A Fibration of Fermat Surface in Characteristic

1. Introduction

Let k be an algebraically closed field and let S be the hypersurface in $P_k^3$ defined by the equation

$$z_0^5-z_1^5=z_2^5-z_3^5,$$

(1)

where $z_0,z_1,z_2,z_3$ are the homogeneous coordinates.

The morphism $f:S\to P_k^1$ is defined as follows:

$$f:(z_0:z_1:z_2:z_3)\mapsto \left\{\begin{array}{c}\hfill z_2^4/z_0^4,\mathrm{if}{z}_0=z_1\mathrm{and}{z}_2=z_3,\\ \hfill (z_0-z_1)/(z_2-z_3),\mathrm{otherwise}.\end{array}\right.$$

(2)

When k is the complex numbers C, f is a fibration. It is well-known that the general fiber of f is a smooth curve. (see [1], Chapter III, Corollary 10.7). In [2,3,4], they investigate the fibration $f:S\to P_C^1$. For example, the general fiber of f is a smooth curve of genus 3. On the other hand, let $S_{\lambda }$ denote the fiber over a point $\lambda \in P_C^1$, they obtain that $S_{\lambda }$ is a singular fiber if and only if $\lambda$ belongs to the following set

$$E:=\left\{\lambda \right|{\lambda }^5=-1/4,1,or-4\}\cup \{0,\infty \}.$$

(3)

However, when k is an algebraically closed field of positive characteristic, the situation will be different. The general fiber of a fibration may be singular (see [5]).

In this paper, k will be an algebraically closed field of characteristic 2. Although f remains a fibration, the general fiber of f is no longer smooth. In Section 2, we recall some necessary preliminaries. In Section 3, we investigate the geometric properties of fiber of f. We obtain that the following theorem.

Theorem 1.

$(\mathrm{see}\mathrm{Proposition}1,\mathrm{Proposition}2)$. Let k be an algebraically closed field of characteristic 2 and let S be a Fermat surface of degree 5 defined by the equation $\left(1\right)$. Let $f:S\to P_k^1$ be a fibration defined by $\left(2\right)$. We obtain that

(i) for any $\lambda \in k$, the fiber $S_{\lambda }$ has arithmetic genus 3.

(ii) the fiber $S_{\lambda }$ is reducible if and only if

$$\lambda \in \left\{c\right|c^{15}=1\}\cup \{0,\infty \}.$$

(4)

(iii) if $\lambda \notin \left\{c\right|c^{15}=1\}\cup \{0,\infty \}$, then the fiber $S_{\lambda }$ is a rational curve with a unique singular point $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}}:{\lambda }^{{\displaystyle \frac{1}{4}}})$.

In Section 4, we investigate the fibration-preserving automorphisms group of f. More specifically, let

$$G:=\{\sigma \in {\mathrm{Aut}}_k\left(S\right)|\exists \phi \in {\mathrm{Aut}}_k\left(P_k^1\right)\mathrm{such}\mathrm{that}f\circ \sigma =\phi \circ f\}$$

(5)

and ${\mathrm{Aut}}_{P_k^1}\left(S\right):=\{\sigma \in G|f·\sigma =f\}$. We have the following theorem.

Theorem 2.

$(\mathrm{see}\mathrm{Theorem}5,\mathrm{Theorem}6)$. Let H be the Sylow p-group of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$. We have

(i) for any $\sigma \in H$, there are some elements $a,b,c$ and d in k such that

$$\sigma =\left(\begin{array}{cccc}a& 1+a& b& b\\ 1+a& a& b& b\\ c& c& d& 1+d\\ c& c& 1+d& d\end{array}\right),$$

(6)

$a^4=a$, $c=b^4=c^{16}$ and $d^4=d$. Conversely, if σ is of the form as in $\left(6\right)$, then $\sigma \in H$.

(ii) the order of H is 256.

(iii) ${\mathrm{Aut}}_{P_k^1}\left(S\right)=H$.

2. Preliminaries

In this section, let k be an algebraically closed field. Let S be a smooth projective surface, and C a smooth projective curve over k.

2.1. Fibration

We say $f:S\to C$ is a fibration if f is a surjective morphism and $f_{*}{\mathcal{O}}_S={\mathcal{O}}_C$.

Definition 1.

Let $f:S\to C$ be a fibration and let $c\in C$ be a point. Let $k\left(c\right)$ be the residue field of c. Then the fiber of f over the point c is defined to be the fiber product $S_c:=S{\times }_C\mathrm{Spec}k\left(c\right)$.

It is a fact that the topological space of $S_c$ is homeomorphic to $f^{-1}\left(c\right)$ with the induced topology. (see [1], Chapter II, exercise 3.10)

2.2. Automorphism Group of $P_k^1$

Let ${\mathrm{GL}}_2\left(k\right)$ be the group of invertible $2\times 2$ matrices with entries in k, and ${\mathrm{PGL}}_2\left(k\right):={\mathrm{GL}}_2\left(k\right)/k^{*}$. It is known that ${\mathrm{PGL}}_2\left(k\right)\cong {\mathrm{Aut}}_k\left(P_k^1\right)$.

Theorem 3.

([6], Lemma 3.2 ). Let $s\in {\mathrm{PGL}}_2\left(k\right)$ be nontrivial and of finite order. Then $s^p=\mathrm{id}$ if and only if s has a unique fixed point in $P_k^1$.

Corollary 1.

Let H be a finite subgroup of ${\mathrm{PGL}}_2\left(k\right)$. If H fixes at least two points of $P_k^1$, then H is a cyclic group with order prime to p.

Proof. 

For any $s\in H$, if H fixes at least two points of $P_k^1$, then s fixes at least two points of $P_k^1$. By Theorem 3, the order of s is prime to p. This implies that the order of H is prime to p. Furthermore, H is a cyclic group (refer to [6], Proposition 4.5). □

Theorem 4.

([6], Proposition 4.7 ). If D is a finite subgroup of ${\mathrm{PGL}}_2\left(k\right)$, then D fixes a unique point of $P_k^1$ if and only if $D\cong H⋊M$, where $H\subseteq D$ is a nontrivial Sylow p-group and $M\subseteq D$ is a cyclic group with order prime to p.

2.3. Automorphism Group of Fermat Surface of Degree 5

Let k be an algebraically closed field of characteristic $p=2$ and $q:=p^2$. Let S be the Fermat surface defined by the Equation (1). Then

$${\mathrm{Aut}}_k\left(S\right)=PU(4,q):=\{\left(a_{ij}\right)\in GL(4,F_{q^2})|{\left(a_{ij}\right)}^t\left(a_{ij}^4\right)=I\}/\left\{scalars\right\}$$

(7)

(see [7], page 100, Theorem). Furthermore, one obtains that the order of $PU(4,q)$ is $4^6·(4^2-1)(4^3+1)(4^4-1)$ (see [8], page 8, Table 1).

3. The Fiber of f

In this section, let k be an algebraically closed field of characteristic $p=2$ and let S be the Fermat surface defined by the Equation (1). Note that the characteristic of k is 2, so $a+b=a-b$ for any $a,b\in k$.

3.1. Reducible Fiber

Proposition 1.

For any $\lambda \in P_k^1$, the fiber $S_{\lambda }$ is reducible if and only if

$$\lambda \in \left\{c\right|c^{15}=1\}\cup \{0,\infty \}.$$

(8)

Proof. 

When $\lambda \in \{0,\infty \}$, we obtain that $S_{\lambda }$ is a union of four projective lines meeting in a point (see 3.1.1).

Now, let us suppose $\lambda \notin \{0,\infty \}$. Let

$$F(z_0,z_1,z_2):=z_2^4+{\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3·z_2+(\lambda +{\displaystyle \frac{1}{{\lambda }^4}}){(z_0+z_1)}^4+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3).$$

(9)

By definition, $(z_0:z_1:z_2:z_3)\in f^{-1}\left(\lambda \right)$ if and only if

$$\left\{\begin{array}{c}\hfill F(z_0,z_1,z_2)=0,\\ \hfill z_0+z_1=\lambda (z_2+z_3)\end{array}\right.$$

(10)

Therefore, if $(z_0:z_1:z_2:z_3)\in f^{-1}\left(\lambda \right)$, then $F(z_0,z_1,z_2)=0$.

If $F(z_0,z_1,z_2)$ is reducible, then either

$$F(z_0,z_1,z_2)=(a_0+a_1{z}_2+a_2{z}_2^2+z_2^3)(b_0+z_2)$$

(11)

for some $a_0,a_1,a_2,b_0\in k[z_0,z_1]$, or

$$F(z_0,z_1,z_2)=({\alpha }_0+{\alpha }_1{z}_2+z_2^2)({\beta }_0+{\beta }_1{z}_2+z_2^2)$$

(12)

for some ${\alpha }_0,{\alpha }_1,{\beta }_0,{\beta }_1\in k[z_0,z_1]$.

In 3.1.2 and 3.1.3, we obtain that the necessary condition for $S_{\lambda }$ to be reducible is that ${\lambda }^{15}=1$. In 3.1.4, we obtain that $S_{\lambda }$ is still a union of four projective lines meeting in a point if ${\lambda }^{15}=1$.

In summary, the proof is complete. □

3.1.1. Case $\lambda \in \{0,\infty \}$

Let $E:=\{(z_0:z_1:z_2:z_3)\in S|z_0=z_1,z_2\ne z_3\}$.

By definition, $f^{-1}\left(0\right)=E\cup \left\{(1:1:0:0)\right\}$. More specifically, if $(z_0:z_1:z_2:z_3)\in E$, then $z_2^5+z_3^5=z_0^5+z_1^5=0$. Hence $0=z_2^5+z_3^5=(z_2+z_3)(z_2^4+z_2^3{z}_3+z_2^2{z}_3^2+z_2{z}_3^3+z_3^4)$. Combining this with the inequation that $z_2\ne z_3$, we obtain $z_2^4+z_2^3{z}_3+z_2^2{z}_3^2+z_2{z}_3^3+z_3^4=0$. Hence

$$E=\{(z_0:z_0:z_2:z_3)\in P_k^3|z_2^4+z_2^3{z}_3+z_2^2{z}_3^2+z_2{z}_3^3+z_3^4=0\mathrm{and}{z}_2\ne z_3\}.$$

(13)

Let $E^{\prime }:=\{(z_0:z_0:z_2:z_3)\in P_k^3|z_2^4+z_2^3{z}_3+z_2^2{z}_3^2+z_2{z}_3^3+z_3^4=0\}$. Note that $E^{\prime }=E\cup \left\{(1:1:0:0)\right\}$, so $f^{-1}\left(0\right)=E^{\prime }$. Let $\{{\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4\}$ be the solution set of the equation $z^4+z^3+z^2+z=1$. We have $z_2^4+z_2^3{z}_3+z_2^2{z}_3^2+z_2{z}_3^3+z_3^4=(z_2+{\theta }_1{z}_3)(z_2+{\theta }_2{z}_3)(z_2+{\theta }_3{z}_3)(z_2+{\theta }_4{z}_3)$. Hence

$${\displaystyle f^{-1}\left(0\right)=\bigcup _{i=1}^4\{(z_0:z_0:z_2:z_3)\in P_k^3|z_2+{\theta }_i{z}_3=0\}}$$

(14)

Therefore, $f^{-1}\left(0\right)$ is reducible and it has a unique singular point $(1:1:0:0)$. See the figure below.

Similarly,

$${\displaystyle f^{-1}(\infty )=\bigcup _{i=1}^4\{(z_0:z_1:z_2:z_2)\in P_k^3|z_0+{\theta }_i{z}_1=0\}.}$$

(15)

It is reducible and it has a unique singular point $(0:0:1:1)$.

3.1.2. Case $F(z_0,z_1,z_2)=(a_0+a_1{z}_2+a_2{z}_2^2+z_2^3)(b_0+z_2)$

Suppose $\lambda \notin \{0,\infty \}$. Note that

$$\begin{array}{cc}\hfill F(z_0,z_1,z_2)& =(a_0+a_1{z}_2+a_2{z}_2^2+z_2^3)(b_0+z_2)\hfill \\ \hfill & =z_2^4+(b_0+a_2)z_2^3+(b_0{a}_2+a_1)z_2^2+(b_0{a}_1+a_0)z_2+b_0{a}_0.\hfill \end{array}$$

(16)

Combining this with (9), we obtain

$$\left\{\begin{array}{c}\hfill a_2=b_0,\\ \hfill a_1=b_0{a}_2=b_0^2,\\ \hfill a_0=b_0{a}_1+{\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3=b_0^3+{\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3,\\ \hfill a_0{b}_0=(\lambda +{\displaystyle \frac{1}{{\lambda }^4}})(z_0^4+z_1^4)+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3).\end{array}\right.$$

(17)

Hence $a_0{b}_0=b_0·[b_0^3+{\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3]=b_0^4+{\displaystyle \frac{1}{{\lambda }^3}}{b}_0{(z_0+z_1)}^3$. We may assume $b_0=c_0{z}_0+c_1{z}_1$, where $c_0,c_1\in k$. Thus

$$\begin{array}{cc}\hfill a_0{b}_0& =b_0^4+{\displaystyle \frac{b_0}{{\lambda }^3}}{(z_0+z_1)}^3\hfill \\ & =(c_0^4+{\displaystyle \frac{c_0}{{\lambda }^3}})z_0^4+(c_1^4+{\displaystyle \frac{c_1}{{\lambda }^3}})z_1^4+{\displaystyle \frac{c_0+c_1}{{\lambda }^3}}(z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3).\hfill \end{array}$$

(18)

Combining this with (17), we obtain

$$\begin{array}{c}\hfill (\lambda +{\displaystyle \frac{1}{{\lambda }^4}})(z_0^4+z_1^4)+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3)\\ \hfill =(c_0^4+{\displaystyle \frac{c_0}{{\lambda }^3}})z_0^4+(c_1^4+{\displaystyle \frac{c_1}{{\lambda }^3}})z_1^4+{\displaystyle \frac{c_0+c_1}{{\lambda }^3}}(z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3).\end{array}$$

(19)

Therefore,

$$\left\{\begin{array}{c}\hfill c_0^4+{\displaystyle \frac{c_0}{{\lambda }^3}}=c_1^4+{\displaystyle \frac{c_1}{{\lambda }^3}}=\lambda +{\displaystyle \frac{1}{{\lambda }^4}},\\ \hfill {\displaystyle \frac{1}{{\lambda }^3}}(c_0+c_1)=\lambda .\end{array}\right.$$

(20)

This implies that $c_0+c_1={\lambda }^4$, hence $0=c_0^4+{\displaystyle \frac{c_0}{{\lambda }^3}}+c_1^4+{\displaystyle \frac{c_1}{{\lambda }^3}}={\lambda }^{16}+\lambda$. Therefore, ${\lambda }^{15}=1$.

3.1.3. Case $F(z_0,z_1,z_2)=({\alpha }_0+{\alpha }_1{z}_2+z_2^2)({\beta }_0+{\beta }_1{z}_2+z_2^2)$

Suppose $\lambda \notin \{0,\infty \}$. Note that

$$\begin{array}{cc}\hfill F(z_0,z_1,z_2)& =({\alpha }_0+{\alpha }_1{z}_2+z_2^2)({\beta }_0+{\beta }_1{z}_2+z_2^2)\hfill \\ & =z_2^4+({\alpha }_1+{\beta }_1)z_2^3+({\alpha }_0+{\beta }_0+{\alpha }_1{\beta }_1)z_2^2+({\alpha }_0{\beta }_1+{\alpha }_1{\beta }_0)z_2+{\alpha }_0{\beta }_0.\hfill \end{array}$$

(21)

Combining this with (9), we obtain

$$\left\{\begin{array}{c}\hfill {\alpha }_1={\beta }_1,\\ {\alpha }_0={\alpha }_1{\beta }_1+{\beta }_0={\beta }_1^2+{\beta }_0,\hfill \\ {\alpha }_0{\beta }_1+{\alpha }_1{\beta }_0=({\beta }_1^2+{\beta }_0){\beta }_1+{\beta }_0{\beta }_1={\beta }_1^3={\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3,\hfill \\ {\alpha }_0{\beta }_0={\beta }_0{\beta }_1^2+{\beta }_0^2=(\lambda +{\displaystyle \frac{1}{{\lambda }^4}})(z_0^4+z_1^4)+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3).\hfill \end{array}\right.$$

(22)

By (22), ${\beta }_1^3={\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3$. Therefore, we may assume ${\beta }_1={\displaystyle \frac{\theta }{\lambda }}(z_0+z_1)$, where $\theta \in \{z\in k|z^3=1\}$. We may assume ${\beta }_0=c_0{z}_0^2+c_1{z}_0{z}_1+c_2{z}_1^2$, where $c_0,c_1,c_2\in k$. Thus

$$\begin{array}{cc}\hfill {\alpha }_0{\beta }_0& ={\beta }_0{\beta }_1^2+{\beta }_0^2\hfill \\ \hfill & =(c_0{z}_0^2+c_1{z}_0{z}_1+c_2{z}_1^2)·{\displaystyle \frac{{\theta }^2}{{\lambda }^2}}{(z_0+z_1)}^2+{(c_0{z}_0^2+c_1{z}_0{z}_1+c_2{z}_1^2)}^2\hfill \\ \hfill & =(c_0^2+{\displaystyle \frac{c_0{\theta }^2}{{\lambda }^2}})z_0^4+(c_2^2+{\displaystyle \frac{c_2{\theta }^2}{{\lambda }^2}})z_1^4+{\displaystyle \frac{c_1{\theta }^2}{{\lambda }^2}}(z_0^3{z}_1+z_0{z}_1^3)+[{\displaystyle \frac{(c_0+c_2){\theta }^2}{{\lambda }^2}}+c_1^2]z_0^2{z}_1^2.\hfill \end{array}$$

(23)

Combining this with (22), we obtain

$$\left\{\begin{array}{c}\hfill c_0^2+{\displaystyle \frac{c_0{\theta }^2}{{\lambda }^2}}=c_2^2+{\displaystyle \frac{c_2{\theta }^2}{{\lambda }^2}}=\lambda +{\displaystyle \frac{1}{{\lambda }^4}},\\ \hfill {\displaystyle \frac{c_1{\theta }^2}{{\lambda }^2}}={\displaystyle \frac{(c_0+c_2){\theta }^2}{{\lambda }^2}}+c_1^2=\lambda .\end{array}\right.$$

(24)

Hence $0=c_0^2+{\displaystyle \frac{c_0{\theta }^2}{{\lambda }^2}}+c_2^2+{\displaystyle \frac{c_2{\theta }^2}{{\lambda }^2}}=(c_0+c_2)(c_0+c_2+{\displaystyle \frac{{\theta }^2}{{\lambda }^2}})$. This implies that either $c_0+c_2=0$ or $c_0+c_2+{\displaystyle \frac{{\theta }^2}{{\lambda }^2}}=0$.

If $c_0+c_2=0$, then $\lambda ={\displaystyle \frac{c_1{\theta }^2}{{\lambda }^2}}={\displaystyle \frac{(c_0+c_2){\theta }^2}{{\lambda }^2}}+c_1^2=c_1^2$ by (24). It follows that ${\lambda }^3=c_1{\theta }^2$ and $\lambda =c_1^2$. Note that ${\theta }^3=1$, so ${\lambda }^6=c_1^2{\theta }^4=\lambda \theta$. Hence ${\lambda }^{15}={\theta }^3=1$.

If $c_0+c_2+{\displaystyle \frac{{\theta }^2}{{\lambda }^2}}=0$, then $\lambda ={\displaystyle \frac{c_1{\theta }^2}{{\lambda }^2}}={\displaystyle \frac{(c_0+c_2){\theta }^2}{{\lambda }^2}}+c_1^2={\displaystyle \frac{{\theta }^4}{{\lambda }^2}}+c_1^2={\displaystyle \frac{\theta }{{\lambda }^2}}+c_1^2$ by (24). In other words, we have $\lambda ={\displaystyle \frac{c_1{\theta }^2}{{\lambda }^2}}={\displaystyle \frac{\theta }{{\lambda }^2}}+c_1^2$. This implies that $c_1={\displaystyle \frac{{\lambda }^3}{{\theta }^2}}$, hence $c_1^2={\displaystyle \frac{{\lambda }^6}{{\theta }^4}}={\displaystyle \frac{{\lambda }^6}{\theta }}$. Substituting this into $\lambda ={\displaystyle \frac{\theta }{{\lambda }^2}}+c_1^2$, we obtain ${\lambda }^{10}+{\lambda }^5\theta +{\theta }^2=0$. Note that $0={\theta }^3+1=(\theta +1)({\theta }^2+\theta +1)$. If $\theta =1$, then ${\lambda }^{10}+{\lambda }^5+1=0$, and hence ${\lambda }^{15}+1=({\lambda }^5+1)({\lambda }^{10}+{\lambda }^5+1)=0$. Otherwise ${\theta }^2+\theta +1=0$, as ${\lambda }^{10}+{\lambda }^5\theta +{\theta }^2=0$, we obtain that either ${\lambda }^5=1$ or ${\lambda }^5=1+\theta$. Therefore, ${\lambda }^{15}=1$.

3.1.4. Case ${\lambda }^{15}=1$

Note that ${\lambda }^{15}+1=({\lambda }^5+1)({\lambda }^{10}+{\lambda }^5+1)$. If ${\lambda }^{15}+1=0$, then we obtain that either ${\lambda }^5+1=0$ or ${\lambda }^{10}+{\lambda }^5+1=0$.

(i) Suppose ${\lambda }^5+1=0$. Let $\{{\theta }_0,{\theta }_1\}$ be the solution set of the equation $z^2+z+1=0$. Then we have

$$F(z_0,z_1,z_2)=(z_2+{\lambda }^4{z}_0)(z_2+{\lambda }^4{z}_1)(z_2+{\lambda }^4{\theta }_0{z}_0+{\lambda }^4{\theta }_1{z}_1)(z_2+{\lambda }^4{\theta }_1{z}_0+{\lambda }^4{\theta }_0{z}_1).$$

(25)

Hence

$${\displaystyle f^{-1}\left(\lambda \right)=\bigcup _{i=1}^4\{(z_0:z_1:z_2:z_3)\in P_k^3|F_i=0\mathrm{and}{z}_0+z_1=\lambda (z_2+z_3)\},}$$

(26)

where $F_1:=z_2+{\lambda }^4{z}_0,F_2:=z_2+{\lambda }^4{z}_1,F_3:=z_2+{\lambda }^4{\theta }_0{z}_0+{\lambda }^4{\theta }_1{z}_1$, and $F_4:=z_2+{\lambda }^4{\theta }_1{z}_0+{\lambda }^4{\theta }_0{z}_1$. Therefore, $f^{-1}\left(\lambda \right)$ is reducible and it has a unique singular point $(1:1:{\lambda }^4:{\lambda }^4)$.

(ii) Suppose ${\lambda }^{10}+{\lambda }^5+1=0$. Let $\{{\delta }_0,{\delta }_1\}$ be the solution set of the equation $z^2+{\lambda }^{9}z+{\lambda }^{13}=0$. Let $F_1:=z_2+{\delta }_0(z_0+z_1)+{\lambda }^4{z}_1,F_2:=z_2+{\delta }_1(z_0+z_1)+{\lambda }^4{z}_1,F_3:=z_2+{\delta }_0(z_0+z_1)+{\lambda }^4{z}_0$, and $F_4:=z_2+{\delta }_1(z_0+z_1)+{\lambda }^4{z}_0$. Then we have $F(z_0,z_1,z_2)={\prod }_{i=1}^{i=4}{F}_i$. Hence

$${\displaystyle f^{-1}\left(\lambda \right)=\bigcup _{i=1}^4\{(z_0:z_1:z_2:z_3)\in P_k^3|F_i=0\mathrm{and}{z}_0+z_1=\lambda (z_2+z_3)\}.}$$

(27)

Therefore, $f^{-1}\left(\lambda \right)$ is reducible and it has a unique singular point $(1:1:{\lambda }^4:{\lambda }^4)$.

3.2. Irreducible Fiber

By Proposition 1, if $\lambda \notin \left\{c\right|c^{15}=1\}\cup \{0,\infty \}$, then the fiber $S_{\lambda }$ is irreducible. $S_{\lambda }$ is defined by

$$\left\{\begin{array}{c}\hfill z_0+z_1=\lambda (z_2+z_3)\\ \hfill z_2^4+{\displaystyle \frac{1}{{\lambda }^3}}{z}_2{(z_0+z_1)}^3+(\lambda +{\displaystyle \frac{1}{{\lambda }^4}})(z_0^4+z_1^4)+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3)=0\end{array}\right.$$

(28)

Hence $S_{\lambda }$ is isomorphic to the curve $C_{\lambda }$ in $P_k^2$, where $C_{\lambda }$ is defined by the equation

$$h_{\lambda }:=z_2^4+{\displaystyle \frac{1}{{\lambda }^3}}{z}_2{(z_0+z_1)}^3+(\lambda +{\displaystyle \frac{1}{{\lambda }^4}})(z_0^4+z_1^4)+\lambda (z_0^3{z}_1+z_0^2{z}_1^2+z_0{z}_1^3)=0.$$

(29)

Proposition 2.

(i) If $\lambda \notin \left\{c\right|c^{15}=1\}\cup \{0,\infty \}$, then the fiber $S_{\lambda }$ is a rational curve with a unique singular point $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}}:{\lambda }^{{\displaystyle \frac{1}{4}}})$.

(ii) For any λ, the arithmetic genus $p_a\left(S_{\lambda }\right)$ is equal to 3.

Proof. (i) Let us consider the curve $C_{\lambda }$ in $P_k^2$. We have

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial h_{\lambda }}{\partial z_2}}={\displaystyle \frac{1}{{\lambda }^3}}{(z_0+z_1)}^3\\ \hfill {\displaystyle \frac{\partial h_{\lambda }}{\partial z_1}}=({\displaystyle \frac{1}{{\lambda }^3}}{z}_2+\lambda z_0){(z_0+z_1)}^2\\ \hfill {\displaystyle \frac{\partial h_{\lambda }}{\partial z_0}}=({\displaystyle \frac{1}{{\lambda }^3}}{z}_2+\lambda z_1){(z_0+z_1)}^2\end{array}\right.$$

(30)

By Jacobian criterion, $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}})$ is the unique singular point of $C_{\lambda }$. In other words, $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}}:{\lambda }^{{\displaystyle \frac{1}{4}}})$ is the unique singular point of $S_{\lambda }$.

Note that the arithmetic genus $p_a\left(C_{\lambda }\right)={\displaystyle \frac{(4-1)(4-2)}{2}}=3$, and $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}})$ is a triple singular point of $C_{\lambda }$. Hence the normalization of $C_{\lambda }$ is a rational curve. Since $S_{\lambda }$ is isomorphic to $C_{\lambda }$, we obtain that $S_{\lambda }$ is a rational curve with a unique singular point $(1:1:{\lambda }^{{\displaystyle \frac{1}{4}}}:{\lambda }^{{\displaystyle \frac{1}{4}}})$.

(ii) According to chapter III, corollary 9.10 in [1], the arithmetic genus of $S_{\lambda }$ are all independent of $\lambda$. □

4. Subgroups of ${\mathrm{Aut}}_k\left(S\right)$

In this section, let k be an algebraically closed field of characteristic $p=2$ and $q:=p^2$. Let S be the Fermat surface defined by the Equation (1). Note that ${\mathrm{Aut}}_k\left(S\right)=PU(4,q^2)$. Denoting $G:=\{\sigma \in {\mathrm{Aut}}_k\left(S\right)|\exists \phi \in {\mathrm{Aut}}_k\left(P_k^1\right)\mathrm{such}\mathrm{that}f\circ \sigma =\phi \circ f\}$ and ${\mathrm{Aut}}_{P_k^1}\left(S\right):=\{\sigma \in G|f\circ \sigma =f\}$.

For any $\sigma =\left(a_{ij}\right)\in {\mathrm{Aut}}_{P_k^1}\left(S\right)$, as $\sigma$ preserves the fiber of f, we have $\sigma \left(f^{-1}\left(0\right)\right)=f^{-1}\left(0\right)$. In particular, $\sigma$ fixes the unique singular point $(1:1:0:0)$ of $f^{-1}\left(0\right)$. Hence

$$\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)\left(\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right)$$

(31)

We obtain that

$$\left\{\begin{array}{c}\hfill a_{11}+a_{12}=1\\ \hfill a_{21}+a_{22}=1\\ \hfill a_{31}+a_{32}=0\\ \hfill a_{41}+a_{42}=0\end{array}\right.$$

(32)

Similarly, $\sigma$ fixes the unique singular point $(0:0:1:1)$ of $f^{-1}(\infty )$. Hence

$$\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right)$$

(33)

We obtain that

$$\left\{\begin{array}{c}\hfill a_{13}+a_{14}=0\\ \hfill a_{23}+a_{24}=0\\ \hfill a_{33}+a_{34}=1\\ \hfill a_{43}+a_{44}=1\end{array}\right.$$

(34)

Combining (32) and (34), we have

$$\sigma =\left(\begin{array}{cccc}{a}_{11}& 1+a_{11}& a_{13}& a_{13}\\ 1+a_{22}& a_{22}& a_{23}& a_{23}\\ a_{31}& a_{31}& a_{33}& 1+a_{33}\\ a_{41}& a_{41}& 1+a_{44}& a_{44}\end{array}\right)$$

(35)

4.1. Sylow p-Group of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$

Let H be the Sylow p-group of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$. Let F be a general fiber of f. We have the following exact sequences:

$$1\to {\mathrm{Aut}}_{P_k^1}\left(S\right)\to {\mathrm{Aut}}_k\left(F\right).$$

(36)

By Proposition 2, F is a rational curve with a unique singular point. Hence ${\mathrm{Aut}}_k\left(F\right)\subset {\mathrm{Aut}}_k\left(P_k^1\right)\cong \mathrm{PGL}(2,k)$. By (36), we may regard ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ as a subgroup of ${\mathrm{Aut}}_k\left(F\right)$ (hence of ${\mathrm{Aut}}_k\left(P_k^1\right)$). By Theorem 3 and Corollary 1, we obtain that ${\sigma }^2=\mathrm{id}$ for any $\sigma \in H$.

Let $\sigma =\left(a_{ij}\right)\in H$. As ${\sigma }^2=\mathrm{id}$, we obtain that $\left(a_{ij}\right)\left(a_{ij}\right)=I$, and hence ${\left(a_{ij}\right)}^t{\left(a_{ij}\right)}^t=I$. On the other hand, $\sigma \in H$, hence $\sigma \in {\mathrm{Aut}}_k\left(S\right)=PU(4,q^2)$. This implies that ${\left(a_{ij}\right)}^t\left(a_{ij}^4\right)=I$. Therefore, we have $\left(a_{ij}^4\right)={\left(a_{ij}\right)}^t$. In other words, we obtain that

$$\left\{\begin{array}{c}\hfill a_{ij}=a_{ji}^4=a_{ij}^{16},\mathrm{if}i\ne j,\\ \hfill a_{ii}=a_{ii}^4,\mathrm{otherwise}.\end{array}\right.$$

(37)

Note that $\sigma$ is also an element of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$, so $\sigma$ has the form of (35). We obtain that ${(1+a_{11})}^4=1+a_{22}$ and $a_{11}^4=a_{11}$. This implies that $a_{11}=a_{11}^4=a_{22}$. Similarly, we have $a_{33}=a_{44}$, $a_{13}=a_{23}$ and $a_{31}=a_{41}$. Therefore, we may assume

$$\sigma =\left(\begin{array}{cccc}a& 1+a& b& b\\ 1+a& a& b& b\\ c& c& d& 1+d\\ c& c& 1+d& d\end{array}\right)$$

(38)

where $a^4=a$, $c=b^4=c^{16}$ and $d^4=d$.

Conversely, as

$$\sigma ·\left(\begin{array}{c}{z}_0\\ z_1\\ z_2\\ z_3\end{array}\right)=\left(\begin{array}{c}a(z_0+z_1)+b(z_2+z_3)+z_1\\ a(z_0+z_1)+b(z_2+z_3)+z_0\\ c(z_0+z_1)+d(z_2+z_3)+z_3\\ c(z_0+z_1)+d(z_2+z_3)+z_2\end{array}\right),$$

(39)

we obtain that $\sigma$ preserves the fiber of f. Hence any element of the p-group of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ is of the form as in (38). We have proved the following theorem.

Theorem 5.

Let H be the Sylow p-group of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$. We have

(i) for any $\sigma \in H$, there are some elements $a,b,c$ and d in k such that

$$\sigma =\left(\begin{array}{cccc}a& 1+a& b& b\\ 1+a& a& b& b\\ c& c& d& 1+d\\ c& c& 1+d& d\end{array}\right),$$

(40)

$a^4=a$, $c=b^4=c^{16}$ and $d^4=d$. Conversely, if σ be of the form as in $\left(40\right)$, then $\sigma \in H$.

(ii) the order of H is 256.

Remark 1.

In particular,

$$\left(\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right)$$

(41)

is a nontrivial element of H.

Remark 2.

Let F be a general fiber of f and let P be the unique singular point of F. By $\left(36\right)$, we may regard ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ as a subgroup of ${\mathrm{Aut}}_k\left(F\right)$. By Corollary 1 and the fact that H is nontrivial, we obtain that ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ fixes a unique point P of F. By Theorem 4, we have ${\mathrm{Aut}}_{P_k^1}\left(S\right)\cong H⋊M$, where M is a cyclic group with order prime to p.

4.2. Subgroups of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ with order prime to p

By Remark 4, we have ${\mathrm{Aut}}_{P_k^1}\left(S\right)\cong H⋊M$, where M is a cyclic group with order prime to p.

Theorem 6.

Let M as above. We obtain that

(i) M is trivial.

(ii) ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ is a p-group of order 256.

Proof. (i) We may assume the order of M is r. Suppose $r\ne 1$. Let $\psi :=\left(a_{ij}\right)$ be a generator of M. Note that ${\psi }^r={\left(a_{ij}\right)}^r=I$ and ${\left(a_{ij}\right)}^t\left(a_{ij}^4\right)=I$, so ${\left(a_{ij}\right)}^{r-1}={\left(a_{ij}^4\right)}^t$. As $\psi$ is an element of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$, we may assume $\psi$ has the form of (35). Therefore,

$${\psi }^{r-1}={\left(\begin{array}{cccc}{a}_{11}^4& 1+a_{11}^4& a_{13}^4& a_{13}^4\\ 1+a_{22}^4& a_{22}^4& a_{23}^4& a_{23}^4\\ a_{31}^4& a_{31}^4& a_{33}^4& 1+a_{33}^4\\ a_{41}^4& a_{41}^4& 1+a_{44}^4& a_{44}^4\end{array}\right)}^t$$

(42)

On the other hand, by Lemma 5, we may assume

$${\psi }^{r-1}=\left(\begin{array}{cccc}{c}_{11}& 1+c_{11}& c_{13}& c_{13}\\ 1+c_{22}& c_{22}& c_{23}& c_{23}\\ c_{31}& c_{31}& c_{33}& 1+c_{33}\\ c_{41}& c_{41}& 1+c_{44}& c_{44}\end{array}\right)$$

(43)

Combining (42) and (43), we obtain that $a_{11}^4=c_{11}$ and $1+a_{22}^4=1+c_{11}$. This implies that $a_{11}^4=c_{11}=a_{22}^4$, hence $a_{11}=a_{22}$. Similarly, we have $a_{33}=a_{44}$, $a_{13}=a_{23}$, and $a_{31}=a_{41}$. Therefore, we may assume

$$\psi =\left(\begin{array}{cccc}a& 1+a& b& b\\ 1+a& a& b& b\\ c& c& d& 1+d\\ c& c& 1+d& d\end{array}\right)$$

(44)

where $a,b,c,d\in k$, so ${\psi }^2=I$. Hence the order of M is 2, which is a contradiction.

Therefore, $r=1$ and M is trivial.

(ii) As M is trivial, we obtain that ${\mathrm{Aut}}_{P_k^1}\left(S\right)=H$ is a p-group. By Theorem 5, the order of ${\mathrm{Aut}}_{P_k^1}\left(S\right)$ is 256. □

Lemma 1.

Let

$$A=\left(\begin{array}{cccc}{a}_{11}& 1+a_{11}& a_{13}& a_{13}\\ 1+a_{22}& a_{22}& a_{23}& a_{23}\\ a_{31}& a_{31}& a_{33}& 1+a_{33}\\ a_{41}& a_{41}& 1+a_{44}& a_{44}\end{array}\right)$$

(45)

and

$$B=\left(\begin{array}{cccc}{b}_{11}& 1+b_{11}& b_{13}& b_{13}\\ 1+b_{22}& b_{22}& b_{23}& b_{23}\\ b_{31}& b_{31}& b_{33}& 1+b_{33}\\ b_{41}& b_{41}& 1+b_{44}& b_{44}\end{array}\right).$$

(46)

Then

$$AB=\left(\begin{array}{cccc}{c}_{11}& 1+c_{11}& c_{13}& c_{13}\\ 1+c_{22}& c_{22}& c_{23}& c_{23}\\ c_{31}& c_{31}& c_{33}& 1+c_{33}\\ c_{41}& c_{41}& 1+c_{44}& c_{44}\end{array}\right)$$

(47)

where

$$\left\{\begin{array}{c}\hfill c_{11}=a_{11}{b}_{11}+(1+a_{11})(1+b_{22})+a_{13}(b_{31}+b_{41}),\\ \hfill c_{13}=a_{11}{b}_{13}+b_{23}(1+a_{11})+a_{13}(1+b_{33}+b_{44}),\\ \hfill c_{22}=a_{22}{b}_{22}+(1+a_{22})(1+b_{11})+a_{23}(b_{31}+b_{41}),\\ \hfill c_{23}=a_{22}{b}_{23}+b_{13}(1+a_{22})+a_{23}(1+b_{33}+b_{44}),\\ \hfill c_{33}=a_{33}{b}_{33}+(1+a_{33})(1+b_{44})+a_{31}(b_{13}+b_{23}),\\ \hfill c_{31}=a_{33}{b}_{31}+b_{41}(1+a_{33})+a_{31}(1+b_{11}+b_{22}),\\ \hfill c_{44}=a_{44}{b}_{44}+(1+a_{44})(1+b_{33})+a_{41}(b_{13}+b_{23}),\\ \hfill c_{41}=a_{44}{b}_{41}+b_{31}(1+a_{44})+a_{41}(1+b_{11}+b_{22}).\end{array}\right.$$

(48)

Remark 3.

By Theorem 6, for any nontrivial element $\sigma \in {\mathrm{Aut}}_k\left(S\right)$ with order prime to p, we obtain that σ does preserve the fiber of f.