## 55.14 Semistable reduction

In this section we carefully define what we mean by semistable reduction.

Example 55.14.1. 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 $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. 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 53.20. 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 31.32. 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 $A[\frac{\mathfrak m}{x}]$.

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.

Lemma 55.14.2. Let $R$ be a discrete valuation ring. Let $X$ be a scheme which is at-worst-nodal of relative dimension $1$ over $R$. Let $x \in X$ be a point of the special fibre of $X$ over $R$. Then there exists a commutative diagram

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

where $R \subset R'$ is an étale extension of discrete valuation rings, the morphism $U \to X$ is étale, the morphism $U \to \mathop{\mathrm{Spec}}(A)$ is étale, there is a point $x' \in U$ mapping to $x$, and

$A = R'[u, v]/(uv) \quad \text{or}\quad A = R'[u, v]/(uv - \pi ^ n)$

where $n \geq 0$ and $\pi \in R'$ is a uniformizer.

Proof. We have already proved this lemma in much greater generality, see Algebraic Curves, Lemma 53.20.12. All we have to do here is to translate the statement given there into the statement given above.

First, if the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is smooth at $x$, then we can find an étale morphism $U \to \mathbf{A}^1_ R = \mathop{\mathrm{Spec}}(R[u])$ for some affine open neighbourhood $U \subset X$ of $x$. This is Morphisms, Lemma 29.36.20. After replacing the coordinate $u$ by $u + 1$ if necessary, we may assume that $x$ maps to a point in the standard open $D(u) \subset \mathbf{A}^1_ R$. Then $D(u) = \mathop{\mathrm{Spec}}(A)$ with $A = R[u, v]/(uv - 1)$ and we see that the result is true in this case.

Next, assume that $x$ is a singular point of the fibre. Then we may apply Algebraic Curves, Lemma 53.20.12 to get a diagram

$\xymatrix{ X \ar[d] & U \ar[rr] \ar[l] \ar[rd] & & W \ar[r] \ar[ld] & \mathop{\mathrm{Spec}}(\mathbf{Z}[u, v, a]/(uv - a)) \ar[d] \\ \mathop{\mathrm{Spec}}(R) & & V \ar[ll] \ar[rr] & & \mathop{\mathrm{Spec}}(\mathbf{Z}[a]) }$

with all the properties mentioned in the statement of the cited lemma. Let $x' \in U$ be the point mapping to $x$ promised by the lemma. First we shrink $V$ to an affine neighbourhood of the image of $x'$. Say $V = \mathop{\mathrm{Spec}}(R')$. Then $R \to R'$ is étale. Since $R$ is a discrete valuation ring, we see that $R'$ is a finite product of quasi-local Dedekind domains (use More on Algebra, Lemma 15.44.4). Hence (for example using prime avoidance) we find a standard open $D(f) \subset V = \mathop{\mathrm{Spec}}(R')$ containing the image of $x'$ such that $R'_ f$ is a discrete valuation ring. Replacing $R'$ by $R'_ f$ we reach the situation where $V = \mathop{\mathrm{Spec}}(R')$ with $R \subset R'$ an étale extension of discrete valuation rings (extensions of discrete valuation rings are defined in More on Algebra, Definition 15.111.1).

The morphism $V \to \mathop{\mathrm{Spec}}(\mathbf{Z}[a])$ is determined by the image $h$ of $a$ in $R'$. Then $W = \mathop{\mathrm{Spec}}(R'[u, v]/(uv - h))$. Thus the lemma holds with $A = R'[u, v]/(uv - h)$. If $h = 0$ then we clearly obtain the first case mentioned in the lemma. If $h \not= 0$ then we may write $h = \epsilon \pi ^ n$ for some $n \geq 0$ where $\epsilon$ is a unit of $R'$. Changing coordinates $u_{new} = \epsilon u$ and $v_{new} = v$ we obtain the second isomorphism type of $A$ listed in the lemma. $\square$

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.

A slightly stronger statement (also true) would be that no matter how you blow up in singular points you eventually end up with a resolution and all the intermediate blowups are at-worst-nodal of relative dimension $1$ over $R$.

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 10.163.10). 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. $\square$

Lemma 55.14.4. Let $R$ be a discrete valuation ring with fraction field $K$ and residue field $k$. Assume $X \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$ over $R$. Let $X \to X'$ be the contraction of an exceptional curve $E \subset X$ of the first kind. Then $X'$ is at-worst-nodal of relative dimension $1$ over $R$.

Proof. Namely, let $x' \in X'$ be the image of $E$. Then the only issue is to see that $X' \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$ in a neighbourhood of $x'$. The closed fibre of $X \to \mathop{\mathrm{Spec}}(R)$ is reduced, hence $\pi \in R$ vanishes to order $1$ on $E$. This immediately implies that $\pi$ viewed as an element of $\mathfrak m_{x'} \subset \mathcal{O}_{X', x'}$ but is not in $\mathfrak m_{x'}^2$. Since $\mathcal{O}_{X', x'}$ is regular of dimension $2$ (by definition of contractions in Resolution of Surfaces, Section 54.16), this implies that $\mathcal{O}_{X'_ k, x'}$ is regular of dimension $1$ (Algebra, Lemma 10.106.3). On the other hand, the curve $E$ has to meet at least one other component, say $C$ of the closed fibre $X_ k$. Say $x \in E \cap C$. Then $x$ is a node of the special fibre $X_ k$ and hence $\kappa (x)/k$ is finite separable, see Algebraic Curves, Lemma 53.19.7. Since $x \mapsto x'$ we conclude that $\kappa (x')/k$ is finite separable. By Algebra, Lemma 10.140.5 we conclude that $X'_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth in an open neighbourhood of $x'$. Combined with flatness, this proves that $X' \to \mathop{\mathrm{Spec}}(R)$ is smooth in a neighbourhood of $x'$ (Morphisms, Lemma 29.34.14). This finishes the proof as a smooth morphism of relative dimension $1$ is at-worst-nodal of relative dimension $1$ (Algebraic Curves, Lemma 53.20.3). $\square$

Lemma 55.14.5. Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. The following are equivalent

1. there exists a proper model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$,

2. there exists a minimal model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$, and

3. any minimal model of $C$ is at-worst-nodal of relative dimension $1$ over $R$.

Proof. To make sense out of this statement, recall that a minimal model is defined as a regular proper model without exceptional curves of the first kind (Definition 55.8.4), that minimal models exist (Proposition 55.8.6), and that minimal models are unique if the genus of $C$ is $> 0$ (Lemma 55.10.1). Keeping this in mind the implications (2) $\Rightarrow$ (1) and (3) $\Rightarrow$ (2) are clear.

Assume (1). Let $X$ be a proper model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$. Applying Lemma 55.14.3 we see that we may assume $X$ is regular as well. Let

$X = X_ m \to X_{m - 1} \to \ldots \to X_1 \to X_0$

be as in Lemma 55.8.5. By Lemma 55.14.4 and induction this implies $X_0$ is at-worst-nodal of relative dimension $1$ over $R$.

To finish the proof we have to show that (2) implies (3). This is clear if the genus of $C$ is $> 0$, since then the minimal model is unique (see discussion above). On the other hand, if the minimal model is not unique, then the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is smooth for any minimal model as its special fibre will be isomorphic to $\mathbf{P}^1_ k$ by Lemma 55.12.4. $\square$

Definition 55.14.6. Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. We say that $C$ has semistable reduction if the equivalent conditions of Lemma 55.14.5 are satisfied.

Lemma 55.14.7. Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. The following are equivalent

1. there exists a proper smooth model for $C$,

2. there exists a minimal model for $C$ which is smooth over $R$,

3. any minimal model is smooth over $R$.

Proof. If $X$ is a smooth proper model, then the special fibre is connected (Lemma 55.9.4) and smooth, hence irreducible. This immediately implies that it is minimal. Thus (1) implies (2). To finish the proof we have to show that (2) implies (3). This is clear if the genus of $C$ is $> 0$, since then the minimal model is unique (Lemma 55.10.1). On the other hand, if the minimal model is not unique, then the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is smooth for any minimal model as its special fibre will be isomorphic to $\mathbf{P}^1_ k$ by Lemma 55.12.4. $\square$

Definition 55.14.8. Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. We say that $C$ has good reduction if the equivalent conditions of Lemma 55.14.7 are satisfied.

Comment #5935 by Morten on

By symmetry the case of the algebra A[𝔪y] is the same as the case of A[𝔪y].

Should one y be replaced by x?

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