Proof. Direct calculation. ◻

Proof. We first produce the stable limit of the Craighero–Gattazzo quintic  $X$ in characteristic seven. Let ${\mathcal{X}}^{0}$ be the generic fiber of ${\mathcal{X}}$ given by (2.3). Consider the family $\hat{{\mathcal{X}}}\rightarrow \operatorname{Spec}{\mathcal{R}}$ given by equations

$$\begin{eqnarray}(f_{1}w^{2}+f_{3}w+f_{5}+\text{h.o.t.}=0,\;f_{2}=7w)\subset \mathbb{P}_{[x:y:z:t:w]}^{4}(1,1,1,1,2)_{{\mathcal{R}}}\end{eqnarray}$$

obtained by substituting $f_{2}$ for $7w$ in the first three terms of (2.3) and dividing by  $343$ . Here, and throughout, ‘h.o.t.’ refers to higher-order terms with respect to the $7$ -adic valuation. The generic fiber of $\hat{{\mathcal{X}}}$ is clearly isomorphic to ${\mathcal{X}}^{0}$ .

The special fiber $\hat{{\mathcal{X}}}_{0}$ is given by

$$\begin{eqnarray}(f_{1}w^{2}+f_{3}w+f_{5}=0,\;f_{2}=0)\subset \mathbb{P}_{[x:y:z:t:w]}^{4}(1,1,1,1,2)_{k}.\end{eqnarray}$$

We claim that it is isomorphic to the surface $Z^{\prime }$ obtained by blowing down four elliptic $(-1)$ -curves on $S_{0}$ to $\tilde{E}_{8}$ -singularities.

The point $(0:0:0:0:1)$ is an isolated singularity with equation, in a local chart,

$$\begin{eqnarray}(f_{1}+f_{3}+f_{5}=0,\;f_{2}=0)\subset {\textstyle \frac{1}{2}}(1,1,1,1).\end{eqnarray}$$

The singularity is formally isomorphic to

$$\begin{eqnarray}(xy=z^{2})\subset {\textstyle \frac{1}{2}}(1,1,1)_{k},\end{eqnarray}$$

which has a $(-4)$ -curve as the resolution graph. Moreover, the equation of the whole family $\hat{{\mathcal{X}}}$ near this point is formally isomorphic to

$$\begin{eqnarray}(xy=z^{2}+7)\subset {\textstyle \frac{1}{2}}(1,1,1)_{{\mathcal{R}}}.\end{eqnarray}$$

Next we analyze $\hat{{\mathcal{X}}}_{0}$ away from $t_{0}=(0:0:0:0:1)$ . We use the generically two-to-one map $\unicode[STIX]{x1D70B}:S_{0}\setminus \{t_{0}\}\rightarrow Q$ given by $[x:y:z:t:w]\rightarrow [x:y:z:t]$ . Away from $\unicode[STIX]{x1D6E5}=L\cap Q$ , $\unicode[STIX]{x1D70B}$  is a double cover branched along $(f_{3}^{2}-4f_{1}f_{5}=0)=B_{1}\cup B_{2}$ . Thus, it can be identified with $Z^{\prime }\setminus (\unicode[STIX]{x1D6E5}_{2}\cup N_{1}\cup N_{2})$ . Over $\unicode[STIX]{x1D6E5}$ , but away from $t_{0}$ (which includes $Q_{1}$ and $Q_{2}$ ) the map $\unicode[STIX]{x1D70B}$ is one-to-one. The preimages of $Q_{1}$ and $Q_{2}$ are lines (with coordinate $w$ ). The preimages of the other four points where $f_{3}=0$ are empty; in Figure 2 these are the points where $B_{1}$ and $B_{2}$ are tangent to $\unicode[STIX]{x1D6E5}$ . It follows that $\hat{{\mathcal{X}}}_{0}$ and $Z^{\prime }$ are normal surfaces isomorphic in codimension one, and therefore isomorphic.

It remains to note that the family $\hat{{\mathcal{X}}}$ has $\tilde{E}_{8}$ singularities along the sections $(1:0:0:0:0)$ , $(0:1:0:0:0)$ , $(0:0:1:0:0)$ , and $(0:0:0:1:0)$ . Resolving them gives a family ${\mathcal{S}}\rightarrow \operatorname{Spec}{\mathcal{R}}$ with special fiber $S_{0}$ and generic fiber (after pulling back to $\operatorname{Spec}\mathbb{C}$ ) the Craighero–Gattazzo surface  $S$ .◻

Proof. We follow [Reference RanaRan14, 4.8, 4.10] closely. It suffices to show that

(3.3) $$\begin{eqnarray}H^{2}(W,T_{W}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+N_{1})))=0.\end{eqnarray}$$

Indeed, if this is the case, then Serre duality implies

$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle H^{0}(W,\unicode[STIX]{x1D6FA}_{W}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+N_{1}))(K_{W}))\nonumber\\ \displaystyle & = & \displaystyle H^{0}(Y,\unicode[STIX]{x1D70F}_{\ast }[\unicode[STIX]{x1D6FA}_{W}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+N_{1}))(G_{1}+\cdots +G_{4})](K_{Y}))\nonumber\end{eqnarray}$$

(by Lemma 3.3)

$$\begin{eqnarray}=H^{0}(Y,\unicode[STIX]{x1D6FA}_{Y}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+N_{1}))(K_{Y}))=H^{2}(Y,T_{Y}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+N_{1})))^{\vee }.\end{eqnarray}$$

Arguing as in [Reference RanaRan14, 4.8], (3.3) will follow if we can show that

(3.4) $$\begin{eqnarray}H^{2}(W,T_{W}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1})))_{-}=0,\end{eqnarray}$$

and

(3.5) $$\begin{eqnarray}H^{2}(W,T_{W}(-\text{log}(N_{1})))_{+}=0,\end{eqnarray}$$

where $+/-$ denotes the symmetric/skew-symmetric part with respect to the $\unicode[STIX]{x1D707}_{2}$ -action on the double cover. Explicitly, and using Serre duality multiple times, if $\unicode[STIX]{x1D6FC}\in H^{0}(W,\unicode[STIX]{x1D6FA}_{W}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+N_{1}))(K))$ , then since

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{W}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+N_{1}))(K)\subset \unicode[STIX]{x1D6FA}_{W}^{1}(\log (\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1}))(K)\end{eqnarray}$$

the one-form $\unicode[STIX]{x1D6FC}$ must be invariant. But $\unicode[STIX]{x1D707}_{2}$ interchanges $\unicode[STIX]{x1D6E5}_{1}$ and $\unicode[STIX]{x1D6E5}_{2}$ , so that $\unicode[STIX]{x1D6FC}$ does not have a pole along $\unicode[STIX]{x1D6E5}_{1}$ . Thus, $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6FA}_{W}^{1}(\log N_{1})(K)$ is an invariant one-form. Equation (3.5) implies that $\unicode[STIX]{x1D6FC}=0$ .

Proof of (3.4).

At each of the points $Q_{3},\ldots ,Q_{6}$ (the remaining points of $B_{i}\cap \unicode[STIX]{x1D6E5}$ ) we blowup twice to obtain a surface $\mathbb{P}_{1}$ where $\unicode[STIX]{x1D6E5}$ and $B_{i}$ have normal crossings. Let $\bar{C}_{i}$ , $\bar{F}_{i}$ , $i=3,\ldots ,6$ be the exceptional divisors of these blowups, so that on $\mathbb{P}_{1}$ we have $\bar{C}_{i}^{2}=-2$ and $\bar{F}_{i}^{2}=-1$ . Let $\unicode[STIX]{x1D70E}^{\prime }:\mathbb{P}_{1}\rightarrow Q$ be the composition of these blowups, and let $f:W_{1}\rightarrow \mathbb{P}_{1}$ be the double cover branched over $B_{1}+B_{2}+\sum \bar{G}_{i}+\sum \bar{C}_{i}$ .

The surface $W_{1}$ contains $(-1)$ -curves $C_{i}$ and $(-2)$ -curves $F_{i}$ which contract to give the surface $W$ . By the $(-1)$ - and $(-2)$ -curve principles [Reference Park, Shin and UrzúaPSU13, Proposition 4.3, Theorem 4.4] (here we only need the $(-1)$ -curve principle), we have

$$\begin{eqnarray}H^{2}(W_{1},T_{W_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1})))\simeq H^{2}(W,T_{W}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1}))).\end{eqnarray}$$

Note that the double cover $f$ is defined by (see (3.2))

$$\begin{eqnarray}B_{1}+B_{2}+\sum \bar{G}_{i}+\sum \bar{C}_{i}\sim 2L,\end{eqnarray}$$

where

$$\begin{eqnarray}L\sim 3{\unicode[STIX]{x1D70E}^{\prime }}^{\ast }({\mathcal{O}}_{Q}(1,1))-\sum \bar{G}_{i}-3\sum \bar{E}_{i}-\sum \bar{N}_{i}-\sum \bar{F}_{i}.\end{eqnarray}$$

Also we have

$$\begin{eqnarray}K_{\mathbb{P}_{1}}=-2{\unicode[STIX]{x1D70E}^{\prime }}^{\ast }({\mathcal{O}}_{Q}(1,1))+\sum \bar{N}_{i}+\sum \bar{G}_{i}+2\sum \bar{E}_{i}+\sum \bar{C}_{i}+2\sum \bar{F}_{i},\end{eqnarray}$$

and so

$$\begin{eqnarray}K_{\mathbb{P}_{1}}+L\sim {\unicode[STIX]{x1D70E}^{\prime }}^{\ast }({\mathcal{O}}_{Q}(1,1))-\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}.\end{eqnarray}$$

By Lemma 3.2, we have

$$\begin{eqnarray}f_{\ast }(T_{W_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1})))_{-}=T_{\mathbb{P}_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}+\bar{N}_{1}))(-L).\end{eqnarray}$$

By Serre duality, it suffices to prove vanishing of

$$\begin{eqnarray}H^{0}(\mathbb{P}_{1},\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}(\log (\unicode[STIX]{x1D6E5}+\bar{N}_{1}))(K_{\mathbb{P}_{1}}+L)),\end{eqnarray}$$

or

$$\begin{eqnarray}H^{0}\biggl(\mathbb{P}_{1},\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}(\log (\unicode[STIX]{x1D6E5}+\bar{N}_{1}))\biggl({\unicode[STIX]{x1D70E}^{\prime }}^{\ast }({\mathcal{O}}_{Q}(1,1))-\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}\biggr)\biggr).\end{eqnarray}$$

By Lemma 3.3, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D70E}_{\ast }^{\prime }\biggl(\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}(\log (\unicode[STIX]{x1D6E5}+\bar{N}_{1}))\biggl(\unicode[STIX]{x1D70E}^{\prime \ast }({\mathcal{O}}_{Q}(1,1))-\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \subset \unicode[STIX]{x1D70E}_{\ast }^{\prime }\biggl(\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}\biggl(\unicode[STIX]{x1D6E5}+\bar{N}_{1}+\unicode[STIX]{x1D70E}^{\prime \ast }({\mathcal{O}}_{Q}(1,1))-\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D70E}_{\ast }^{\prime }\biggl(\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}\biggl(\unicode[STIX]{x1D70E}^{\prime \ast }(\unicode[STIX]{x1D6E5})+\unicode[STIX]{x1D70E}^{\prime \ast }({\mathcal{O}}_{Q}(1,1))-\bar{N}_{2}-\sum \bar{E}_{i}-\sum \bar{F}_{i}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \subset \unicode[STIX]{x1D6FA}_{Q}^{1}\otimes {\mathcal{O}}_{Q}(2,2)\otimes {\mathcal{I}}_{Q_{2}}\bigotimes _{i=1}^{4}{\mathcal{I}}_{P_{i}}.\nonumber\end{eqnarray}$$

Since $\unicode[STIX]{x1D6FA}_{Q}^{1}={\mathcal{O}}_{Q}(-2,0)\oplus {\mathcal{O}}_{Q}(0,-2)$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{Q}^{1}\otimes {\mathcal{O}}_{Q}(2,2)={\mathcal{O}}_{Q}(0,2)\oplus {\mathcal{O}}_{Q}(2,0).\end{eqnarray}$$

Thus, any global section of $\unicode[STIX]{x1D6FA}_{Q}^{1}\otimes {\mathcal{O}}_{Q}(2,2)\otimes {\mathcal{I}}_{Q_{2}}\bigotimes _{i=1}^{4}{\mathcal{I}}_{P_{i}}$ is a global section of ${\mathcal{O}}_{Q}(0,2)\oplus {\mathcal{O}}_{Q}(2,0)$ vanishing at the points $Q_{2},P_{1},\ldots ,P_{4}$ . Since these points are in three distinct horizontal and vertical fibers of $Q$ , any such global section must be zero. This completes the proof of (3.4).

Lemma 3.2. Let $Y$ be a smooth projective surface defined over an algebraically closed field of characteristic $\neq 2$ . Let $f:X\rightarrow Y$ be a double cover with a smooth branch divisor $B\subset Y$ . Let $C=f^{-1}(D)$ be the preimage of a smooth curve $D$ on $Y$ , and suppose that $D$ intersects $B$ transversally. Then

$$\begin{eqnarray}f_{\ast }(\unicode[STIX]{x1D6FA}_{X}^{1}(\log C))=\unicode[STIX]{x1D6FA}_{Y}^{1}(\log (D))\oplus \unicode[STIX]{x1D6FA}_{Y}^{1}(\log (D+B))(-L)\end{eqnarray}$$

and

$$\begin{eqnarray}f_{\ast }(T_{X}(-\text{log}\,C))=T_{Y}(-\text{log}(D+B))\oplus T_{Y}(-\text{log}(D))(-L),\end{eqnarray}$$

where $B\sim 2L$ . Moreover, these decompositions break the sheaves into their invariant and anti-invariant subspaces under the action of $\unicode[STIX]{x1D707}_{2}$ by deck transformations.

Proof of (3.5). Note that we have the short exact sequence

$$\begin{eqnarray}0\rightarrow T_{W}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1}))\rightarrow T_{W}(-\text{log},N_{1})\rightarrow {\mathcal{N}}_{\unicode[STIX]{x1D6E5}_{1}/W}\oplus {\mathcal{N}}_{\unicode[STIX]{x1D6E5}_{2}/W}\rightarrow 0.\end{eqnarray}$$

Since $H^{2}(W,{\mathcal{N}}_{\unicode[STIX]{x1D6E5}_{1}/W}\oplus {\mathcal{N}}_{\unicode[STIX]{x1D6E5}_{2}/W})=0$ , it suffices to prove that

$$\begin{eqnarray}H^{2}(W,T_{W}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1})))_{+}=0.\end{eqnarray}$$

This part is more delicate and the proof occupies the rest of the section.

By Lemma 3.2, we have

$$\begin{eqnarray}f_{\ast }(T_{W_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}_{1}+\unicode[STIX]{x1D6E5}_{2}+N_{1})))_{+}=T_{\mathbb{P}_{1}}\biggl(-\text{log}\biggl(\unicode[STIX]{x1D6E5}+\bar{N}_{1}+B_{1}+B_{2}+\sum \bar{G}_{i}+\sum \bar{C}_{i}\biggr)\biggr).\end{eqnarray}$$

Again applying the $(-1)$ and $(-2)$ -curve principles, it suffices to show that

$$\begin{eqnarray}H^{2}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}+B_{1}+B_{2})))=0.\end{eqnarray}$$

To begin with, we claim that

(3.6) $$\begin{eqnarray}H^{2}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(B_{1}+B_{2})))=0.\end{eqnarray}$$

Because $B_{1}$ and $B_{2}$ have simple normal crossings after contracting the curves $\bar{C}_{i}$ and $\bar{F}_{i}$ , it suffices to show that

$$\begin{eqnarray}H^{2}(\mathbb{P},T_{\mathbb{ P}}(-\text{log}(B_{1}+B_{2})))=0\end{eqnarray}$$

or equivalently (by Serre duality)

$$\begin{eqnarray}H^{0}\biggl(\mathbb{P},\unicode[STIX]{x1D6FA}_{\mathbb{ P}}^{1}(\log (B_{1}+B_{2}))\biggl(-2D+\sum \bar{N}_{i}+\sum \bar{G}_{i}+2\sum \bar{E}_{i}\biggr)\biggr)=0.\end{eqnarray}$$

Letting ${\mathcal{F}}={\mathcal{O}}_{\mathbb{P}}(-2D+\sum \bar{N}_{i}+\sum \bar{G}_{i}+2\sum \bar{E}_{i})$ , we have the short exact sequence

$$\begin{eqnarray}0\rightarrow \unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}\otimes {\mathcal{F}}\rightarrow \unicode[STIX]{x1D6FA}_{\mathbb{ P}}^{1}(\log (B_{1}+B_{2}))\otimes {\mathcal{F}}\rightarrow ({\mathcal{O}}_{B_{1}}\oplus {\mathcal{O}}_{B_{2}})\otimes {\mathcal{F}}\rightarrow 0.\end{eqnarray}$$

The products $B_{j}\cdot {\mathcal{F}}=-4<0$ for $j=1,2$ and thus

$$\begin{eqnarray}H^{0}(({\mathcal{O}}_{B_{1}}\oplus {\mathcal{O}}_{B_{2}})\otimes {\mathcal{F}})=0.\end{eqnarray}$$

The projection formula and Lemma 3.3 give

$$\begin{eqnarray}H^{0}(\mathbb{P}_{1},\unicode[STIX]{x1D6FA}_{\mathbb{P}_{1}}^{1}\otimes {\mathcal{F}})\simeq H^{0}(Q,\unicode[STIX]{x1D6FA}_{Q}^{1}(-2D)).\end{eqnarray}$$

The sheaf $\unicode[STIX]{x1D6FA}_{Q}^{1}(-2D)={\mathcal{O}}_{Q}(-4,-2)\oplus {\mathcal{O}}_{Q}(-2,-4)$ has no global sections, completing the proof of claim (3.6).

Now consider the short exact sequence

$$\begin{eqnarray}0\rightarrow T_{\mathbb{P}_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}+B_{1}+B_{2}))\rightarrow T_{\mathbb{P}_{1}}(-\text{log}(B_{1}+B_{2}))\rightarrow {\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}}\rightarrow 0.\end{eqnarray}$$

By claim (3.6), vanishing of $H^{2}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}+B_{1}+B_{2})))$ will be complete once we show that the map

(3.7) $$\begin{eqnarray}H^{1}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(B_{1}+B_{2})))\rightarrow H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})\end{eqnarray}$$

is surjective. We identify $H^{1}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(B_{1}+B_{2})))$ with the space of first-order infinitesimal deformations of $\mathbb{P}_{1}$ which contain an embedded first-order deformation of $B_{1}\cup B_{2}$ . We identify $H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})$ with the space of obstructions to deforming $\unicode[STIX]{x1D6E5}$ in $\mathbb{P}_{1}$ . Thus, the map (3.7) factors through the natural map

(3.8) $$\begin{eqnarray}H^{1}(\mathbb{P}_{1},T_{\mathbb{P}_{1}})\rightarrow H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})\end{eqnarray}$$

which sends an infinitesimal first-order deformation of $\mathbb{P}_{1}$ to the obstruction to deforming $\unicode[STIX]{x1D6E5}$ in this first-order deformation of $\mathbb{P}_{1}$ . We have to show that given any such obstruction, there is a deformation of the pair $(\mathbb{P}_{1},B_{1}+B_{2})$ that maps to the given obstruction.

Recall that $\mathbb{P}_{1}$ is obtained from $\mathbb{P}^{1}\times \mathbb{P}^{1}$ by blowing up once at each of $Q_{1},\ldots ,Q_{6};P_{1},\ldots ,P_{4}$ , and again at each of $Q_{3},\ldots ,Q_{6}$ in the direction of the proper transform of $\unicode[STIX]{x1D6E5}$ and at each of $P_{1},\ldots ,P_{4}$ in the direction of tangent cone of $B_{1}\cup B_{2}$ . We denote by $\unicode[STIX]{x1D70E}_{2}:\mathbb{P}_{1}\rightarrow \tilde{Q}$ the ‘intermediate’ blowup, i.e. the map which contracts the last eight $(-1)$ -curves on $\mathbb{P}_{1}$ .

We have the following exact sequence of sheaves on $\tilde{Q}$

$$\begin{eqnarray}0\rightarrow (\unicode[STIX]{x1D70E}_{2})_{\ast }T_{\mathbb{P}_{1}}\rightarrow T_{\tilde{Q}}\rightarrow \bigoplus _{i=3}^{6}k_{Q_{i}}^{2}\oplus \bigoplus _{i=1}^{4}k_{P_{i}}^{2}\rightarrow 0.\end{eqnarray}$$

Looking at the corresponding exact sequence in cohomology, we see that every infinitesimal first-order deformation of $\mathbb{P}_{1}$ arises from either an infinitesimal first-order deformation of $\tilde{Q}$ (corresponding to an element of $H^{1}(\tilde{Q},T_{\tilde{Q}})$ ) or from an infinitesimal first-order deformation of the points $Q_{3},\ldots ,Q_{6},P_{1},\ldots ,P_{4}$ on $\tilde{Q}$ , or both. This latter space is isomorphic to a vector space $V=(k^{2})^{8}$ . We note that $V$ has a linear subspace $V_{1}\simeq k^{8}$ corresponding to infinitesimal first-order deformations of the points $Q_{3},\ldots ,Q_{6}$ , $P_{1},\ldots ,P_{4}$ to points along the exceptional divisors of $\unicode[STIX]{x1D70E}_{2}$ , i.e. changing the tangent direction of the infinitely-near blowup.

Similarly, because $\mathbb{P}^{1}\times \mathbb{P}^{1}$ is rigid, every first-order infinitesimal deformation of $\tilde{Q}$ arises from a first-order infinitesimal deformation of the points $Q_{1},\ldots ,Q_{6};P_{1},\ldots ,P_{4}$ in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ . This latter deformation space is isomorphic to the vector space $W=(k^{2})^{10}$ . Thus, we have short exact sequences

$$\begin{eqnarray}\displaystyle & \displaystyle 0\rightarrow V\rightarrow H^{1}(\mathbb{P}_{1},T_{\mathbb{P}_{1}})\rightarrow H^{1}(\tilde{Q},T_{\tilde{Q}})\rightarrow 0 & \displaystyle \nonumber\\ \displaystyle & \displaystyle 0\rightarrow H^{0}(\mathbb{P}^{1}\times \mathbb{P}^{1},T_{\mathbb{P}^{1}\times \mathbb{P}^{1}})\rightarrow W\rightarrow H^{1}(\tilde{Q},T_{\tilde{Q}})\rightarrow 0 & \displaystyle \nonumber\end{eqnarray}$$

signifying that every first-order infinitesimal deformation of $\mathbb{P}_{1}$ , and therefore of $(\mathbb{P}_{1},B_{1}\cup B_{2})$ , arises from a first-order infinitesimal deformation of the points $Q_{1},\ldots ,Q_{6}$ ; $P_{1},\ldots ,P_{4}$ in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ (i.e. an element of $W$ ) or a first-order deformation of $Q_{3},\ldots ,Q_{6},P_{1},\ldots ,P_{4}$ in $\tilde{Q}$ , or both.

We note that (3.8), and even $V_{1}\rightarrow H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})$ , is surjective, i.e. each obstruction in $H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})$ arises from a first-order infinitesimal deformation of $Q_{1},\ldots ,Q_{6}$ and the tangent directions of $Q_{3},\ldots ,Q_{6}$ in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ that fails to induce a first-order embedded deformation of $\unicode[STIX]{x1D6E5}$ .

To show that the map (3.7) is surjective, it suffices to show that for each deformation type listed, there exists an equisingular deformation of $B_{1}\cup B_{2}$ in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ which passes through the points to which $Q_{1},\ldots ,Q_{6}$ deform and which has the desired tangent direction at each point.

To begin, let us choose bi-homogeneous coordinates $((\unicode[STIX]{x1D6FC}:\unicode[STIX]{x1D6FC}^{\prime }),(\unicode[STIX]{x1D6FD}:\unicode[STIX]{x1D6FD}^{\prime }))$ on $Q=\mathbb{P}^{1}\times \mathbb{P}^{1}$ so that $\unicode[STIX]{x1D6FC}=x/y=-t/z$ and $\unicode[STIX]{x1D6FD}=x/t=-y/z$ . Let $g_{1}$ and $g_{2}$ be the equations (bihomogeneous of degree $(3,3)$ ) of $B_{1}$ and $B_{2}$ , respectively. Referring to (2.4) and (2.5), we have

$$\begin{eqnarray}\displaystyle g_{1} & = & \displaystyle -\unicode[STIX]{x1D6FC}{\unicode[STIX]{x1D6FC}^{\prime }}^{2}\unicode[STIX]{x1D6FD}^{3}+3\unicode[STIX]{x1D6FC}^{2}\unicode[STIX]{x1D6FC}^{\prime }\unicode[STIX]{x1D6FD}^{2}\unicode[STIX]{x1D6FD}^{\prime }-3\unicode[STIX]{x1D6FD}^{2}\unicode[STIX]{x1D6FD}^{\prime }{\unicode[STIX]{x1D6FC}^{\prime }}^{3}-3\unicode[STIX]{x1D6FC}{\unicode[STIX]{x1D6FC}^{\prime }}^{2}\unicode[STIX]{x1D6FD}{\unicode[STIX]{x1D6FD}^{\prime }}^{2}+3\unicode[STIX]{x1D6FC}^{3}\unicode[STIX]{x1D6FD}{\unicode[STIX]{x1D6FD}^{\prime }}^{2}+\unicode[STIX]{x1D6FC}^{2}\unicode[STIX]{x1D6FC}^{\prime }{\unicode[STIX]{x1D6FD}^{\prime }}^{3}\nonumber\\ \displaystyle g_{2} & = & \displaystyle -\unicode[STIX]{x1D6FD}{\unicode[STIX]{x1D6FD}^{\prime }}^{2}{\unicode[STIX]{x1D6FC}^{\prime }}^{3}-3\unicode[STIX]{x1D6FC}{\unicode[STIX]{x1D6FC}^{\prime }}^{2}\unicode[STIX]{x1D6FD}^{2}\unicode[STIX]{x1D6FD}^{\prime }+3\unicode[STIX]{x1D6FC}{\unicode[STIX]{x1D6FC}^{\prime }}^{2}{\unicode[STIX]{x1D6FD}^{\prime }}^{3}-3\unicode[STIX]{x1D6FD}{\unicode[STIX]{x1D6FD}^{\prime }}^{2}\unicode[STIX]{x1D6FC}^{2}\unicode[STIX]{x1D6FC}^{\prime }+3\unicode[STIX]{x1D6FD}^{3}\unicode[STIX]{x1D6FC}^{2}\unicode[STIX]{x1D6FC}^{\prime }-\unicode[STIX]{x1D6FD}^{2}\unicode[STIX]{x1D6FD}^{\prime }\unicode[STIX]{x1D6FC}^{3}.\nonumber\end{eqnarray}$$

Global first-order deformations $\tilde{B}_{1}$ and $\tilde{B}_{2}$ of $B_{1}$ and $B_{2}$ are given by equations

$$\begin{eqnarray}g_{1}+\unicode[STIX]{x1D700}\bar{g_{1}}=g_{1}+\unicode[STIX]{x1D700}\mathop{\sum }_{0\leqslant i,j\leqslant 3}a_{ij}\unicode[STIX]{x1D6FC}^{i}{\unicode[STIX]{x1D6FC}^{\prime }}^{3-i}\unicode[STIX]{x1D6FD}^{j}{\unicode[STIX]{x1D6FD}^{\prime }}^{3-j}\end{eqnarray}$$

and

$$\begin{eqnarray}g_{2}+\unicode[STIX]{x1D700}\bar{g_{2}}=g_{2}+\unicode[STIX]{x1D700}\mathop{\sum }_{0\leqslant i,j\leqslant 3}b_{ij}\unicode[STIX]{x1D6FC}^{i}{\unicode[STIX]{x1D6FC}^{\prime }}^{3-i}\unicode[STIX]{x1D6FD}^{j}{\unicode[STIX]{x1D6FD}^{\prime }}^{3-j},\end{eqnarray}$$

respectively. In order to describe equisingular first-order deformations of $B_{1}\cup B_{2}$ , we move the singularities of $B_{1}$ and $B_{2}$ at $P_{1},\ldots ,P_{4}$ to the points $(\unicode[STIX]{x1D700}c_{1},\unicode[STIX]{x1D700}d_{1}),\ldots ,(\unicode[STIX]{x1D700}c_{4},\unicode[STIX]{x1D700}d_{4})$ , given in local coordinates on $U_{1},\ldots ,U_{4}\subset Q$ , respectively, where

$$\begin{eqnarray}\displaystyle & \displaystyle U_{1}=\{\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}=1\},\quad U_{2}=\{\unicode[STIX]{x1D6FC}^{\prime }=\unicode[STIX]{x1D6FD}=1\}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle U_{3}=\{\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}^{\prime }=1\},\quad U_{4}=\{\unicode[STIX]{x1D6FC}^{\prime }=\unicode[STIX]{x1D6FD}^{\prime }=1\}. & \displaystyle \nonumber\end{eqnarray}$$

To simplify calculations, we change coordinates on $U_{1},\ldots ,U_{4}$ , so that the points $(\unicode[STIX]{x1D700}c_{1},\unicode[STIX]{x1D700}d_{1}),\ldots ,(\unicode[STIX]{x1D700}c_{4},\unicode[STIX]{x1D700}d_{4})$ are at the origin.

Letting $g_{ij}+\unicode[STIX]{x1D700}\bar{g}_{ij}$ be the degree $j$ part of the equation $g_{i}+\unicode[STIX]{x1D700}\bar{g}_{i}$ with respect to the new coordinates, we have the following conditions. These ensure that $B_{1}\cup B_{2}$ maintains the singularities, with possibly different tangent cones, at the points to which $P_{1},\ldots ,P_{4}$ deform. For simplicity we use the same notation for $P_{1},\ldots ,P_{4}$ and the points to which they deform:

1. (1) $g_{10}+\unicode[STIX]{x1D700}\bar{g}_{10}=0$ on each $U_{i}$ ; this forces $\tilde{B}_{1}$ to pass through $P_{1},\ldots ,P_{4}$ ;

2. (2) $g_{11}+\unicode[STIX]{x1D700}\bar{g}_{11}=0$ on $U_{1}$ , $U_{4}$ ; this forces $\tilde{B}_{1}$ to be singular at $P_{1}$ and $P_{4}$ ;

3. (3) $g_{12}+\unicode[STIX]{x1D700}\bar{g}_{12}=(m+m_{1}\unicode[STIX]{x1D700})(g_{21}+\unicode[STIX]{x1D700}\bar{g}_{21})^{2}$ , for some constants $m,m_{1}$ , on $U_{1}$ , $U_{4}$ (where $m,m_{1}$ may differ on $U_{1}$ , $U_{4}$ ); this forces the tangent cones of $\tilde{B}_{1}$ at $P_{1}$ and $P_{4}$ to be the same as those of $\tilde{B}_{2}$ at $P_{1}$ and $P_{4}$ ;

4. (4) $g_{13}+\unicode[STIX]{x1D700}\bar{g}_{13}=(g_{21}+\unicode[STIX]{x1D700}\bar{g}_{21})(h+\unicode[STIX]{x1D700}h_{1})$ , where $h$ and $h_{1}$ are quadratic forms; by Lemma 3.6, this forces $\tilde{B}_{1}$ to have tacnodes at the points $P_{1}$ and $P_{4}$ ;

5. (5) $g_{20}+\unicode[STIX]{x1D700}\bar{g}_{20}=0$ on each $U_{i}$ ; this forces and $\tilde{B}_{2}$ to pass through $P_{1},\ldots ,P_{4}$ ;

6. (6) $g_{21}+\unicode[STIX]{x1D700}\bar{g}_{21}=0$ on $U_{2}$ , $U_{3}$ ; this forces $\tilde{B}_{2}$ to be singular at $P_{2}$ and $P_{3}$ ;

7. (7) $g_{22}+\unicode[STIX]{x1D700}\bar{g}_{22}=(n+n_{1}\unicode[STIX]{x1D700})(g_{11}+\unicode[STIX]{x1D700}\bar{g}_{11})^{2}$ , for some constants $n,n_{1}$ , on $U_{2}$ , $U_{3}$ (where $n,n_{1}$ may differ on $U_{2}$ , $U_{3}$ ); this forces the tangent cones of $\tilde{B}_{2}$ at $P_{2}$ and $P_{3}$ to be the same as those of $\tilde{B}_{1}$ at $P_{2}$ and $P_{3}$ ;

8. (8) $g_{23}+\unicode[STIX]{x1D700}\bar{g}_{23}=(g_{11}+\unicode[STIX]{x1D700}\bar{g}_{11})(h+\unicode[STIX]{x1D700}h_{1})$ , where $h$ and $h_{1}$ are quadratic forms; by Lemma 3.6, this forces $\tilde{B}_{2}$ to have tacnodes at the points $P_{2}$ and $P_{3}$ .

Returning to original coordinates, and after simple algebraic manipulations, this gives the following system of $28$ linear equations in $c_{i},d_{i},a_{ij},b_{ij}$ (four blocks for four charts).

Equations 3.5.

$$\begin{eqnarray}\displaystyle & \displaystyle a_{33}=0 & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{33}=d_{1}-3c_{1} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{32}=-3c_{1}-6d_{1} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{23}=2c_{1}-3d_{1} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{22}=a_{31}+b_{23}+3b_{32} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{13}=2a_{31}-2b_{32}+4b_{23} & \displaystyle \nonumber\\ \displaystyle & \displaystyle 2c_{1}-d_{1}+3a_{12}+a_{03}+2a_{21}-a_{30}=0 & \displaystyle \nonumber\end{eqnarray}$$

$$\begin{eqnarray}\displaystyle & \displaystyle a_{30}=-c_{2}-3d_{2} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{30}=0 & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{31}=3c_{2}+2d_{2} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{20}=3d_{2}+c_{2} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{32}=2b_{10}+4a_{31}+2a_{20} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{21}=6b_{10}+6a_{31}+3a_{20} & \displaystyle \nonumber\\ \displaystyle & \displaystyle 5c_{2}+3d_{2}+5b_{00}+3b_{11}+6b_{22}+5b_{33}=0 & \displaystyle \nonumber\end{eqnarray}$$

$$\begin{eqnarray}\displaystyle & \displaystyle a_{03}=c_{3}+3d_{3} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{03}=0 & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{02}=2d_{3}+3c_{3} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{13}=3d_{3}+c_{3} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{01}=5a_{13}+2b_{23}+3a_{02} & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{12}=4a_{13}+6b_{23}+a_{02} & \displaystyle \nonumber\\ \displaystyle & \displaystyle c_{3}+2d_{3}-3b_{11}+b_{33}+2b_{22}+b_{00}=0 & \displaystyle \nonumber\end{eqnarray}$$

$$\begin{eqnarray}\displaystyle & \displaystyle a_{00}=0 & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{00}=d_{4}-3c_{4} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{01}=3c_{4}+6d_{4} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{10}=3d_{4}-2c_{4} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{20}=2a_{02}+2b_{01}+3b_{10} & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{11}=a_{02}+4b_{01}+6b_{10} & \displaystyle \nonumber\\ \displaystyle & \displaystyle 4c_{4}+5d_{4}+2a_{03}+3a_{12}+a_{21}+5a_{30}=0. & \displaystyle \nonumber\end{eqnarray}$$

Next, we determine all additional conditions on $a_{ij}$ , $b_{ij}$ , $c_{i}$ , $d_{i}$ which ensure that $\tilde{B}_{1}$ and $\tilde{B}_{2}$ pass through the points to which $Q_{1},\ldots ,Q_{6}$ deform, with the desired multiplicities at each point. To do so, we look in the chart $U_{4}$ . Here, the equation of $\unicode[STIX]{x1D6E5}$ is

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}(1+\unicode[STIX]{x1D6FD})+\unicode[STIX]{x1D6FD}-1=0.\end{eqnarray}$$

Solving for $\unicode[STIX]{x1D6FC}$ gives

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\frac{1-\unicode[STIX]{x1D6FD}}{1+\unicode[STIX]{x1D6FD}}.\end{eqnarray}$$

Thus, the points at which $\unicode[STIX]{x1D6E5}$ intersects $B_{1}$ and $B_{2}$ are the roots of the following polynomials:

$$\begin{eqnarray}(\unicode[STIX]{x1D6FD}^{2}+1)(\unicode[STIX]{x1D6FD}^{2}+4\unicode[STIX]{x1D6FD}+6)^{2}\end{eqnarray}$$

and

$$\begin{eqnarray}(\unicode[STIX]{x1D6FD}^{2}+1)(\unicode[STIX]{x1D6FD}^{2}+6\unicode[STIX]{x1D6FD}+6)^{2}.\end{eqnarray}$$

This gives the six points at which $B_{1}$ and $B_{2}$ intersect $\unicode[STIX]{x1D6E5}$ :

$$\begin{eqnarray}\displaystyle & \displaystyle Q_{1}=(-i,i),\quad Q_{2}=(i,-i), & \displaystyle \nonumber\\ \displaystyle & \displaystyle Q_{3}=(3-5i,-2+4i),\quad Q_{4}=(3+5i,-2-4i), & \displaystyle \nonumber\\ \displaystyle & \displaystyle Q_{5}=(-5+4i,-3+5i),\quad Q_{6}=(-5-4i,-3-5i), & \displaystyle \nonumber\end{eqnarray}$$

where $i^{2}+1=0~\text{mod}~7$ .

The intersections of $\bar{g}_{1}=0$ and $\bar{g}_{2}=0$ with $\unicode[STIX]{x1D6E5}$ are given by the zeros of the following polynomials:

$$\begin{eqnarray}\displaystyle {\hat{g}}_{1} & = & \displaystyle (1+\unicode[STIX]{x1D6FD})^{3}(a_{00}+a_{01}\unicode[STIX]{x1D6FD}+a_{02}\unicode[STIX]{x1D6FD}^{2}+a_{03}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1+\unicode[STIX]{x1D6FD})^{2}(1-\unicode[STIX]{x1D6FD})(a_{10}+a_{11}\unicode[STIX]{x1D6FD}+a_{12}\unicode[STIX]{x1D6FD}^{2}+a_{13}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1+\unicode[STIX]{x1D6FD})(1-\unicode[STIX]{x1D6FD})^{2}(a_{20}+a_{21}\unicode[STIX]{x1D6FD}+a_{22}\unicode[STIX]{x1D6FD}^{2}+a_{23}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1-\unicode[STIX]{x1D6FD})^{3}(a_{30}+a_{31}\unicode[STIX]{x1D6FD}+a_{32}\unicode[STIX]{x1D6FD}^{2}+a_{33}\unicode[STIX]{x1D6FD}^{3})\nonumber\end{eqnarray}$$

$$\begin{eqnarray}\displaystyle {\hat{g}}_{2} & = & \displaystyle (1+\unicode[STIX]{x1D6FD})^{3}(b_{00}+b_{01}\unicode[STIX]{x1D6FD}+b_{02}\unicode[STIX]{x1D6FD}^{2}+b_{03}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1+\unicode[STIX]{x1D6FD})^{2}(1-\unicode[STIX]{x1D6FD})(b_{10}+b_{11}\unicode[STIX]{x1D6FD}+b_{12}\unicode[STIX]{x1D6FD}^{2}+b_{13}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1+b)(1-\unicode[STIX]{x1D6FD})^{2}(b_{20}+b_{21}\unicode[STIX]{x1D6FD}+b_{22}\unicode[STIX]{x1D6FD}^{2}+b_{23}\unicode[STIX]{x1D6FD}^{3})\nonumber\\ \displaystyle & & \displaystyle +\,(1-\unicode[STIX]{x1D6FD})^{3}(b_{30}+b_{31}\unicode[STIX]{x1D6FD}+b_{32}\unicode[STIX]{x1D6FD}^{2}+b_{33}\unicode[STIX]{x1D6FD}^{3}).\nonumber\end{eqnarray}$$

Using these equations, we obtain eight additional linear equations in $a_{ij}$ , $b_{ij}$ , $c_{i}$ , $d_{i}$ . These ensure that $\tilde{B}_{1}$ and $\tilde{B}_{2}$ pass through $Q_{1},Q_{2}$ , that $\tilde{B}_{1}$ passes through $Q_{3},Q_{4}$ , and that $\tilde{B}_{2}$ passes through $Q_{5},Q_{6}$ . Note that each restriction arises from setting $\unicode[STIX]{x1D6FD}$ equal to $i$ , $-i$ , $-2+4i$ , $-2-4i$ , $-3+5i$ , or $-3-5i$ in the appropriate equation.

(B1Q1)

$$\begin{eqnarray}\displaystyle & & \displaystyle (3c_{1}-3c_{4}+3d_{1}+d_{4}-2a_{20}-2a_{21}-a_{31}+3a_{02}-a_{12}+3a_{03}-2b_{10}+3b_{32}+2b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{1}-3c_{4}+3d_{1}+d_{4}+2a_{20}-2a_{21}+a_{31}-3a_{02}-a_{12}+3a_{03}+2b_{10}-3b_{32}-2b_{23}=0\nonumber\end{eqnarray}$$

(B1Q2)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-3c_{1}+3c_{4}-3d_{1}-d_{4}+2a_{20}+2a_{21}+a_{31}-3a_{02}+a_{12}-3a_{03}+2b_{10}-3b_{32}-2b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{1}-3c_{4}+3d_{1}+d_{4}+2a_{20}-2a_{21}+a_{31}-3a_{02}-a_{12}+3a_{03}+2b_{10}-3b_{32}-2b_{23}=0\nonumber\end{eqnarray}$$

(B2Q1)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-3c_{2}-c_{3}-3d_{3}-a_{20}-3a_{31}-a_{02}-2b_{11}+b_{22}+3b_{32}+b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{2}+c_{3}+3d_{3}-a_{20}-3a_{31}-a_{02}+2b_{11}-b_{22}+3b_{32}+b_{23}=0\nonumber\end{eqnarray}$$

(B2Q2)

$$\begin{eqnarray}\displaystyle & & \displaystyle (3c_{2}+c_{3}+3d_{3}+a_{20}+3a_{31}+a_{02}+2b_{11}-b_{22}-3b_{32}-b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{2}+c_{3}+3d_{3}-a_{20}-3a_{31}-a_{02}+2b_{11}-b_{22}+3b_{32}+b_{23}=0\nonumber\end{eqnarray}$$

(B1Q3)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-c_{4}+2d_{1}-2d_{4}+3a_{20}+3a_{21}-a_{31}+3a_{12}+2b_{10}-2b_{32}-3b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{1}-c_{4}-3d_{1}-2d_{4}-2a_{20}+3a_{21}-3a_{31}+a_{02}+a_{12}-a_{03}-3b_{10}-3b_{32}+b_{23}=0\nonumber\end{eqnarray}$$

(B1Q4)

$$\begin{eqnarray}\displaystyle & & \displaystyle (c_{4}-2d_{1}+2d_{4}-3a_{20}-3a_{21}+a_{31}-3a_{12}-2b_{10}+2b_{32}+3b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,3c_{1}-c_{4}-3d_{1}-2d_{4}-2a_{20}+3a_{21}-3a_{31}+a_{02}+a_{12}-a_{03}-3b_{10}-3b_{32}+b_{23}=0\nonumber\end{eqnarray}$$

(B2Q5)

$$\begin{eqnarray}\displaystyle & & \displaystyle (c_{1}+3c_{2}+2c_{3}+2d_{1}-d_{3}+a_{20}-3a_{31}+2a_{02}-b_{10}+3b_{11}-b_{22}+b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,2c_{1}-2c_{2}+2c_{3}-3d_{1}+3d_{2}-d_{3}-2a_{20}-a_{31}+3a_{02}+b_{10}+3b_{11}-3b_{22}+b_{32}-3b_{23}=0\nonumber\end{eqnarray}$$

(B2Q6)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-c_{1}-3c_{2}-2c_{3}-2d_{1}+d_{3}-a_{20}+3a_{31}-2a_{02}+b_{10}-3b_{11}+b_{22}-b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad +\,2c_{1}-2c_{2}+2c_{3}-3d_{1}+3d_{2}-d_{3}-2a_{20}-a_{31}+3a_{02}+b_{10}+3b_{11}-3b_{22}+b_{32}-3b_{23}=0.\nonumber\end{eqnarray}$$

Taking the derivatives of ${\hat{g}}_{1}$ and ${\hat{g}}_{2}$ with respect to $\unicode[STIX]{x1D6FD}$ and setting $\unicode[STIX]{x1D6FD}$ equal to $-2+4i$ , $-2-4i$ , $-3+5i$ , or $-3-5i$ as appropriate gives the final four linear equations in $a_{ij},b_{ij},c_{i},d_{i}$ . These ensure that $\tilde{B}_{1}$ and $\tilde{B}_{2}$ are tangent to $\unicode[STIX]{x1D6E5}$ at $Q_{3},Q_{4}$ and $Q_{5},Q_{6}$ , respectively.

(dB1Q3)

$$\begin{eqnarray}\displaystyle & & \displaystyle (3c_{1}-2c_{4}-3d_{1}+3d_{4}+a_{21}+2a_{31}+2a_{02}+a_{03}-b_{10}-2b_{32})i\nonumber\\ \displaystyle & & \displaystyle \quad -\,c_{1}+c_{4}-2d_{1}+2d_{4}+2a_{20}+a_{21}+3a_{02}+a_{03}-3b_{10}-2b_{32}-b_{23}=0\nonumber\end{eqnarray}$$

(dB1Q4)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-3c_{1}+2c_{4}+3d_{1}-3d_{4}-a_{21}-2a_{31}-2a_{02}-a_{03}+b_{10}+2b_{32})i\nonumber\\ \displaystyle & & \displaystyle \quad -\,c_{1}+c_{4}-2d_{1}+2d_{4}+2a_{20}+a_{21}+3a_{02}+a_{03}-3b_{10}-2b_{32}-b_{23}=0\nonumber\end{eqnarray}$$

(dB2Q5)

$$\begin{eqnarray}\displaystyle & & \displaystyle (c_{1}-3c_{2}-c_{3}+2d_{1}+d_{2}-3d_{3}+3a_{20}+2a_{31}+3a_{02}+2b_{10}-2b_{11}-3b_{22}-2b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad -\,c_{2}-3c_{3}-3d_{2}-2d_{3}-3a_{31}-3a_{02}-b_{10}-b_{22}-3b_{32}=0\nonumber\end{eqnarray}$$

(dB2Q6)

$$\begin{eqnarray}\displaystyle & & \displaystyle (-c_{1}+3c_{2}+c_{3}-2d_{1}-d_{2}+3d_{3}-3a_{20}-2a_{31}-3a_{02}-2b_{10}+2b_{11}+3b_{22}+2b_{23})i\nonumber\\ \displaystyle & & \displaystyle \quad -\,c_{2}-3c_{3}-3d_{2}-2d_{3}-3a_{31}-3a_{02}-b_{10}-b_{22}-3b_{32}=0.\nonumber\end{eqnarray}$$

Consider a basis element in $H^{1}(\mathbb{P}_{1},{\mathcal{N}}_{\unicode[STIX]{x1D6E5}/\mathbb{P}_{1}})$ corresponding via Lemma 3.4 to some deformation of the points $P_{1},\ldots ,P_{4},Q_{1},\ldots ,Q_{6}$ together with the tangent directions of $P_{1},\ldots ,P_{4},Q_{4},\ldots ,Q_{6}$ in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ . There are two cases, as in Lemma 3.4.

Consider for example the basis element $I_{1}$ . The existence of an equisingular deformation of $(\mathbb{P}_{1},B_{1}\cup B_{2})$ mapping to $I_{1}$ is equivalent to the existence of $a_{ij},b_{ij},c_{i},d_{i}$ which satisfy (3.5), as well as B1Q1, B1Q2, B2Q1, B2Q2, B1Q3, dB1Q3, B1Q4-1, B2Q5, B2Q6. Here, we use Lemma 3.7.

Next we consider the basis element $I_{4}$ . The existence of an equisingular deformation of $(\mathbb{P}_{1},B_{1}\cup B_{2})$ mapping to $I_{4}$ is equivalent to the existence of $a_{ij},b_{ij},c_{i},d_{i}$ which satisfy (3.5), as well as B1Q1, B1Q2, B2Q1, B2Q2, B1Q3, dB1Q3, dB1Q3-1, B2Q4, B2Q5, B2Q6. Here, we use Lemma 3.7.

Thus, each of the seven basis elements corresponds to finding a non-trivial solution of a large system of linear equations. As working with such large matrices is unwieldy, we use Macaulay2 to check this (see the code included in the Appendix). In each case, we find that solutions indeed form either a three- or four-dimensional vector space, depending on the basis element, completing the proof. ◻

Proof. Completing the square, the equation becomes $((h_{1}+{\textstyle \frac{1}{2}}h_{2})^{2}+\text{h.o.t.}=0)\subset \mathbb{A}^{2}$ . Letting $h=h_{1}+{\textstyle \frac{1}{2}}h_{2}$ , the singularity becomes $(h^{2}+\text{h.o.t.}=0)\subset \mathbb{A}^{2}$ . As there are no terms of degree $3$ , this is a tacnode.◻

Proof. Suppose $\bar{g}(0,0)=0$ . We have to show that there exists $x_{0}$ with

$$\begin{eqnarray}g(\unicode[STIX]{x1D700}x_{0},0)+\unicode[STIX]{x1D700}\bar{g}(\unicode[STIX]{x1D700}x_{0},0)=0\end{eqnarray}$$

and that

$$\begin{eqnarray}\frac{d}{dx}(g(\unicode[STIX]{x1D700}x_{0},0)+\unicode[STIX]{x1D700}\bar{g}(\unicode[STIX]{x1D700}x_{0},0))=0.\end{eqnarray}$$

Taking the Taylor expansion of these with respect to $\unicode[STIX]{x1D700}$ , the first of these obviously holds. The second holds for

$$\begin{eqnarray}x_{0}=\frac{-\bar{g}_{1}^{\prime }(0,0)}{g_{1}^{\prime \prime }(0,0)}.\end{eqnarray}$$

◻

Proof. By 3.1, $\mathbb{P}$ has a fibration $\mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ with connected fibers such that the general fiber is smooth of genus one; see [Reference BǎdescuBǎd01, § 7]. Moreover, the $I_{8}$ fiber $\sum _{i=1}^{4}A_{i}+\sum _{i=1}^{4}\bar{G}_{i}$ has multiplicity three. Thus, this elliptic fibration is a Halphen surface of index three (after one blows down $\bar{N}_{1}$ and $\bar{N}_{2}$ ); see [Reference Cossec and DolgachevCD89, ch. V, Theorem 5.6.1].◻

Proof. We denote by $\unicode[STIX]{x1D6FC}$ the composition $W\rightarrow \mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ . Since this is a projective morphism, we have a Stein factorization for $\unicode[STIX]{x1D6FC}$ , i.e. maps $\unicode[STIX]{x1D6FD}:W\rightarrow C$ with connected fibers and $\unicode[STIX]{x1D6FE}:C\rightarrow \mathbb{P}_{k}^{1}$ a finite morphism such that $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FE}\circ \unicode[STIX]{x1D6FD}$ . Note that the multiplicity of the fiber $B_{1}+B_{2}+\bar{N}_{1}+\bar{N}_{2}$ of $\mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ is $1$ , and so $\unicode[STIX]{x1D6FE}:C\rightarrow \mathbb{P}_{k}^{1}$ is a finite separable morphism. Note also that the fibers $B_{1}+B_{2}+\bar{N}_{1}+\bar{N}_{2}$ and $I_{8}$ in $\mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ pull back to connected fibers of $\unicode[STIX]{x1D6FC}$ with multiplicities two and three, respectively. Since these multiplicities are coprime, we must have that the degree of $\unicode[STIX]{x1D6FE}$ is one, and so $\unicode[STIX]{x1D6FE}$ is an isomorphism. In this way $\unicode[STIX]{x1D6FC}$ has connected fibers. In addition, since it has two multiple fibers, the Kodaira dimension of $Y$ is non-negative [Reference Cossec and DolgachevCD89].

The double cover $W\rightarrow \mathbb{P}$ induces a connected étale cover between the non-multiple fibers of $\unicode[STIX]{x1D6FC}$ . Note that $\mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ can only have irreducible singular fibers apart from $B_{1}+B_{2}+\bar{N}_{1}+\bar{N}_{2}$ and $I_{8}$ , because the Picard number of $\mathbb{P}$ is $12$ . Therefore, we can have either two $I_{1}$ or one II as extra singular fibers. But a fiber of type II is étale simply connected, and so it does not have a connected étale cover of degree two. Thus, $\mathbb{P}\rightarrow \mathbb{P}_{k}^{1}$ has precisely two extra $I_{1}$ singular fibers, and their pre-images under $W\rightarrow \mathbb{P}$ give two $I_{2}$ reduced fibers for $\unicode[STIX]{x1D6FC}$ . This elliptic fibration induces a relatively minimal elliptic fibration $Y\rightarrow \mathbb{P}_{k}^{1}$ , after we blow-down the curves $G_{1},\ldots ,G_{4}$ .

Using well-known facts on double covers, one can easily verify that $K_{Y}^{2}=0$ , $\unicode[STIX]{x1D712}({\mathcal{O}}_{Y})=1$ , and

(4.1) $$\begin{eqnarray}p_{g}(Y)=h^{2}(-L)=h^{0}(K_{\mathbb{ P}}+L)=0,\end{eqnarray}$$

where

$$\begin{eqnarray}L=3\unicode[STIX]{x1D70E}^{\ast }(\unicode[STIX]{x1D6E5})-\mathop{\sum }_{i=1}^{4}\bar{E}_{i}-2\mathop{\sum }_{i=1}^{4}\bar{G}_{i}-\bar{N}_{1}-\bar{N}_{2}\end{eqnarray}$$

is the line bundle defining the double cover $\unicode[STIX]{x1D70B}^{\prime }$ . Thus, $q(Y)=0$ .◻

Proof. The Boyd surface $Y$ is the minimal resolution of the surface $S_{0}$ , which has log terminal singularities. Therefore, it suffices to show that $K_{Y}$ is nef, which follows from (4.2).◻

Proof. We can substitute $T$ with an affine connected component $\operatorname{Spec}A$ of  $o$ . By [Reference MumfordMum70, p. 56], it suffices to prove that $T_{K}$ is geometrically connected. Since it is smooth over $\operatorname{Spec}K$ and has a $K$ -point $\unicode[STIX]{x1D70E}_{1}(\unicode[STIX]{x1D702})$ , it suffices to prove that it is connected. Arguing by contradiction, suppose it is disconnected. Then $H^{0}(T_{K},{\mathcal{O}}_{T_{K}})$ contains a non-trivial idempotent $e$ . Let $\unicode[STIX]{x1D70B}\in {\mathcal{R}}$ be a uniformizer. Since $T$ is flat over $\operatorname{Spec}{\mathcal{R}}$ , $\unicode[STIX]{x1D70B}$ is not a zero-divisor in $A$ , and so  $e\in A[1/\unicode[STIX]{x1D70B}]$ . Let $n$ be the minimal non-negative integer such that $e$ can be written as $a/\unicode[STIX]{x1D70B}^{n}$ with $a\in A$ . Then $a^{2}=\unicode[STIX]{x1D70B}^{n}a$ . Since $T$ is smooth over $\operatorname{Spec}{\mathcal{R}}$ , its special fiber is reduced. It follows that $n=0$ because otherwise $a^{2}=0\operatorname{mod}(\unicode[STIX]{x1D70B})$ and therefore $a=0\operatorname{mod}(\unicode[STIX]{x1D70B})$ , which implies that $n$ is not minimal. So $e\in A$ , which contradicts the connectedness of $T$ .◻

Proof. By [Reference SchlessingerSch68], $F_{\unicode[STIX]{x1D701}_{0}}$ admits a hull, and by (5.1) we can assume that the hull is induced by an element $\bar{\unicode[STIX]{x1D701}}\in F_{\unicode[STIX]{x1D701}_{0}}(\operatorname{Spec}{\mathcal{H}})$ , where $({\mathcal{H}},\mathfrak{m})$ is a complete local Noetherian ${\mathcal{R}}$ -algebra with residue field $k$ . By Artin’s algebraization theorem [Reference ArtinArt69, Theorem 1.6], there exists an ${\mathcal{R}}$ -scheme of finite type $T$ , a closed $k$ -point $o\in T$ , an element $\unicode[STIX]{x1D701}\in F_{\unicode[STIX]{x1D701}_{0}}(T,o)$ , and an isomorphism $\unicode[STIX]{x1D70E}:\hat{{\mathcal{O}}}_{T,o}\rightarrow {\mathcal{H}}$ such that $F(\unicode[STIX]{x1D70E})\unicode[STIX]{x1D701}$ and $\bar{\unicode[STIX]{x1D701}}$ agree on ${\mathcal{H}}/\mathfrak{m}^{n}$ for all $n\geqslant 1$ . By (5.1), in fact $F(\unicode[STIX]{x1D70E})\unicode[STIX]{x1D701}=\bar{\unicode[STIX]{x1D701}}$ .

Since $F_{\unicode[STIX]{x1D701}_{0}}$ is smooth, $T\rightarrow \operatorname{Spec}{\mathcal{R}}$ is formally smooth at $o$ , and therefore we can assume that $T$ is a smooth ${\mathcal{R}}$ -scheme after shrinking it if necessary.

Since ${\mathcal{R}}$ is complete, we can find sections $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}:\operatorname{Spec}{\mathcal{R}}\rightarrow T$ such that $F(\unicode[STIX]{x1D70E}_{i})(\unicode[STIX]{x1D701})$ and $\unicode[STIX]{x1D6F4}_{i}$ agree on ${\mathcal{R}}/\mathfrak{n}^{n}$ for any  $n\geqslant 1$ , where $\mathfrak{n}\subset {\mathcal{R}}$ is the maximal ideal. By (5.1), $F(\unicode[STIX]{x1D70E}_{i})(\unicode[STIX]{x1D701})=\unicode[STIX]{x1D6F4}_{i}$ . It remains to apply Lemma 5.1.◻

Proof. Since ${\mathcal{K}}_{{\mathcal{R}}}^{\unicode[STIX]{x1D714}}$ is an algebraic ${\mathcal{R}}$ -stack, its associated set-valued functor $\operatorname{Def}_{{\mathcal{R}}}^{\mathbb{Q}G}$ satisfies the conditions of Lemma 5.3 by Artin’s criterion [Reference ArtinArt74].◻

Proof. This is well-known but we sketch a proof for completeness. Let $\mathfrak{m}\subset {\mathcal{R}}$ be the maximal ideal and let ${\mathcal{R}}_{n}={\mathcal{R}}/\mathfrak{m}^{n+1}$ for each $n=0,1,\ldots .$ We first lift $(Y;C_{1},\ldots ,C_{r})$ to a scheme and a collection of subschemes flat over $\operatorname{Spec}{\mathcal{R}}_{n}$ for each $n$ by induction on $n$ . So assume we already have a lift $(Y^{n};C_{1}^{n},\ldots ,C_{r}^{n})$ to $\operatorname{Spec}{\mathcal{R}}_{n}$ . We have an exact sequence

(5.2) $$\begin{eqnarray}0\rightarrow T_{Y}(-\text{log}(C_{1}+\cdots +C_{r}))\rightarrow T_{Y}\rightarrow i_{1\ast }N_{C_{1}/Y}\oplus \cdots \oplus i_{r\ast }N_{C_{r}/Y}\rightarrow 0\end{eqnarray}$$

of sheaves on $Y$ , where $i_{j}:C_{j}\rightarrow Y$ denotes the embedding for each  $j$ . Since $H^{2}(Y,T_{Y}(-\text{log}(C_{1}+\cdots +C_{r})))=0$ , we have $H^{2}(Y,T_{Y})=0$ as well. Therefore, we can lift $Y^{n}$ to a scheme $Y^{n+1}$ flat (and then automatically smooth and proper) over $\operatorname{Spec}{\mathcal{R}}_{n+1}$ . Moreover, all possible lifts form an affine space with underlying vector space $H^{1}(Y,T_{Y})$ . Since

$$\begin{eqnarray}H^{1}(Y,T_{Y})\rightarrow H^{1}(C_{1},N_{C_{1}/Y})\oplus \cdots \oplus H^{1}(C_{r},N_{C_{r}/Y})\end{eqnarray}$$

is surjective by $H^{2}(Y,T_{Y}(-\text{log}(C_{1}+\cdots +C_{r})))=0$ , we can choose a lift such that the corresponding class in $H^{1}(C_{i},N_{C_{i}/Y})$ vanishes for each  $i$ . This class can be interpreted as an obstruction to lifting $C_{i}^{n}\subset Y^{n}$ to a subscheme $C_{i}^{n+1}\subset Y^{n+1}$ flat over $\operatorname{Spec}{\mathcal{R}}_{n+1}$ . So we can lift all $C_{i}$ to subschemes $C_{i}^{n+1}\subset Y^{n+1}$ flat (and automatically smooth and proper) over $\operatorname{Spec}{\mathcal{R}}_{n+1}$ . The projective limit $\hat{{\mathcal{Y}}}=\mathop{\lim }_{\leftarrow }Y^{n}$ is a formal scheme smooth and proper over $\operatorname{Spf}{\mathcal{R}}$ . The projective limits $\hat{{\mathcal{C}}}_{i}=\mathop{\lim }_{\leftarrow }C_{i}^{n}$ for $i=1,\ldots ,n$ are closed formal subschemes smooth and proper over $\operatorname{Spf}{\mathcal{R}}$ .

Since $H^{2}(Y,{\mathcal{O}}_{Y})=0$ , we can lift any ample invertible sheaf on $Y$ to an (automatically ample) invertible sheaf on $\hat{{\mathcal{Y}}}$ . By Grothendieck’s existence theorem [Reference Grothendieck and DieudonnéEGAIII, 5.4.5], there exists a scheme ${\mathcal{Y}}$ projective and flat (and then automatically smooth) over $\operatorname{Spec}{\mathcal{R}}$ such that $\hat{{\mathcal{Y}}}$ is a completion of its special fiber. By [Reference Grothendieck and DieudonnéEGAIII, 5.1.8], there exist closed subschemes ${\mathcal{C}}_{1},\ldots ,{\mathcal{C}}_{r}\subset {\mathcal{Y}}$ such that $\hat{{\mathcal{C}}}_{1},\ldots ,\hat{{\mathcal{C}}}_{r}$ are completions of their special fibers. They are flat (and automatically smooth and proper) over $\operatorname{Spec}{\mathcal{R}}$ .◻

Notation 5.6. We revert to the notation of the previous sections; in particular, ${\mathcal{R}}$ will denote the ring of Witt vectors of an algebraically closed field $k$ of characteristic seven. We denote by $Y$ the Boyd surface over $k$ . The $(-4)$ -curve $\unicode[STIX]{x1D6E5}_{1}$ and the $(-2)$ -curve $N_{1}$ of $Y$ intersect transversally and in one point.

Proof. This follows from Theorem 3.1, (4.1), preservation of intersection numbers, and Lemma 5.5.◻

Proof. In one direction, we obtain ${\mathcal{Y}}$ by blowing up $\unicode[STIX]{x1D6F4}$ . In the opposite direction, since the question is étale-local on $S$ and ${\mathcal{X}}$ , we can assume that ${\mathcal{X}}$ and $S$ are spectra of Henselian local rings. By [Reference Lee and NakayamaLN13, Theorem 2.13], it suffices to find relative Cartier divisors $D_{1}$ and $D_{2}$ of ${\mathcal{X}}\rightarrow S$ such that their scheme-theoretic intersections with $\mathbb{P}$ are disjoint sections of the $\mathbb{P}^{1}$ -bundle. As in the proof of [Reference Lee and NakayamaLN13, Theorem 2.11], their existence follows from surjectivity of $\operatorname{Pic}{\mathcal{X}}\rightarrow \operatorname{Pic}\mathbb{P}_{s}^{1}$ (see [Reference Grothendieck and DieudonnéEGAIV, Corollary 21.9.12]), where $s\in S$ is the closed point. Finally, $\mathbb{X}_{S^{\prime }}$ (being toric) and hence ${\mathcal{X}}$ satisfy the Kollár condition.◻

Proof. This follows from the fact that $R^{1}\unicode[STIX]{x1D6FC}_{\ast }({\mathcal{O}}_{Y})=0$ as in [Reference WahlWah76] (where the equi-characteristic local case is worked out). Specifically, let $\hat{{\mathcal{Y}}}$ be the formal completion of the special fiber in ${\mathcal{Y}}$ . Let $\hat{\bar{{\mathcal{Y}}}}$ be a formal scheme with underlying topological space $S_{0}$ and sheaf of rings $\unicode[STIX]{x1D6FC}_{\ast }{\mathcal{O}}_{\hat{{\mathcal{Y}}}}$ . The vanishing of $R^{1}\unicode[STIX]{x1D6FC}_{\ast }({\mathcal{O}}_{Y})$ implies that $\hat{\bar{{\mathcal{Y}}}}$ is flat over $\operatorname{Spf}{\mathcal{R}}$ by [Reference WahlWah76, 0.4.4]. Since $H^{2}(S_{0},{\mathcal{O}}_{S_{0}})=0$ and $S_{0}$ is projective, $\hat{\bar{{\mathcal{Y}}}}$ carries an ample line bundle, and therefore is a formal fiber of a scheme $\bar{{\mathcal{Y}}}$ projective and flat over $\operatorname{Spec}{\mathcal{R}}$ , by Grothendieck’s existence theorem [Reference Grothendieck and DieudonnéEGAIII, 5.4.5]. Since the formal fiber functor is fully faithful [Reference Grothendieck and DieudonnéEGAIII, 5.4.1], the morphism $\hat{{\mathcal{Y}}}\rightarrow \hat{\bar{{\mathcal{Y}}}}$ is induced by the morphism $\unicode[STIX]{x1D6FC}:{\mathcal{Y}}\rightarrow \bar{{\mathcal{Y}}}$ . The rest follows from Lemma 5.9.◻

Proof. It suffices to prove that the special fiber of ${\mathcal{K}}_{{\mathcal{R}}}^{\unicode[STIX]{x1D714}}$ , i.e. the algebraic stack of Kollár families over $k$ , is smooth at $S_{0}\rightarrow \operatorname{Spec}k$ . There are several ways to deduce this from Theorem 3.1. One is to use the theory of index one covers as in [Reference HackingHac04, § 3] (which assumes characteristic zero but in our case this is not important because the index of the singularity two is not divisible by the characteristic seven). One can also mimic calculations in [Reference HackingHac04] in the setting of [Reference Abramovich, Hassett, Faber, van der Geer and LooijengaAH11]. Finally, one can apply [Reference WahlWah81, Proposition 6.4] (or [Reference Lee and NakayamaLN13, Theorem 4.6]), which shows that the morphism of deformation functors of artinian rings $\operatorname{Def}X\rightarrow \operatorname{Def}^{\text{loc}}X$ is smooth and that local $\mathbb{Q}$ -Gorenstein deformations of a $\frac{1}{4}(1,1)$ -singularity are unobstructed.◻

Proof. This follows from Theorem 5.4 and Lemmas 5.10, 2.4, and 5.11. ◻

Porism 5.13. The Craighero–Gattazzo surface is unobstructed and its local moduli space is smooth of dimension eight.

Proof. Since the stack of Kollár families ${\mathcal{K}}_{{\mathcal{R}}}^{\unicode[STIX]{x1D714}}$ is ${\mathcal{R}}$ -smooth at $S_{0}\rightarrow \operatorname{Spec}k$ , the stack ${\mathcal{K}}_{\mathbb{C}}^{\unicode[STIX]{x1D714}}$ is $\mathbb{C}$ -smooth at $S\rightarrow \operatorname{Spec}\mathbb{C}$ . But in the neighborhood of a smooth surface such as $S$ , ${\mathcal{K}}_{\mathbb{C}}^{\unicode[STIX]{x1D714}}$ can be identified with the Deligne–Mumford stack of Gieseker families of canonically polarized surfaces with canonical singularities.◻

Proof. We first claim that

(6.1) $$\begin{eqnarray}\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(D_{0})=1.\end{eqnarray}$$

We are going to use that $\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(S)=1$ (see [Reference Dolgachev and WernerDW99]). Since this is the only fact about $S$ that we need, we can shrink the curve $U$ from Theorem 5.12 and without loss of generality assume that $U$ is a complex disc. Since ${\mathcal{S}}$ contracts onto $D_{0}$ , we have $\unicode[STIX]{x1D70B}_{1}({\mathcal{S}})=\unicode[STIX]{x1D70B}_{1}(D_{0})$ . Now using the same argument as in [Reference XiaoXia91, p. 601], we have an exact sequence

$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{1}(S)\rightarrow \unicode[STIX]{x1D70B}_{1}({\mathcal{S}})\rightarrow \unicode[STIX]{x1D70B}_{1}(U)\rightarrow 1,\end{eqnarray}$$

and so $\unicode[STIX]{x1D70B}_{1}(S)$ surjects onto $\unicode[STIX]{x1D70B}_{1}(D_{0})$ . The right exactness of profinite completions [Reference Ribes and ZalesskiiRZ10, Proposition 3.2.5] implies that $\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(S)$ surjects onto $\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(D_{0})$ , which implies (6.1). Alternatively, surjectivity of $\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(S)\rightarrow \unicode[STIX]{x1D70B}_{1}^{\text{alg}}(D_{0})$ follows from the Grothendieck’s specialization theorem [Reference Grothendieck and RaynaudSGA1, Corollary 2.3].

We have $K_{D}^{2}=0$ . By Lemma 4.3 and Corollary 5.12, $K_{D_{0}}$ is nef. Therefore, $D$ is not rational. Indeed, if $D$ is rational, then by Riemann–Roch $h^{0}(D,-K_{D})\geqslant 1$ and so $-K_{D}\sim E\geqslant 0$ . Since $K_{D}\cdot \unicode[STIX]{x1D6E4}=2$ , we have $\unicode[STIX]{x1D6E4}\subset E$ . We know that $f^{\ast }(2K_{D_{0}})\sim -2E+\unicode[STIX]{x1D6E4}$ where $f:D\rightarrow D_{0}$ is the minimal resolution. But $E\neq \unicode[STIX]{x1D6E4}$ , and so $f^{\ast }(2K_{D_{0}})$ cannot be nef. Also, the Kodaira dimension of $D$ cannot be zero because of the Enriques classification and $K_{D}\cdot \unicode[STIX]{x1D6E4}=2$ , and cannot be two because of Kawamata’s argument [Reference KawamataKaw92] (see [Reference RanaRan14, Lemma 2.4]). Therefore, the Kodaira dimension is one, and so $D$ is an elliptic fibration over $\mathbb{P}^{1}$ (since $q(D)=0$ ).

Say we have $r$ multiple fibers of multiplicities $m_{1},\ldots ,m_{r}$ . By [Reference XiaoXia91, p. 601],

$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{1}(D)\simeq \langle a_{1},\ldots ,a_{r}:a_{1}\cdots a_{r}=a_{1}^{m_{1}}=\cdots =a_{r}^{m_{r}}=1\rangle .\end{eqnarray}$$

But this group is residually finite (see [Reference Lyndon and SchuppLS77, p. 126] and [Reference Lyndon and SchuppLS77, p. 141 last paragraph]). We also have $\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(D)=\unicode[STIX]{x1D70B}_{1}^{\text{alg}}(D_{0})$ (see [Reference KollárKol93]), and so by the above we get $\unicode[STIX]{x1D70B}_{1}(D)=1$ . This implies that there are only two multiple fibers $m_{1}F_{1},m_{2}F_{2}$ with coprime multiplicities $m_{1},m_{2}$ . Let $F$ be a general fiber of $D\rightarrow \mathbb{P}^{1}$ , and let $\unicode[STIX]{x1D6E4}\cdot F=d$ . Then, since $K_{D}\sim -F+(m_{1}-1)F_{1}+(m_{2}-1)F_{2}$ , we have $\unicode[STIX]{x1D6E4}\cdot K_{D}=d-d/m_{1}-d/m_{2}=2$ . In addition, since $\unicode[STIX]{x1D6E4}\cdot F_{1}=d/m_{1}$ and $\unicode[STIX]{x1D6E4}\cdot F_{2}=d/m_{2}$ , we have $d=\unicode[STIX]{x1D706}m_{1}m_{2}$ , and so $\unicode[STIX]{x1D706}(m_{1}m_{2}-m_{1}-m_{2})=2$ . The only possible solutions, up to permuting one and two, are $\unicode[STIX]{x1D706}=2$ , $m_{1}=2$ , $m_{2}=3$ .◻

Proof. Here we use the method of [Reference Lee and ParkLP07], which applies Van Kampen’s theorem and the Milnor fiber of the $\mathbb{Q}$ -Gorenstein smoothing of $\frac{1}{4}(1,1)$ . We only need $\unicode[STIX]{x1D70B}_{1}(D\setminus \unicode[STIX]{x1D6E4})=1$ . By Van Kampen’s theorem, we have $\unicode[STIX]{x1D70B}_{1}(D)\simeq \unicode[STIX]{x1D70B}_{1}(D\setminus \unicode[STIX]{x1D6E4})/\overline{\langle \unicode[STIX]{x1D6FC}\rangle }$ where $\unicode[STIX]{x1D6FC}$ is a loop around $\unicode[STIX]{x1D6E4}$ , and $\overline{\langle \unicode[STIX]{x1D6FC}\rangle }$ is the smallest normal subgroup of $\unicode[STIX]{x1D70B}_{1}(D\setminus \unicode[STIX]{x1D6E4})$ containing $\langle \unicode[STIX]{x1D6FC}\rangle$ . We can and do consider $\unicode[STIX]{x1D6FC}$ as given by a loop around $N$ , since $N$ and $\unicode[STIX]{x1D6E4}$ intersect transversally. As $N\cdot \unicode[STIX]{x1D6E4}=1$ , the set $N^{\prime }:=N\cap (D\setminus \unicode[STIX]{x1D6E4})$ is simply connected, and so $\unicode[STIX]{x1D6FC}\subset N^{\prime }\subset D\setminus \unicode[STIX]{x1D6E4}$ is homotopically trivial. Therefore, $\overline{\langle \unicode[STIX]{x1D6FC}\rangle }=1$ , and so $\unicode[STIX]{x1D70B}_{1}(D\setminus \unicode[STIX]{x1D6E4})=1$ since by Proposition 6.1 we have $\unicode[STIX]{x1D70B}_{1}(D)=1$ . After this, one directly applies [Reference Lee and ParkLP07, pp. 493 and 499].◻

Proof. In § 3, the main point was to prove that

$$\begin{eqnarray}H^{2}(\mathbb{P}_{1},T_{\mathbb{P}_{1}}(-\text{log}(\unicode[STIX]{x1D6E5}+B_{1}+B_{2})))=0.\end{eqnarray}$$

By applying the $(-1)$ and $(-2)$ principles as before, we have

$$\begin{eqnarray}H^{2}\biggl(\mathbb{P}_{1},T_{\mathbb{P}_{1}}\biggl(-\text{log}\biggl(\unicode[STIX]{x1D6E5}+B_{1}+B_{2}+\sum \bar{N}_{i}+\sum \bar{G}_{i}+\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}\biggr)\biggr)\biggr)=0.\end{eqnarray}$$

By Lemma 5.5, preservation of intersection numbers, and $H^{2}(\mathbb{P}_{1},{\mathcal{O}}_{\mathbb{P}_{1}})=0$ , we have that the configuration of curves $\unicode[STIX]{x1D6E5}+B_{1}+B_{2}+\sum \bar{N}_{i}+\sum \bar{G}_{i}+\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}$ exists in $\mathbb{P}_{1}$ over $\mathbb{C}$ . We will use the same notation as in characteristic seven. Then, by contracting $\sum \bar{N}_{i}+\sum \bar{G}_{i}+\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}$ , we obtain curves $\unicode[STIX]{x1D6E5}+B_{1}+B_{2}$ in $\mathbb{P}_{\mathbb{C}}^{1}\times \mathbb{P}_{\mathbb{C}}^{1}$ with the corresponding singularities. In this way, we can check that $\unicode[STIX]{x1D6E5}\sim (1,1)$ and $B_{i}\sim (3,3)$ in Pic $(\mathbb{P}^{1}\times \mathbb{P}^{1})$ . Note that the two singularities of $B_{1}$ and the two singularities of $B_{2}$ may not be located at the special position we had in characteristic seven. Let us call these points $P_{1},\ldots ,P_{4}$ as before.

The linear system $|{\mathcal{O}}(2,2)|$ contains a member, which we call $\unicode[STIX]{x1D6E4}$ , that passes through $P_{1},\ldots ,P_{4}$ with the direction of the tangent cone to $B_{1}\cup B_{2}$ . Indeed, $\unicode[STIX]{x1D6E4}$ exists because $h^{0}(\mathbb{P}^{1}\times \mathbb{P}^{1},{\mathcal{O}}(2,2))=9$ and passing through four points with four given directions imposes eight conditions.

Then one easily checks in $\mathbb{P}_{1}$ that

(7.1) $$\begin{eqnarray}B_{1}+B_{2}+2\bar{N}_{1}+2\bar{N}_{2}\sim 3\unicode[STIX]{x1D6E4}\end{eqnarray}$$

as well as

(7.2) $$\begin{eqnarray}B_{1}+B_{2}+\mathop{\sum }_{i=1}^{4}\bar{G}_{i}\sim 2\biggl(3\unicode[STIX]{x1D70E}^{\ast }(\unicode[STIX]{x1D6E5})-\bar{N}_{1}-\bar{N}_{2}-3\mathop{\sum }_{i=1}^{4}\bar{E}_{i}-\mathop{\sum }_{i=1}^{4}\bar{G}_{i}\biggr).\end{eqnarray}$$

In this way, (7.1) gives an elliptic fibration $\mathbb{P}_{1}\rightarrow \mathbb{P}_{\mathbb{C}}^{1}$ with one multiple fiber $\unicode[STIX]{x1D6E4}$ of multiplicity three, and (7.2) gives a double cover $W\rightarrow \mathbb{P}_{1}$ of $\mathbb{P}_{1}$ branched along $B_{1}+B_{2}+\sum _{i=1}^{4}\bar{G}_{i}$ , just as before. Again the pre-images of $\bar{G}_{1},\ldots ,\bar{G}_{4}$ give $(-1)$ -curves in $W$ , which we contract to obtain a surface $D$ . Using the standard formulas for double covers, as before, we get $K_{D}^{2}=0$ and $\unicode[STIX]{x1D712}({\mathcal{O}}_{D})=1$ . Also, we can directly compute $p_{g}(D)=0$ using the defining line bundle of the double cover, and so $q(D)=0$ . The pull-back of the elliptic fibration $\mathbb{P}_{1}\rightarrow \mathbb{P}_{\mathbb{C}}^{1}$ gives an elliptic fibration $D\rightarrow \mathbb{P}_{\mathbb{C}}^{1}$ with two multiple fibers: the pre-images of $\unicode[STIX]{x1D6E4}$ and $B_{1}+B_{2}+\bar{N}_{1}+\bar{N}_{2}$ , with multiplicities three and two, respectively. The two $(-4)$ curves are pre-images of $\unicode[STIX]{x1D6E5}$ .◻

Proof. In characteristic seven, we have two genus two fibrations on the Boyd surface induced by the two rulings in $\mathbb{P}^{1}\times \mathbb{P}^{1}$ . We first want to find out the singular fibers of these fibrations. For that, we need to look at the induced morphisms $B_{i}\subset \mathbb{P}\rightarrow \mathbb{P}^{1}\times \mathbb{P}^{1}\rightarrow \mathbb{P}^{1}$ for each $i$ and for each ruling.

Using (2.4) and (2.5) of $B_{1}$ and $B_{2}$ , respectively, we obtain that, for the ruling $\unicode[STIX]{x1D6FD}=x/t$ , the morphism $B_{1}\rightarrow \mathbb{P}^{1}$ has branch points at $\unicode[STIX]{x1D6FD}$ satisfying $(\unicode[STIX]{x1D6FD}^{2}+1)^{2}=0$ , and the morphism $B_{2}\rightarrow \mathbb{P}^{1}$ is branched at $\unicode[STIX]{x1D6FD}$ satisfying $\unicode[STIX]{x1D6FD}^{4}+4\unicode[STIX]{x1D6FD}^{2}+1=0$ . One verifies that in the first case, the points of ramification are $Q_{1}=(-i,i)$ and $Q_{2}=(i,-i)$ , and $B_{1}$ is tangent to the ruling with flex points at $Q_{1}$ and $Q_{2}$ . For the second ruling, the roles of $B_{1}$ and $B_{2}$ are interchanged in relation to ramification, and $B_{2}$ is tangent to the ruling with flex points at $Q_{1}$ and $Q_{2}$ for $B_{2}$ .

Using the previous observations on the ramification points of $B_{1}\rightarrow \mathbb{P}^{1}$ and $B_{2}\rightarrow \mathbb{P}^{1}$ , we obtain the following singular fibers for the genus two fibrations $Y\rightarrow \mathbb{P}^{1}$ (we take it from one ruling, the other is analogous).

1. (1) Two reduced singular fibers consisting of $E_{1}\cup A_{1}\cup E_{4}$ and $E_{2}\cup A_{2}\cup E_{3}$ where $E_{i}$ are disjoint elliptic $(-1)$ curves, and $A_{i}$ are $(-2)$ rational curves, each intersecting two $E_{j}$ at one nodal point.

2. (2) Two reduced singular fibers over $\unicode[STIX]{x1D6FD}=i,-i$ consisting of one nodal rational curve together with $N_{1}$ , and another rational nodal curve with $N_{2}$ . Each of the $N_{i}$ passes through the corresponding node, forming a simple triple point for the fiber.

3. (3) Four reduced singular curves, each consisting of a nodal curve whose resolution is an elliptic curve.

We claim that there exists a lifting of this Dolgachev surface to characteristic zero as in Theorem 7.1 such that case (2) is eliminated. In other words, we have to construct a lifting of $\mathbb{P}_{1}$ together with the curves $\unicode[STIX]{x1D6E5}+B_{1}+B_{2}+\sum \bar{N}_{i}+\sum \bar{G}_{i}+\sum \bar{E}_{i}+\sum \bar{C}_{i}+\sum \bar{F}_{i}$ such that the flex ramification points for $B_{1}\rightarrow \mathbb{P}^{1}$ disappear, becoming simple ramification for a degree three morphism $B_{1}\rightarrow \mathbb{P}_{\mathbb{C}}^{1}$ . Using the Macaulay2 code in the Appendix, we show the existence of a first other deformation of that type. This together with unobstructed deformations, as in the remark above, gives a lifting to $\operatorname{Spec}{\mathcal{R}}$ such that, over the generic point, the curve $B_{1}$ is not flex with respect to any ruling. In this way, at least for one ruling, the corresponding genus two fibration on the complex $2,3$ Dolgachev surface has only singular fibers which are reduced and with nodes as singularities, i.e. it is a Lefschetz fibration.◻