (f_2, f_3)^2$$ otherwise there would be a relation between the classes of $z_2$ and $z_3$ in $\overline{\mathfrak m}'/(\overline{\mathfrak m}')^2$. The claim now follows from Lemma 50.12.2.

Resolution in case I. By the claim any singular point of $X$ is $\kappa$-rational. Pick such a singular point $p$. We may choose our $x_1, x_2, x_3 \in \mathfrak m$ such that $p$ lies on the chart described above and has coordinates $y_2 = y_3 = 0$. Since it is a singular point arguing as in the proof of the claim we find that $q(1, y_2, y_3) \in (y_2, y_3)^2$. Thus we can choose $a_{11} = a_{12} = a_{13} = 0$ and $q(t_1, t_2, t_3) = q(t_2, t_3)$. It follows that $$E = V(q) \subset \mathbf{P}^1_\kappa$$ either is the union of two distinct lines meeting at $p$ or is a degree $2$ curve with a unique $\kappa$-rational point (small detail omitted; use that $q$ is not a square of a linear form up to a scalar). In both cases we conclude that $X$ has a unique singular point $p$ which is $\kappa$-rational. We need a bit more information in this case. First, looking at higher terms in the expression above, we find that $\overline{a}_{111} = 0$ because $p$ is singular. Then we can write $a_{111} = b_{111} x_1 \bmod (x_2, x_3)$ for some $b_{111} \in A$. Then the quadratic form at $p$ for the generators $x_1, y_2, y_3$ of $\mathfrak m'$ is $$q' = \overline{b}_{111} t_1^2 + \overline{a}_{112} t_1 t_2 + \overline{a}_{113} t_1 t_3 + \overline{a}_{22} t_2^2 + \overline{a}_{23} t_2 t_3 + \overline{a}_{33} t_3^2$$ We see that $E' = V(q')$ intersects the line $t_1 = 0$ in either two points or one point of degree $2$. We conclude that $p$ lies in case I.

Suppose that the blowing up $X' \to X$ of $X$ at $p$ again has a singular point $p'$. Then we see that $p'$ is a $\kappa$-rational point and we can blow up to get $X'' \to X'$. If this process does not stop we get a sequence of blowings up $$\mathop{\mathrm{Spec}}(A) \leftarrow X \leftarrow X' \leftarrow X'' \leftarrow \ldots$$ We want to show that Lemma 50.10.1 applies to this situation. To do this we have to say something about the choice of the element $x_1$ of $\mathfrak m$. Suppose that $A$ is in case I and that $X$ has a singular point. Then we will say that $x_1 \in \mathfrak m$ is a good coordinate if for any (equivalently some) choice of $x_2, x_3$ the quadratic form $q(t_1, t_2, t_3)$ has the property that $q(0, t_2, t_3)$ is not a scalar times a square. We have seen above that a good coordinate exists. If $x_1$ is a good coordinate, then the singular point $p \in E$ of $X$ does not lie on the hypersurface $t_1 = 0$ because either this does not have a rational point or if it does, then it is not singular on $X$. Observe that this is equivalent to the statement that the image of $x_1$ in $\mathcal{O}_{X, p}$ cuts out the exceptional divisor $E$. Now the computations above show that if $x_1$ is a good coordinate for $A$, then $x_1 \in \mathfrak m'\mathcal{O}_{X, p}$ is a good coordinate for $p$. This of course uses that the notion of good coordinate does not depend on the choice of $x_2$, $x_3$ used to do the computation. Hence $x_1$ maps to a good coordinate at $p'$, $p''$, etc. Thus Lemma 50.10.1 applies and our sequence of blowing ups comes from a nonsingular arc $A \to R$. Then the map $A^\wedge \to R$ is a surjection. Since the completion of $A$ is normal, we conclude by Lemma 50.10.2 that after a finite number of blowups $$\mathop{\mathrm{Spec}}(A^\wedge) \leftarrow X^\wedge \leftarrow (X')^\wedge \leftarrow \ldots$$ the resulting scheme $(X^{(n)})^\wedge$ is regular. Since $(X^{(n)})^\wedge \to X^{(n)}$ induces isomorphisms on complete local rings (Lemma 50.11.1) we conclude that the same is true for $X^{(n)}$.

Resolution in case II. Here we have $$x_3^2 = \sum a_{ijk}x_ix_jx_k$$ in $A$ for some choice of generators $x_1, x_2, x_3$ of $\mathfrak m$. Then $q = t_3^2$ and $E = 2C$ where $C$ is a line. Recall that in $A'$ we get $$y_3^2 = x_1(\sum a_{ijk} y_iy_jy_k)$$ Since we know that $X$ is normal, we get a discrete valuation ring $\mathcal{O}_{X, \xi}$ at the generic point $\xi$ of $C$. The element $y_3 \in A'$ maps to a uniformizer of $\mathcal{O}_{X, \xi}$. Since $x_1$ scheme theoretically cuts out $E$ which is $C$ with multiplicity $2$, we see that $x_1$ is a unit times $y_3^2$ in $\mathcal{O}_{X, \xi}$. Looking at our equality above we conclude that $$h(y_2) = \overline{a}_{111} + \overline{a}_{112} y_2 + \overline{a}_{122} y_2^2 + \overline{a}_{222} y_2^3$$ must be nonzero in the residue field of $\xi$. Now, suppose that $p \in C$ defines a singular point. Then $y_3$ is zero at $p$ and $p$ must correspond to a zero of $h$ by the reasoning used in proving the claim above. If $h$ does not have a double zero at $p$, then the quadratic form $q'$ at $p$ is not a square and we conclude that $p$ falls in case I which we have treated above1. Since the degree of $h$ is $3$ we get at most one singular point $p \in C$ falling into case II which is moreover $\kappa$-rational. After changing our choice of $x_1, x_2, x_3$ we may assume this is the point $y_2 = y_3 = 0$. Then $h = \overline{a}_{122} y_2^2 + \overline{a}_{222} y_2^3$. Moreover, it still has to be the case that $\overline{a}_{113} = 0$ for the quadratic form $q'$ to have the right shape. Thus the local ring $\mathcal{O}_{X, p}$ defines a singularity as in the next paragraph.

The final case we treat is the case where we can choose our generators $x_1, x_2, x_3$ of $\mathfrak m$ such that $$x_3^2 + x_1(a x_2^2 + b x_2x_3 + c x_3^2) \in \mathfrak m^4$$ for some $a, b, c \in A$. This is a subclass of case II. If $\overline{a} = 0$, then we can write $a = a_1 x_1 + a_2 x_2 + a_3 x_3$ and we get after blowing up $$y_3^2 + x_1(a_1 x_1 y_2^2 + a_2 x_1 y_2^3 + a_3 x_1 y_2^2 y_3 + b y_2 y_3 + c y_3^2) = x_1^2 (\sum a_{ijkl}y_iy_jy_ky_l)$$ This means that $X$ is not normal2 a contradiction. By the result of the previous paragraph, if the blowup $X$ has a singular point $p$ which falls in case II, then there is only one and it is $\kappa$-rational. Computing the affine blowup algebras $A[\frac{\mathfrak m}{x_2}]$ and $A[\frac{\mathfrak m}{x_3}]$ the reader easily sees that $p$ cannot be contained the corresponding opens of $X$. Thus $p$ is in the spectrum of $A[\frac{\mathfrak m}{x_1}]$. Doing the blowing up as before we see that $p$ must be the point with coordinates $y_2 = y_3 = 0$ and the new equation looks like $$y_3^2 + x_1(a y_2^2 + b y_2 y_3 + c y_3^2) \in (\mathfrak m')^4$$ which has the same shape as before and has the property that $x_1$ defines the exceptional divisor. Thus if the process does not stop we get an infinite sequence of blowups and on each of these $x_1$ defines the exceptional divisor in the local ring of the singular point. Thus we can finish the proof using Lemmas 50.10.1 and 50.10.2 and the same reasoning as before.

Lemma 50.12.3. Let $(A, \mathfrak m, \kappa)$ be a local normal Nagata domain of dimension $2$ which defines a rational singularity, whose completion is normal, and which is Gorenstein. Then there exists a finite sequence of blowups in singular closed points $$X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = \mathop{\mathrm{Spec}}(A)$$ such that $X_n$ is regular and such that each intervening schemes $X_i$ is normal with finitely many singular points of the same type.

Proof. This is exactly what was proved in the discussion above. $\square$

1. The maximal ideal at $p$ in $A'$ is generated by $y_3, x_1$ and a third element $g$ whose image in $\kappa[y_2]$ is the prime divisor of $h$ corresponding to $p$. If this prime divisor doesn't divide $h$ twice, then we see that the quadratic form at $p$ looks like $$y_3^2 - x_1((something)x_1 + (something)y_3 + (unit)g)$$ and this can never be a square in $\kappa[y_3, x_1, g]$.
2. Namely, the equation shows that you get something singular along the $1$-dimensional locus $x_1 = y_3 = 0$ which cannot happen for a normal surface.

The code snippet corresponding to this tag is a part of the file resolve.tex and is located in lines 3029–3488 (see updates for more information).

\section{Rational double points}
\label{section-rational-double-points}

\noindent
In Section \ref{section-rational-singularities}
we argued that resolution of $2$-dimensional
rational singularities reduces to the Gorenstein case.
A Gorenstein rational surface singularity is a rational double point.
We will resolve them by explicit computations.

\medskip\noindent
According to the discussion in Examples, Section \ref{examples-section-bad}
there exists a normal Noetherian local domain $A$ whose completion
is isomorphic to $\mathbf{C}[[x, y, z]]/(z^2)$. In this case one could
say that $A$ has a rational double point singularity, but on the other
hand, $\Spec(A)$ does not have a resolution of singularities.
This kind of behaviour cannot occur if $A$ is a Nagata ring, see
Algebra, Lemma \ref{algebra-lemma-local-nagata-domain-analytically-unramified}.

\medskip\noindent
However, it gets worse as there exists a local normal Nagata domain $A$
whose completion is $\mathbf{C}[[x, y, z]]/(yz)$ and another whose
completion is $\mathbf{C}[[x, y, z]]/(y^2 - z^3)$. This is Example 2.5 of
\cite{Nishimura-few}. This is why we need to assume the completion of
our ring is normal in this section.

\begin{situation}
\label{situation-rational-double-point}
Here $(A, \mathfrak m, \kappa)$ be a Nagata local normal domain of
dimension $2$ which defines a rational singularity, whose completion
is normal, and which is Gorenstein. We assume $A$ is not regular.
\end{situation}

\noindent
The arguments in this section will show that repeatedly blowing
up singular points resolves $\Spec(A)$ in this situation. We will
need the following lemma in the course of the proof.

\begin{lemma}
\label{lemma-issquare}
Let $\kappa$ be a field. Let $I \subset \kappa[x, y]$ be an ideal. Let
$$a + b x + c y + d x^2 + exy + f y^2 \in I^2$$
for some $a, b, c, d, e, f \in k$ not all zero. If the colength
of $I$ in $\kappa[x, y]$ is $> 1$, then
$a + b x + c y + d x^2 + exy + f y^2 = j(g + hx + iy)^2$
for some $j, g, h, i \in \kappa$.
\end{lemma}

\begin{proof}
Consider the partial derivatives $b + 2dx + ey$ and
$c + ex + 2fy$. By the Leibniz rules these are contained in $I$.
If one of these is nonzero, then after a linear change of coordinates,
i.e., of the form $x \mapsto \alpha + \beta x + \gamma y$ and
$y \mapsto \delta + \epsilon x + \zeta y$, we may assume
that $x \in I$. Then we see that $I = (x)$ or $I = (x, F)$ with
$F$ a monic polynomial of degree $\geq 2$ in $y$.
In the first case the statement is clear. In the second case
observe that we can write any element in $I^2$ in the form
$$A(x, y) x^2 + B(y) x F + C(y) F^2$$
for some $A(x, y) \in \kappa[x, y]$ and $B, C \in \kappa[y]$.
Thus
$$a + b x + c y + d x^2 + exy + f y^2 = A(x, y) x^2 + B(y) x F + C(y) F^2$$
and by degree reasons we see that $B = C = 0$ and $A$ is a constant.

\medskip\noindent
To finish the proof we need to deal with the case that both
partial derivatives are zero. This can only happen in characteristic $2$
and then we get
$$a + d x^2 + f y^2 \in I^2$$
We may assume $f$ is nonzero (if not, then switch the roles of $x$ and $y$).
After dividing by $f$ we obtain the case where the characteristic of
$\kappa$ is $2$ and
$$a + d x^2 + y^2 \in I^2$$
If $a$ and $d$ are squares in $\kappa$, then we are done. If not,
then there exists a derivation $\theta : \kappa \to \kappa$ with
$\theta(a) \not = 0$ or $\theta(d) \not = 0$, see
Algebra, Lemma \ref{algebra-lemma-derivative-zero-pth-power}.
We can extend this to a derivation of $\kappa[x, y]$ by setting
$\theta(x) = \theta(y) = 0$. Then we find that
$$\theta(a) + \theta(d) x^2 \in I$$
The case $\theta(d) = 0$ is absurd. Thus we may assume
that $\alpha + x^2 \in I$ for some $\alpha \in \kappa$.
Combining with the above we find that $a + \alpha d + y^2 \in I$.
Hence
$$J = (\alpha + x^2, a + \alpha d + y^2) \subset I$$
with codimension at most $2$. Observe that
$J/J^2$ is free over $\kappa[x, y]/J$ with basis
$\alpha + x^2$ and $a + \alpha d + y^2$.
Thus $a + d x^2 + y^2 = 1 \cdot (a + \alpha d + y^2) + d \cdot (\alpha + x^2) \in I^2$
implies that the inclusion $J \subset I$ is strict.
Thus we find a nonzero element of the form $g + hx + iy + jxy$ in $I$.
If $j = 0$, then $I$ contains a linear form and we can
conclude as in the first paragraph. Thus $j \not = 0$
and $\dim_\kappa(I/J) = 1$ (otherwise we could find
an element as above in $I$ with $j = 0$).
We conclude that $I$ has the form
$(\alpha + x^2, \beta + y^2, g + hx + iy + jxy)$
with $j \not = 0$ and has colength $3$.
In this case $a + dx^2 + y^2 \in I^2$ is impossible.
This can be shown by a direct computation, but we prefer to argue
as follows. Namely, to prove this statement we may assume that
$\kappa$ is algebraically closed. Then we can do a coordinate
change $x \mapsto \sqrt{\alpha} + x$ and $y \mapsto \sqrt{\beta} + y$
and assume that $I = (x^2, y^2, g' + h'x + i'y + jxy)$ with the same $j$.
Then $g' = h' = i' = 0$ otherwise the colength of $I$ is not $3$.
Thus we get $I = (x^2, y^2, xy)$ and the result is clear.
\end{proof}

\noindent
Let $(A, \mathfrak m, \kappa)$ be as in
Situation \ref{situation-rational-double-point}.
Let $X \to \Spec(A)$ be the blowing up of $\mathfrak m$ in $\Spec(A)$.
By Lemma \ref{lemma-blow-up-normal-rational} we see that $X$ is normal.
All singularities of $X$ are rational singularities
by Lemma \ref{lemma-rational-propagates}.
Since $\omega_A = A$ we see from Lemma \ref{lemma-dualizing-blow-up-rational}
that $\omega_X \cong \mathcal{O}_X$ (see discussion in
Remark \ref{remark-dualizing-setup} for conventions).
Thus all singularities of $X$ are Gorenstein.
Moreover, the local rings of $X$ at closed point have
normal completions by Lemma \ref{lemma-blowup-still-good}.
In other words, by blowing up $\Spec(A)$ we obtain a normal
surface $X$ whose singular points are as in
Situation \ref{situation-rational-double-point}.
We will use this below without further mention.
(Note: we will see in the course of the discussion below
that there are finitely many of these singular points.)

\medskip\noindent
Let $E \subset X$ be the exceptional divisor. We have
$\omega_E = \mathcal{O}_E(-1)$ by Lemma \ref{lemma-dualizing-blow-up-rational}.
By Lemma \ref{lemma-cohomology-blow-up-rational} we have
$\kappa = H^0(E, \mathcal{O}_E)$.
Thus $E$ is a Gorenstein curve and by Riemann-Roch as discussed in
Algebraic Curves, Section \ref{curves-section-Riemann-Roch}
we have
$$\chi(E, \mathcal{O}_E) = 1 - g = -(1/2) \deg(\omega_E) = (1/2)\deg(\mathcal{O}_E(1))$$
where $g = \dim_\kappa H^1(E, \mathcal{O}_E) \geq 0$.
Since $\deg(\mathcal{O}_E(1))$ is positive
by Varieties, Lemma
\ref{varieties-lemma-ampleness-in-terms-of-degrees-components}
we find that $g = 0$ and $\deg(\mathcal{O}_E(1)) = 2$. It follows that
we have
$$\dim_\kappa (\mathfrak m^n/\mathfrak m^{n + 1}) = 2n + 1$$
by Lemma \ref{lemma-cohomology-blow-up-rational} and Riemann-Roch
on $E$.

\medskip\noindent
Choose $x_1, x_2, x_3 \in \mathfrak m$ which map to a basis of
$\mathfrak m/\mathfrak m^2$. Because
$\dim_\kappa(\mathfrak m^2/\mathfrak m^3) = 5$
the images of $x_i x_j$, $i \geq j$ in this $\kappa$-vector space
satisfy a relation. In other words, we can find $a_{ij} \in A$,
$i \geq j$, not all contained in $\mathfrak m$, such that
$$a_{11} x_1^2 + a_{12} x_1x_2 + a_{13}x_1x_3 + a_{22} x_2^2 + a_{23} x_2x_3 + a_{33} x_3^2 = \sum a_{ijk} x_ix_jx_k$$
for some $a_{ijk} \in A$ where $i \leq j \leq k$. Denote
$a \mapsto \overline{a}$ the map $A \to \kappa$.
$q = \sum \overline{a}_{ij} t_i t_j \in \kappa[t_1, t_2, t_3]$
is well defined up to multiplication by an element of $\kappa^*$
by our choices. If during the course of our arguments we find
that $\overline{a}_{ij} = 0$ in $\kappa$,
then we can subsume the term $a_{ij} x_i x_j$ in the right
hand side and assume $a_{ij} = 0$; this operation changes the $a_{ijk}$
but not the other $a_{i'j'}$.

\medskip\noindent
The blowing up is covered by $3$ affine charts corresponding to
the variables'' $x_1, x_2, x_3$. By symmetry it suffices to study
one of the charts. To do this let
$$A' = A[\mathfrak m/x_1]$$
be the affine blowup algebra (as in
Algebra, Section \ref{algebra-section-blow-up}).
Since $x_1, x_2, x_3$ generate $\mathfrak m$ we see that $A'$
is generated by $y_2 = x_2/x_1$ and $y_3 = x_3/x_1$ over $A$.
We will occasionally use $y_1 = 1$ to simplify formulas.
Moreover, looking at our relation above we find that
$$a_{11} + a_{12} y_2 + a_{13} y_3 + a_{22} y_2^2 + a_{23} y_2y_3 + a_{33} y_3^2 = x_1 (\sum a_{ijk} y_iy_jy_k)$$
in $A'$. Recall that $x_1 \in A'$ defines the exceptional divisor $E$
on our affine open of $X$
which is therefore scheme theoretically given by
$$\kappa[y_2, y_3]/ (\overline{a}_{11} + \overline{a}_{12} y_2 + \overline{a}_{13} y_3 + \overline{a}_{22} y_2^2 + \overline{a}_{23} y_2y_3 + \overline{a}_{33} y_3^2)$$
In other words,
$E \subset \mathbf{P}^2_\kappa = \text{Proj}(\kappa[t_1, t_2, t_3])$ is
the zero scheme of the quadratic form $q$ introduced above.

\medskip\noindent
The quadratic form $q$ is an important invariant of the singularity
defined by $A$. Let us say we are in
{\bf case II} if $q$ is a square of a linear form times
an element of $\kappa^*$ and in {\bf case I} otherwise.
Observe that we are in case II exactly if, after
changing our choice of $x_1, x_2, x_3$, we have
$$x_3^2 = \sum a_{ijk}x_ix_jx_k$$
in the local ring $A$.

\medskip\noindent
Let $\mathfrak m' \subset A'$ be a maximal ideal lying over $\mathfrak m$
with residue field $\kappa'$. In other words, $\mathfrak m'$ corresponds
to a closed point $p \in E$ of the exceptional divisor. Recall that the
surjection
$$\kappa[y_2, y_3] \to \kappa'$$
has kernel generated by two elements $f_2, f_3 \in \kappa[y_2, y_3]$
(see for example Algebra, Example \ref{algebra-example-spec-kxy}
or the proof of
Algebra, Lemma \ref{algebra-lemma-dim-affine-space}).
Let $z_2, z_3 \in A'$ map to $f_2, f_3$ in $\kappa[y_2, y_3]$.
Then we see that $\mathfrak m' = (x_1, z_2, z_3)$ because
$x_2$ and $x_3$ become divisible by $x_1$ in $A'$.

\medskip\noindent
{\bf Claim.} If $X$ is singular at $p$, then $\kappa' = \kappa$ or we are
in case II. Namely, if $A'_{\mathfrak m'}$
is singular, then $\dim_{\kappa'} \mathfrak m'/(\mathfrak m')^2 = 3$
which implies that
$\dim_{\kappa'} \overline{\mathfrak m}'/(\overline{\mathfrak m}')^2 = 2$
where $\overline{m}'$ is the maximal ideal of
$\mathcal{O}_{E, p} = \mathcal{O}_{X, p}/x_1\mathcal{O}_{X, p}$.
This implies that
$$q(1, y_2, y_3) = \overline{a}_{11} + \overline{a}_{12} y_2 + \overline{a}_{13} y_3 + \overline{a}_{22} y_2^2 + \overline{a}_{23} y_2y_3 + \overline{a}_{33} y_3^2 \in (f_2, f_3)^2$$
otherwise there would be a relation between the classes of $z_2$
and $z_3$ in $\overline{\mathfrak m}'/(\overline{\mathfrak m}')^2$.
The claim now follows from Lemma \ref{lemma-issquare}.

\medskip\noindent
Resolution in case I. By the claim any
singular point of $X$ is $\kappa$-rational. Pick such a singular point $p$.
We may choose our $x_1, x_2, x_3 \in \mathfrak m$ such that $p$
lies on the chart described above and has coordinates $y_2 = y_3 = 0$.
Since it is a singular point arguing as in the proof of the claim
we find that $q(1, y_2, y_3) \in (y_2, y_3)^2$.
Thus we can choose $a_{11} = a_{12} = a_{13} = 0$ and
$q(t_1, t_2, t_3) = q(t_2, t_3)$. It follows that
$$E = V(q) \subset \mathbf{P}^1_\kappa$$
either is the union of two distinct lines meeting at $p$
or is a degree $2$ curve with a unique $\kappa$-rational point
(small detail omitted; use that $q$ is not a square of a linear
form up to a scalar).
In both cases we conclude that $X$ has a unique singular point $p$
which is $\kappa$-rational. We need a bit more information in this
case. First, looking at higher terms in the expression above, we
find that $\overline{a}_{111} = 0$ because $p$ is singular.
Then we can write $a_{111} = b_{111} x_1 \bmod (x_2, x_3)$
for some $b_{111} \in A$. Then
the quadratic form at $p$ for the generators
$x_1, y_2, y_3$ of $\mathfrak m'$ is
$$q' = \overline{b}_{111} t_1^2 + \overline{a}_{112} t_1 t_2 + \overline{a}_{113} t_1 t_3 + \overline{a}_{22} t_2^2 + \overline{a}_{23} t_2 t_3 + \overline{a}_{33} t_3^2$$
We see that $E' = V(q')$ intersects the line $t_1 = 0$ in either
two points or one point of degree $2$. We conclude that $p$
lies in case I.

\medskip\noindent
Suppose that the blowing up $X' \to X$ of $X$ at $p$ again has a
singular point $p'$. Then we see that $p'$ is a $\kappa$-rational
point and we can blow up to get $X'' \to X'$. If this process
does not stop we get a sequence of blowings up
$$\Spec(A) \leftarrow X \leftarrow X' \leftarrow X'' \leftarrow \ldots$$
We want to show that
Lemma \ref{lemma-sequence-blowups}
applies to this situation. To do this we have to say something
about the choice of the element $x_1$ of $\mathfrak m$.
Suppose that $A$ is in case I and that $X$ has a singular point.
Then we will say that $x_1 \in \mathfrak m$ is a {\it good coordinate}
if for any (equivalently some) choice of $x_2, x_3$ the quadratic
form $q(t_1, t_2, t_3)$ has the property that $q(0, t_2, t_3)$
is not a scalar times a square. We have seen above that a good
coordinate exists. If $x_1$ is a good coordinate, then the
singular point $p \in E$ of $X$ does not lie on the hypersurface
$t_1 = 0$ because either this does not have a rational point or
if it does, then it is not singular on $X$. Observe that this
is equivalent to the
statement that the image of $x_1$ in $\mathcal{O}_{X, p}$ cuts
out the exceptional divisor $E$. Now the computations above show
that if $x_1$ is a good coordinate for $A$, then
$x_1 \in \mathfrak m'\mathcal{O}_{X, p}$ is a good coordinate
for $p$. This of course uses that the notion of good coordinate
does not depend on the choice of $x_2$, $x_3$ used to do the
computation. Hence $x_1$ maps to a good coordinate at
$p'$, $p''$, etc. Thus
Lemma \ref{lemma-sequence-blowups}
applies and our sequence of blowing
ups comes from a nonsingular arc $A \to R$.
Then the map $A^\wedge \to R$ is a surjection.
Since the completion of $A$ is normal, we conclude by
Lemma \ref{lemma-sequence-blowups-along-arc-becomes-nonsingular}
that after a finite number of blowups
$$\Spec(A^\wedge) \leftarrow X^\wedge \leftarrow (X')^\wedge \leftarrow \ldots$$
the resulting scheme $(X^{(n)})^\wedge$ is regular. Since
$(X^{(n)})^\wedge \to X^{(n)}$ induces isomorphisms on complete
local rings (Lemma \ref{lemma-iso-completions}) we conclude
that the same is true for $X^{(n)}$.

\medskip\noindent
Resolution in case II. Here we have
$$x_3^2 = \sum a_{ijk}x_ix_jx_k$$
in $A$ for some choice of generators $x_1, x_2, x_3$ of $\mathfrak m$.
Then $q = t_3^2$ and $E = 2C$ where $C$ is a line.
Recall that in $A'$ we get
$$y_3^2 = x_1(\sum a_{ijk} y_iy_jy_k)$$
Since we know that $X$ is normal, we get a discrete valuation
ring $\mathcal{O}_{X, \xi}$ at the generic point $\xi$ of $C$.
The element $y_3 \in A'$ maps to a uniformizer of $\mathcal{O}_{X, \xi}$.
Since $x_1$ scheme theoretically cuts out $E$
which is $C$ with multiplicity $2$, we see that
$x_1$ is a unit times $y_3^2$ in $\mathcal{O}_{X, \xi}$. Looking
at our equality above we conclude that
$$h(y_2) = \overline{a}_{111} + \overline{a}_{112} y_2 + \overline{a}_{122} y_2^2 + \overline{a}_{222} y_2^3$$
must be nonzero in the residue field of $\xi$.
Now, suppose that $p \in C$ defines a singular point.
Then $y_3$ is zero at $p$ and $p$ must correspond to a
zero of $h$ by the reasoning used in proving the claim above.
If $h$ does not have a double zero at $p$, then the quadratic
form $q'$ at $p$ is not a square and we conclude that $p$
falls in case I which we have treated above\footnote{The maximal
ideal at $p$ in $A'$ is generated by $y_3, x_1$ and a third element
$g$ whose image in $\kappa[y_2]$ is the prime divisor of $h$
corresponding to $p$. If this prime divisor doesn't divide $h$
twice, then we see that the quadratic form at $p$ looks like
$$y_3^2 - x_1((something)x_1 + (something)y_3 + (unit)g)$$
and this can never be a square in $\kappa[y_3, x_1, g]$.}.
Since the degree of $h$ is $3$ we
get at most one singular point $p \in C$ falling into case II
which is moreover $\kappa$-rational. After changing our
choice of $x_1, x_2, x_3$ we may assume this is the point
$y_2 = y_3 = 0$.
Then $h = \overline{a}_{122} y_2^2 + \overline{a}_{222} y_2^3$.
Moreover, it still has to be the case that
$\overline{a}_{113} = 0$ for the quadratic form $q'$ to have
the right shape.
Thus the local ring $\mathcal{O}_{X, p}$ defines a singularity
as in the next paragraph.

\medskip\noindent
The final case we treat is the case where we can choose our generators
$x_1, x_2, x_3$ of $\mathfrak m$ such that
$$x_3^2 + x_1(a x_2^2 + b x_2x_3 + c x_3^2) \in \mathfrak m^4$$
for some $a, b, c \in A$. This is a subclass of case II. If
$\overline{a} = 0$, then we can write
$a = a_1 x_1 + a_2 x_2 + a_3 x_3$ and we get after blowing up
$$y_3^2 + x_1(a_1 x_1 y_2^2 + a_2 x_1 y_2^3 + a_3 x_1 y_2^2 y_3 + b y_2 y_3 + c y_3^2) = x_1^2 (\sum a_{ijkl}y_iy_jy_ky_l)$$
This means that $X$ is not normal\footnote{Namely, the
equation shows that you get something singular along the
$1$-dimensional locus $x_1 = y_3 = 0$ which cannot happen
for a normal surface.} a contradiction. By the result
of the previous paragraph, if the blowup $X$
has a singular point $p$ which falls in case II, then
there is only one and it is $\kappa$-rational.
Computing the affine blowup algebras
$A[\frac{\mathfrak m}{x_2}]$ and $A[\frac{\mathfrak m}{x_3}]$
the reader easily sees that $p$ cannot be contained
the corresponding opens of $X$. Thus $p$ is in the spectrum
of $A[\frac{\mathfrak m}{x_1}]$. Doing the blowing up as before we see that
$p$ must be the point with coordinates $y_2 = y_3 = 0$ and the new
equation looks like
$$y_3^2 + x_1(a y_2^2 + b y_2 y_3 + c y_3^2) \in (\mathfrak m')^4$$
which has the same shape as before and has the property
that $x_1$ defines the exceptional divisor. Thus if the process
does not stop we get an infinite sequence of blowups and on
each of these $x_1$ defines the exceptional divisor in the
local ring of the singular point. Thus we can
finish the proof using
Lemmas \ref{lemma-sequence-blowups} and
\ref{lemma-sequence-blowups-along-arc-becomes-nonsingular}
and the same reasoning as before.

\begin{lemma}
\label{lemma-resolve-rational-double-points}
Let $(A, \mathfrak m, \kappa)$ be a local normal Nagata
domain of dimension $2$ which defines a rational singularity,
whose completion is normal, and which is Gorenstein.
Then there exists a finite sequence of blowups in
singular closed points
$$X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = \Spec(A)$$
such that $X_n$ is regular and such that each intervening
schemes $X_i$ is normal with finitely many singular points
of the same type.
\end{lemma}

\begin{proof}
This is exactly what was proved in the discussion above.
\end{proof}

Comment #3008 by azrt on November 26, 2017 a 6:28 am UTC

> The element $y_3 \in A'$ maps to a uniformizer of $\mathcal O_{X,\zeta}$

Could you explain this part, please?

>we see that $x_1$ is a unit times $y_3$ Presumably, it should "a unit times y_3^2".

Comment #3009 by azrt on November 26, 2017 a 9:10 am UTC

Sorry for being dumb, but "Resolution in Case II" is rather hard to read.

>If $h$ does not have a double zero at $p$, then the quadratic form $q'$ at $p$ is not a square

>Moreover, it still has to be the case $\overline a_{113}=0$ for the quadratic form $q'$ to have the right shape

>This means that $X$ is not normal a contradiction.

Could you elaborate on these points, please? They look confusing.

P.S. I understood the first question in my last comment, so you must not answer it.

Comment #3010 by Johan (site) on November 26, 2017 a 10:25 pm UTC

First, you are right and there is a typo. It should state that $x_1$ is a unit times $y_3^2$ in $\mathcal{O}_{X, \xi}$. Will fix this soon. I think this means everything in your comment #3008 is OK now.

About the first question in comment #3009: the maximal ideal at $p$ in $A'$ is generated by $y_3$, $x_1$ and a third element $g$ whose image in $\kappa[y_2]$ is the prime divisor of $h$ corresponding to $p$. If this prime divisor doesn't divide $h$ twice, then we see that the quadratic form at $p$ looks like $$y_3^2 - x_1((something)x_1 + (something)y_3 + (unit) g)$$ and this can never be a square in $\kappa[y_3, x_1, g]$. Do you agree? (If you do I will add more text to this effect. I also think there may be an easier way to see this.)

Can you try to work out the second question similarly?

As to the third question: I think this is because the equation shows that you get something singular along the $1$-dimensional locus $x_1 = y_3 = 0$ which cannot happen for a normal surface. OK?

Comment #3130 by Johan (site) on February 1, 2018 a 2:40 pm UTC

Made changes as in my comment #3010. See here.

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