We were trying to show that there is no Enriques surface over $\mathbb{Z}$. If $Y \to \operatorname{Spec} \mathbb{Z}$ is an Enriques surface, we proved the following fact.
Theorem 1. The Picard scheme of $Y$ has to be the constant group $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}^{\oplus 10}$.
Using this, we want to get a contradiction. Recall that we had this geometric result.
Proposition 2 (4.2). Let $Y_{\bar{\mathbb{F}}_ 2}$ be an Enriques surface $\mathrm{Pic}_ {Y/\bar{\mathbb{F}}_ 2}^\tau \cong \mathbb{Z}/2\mathbb{Z}$. Then there is a genus-one fibration $\varphi \colon Y_{\bar{\mathbb{F}}_ 2} \to \mathbb{P}_ {\bar{\mathbb{F}}_ 2}^1$ with exactly two multiple fibers $2F_1$ and $2F_2$ (where $2$ is the gcd of the multiplicities of each irreducible component), and $\omega_Y \cong \mathscr{O}(F_1-F_2)$. Moreover, the intersection form on $\mathrm{Num}_ {Y/\bar{\mathbb{F}}_ 2} \cong \mathbb{Z}^{\oplus 10}$ is unimodular (in fact isomorphic to $(\begin{smallmatrix} 0 & 1 \br 1 & 0 \end{smallmatrix}) \oplus E_8$).
Kodaira symbols
What can a fiber of a genus-one fibration $Y \to \mathbb{P}_ \bar{k}^1$ look like? Note that every fiber is geometrically connected, because it is $\mathscr{O}$-connected. Becasue it is a Cartier divison on a smooth variety, we can write it as $$ C = m_1 C_1 + \dotsm + m_r C_r. $$ Then we can make the following observations.
- $C_i \cdot C = 0$
- $C_i \cdot C_j \ge 0$ with strict inequality when they meet
- the self-intersection number $C_i \cdot C_i$ (which is always going to be negative unless $r = 1$) is not $-1$
- $\sum_i m_i (1 - p_a(C_i) + \frac{1}{2} C_i \cdot C_i) = 0$
This means that unless $k = 1$ we would need $p_a(C_i) = 0$ and $C_i \cdot C_i = -2$ for all $i$. Using this, we can classify the possible values of $m_i$ and the graph for the components (up to a scalar).
In the case when $r = 1$, we have
- I0: elliptic curve
- I1: nodal curve
- II: cuspidal curve
and then we have things like
- In: $n$-gon of $\mathbb{P}^1$’s,
- III: two $\mathbb{P}^1$’s meeting tangentially at a double point,
- IV: three $\mathbb{P}^1$’s meeting at a triple point, where they pairwise intersect transversely,
- I0*: four $\mathbb{P}^1$’s branching out from a $\mathbb{P}^1$, with the multiplicity of the middle being $2$,
- IV*: affine $E_6$,
and so on.
Using the triviality of numerical classes
Recall that we know that the scheme of numerical classes of $Y$ is trivial. This actually turns out to put a lot of restriction on the geometry.
Proposition 3 (2.1). Let $Y \to \mathbb{P}_ k^1$ be a genus-one fibration, where $k$ is a perfect field. Assume that $\mathrm{Num}_ {Y/k}$ is a trivial group scheme (which we have becuase even $\mathrm{Pic}_ {Y/k}$ is trivial). Then
- for every closed point $a \in \mathbb{P}_ k^1$ the irreducible components of $Y_a$ are geometrically irreducible,
- if $a$ is not a $k$-point, then $Y_a$ is geometrically irreducible.
The point is that the Galois group action on the set of irreducible components of $Y_a$ must preserve the numerical classes. If you think about it, this is pretty hard. For any $\sigma \in \operatorname{Gal}(\bar{k}/k)$ we should have $\sigma(C_i) \cdot C_i = C_i \cdot C_i \le 0$ and so either $C_i = \sigma(C_i)$ or $C_i \cdot C_i = 0 = \sigma(C_i) \cdot C_i$.
Theorem 4 (5.6). Let $k$ be a perfect field of characteristic $2$ and let $Y/k$ be an Enriques surface with trivial $\mathrm{Pic}_ {Y/k}$. Then
- every $(-2)$-curve $\bar{E} \subseteq Y_{\bar{k}}$ descends to $k$ and moreover $E \cong \mathbb{P}_ k^1$,
- every genus-one fibration $Y_{\bar{k}} \to \mathbb{P}_ \bar{k}^1$ is actually defined over $k$,
- this has exactly two multiple fibers, lying over $k$-points of $\mathbb{P}_ k^1$,
- each multiple fiber is either an ordinary elliptic curve or not semistable.
For the first part, we use that $-2 = \bar{E} \cdot \bar{E} = \sigma(\bar{E}) \cdot \bar{E}$ implies that $\bar{E}$ is stable under the Galois action. This means that we can descend $\bar{E}$ to $E$. On the other hand, becasuse $K$ is numerically trivial, we can see from the Riemann–Roch formula that $p_a(E)-1 = \frac{1}{2} E \cdot (E-K) = -1$ and hence $E$ is a form of $\mathbb{P}^1$. The reason it has a $k$-point is because $E$ defines a primitive numerical class (as $-2$ is square-free) and hence there is an invertible sheaf $\mathscr{N}/Y$ such that $\deg_E(\mathscr{N}) = 1$.
For the second part, we first check that the Galois action sends fibers to fibers. This is becasue most geometric fibers are irreducible, and if we want $\sigma(F) \cdot F = F \cdot F = 0$ then $\sigma(F) \to \mathbb{P}_ \bar{k}^1$ has to avoid a point, hence maps to a point. This shows that we can descend the genus-one fibration to $$ Y \to B, \quad B_\bar{k} \cong \mathbb{P}_ \bar{k}^1 $$ for $B$ some form of $\mathbb{P}^1$. At this point, we choose a half-fiber $\bar{F} \subseteq \bar{Y}$ and consider the line bundle $\mathscr{L}/Y$ that pulls back to $\mathscr{O}_ \bar{Y}(\bar{F})$. If $\bar{F}$ is reducible then it contains a $(-2)$-curve, which must be Galois-stable, and hence we get a $k$-point on $B$ and are done. Otherwise $F$ is an integral curve with $h^0(\mathscr{O}_ F) = h^1(\mathscr{O}_ F) = 1$ and there is an invertible sheaf $\mathscr{N}$ with $\deg_F \mathscr{N} = 1$. This shows that $F$ has a rational point, and so does $B$. This argument also shows that every half-fiber has to live over a rational point, and hence proves the third part as well.
For the last part, we note that for the multiple fiber $F$ we have $\mathscr{O}_ F(F)$ of order two. Now we note that a supersingular elliptic curve doesn’t have nonzero 2-torsion points, and similarly for a semistable curve of genus one.
Combinatorics
At this point, we start analyzing what kind of fibers are possible, using our Kodaira symbols. The way we turn this into a finite problem is using the following observation.
Proposition 5 (7.3). Let $Y / \mathbb{F}_ 2$ be an Enriques surface with trivial $\mathrm{Num}_ {Y/\mathbb{F}_ 2}$. Then $Y(\mathbb{F}_ 2)$ has cardinality $25$.
The main observation is that the Hodge diamond will be concentrated on the main diagonal, and then because all the classes are generated by algebraic cycles over $k$, we have $$ H^{2j}(Y_\bar{k}, \mathbb{Q}_ \ell) \cong \mathbb{Q}_ \ell(-j)^{\oplus b_{2j}}. $$ Using the Lefschetz trace formula, we then see that $$ Y(\mathbb{F}_ 2) = 1 + 10 \cdot 2 + 1 \cdot 2^2 = 25. $$ These should all live over $\mathbb{P}^1(\mathbb{F}_ 2) = \lbrace 0, 1, \infty \rbrace$. Since $25$ is quite a large number compared to $3$, there has to be components with many irreducible components. Now we start analyzing these possibilities and ruling them out. Here is a sample.
Proposition 6 (11.2). It is not possible to have a fiber of type I8 and another fiber of type III and another fiber an elliptic curve $E_4 : y^2+xy=x^3+x$ which is ordinary and has $4$ rational points.
Here, III is two $\mathbb{P}^1$’s with a length-two intersection at a single point, and I8 is a 8-gon of $\mathbb{P}^1$’s. It turns out that the semistable fiber of type I8 has to be simple, so the other two are multiple fibers. Write $$ \varphi^{-1}(0) = 2(D_0 + D_1), \quad \varphi^{-1}(\infty) = C_0 + \dotsb + C_7. $$
What we do is to consider another genus-one fibration $\psi \colon Y \to \mathbb{P}_ {\mathbb{F}_ 2}^1$, with the property that $\varphi^{-1}(c) \cdot \psi^{-1}(d) = 4$. (It doesn’t matter which fiber we choose.) This is a general fact about non-exceptional Enriques surfaces, where we have already seen that all genus-one fibrations descend to $\mathbb{F}_ 2$.
Let us say that $\psi^{-1}(0)$ and $\psi^{-1}(1)$ are double, whereas $\psi^{-1}(\infty)$ is simple. Writing $\psi^{-1}(0) = 2 F = 2 \sum_i m_i \Theta_i$, we see that $$ (D_0 + D_1) \cdot F = 1. $$ Hence we may assume that $F \cdot D_1 = 1$ and $F \cdot D_0 = 0$, so that $\psi(D_0) = \ast$. Now $D_1$ is a section of $\psi$ and so write $D_1 \cdot \Theta_0 = m_0 = 1$ and $D_1 \cdot \Theta_j = 0$ for $j \ge 1$. It also follows that $D_0 \neq \Theta_j$, because we have $D_1 \cdot D_0 = 2$. It follows that $D_0 \cdot \Theta_j \ge 0$ for all $j$, and hence $D_0 \cdot \Theta_j = 0$ for all $j$. In particular, we have $$ \Theta_j \cdot (D_0 + D_1) = 0 $$ for $j \ge 1$, and hence $\Theta_1, \dotsc, \Theta_r$ are contained in fibers of $\varphi$. On the other hand, from $m_0 = 1$ and staring at the diagram, we see that $\Theta_1 + \dotsb + \Theta_r$ is either empty or connected. Because none of them can be elliptic curves, we see that $\Theta_1 + \dotsb + \Theta_r$ is contained in $\varphi^{-1}(\infty)$ and is hence a chain. Because $\psi^{-1}(0)$ has to become a chain of $\mathbb{P}^1$’s after removing one $m_0 = 1$ curve, we see that the half-fiber $\frac{1}{2} \psi^{-1}(0)$ has to be either one of II, III, IV. Combining this we the same statement for $\psi^{-1}(1)$ and doing some more analysis, we see that the fibers of $\psi$ also have to be of the form $$ \psi \colon \mathrm{III} + E_4 + \mathrm{I}_ 8. $$ Playing more around with the diagram gives a contradiction.
At the end of the combinatorics, we are only left with exceptional Enriques surfaces.
Theorem 7. An exceptional Enriques surface over $\mathbb{F}_ 2$ does not lift to $\mathbb{Z}/4\mathbb{Z}$.