The Stacks project

Lemma 55.14.3. Let $R$ be a discrete valuation ring. Let $X$ be a quasi-compact scheme which is at-worst-nodal of relative dimension $1$ with smooth generic fibre over $R$. Then there exists $m \geq 0$ and a sequence

\[ X_ m \to \ldots \to X_1 \to X_0 = X \]

such that

  1. $X_{i + 1} \to X_ i$ is the blowing up of a closed point $x_ i$ where $X_ i$ is singular,

  2. $X_ i \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$,

  3. $X_ m$ is regular.

Proof. Since $X$ is quasi-compact we see that the special fibre $X_ k$ is quasi-compact. Since the singularities of $X_ k$ are at-worst-nodal, we see that $X_ k$ has a finite number of nodes and is otherwise smooth over $k$. As $X \to \mathop{\mathrm{Spec}}(R)$ is flat with smooth generic fibre it follows that $X$ is smooth over $R$ except at the finite number of nodes of $X_ k$ (use Morphisms, Lemma 29.34.14). It follows that $X$ is regular at every point except for possibly the nodes of its special fibre (see Algebra, Lemma 10.162.10). Let $x \in X$ be such a node. Choose a diagram

\[ \xymatrix{ X \ar[d] & U \ar[r] \ar[d] \ar[l] & \mathop{\mathrm{Spec}}(A) \ar[dl] \\ \mathop{\mathrm{Spec}}(R) & \mathop{\mathrm{Spec}}(R') \ar[l] } \]

as in Lemma 55.14.2. Observe that the case $A = R'[u, v]/(uv)$ cannot occur, as this would mean that the generic fibre of $X/R$ is singular (tiny detail omitted). Thus $A = R'[u, v]/(uv - \pi ^ n)$ for some $n \geq 0$. Since $x$ is a singular point, we have $n \geq 2$, see discussion in Example 55.14.1.

After shrinking $U$ we may assume there is a unique point $u \in U$ mapping to $x$. Let $w \in \mathop{\mathrm{Spec}}(A)$ be the image of $u$. We may also assume that $u$ is the unique point of $U$ mapping to $w$. Since the two horizontal arrows are étale we see that $u$, viewed as a closed subscheme of $U$, is the scheme theoretic inverse image of $x \in X$ and the scheme theoretic inverse image of $w \in \mathop{\mathrm{Spec}}(A)$. Since blowing up commutes with flat base change (Divisors, Lemma 31.32.3) we find a commutative diagram

\[ \xymatrix{ X' \ar[d] & U' \ar[l] \ar[d] \ar[r] & W' \ar[d] \\ X & U \ar[l] \ar[r] & \mathop{\mathrm{Spec}}(A) } \]

with cartesian squares where the vertical arrows are the blowing up of $x, u, w$ in $X, U, \mathop{\mathrm{Spec}}(A)$. The scheme $W'$ was described in Example 55.14.1. We saw there that $W'$ at-worst-nodal of relative dimension $1$ over $R'$. Thus $W'$ is at-worst-nodal of relative dimension $1$ over $R$ (Algebraic Curves, Lemma 53.20.7). Hence $U'$ is at-worst-nodal of relative dimension $1$ over $R$ (see Algebraic Curves, Lemma 53.20.8). Since $X' \to X$ is an isomorphism over the complement of $x$, we conclude the same thing is true of $X'/R$ (by Algebraic Curves, Lemma 53.20.8 again).

Finally, we need to argue that after doing a finite number of these blowups we arrive at a regular model $X_ m$. This is rather clear because the “invariant” $n$ decreases by $2$ under the blowup described above, see computation in Example 55.14.1. However, as we want to avoid precisely defining this invariant and establishing its properties, we in stead argue as follows. If $n = 2$, then $W'$ is regular and hence $X'$ is regular at all points lying over $x$ and we have decreased the number of singular points of $X$ by $1$. If $n > 2$, then the unique singular point $w'$ of $W'$ lying over $w$ has $\kappa (w) = \kappa (w')$. Hence $U'$ has a unique singular point $u'$ lying over $u$ with $\kappa (u) = \kappa (u')$. Clearly, this implies that $X'$ has a unique singular point $x'$ lying over $x$, namely the image of $u'$. Thus we can argue exactly as above that we get a commutative diagram

\[ \xymatrix{ X'' \ar[d] & U'' \ar[l] \ar[d] \ar[r] & W'' \ar[d] \\ X' & U' \ar[l] \ar[r] & W' } \]

with cartesian squares where the vertical arrows are the blowing up of $x', u', w'$ in $X', U', W'$. Continuing like this we get a compatible sequence of blowups which stops after $\lfloor n/2 \rfloor $ steps. At the completion of this process the scheme $X^{(\lfloor n/2 \rfloor )}$ will have one fewer singular point than $X$. Induction on the number of singular points completes the proof. $\square$

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