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Tag 0CDC

Example 48.14.1. Let $R$ be a discrete valuation ring with uniformizer $\pi$. Given $n \geq 0$, consider the ring map $$R \longrightarrow A = R[x, y]/(xy - \pi^n)$$ Set $X = \mathop{\rm Spec}(A)$ and $S = \mathop{\rm Spec}(A)$. If $n = 0$, then $X \to S$ is smooth. For all $n$ the morphism $X \to S$ is at-worst-nodal of relative dimension $1$ as defined in Algebraic Curves, Section 46.17. If $n = 1$, then $X$ is regular, but if $n > 1$, then $X$ is not regular as $(x, y)$ no longer generate the maximal ideal $\mathfrak m = (\pi, x, y)$. To ameliorate the situation in case $n > 1$ we consider the blowup $b : X' \to X$ of $X$ in $\mathfrak m$. See Divisors, Section 30.29. By construction $X'$ is covered by three affine pieces corresponding to the blowup algebras $A[\frac{\mathfrak m}{\pi}]$, $A[\frac{\mathfrak m}{x}]$, and $A[\frac{\mathfrak m}{y}]$.

The algebra $A[\frac{\mathfrak m}{\pi}]$ has generators $x' = x/\pi$ and $y' = y/\pi$ and $x'y' = \pi^{n - 2}$. Thus this part of $X'$ is the spectrum of $R[x', y'](x'y' - \pi^{n - 2})$.

The algebra $A[\frac{\mathfrak m}{x}]$ has generators $x$, $u = \pi/x$ subject to the relation $xu - \pi$. Note that this ring contains $y/x = \pi^n/x^2 = u^2\pi^{n - 2}$. Thus this part of $X'$ is regular.

By symmetry the case of the algebra $A[\frac{\mathfrak m}{y}]$ is the same as the case of $A[\frac{\mathfrak m}{y}]$.

Thus we see that $X' \to S$ is at-worst-nodal of relative dimension $1$ and that $X'$ is regular, except for one point which has an affine open neighbourhood exactly as above but with $n$ replaced by $n - 2$. Using induction on $n$ we conclude that there is a sequence of blowing ups in closed points $$X_{\lfloor n/2 \rfloor} \to \ldots \to X_1 \to X_0 = X$$ such that $X_{\lfloor n/2 \rfloor} \to S$ is at-worst-nodal of relative dimension $1$ and $X_{\lfloor n/2 \rfloor}$ is regular.

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\begin{example}
\label{example-blowup}
Let $R$ be a discrete valuation ring with uniformizer $\pi$.
Given $n \geq 0$, consider the ring map
$$R \longrightarrow A = R[x, y]/(xy - \pi^n)$$
Set $X = \Spec(A)$ and $S = \Spec(A)$.
If $n = 0$, then $X \to S$ is smooth.
For all $n$ the morphism $X \to S$ is at-worst-nodal
of relative dimension $1$ as defined in
Algebraic Curves, Section \ref{curves-section-families-nodal}.
If $n = 1$, then $X$ is regular, but if $n > 1$, then $X$ is not
regular as $(x, y)$ no longer generate the maximal ideal
$\mathfrak m = (\pi, x, y)$. To ameliorate the situation
in case $n > 1$ we
consider the blowup $b : X' \to X$ of $X$ in $\mathfrak m$.
See Divisors, Section \ref{divisors-section-blowing-up}.
By construction $X'$ is covered by three affine pieces
corresponding to the blowup algebras $A[\frac{\mathfrak m}{\pi}]$,
$A[\frac{\mathfrak m}{x}]$, and $A[\frac{\mathfrak m}{y}]$.

\medskip\noindent
The algebra $A[\frac{\mathfrak m}{\pi}]$ has generators
$x' = x/\pi$ and $y' = y/\pi$ and $x'y' = \pi^{n - 2}$.
Thus this part of $X'$ is the spectrum of $R[x', y'](x'y' - \pi^{n - 2})$.

\medskip\noindent
The algebra $A[\frac{\mathfrak m}{x}]$ has generators $x$,
$u = \pi/x$ subject to the relation $xu - \pi$. Note that this ring
contains $y/x = \pi^n/x^2 = u^2\pi^{n - 2}$. Thus this part of
$X'$ is regular.

\medskip\noindent
By symmetry the case of the algebra $A[\frac{\mathfrak m}{y}]$ is
the same as the case of $A[\frac{\mathfrak m}{y}]$.

\medskip\noindent
Thus we see that $X' \to S$ is at-worst-nodal of relative dimension $1$
and that $X'$ is regular, except for one point which has an
affine open neighbourhood exactly as above but with $n$ replaced by $n - 2$.
Using induction on $n$ we conclude that there is a sequence of
blowing ups in closed points
$$X_{\lfloor n/2 \rfloor} \to \ldots \to X_1 \to X_0 = X$$
such that $X_{\lfloor n/2 \rfloor} \to S$ is
at-worst-nodal of relative dimension $1$ and
$X_{\lfloor n/2 \rfloor}$ is regular.
\end{example}

Comment #2444 by Bronson Lim on March 5, 2017 a 5:00 am UTC

Should say S = \mathop{\rm Spec}(R).

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