The maximum number of points on a curve of genus eight
over the field of four elements

To obtain a contradiction, suppose such a curve $C$ exists. We will determine the real Weil polynomial (defined below) of $C$, and use this polynomial to show that $C$ must be a triple cover of a certain elliptic curve. Then we will show that no such triple cover can exist.

The Weil polynomial of an abelian variety over a finite field is the characteristic polynomial of its Frobenius endomorphism, and a result of Tate [12, Theorem 1(c), p. 139] says that two abelian varieties over the same finite field $k$ are isogenous to one another (over $k$) if and only if they have the same Weil polynomial. If $A$ is a $g$-dimensional abelian variety over ${\mathbb{F}}_{q}$ with Weil polynomial $f$, then the real Weil polynomial of $A$ is the unique degree-$g$ polynomial $h$ such that $f(x)=x^{g}h(x+q/x)$. We define the Weil polynomial and the real Weil polynomial of a curve over a finite field to be the Weil polynomial and the real Weil polynomial of the curve’s Jacobian, respectively.

Suppose $f$ is the Weil polynomial of an isogeny class of $g$-dimensional abelian varieties over ${\mathbb{F}}_{q}$, and let $\pi_{1},\ldots,\pi_{2g}$ be the complex roots of $f$, listed with multiplicity. For every $n>0$ we define quantities $R_{n}(f)$ and $P_{n}(f)$ by

where $\mu$ is the Möbius function. If $f$ is the Weil polynomial of a curve $C$, then $R_{n}(f)=\#C({\mathbb{F}}_{q^{n}})$ and $P_{n}(f)$ is equal to the number of degree-$n$ places on $C$, and so $P_{n}(f)\geqslant 0$. This gives us a simple test that must be satisfied if an isogeny class of abelian varieties contains a Jacobian: We simply check to see that $P_{n}(f)\geqslant 0$ for all $n$. An isogeny class that fails this test does not contain a Jacobian, while an isogeny class that passes the test may or may not contain a Jacobian.

Lauter [6] sketches an algorithm for enumerating the Weil polynomials $f$ for the isogeny classes of $g$-dimensional abelian varieties over ${\mathbb{F}}_{q}$ with $P_{n}(f)\geqslant 0$ for all $n$ and with $P_{1}(f)$ equal to a given integer $N$. The Jacobian of a genus-$g$ curve over ${\mathbb{F}}_{q}$ with $N$ points will necessarily lie in one of these isogeny classes. The paper [4] and its predecessor [2] provide many methods for showing that certain isogeny classes that pass this initial test do not contain Jacobians; we mention three here that are stated in terms of real Weil polynomials. One criterion, due to Serre [10, p. Se 11 ff.] [11, §II.4], says that if the real Weil polynomial $h$ of an isogeny class can be written as a product of two nontrivial monic factors $h_{1}$ and $h_{2}$ such that the resultant of $h_{1}$ and $h_{2}$ is $\pm 1$, then the isogeny class does not contain a Jacobian (see also [4, Theorem 2.2(a) and Proposition 2.8]). A second criterion [4, Theorem 2.2(b) and Proposition 2.8] says that if we have a factorization $h=h_{1}h_{2}$ where the resultant of $h_{1}$ and $h_{2}$ is equal to $\pm 2$ (or more generally, when the gluing exponent of the pair of isogeny classes associated to $h_{1}$ and $h_{2}$ is equal to $2$), then any curve with real Weil polynomial $h$ is a double cover of a curve whose real Weil polynomial is either $h_{1}$ or $h_{2}$; often, this added information is enough to show that there can be no curve with real Weil polynomial $h$. And finally, if $q$ is a square, say $q=s^{2}$, and if the real Weil polynomial $h$ of an isogeny class can be factored as $h=h_{1}h_{2}$ where $h_{1}$ is a power of $x-2s$ and where $h_{2}(2s)$ is squarefree, then there is no Jacobian with real Weil polynomial $h$ [4, Theorem 3.1]. We will call these three techniques the resultant $1$ argument, the resultant $2$ argument, and the supersingular factor argument, respectively.

Using the Magma programs that accompany [4], which may be found at

http://ewhowe.com/papers/paper35.html,

As the table indicates, all but the first of the $26$ possible real Weil polynomials can be eliminated from consideration by using one of the arguments mentioned above. The entries that are eliminated by the resultant $1$ argument or the supersingular factor argument need no explanation beyond the specification of the splitting $h=h_{1}h_{2}$ that makes the argument work. The entries that are eliminated by the resultant $2$ argument do require some more explanation, to indicate why there would be a problem with a curve with real Weil polynomial $h$ being a double cover of a curve with real Weil polynomial $h_{1}$ or $h_{2}$.

The Riemann–Hurwitz formula shows that a genus-$8$ curve cannot be a double cover of a curve of genus $5$ or larger. That fact is enough to show that for table entries 13, 16, 17, 18, and 24, it is not possible for a curve with real Weil polynomial $h$ to be a double cover of a curve with real Weil polynomial $h_{2}$.

If a genus-$8$ curve with $24$ rational points is a double cover of a curve $D$, then the curve $D$ must have at least $12$ rational points. This shows that for table entries 13, 17, and 18, it is not possible for a curve with real Weil polynomial $h$ to be a double cover of a curve with real Weil polynomial $h_{1}$.

For table entries 16 and 24, there is no curve with real Weil polynomial equal to $h_{1}$, because of the resultant $1$ argument.

The only remaining “resultant $2$” entry on the table is number 6. For that entry, the supersingular factor argument shows there is no curve with real Weil polynomial $h_{2}$. Next, we check that a genus-$4$ curve $D$ with real Weil polynomial $h_{1}$ must have $13$ rational points. We also check that a genus-$8$ curve $C$ with real Weil polynomial $h$ has no places of degree $2$. Therefore, if we have a double cover $C\to D$, we see that $11$ rational points of $D$ split and $2$ ramify, in order to produce the $24$ rational points on $C$. But the ramification of the cover $C\to D$ is necessarily wild, so the Riemann–Hurwitz formula shows that there can be at most one point of $C$ that ramifies in the cover. This contradiction shows that the isogeny class for table entry 6 does not contain a Jacobian.

Thus, the only possible real Weil polynomial for a genus-$8$ curve $C$ over ${\mathbb{F}}_{4}$ with $24$ points is the one given for the first entry in Table 1, namely

Note that we can write $h=h_{1}h_{2}$, where $h_{1}=x+3$ and $h_{2}=x(x+2)^{4}(x+4)^{2}$. Since $h_{1}$ is the real Weil polynomial of the unique elliptic curve $E$ over ${\mathbb{F}}_{4}$ with $8$ points, and since the resultant of $h_{1}$ and $h_{2}$ is $\pm 3$, Propositions 2.5 and 2.8 of [4] show that there exists a degree-$3$ map from $C$ to $E$. Let $\varphi\colon C\to E$ be such a triple cover. The remainder of the proof will focus on this map $\varphi$. First we show that $\varphi$ is not a Galois cover.

Suppose, to obtain a contradiction, that the triple cover $\varphi$ is Galois. Then all ramification in the cover is tame, and the Riemann–Hurwitz formula shows that seven geometric points of $E$ ramify.

The real Weil polynomial for $C$ shows that $C$ has $24$ places of degree $1$ and no places of degree $2$ or $3$. Since $E$ has $8$ places of degree $1$, all of them must split in the extension in order to account for the $24$ degree-$1$ places on $C$, so no degree-$1$ place of $E$ ramifies. No places of $E$ of degree $2$ or $3$ ramify either, because $C$ has no places of degree $2$ or $3$. Therefore, the ramification divisor on $E$ consists of a single place of degree $7$.

Every place of degree $7$ on $C$ lies over a place of degree $7$ on $E$, so the number of degree-$7$ places on $C$ is equal to three times the number of degree-$7$ places on $E$ that split, plus the number of degree-$7$ places on $E$ that ramify. Therefore, the number of degree-$7$ places of $C$ must be $1$ modulo $3$. However, the real Weil polynomial $h$ tells us that $C$ has $2496$ places of degree $7$, and this number is $0$ modulo $3$. This contradiction shows that $\varphi$ cannot be a Galois cover.

Let $k(E)$ and $k(C)$ be the function fields of $E$ and $C$, and view $k(E)$ as a degree-$3$ subfield of $k(C)$ via the embedding $\varphi^{*}$. Let $M$ be the Galois closure of $k(C)$ over $k(E)$, and let $L$ be the quadratic resolvent field. Then we have a field diagram

where $M/k(E)$ is a Galois extension with group $S_{3}$.

Consider how a place $P$ of $E$ decomposes in $M$. There are six possible combinations of decomposition group and inertia group; these six possibilities are listed in Table 2, together with the splitting behavior in $C$ determined by the combination, and the associated contribution to the degrees of the different $\operatorname{{\mathfrak{d}}}_{C}$ of $k(C)/k(E)$ and the different $\operatorname{{\mathfrak{d}}}_{L}$ of $L/k(E)$.

As we noted earlier, $C$ has $24$ places of degree $1$ and no places of degree $2$ or $3$, and every place of degree $1$ on $E$ must split completely in $C$. Each degree-$1$ place $P$ of $E$ must also split completely in $M$, and in particular the places of $M$ over $P$ have residue class field degree $1$. This shows that $L/k(E)$ is not a constant field extension, so $L$ and $M$ are function fields of geometrically irreducible curves over ${\mathbb{F}}_{4}$; let $F$ and $D$ be these curves, so that $L=k(F)$ and $M=k(D)$.

We also see that once again, no place of $E$ of degree $1$, $2$, or $3$ can ramify in $C$. The Riemann–Hurwitz formula says that the degree of $\operatorname{{\mathfrak{d}}}_{C}$ is $14$. Each place $P$ of $E$ that ramifies in $C$ contributes at least $2\deg P\geqslant 8$ to $\operatorname{deg}\operatorname{{\mathfrak{d}}}_{C}$, so we see that exactly one place must ramify. That place $P$ must have degree $7$, and if the ramification is wild then we must have $m_{P}=2$.

Suppose the ramifying place $P$ has inertia group $C_{3}$. Then $L/k(E)$ is unramified, so $M/k(C)$ is unramified, so the curve $D$ has genus $15$. Since every rational point of $E$ splits completely in $D$, we see that $D$ has $48$ rational points. But the Oesterlé bound for a curve of genus $15$ over ${\mathbb{F}}_{4}$ is $37$, a contradiction. Therefore the ramifying place $P$ must have inertia group $C_{2}$.

Since the only ramifying place has inertia group $C_{2}$, Table 2 shows that $\operatorname{deg}\operatorname{{\mathfrak{d}}}_{L}=\operatorname{deg}\operatorname{{\mathfrak{d}}}_{C}=14$, and it follows from the Riemann–Hurwitz formula that the genus of the curve $F$ is $8$.

Since $C$ has no places of degree $2$ or $3$, Table 2 shows that the places of degree $2$ and $3$ on $E$ must have decomposition group $C_{3}$ in $M/k(E)$, so they split in $L/k(E)$. Earlier we showed that the degree-$1$ places of $E$ split completely in $M$, and hence also in $L$. Since $E$ has $8$ degree-$1$ places, $4$ degree-$2$ places, and $16$ degree-$3$ places, it follows that $F$ has $16$ degree-$1$ places, $8$ degree-$2$ places, and $32$ degree-$3$ places.

The Magma code that accompanies [4] includes an implementation of Lauter’s algorithm [6] for enumerating all Weil polynomials $f$ of $g$-dimensional isogeny classes over ${\mathbb{F}}_{q}$ with $P_{n}(f)\geqslant 0$ for all $n$ and with $P_{1}(f)$ equal to a given integer $N$. This can be easily adapted to enumerate all Weil polynomials $f$ of $8$-dimensional isogeny classes over ${\mathbb{F}}_{4}$ with $P_{n}(f)\geqslant 0$ for all $n$ and with $P_{1}(f)=16$, $P_{2}(f)=8$, and $P_{3}(f)=32$. We find that there are $44$ such Weil polynomials; the associated real Weil polynomials are listed in Table 3.