Proof. We work with intersection products on ${\mathcal S}$ . The surfaces ${\mathcal S}_i$ cut ${\mathcal S}$ in divisors $C_i +L_1+\cdots +L_m$ and we are interested in $\Gamma = C_1\cdot C_2$ . Letting H denote a hyperplane section of ${\mathcal S}$ ,

(2.1) $$ \begin{align} \deg \Gamma = (d_1H - (L_1+\cdots+L_m))\cdot (d_2H - (L_1+\cdots+L_m)). \end{align} $$

By classical intersection theory on surfaces,

$$ \begin{align*}H^2 = d,\quad L_i.H = 1 \quad \mathrm{and} \quad L_i^2 = 2-d.\end{align*} $$

The first follows since ${\mathcal S}$ has degree d. For the second, a line and a plane meet at one point, but if the line is contained in ${\mathcal S}$ , the intersection point is also in ${\mathcal S}$ . The third follows from the previous one and the adjunction formula. Indeed, if L is a nonsingular curve of genus g on a surface S and K is the canonical divisor on S, then the adjunction formula gives

$$ \begin{align*}2g-2=L.(L+K)=L^2+L.K.\end{align*} $$

If $L=L_i$ is a line, its genus is zero and we get

$$ \begin{align*}-2=L_i^2+L_i.K.\end{align*} $$

Moreover, the canonical divisor K on a smooth surface in $\mathbb {P}^3$ of degree d satisfies

$$ \begin{align*}K\sim (d-4)H,\end{align*} $$

where H is a hyperplane section. So, we obtain $-2=L_i^2+(d-4)L_i.H$ and, thus, $L_i^2=2-d$ since $L_i.H=1$ . Hence, the right-hand side of (2.1) expands to

$$ \begin{align*} d_1d_2H^2 &- (d_1+d_2)(L_1+\cdots+L_m)\cdot H + (L_1+\cdots+L_m)^2\\ & = dd_1d_2 -m(d_1+d_2) + (L_1+\cdots+L_m)^2. \end{align*} $$

Finally, $L_i\cdot L_j = 0$ or $1$ if $i\ne j$ , depending on whether the lines intersect or not, so

$$ \begin{align*}(L_1+\cdots+L_m)^2 = \sum_{i=1}^m L_i^2 + \sum_{i \ne j} L_i\cdot L_j \le m(2-d)+m(m-1)=m(m+1-d)\end{align*} $$

and

$$ \begin{align*}(L_1+\cdots+L_m)^2 = \sum_{i=1}^m L_i^2 + \sum_{i \ne j} L_i\cdot L_j \ge m(2-d),\end{align*} $$

which combines with (2.1) to prove the lemma.

Proof. In [Reference Voloch, Melles, Brasselet, Kennedy, Lauter and McEwan10], it is proved that the one-dimensional components of X are lines. Hence, the situation here is as in Lemma 2.1, with the surface $S_1$ of degree $d_1 = d + p -1$ and the surface $S_2$ of degree $d_2 =d+2p -2$ . Since we have supposed that the ${\mathbb F}_p$ -lines $L_1,\ldots , L_m$ on S have multiplicity one in X, it follows from Lemma 2.1 that the zero-dimensional subscheme $\Gamma $ of X given by its isolated rational points satisfies

$$ \begin{align*}\deg \Gamma \le d(d + p -1)(d+2p -2) -m(3d +3p-3) + m(m+1).\end{align*} $$

Moreover, as shown in [Reference Voloch, Melles, Brasselet, Kennedy, Lauter and McEwan10, proof of Theorem 2], outside of the lines contained in S, the rational points of S are isolated points of $S\cap S_1\cap S_2$ with multiplicity at least $6$ . So, the rational points of S not on lines have multiplicity at least $6$ in $\Gamma $ and lead to the bound

$$ \begin{align*}\# S({\mathbb F}_p) \le \frac{d(d + p -1)(d+2p -2) -m(3d +3p-3) + m(m+1)}{6} + m(p+1),\end{align*} $$

which can be rewritten as

$$ \begin{align*}\# S({\mathbb F}_p) \le \frac{d(d + p -1)(d+2p -2) -m(3d-7-m)}{6} + \frac{m(p+1)}{2}.\end{align*} $$

Furthermore, Bauer and Rams in [Reference Bauer and Rams2, Theorem 1] show that a smooth surface in ${\mathbb P}^3$ of degree $d>2$ over a field of characteristic 0 or of characteristic $p>d$ contains at most $11d^2-30d+18$ lines. So, the last bound of the theorem follows.

Proof. Following [Reference Andria, Rojas and Mangueira1] (which is ostensibly over the field $\mathbb C$ of complex numbers, but one can see that the proofs work over finite fields), one can prove that S has $3d^2$ lines defined over ${\mathbb F}_p$ , namely the lines with equations

$$ \begin{align*} \begin{cases} \begin{matrix} w = \eta^i x, \\ y = \eta^kz, \cr \end{matrix} \end{cases} \!\!\!\!\!\begin{cases} \begin{matrix} x = \eta^{k+i} z , \\ w = \eta^iy,\cr \end{matrix} \end{cases}\!\!\!\!\! \begin{cases} \begin{matrix} x = v\eta^i y, \\ w = \eta^{k+i}z, \cr \end{matrix} \end{cases} \end{align*} $$

where $\eta $ is a primitive dth root of unity in ${\mathbb F}_p$ and $v\in {\mathbb F}_p$ is such that $v^d=-1$ .

One can prove that already for $S\cap S_1$ (and a posteriori for X), the multiplicity of the lines is one. We show this by verifying that S and $S_1$ intersect transversely outside of a finite set. We work in the affine patch $w=1$ (and the other patches are similar). Then, S has equation $x^d +y^d - z^d =1$ with gradient $(dx^{d-1}, dy^{d-1},-dz^{d-1})$ and $S_1$ has equation $x^{d+q-1} +y^{d+q-1} - z^{d+q-1} =1$ with gradient $((d-1)x^{d+q-2}, (d-1)y^{d+q-2},-(d-1)z^{d+q-2})$ .

The intersection $S\cap S_1$ is not transversal at points where the gradients are proportional. For this to happen, there is a constant $\lambda $ for which $dx^{d-1} = \lambda (d-1)x^{d+q-2}$ and similarly for $y,z$ . So, $x^{q-1} = d/(\lambda (d-1))$ and similarly for $y,z$ . So, $x= \lambda ^{-1/(q-1)}c_x$ , $y= \lambda ^{-1/(q-1)}c_y$ , $z= \lambda ^{-1/(q-1)}c_z$ , where $c_x,c_y,c_z$ come from the same finite set with $c^{q-1} = d/(d-1)$ . However, $(x,y,z) \in S$ , so $\lambda ^{d/(q-1)} = c_x^d+c_y^d+c_z^d$ , so $\lambda $ is also on a finite set, proving that S and $S_1$ intersect transversely outside a finite set.

Proof. Let us remark that if $a\in {\mathbb F}_p$ , then $a^{(p-1)/2}=0$ if and only if $a=0$ , and it is equal to 1 if a is a quadratic residue modulo p and equal to $-1$ otherwise. It is well known that the number of nonzero quadratic residues, and also the number of quadratic nonresidues modulo p, is equal to $(p-1)/2$ . We begin by counting the number of affine rational points on S. When none of the coordinates are zero, we have 6 choices for the positions of the $+1$ , so $6({(p-1)}/{2})^4$ solutions. If two coordinates are zero, then we have also 6 choices for the zero coordinates and then two possibilities for the choices of $+1$ or $-1$ ; thus, $6\times 2 \times ({(p-1)}/{2})^2$ solutions. Adding the point with all coordinates zero, we obtain the number $N_{\mathrm {aff}}$ of affine rational points on S, which is equal to $6({(p-1)}/{2})^4+12 ({(p-1)}/{2})^2+1$ , and then $\sharp S({\mathbb F}_p) = ({N_{\mathrm {aff}}-1})/({p-1})$ , which gives the result.

Let us calculate the degree of $\Gamma $ , the zero-dimensional scheme appearing in the intersection $X:=S\cap S_1\cap S_2$ . By Proposition 3.1, the lines in X have multiplicity one and so we can apply Lemma 2.1 to get

$$ \begin{align*}\deg\Gamma=dd_1d_2 - m(d+d_1+d_2)+2m+ \sum_{i \ne j} L_i\cdot L_j,\end{align*} $$

with $d=(p-1)/2, d_1=3(p-1)/2$ and $d_2=5(p-1)/2$ . Let us scrutinise the intersections of the lines $L_i$ . Since $-1$ is clearly a $(p-1)/2$ th power modulo p, one can use Proposition 3.1 to assert that the lines $L_i$ contained in X are of three types:

$$ \begin{align*} \begin{cases} \begin{matrix} x = ay, \\ z = bw, \cr \end{matrix} \end{cases} \!\!\!\!\!\begin{cases} \begin{matrix} x = a'z , \\ y = b'w,\cr \end{matrix} \end{cases} \!\!\!\!\! \begin{cases} \begin{matrix} x = a' w, \\ y = b'z, \cr \end{matrix} \end{cases}\end{align*} $$

where $a,b$ are quadratic nonresidues modulo p and $a',b'$ are (nonzero) quadratic residues modulo p. Thus, according to Proposition 3.1, the number of lines is $m=3({(p-1)}/{2})^2$ .

If $L_1$ and $L_2$ are two lines of same type given, for example, by equations

$$ \begin{align*} \begin{cases} \begin{matrix} x = a_1y, \\ z = b_1w, \cr \end{matrix} \end{cases} \!\!\!\!\! \begin{cases} \begin{matrix} x = a_2y, \\ z = b_2w,\cr \end{matrix} \end{cases} \end{align*} $$

then $L_1.L_2=1$ if and only if $a_1=a_2$ or $b_1=b_2$ . Hence, in the case of lines $L_i$ and $L_j$ of the same type, the contribution will be

$$ \begin{align*}\sum_{i\not=j, \mathrm{same\ type}}L_i.L_j=3\bigg(\frac{p-1}{2}\bigg)^2\bigg(\bigg(\frac{p-1}{2}-1\bigg)\times 2\bigg)=3\bigg(\frac{p-1}{2}\bigg)^2(p-3).\end{align*} $$

If $L_1$ and $L_2$ are two lines of different types given, for example, by equations

$$ \begin{align*} \begin{cases} \begin{matrix} x = ay, \\ z = bw, \cr \end{matrix} \end{cases} \!\!\!\!\! \begin{cases} \begin{matrix} x = a'z , \\ y = b'w,\cr \end{matrix} \end{cases} \end{align*} $$

then $L_1.L_2=1$ if and only if $ab'=a'b$ . Hence, in the case of lines $L_i$ and $L_j$ of different type, the contribution will be

$$ \begin{align*}\sum_{i\not=j, \mathrm{different\ type}}L_i.L_j=6\bigg(\frac{p-1}{2}\bigg)^3.\end{align*} $$

Finally, we obtain

$$ \begin{align*}\sum_{i\not=j}L_i.L_j=6\bigg(\frac{p-1}{2}\bigg)^2(p-2).\end{align*} $$

One can now compute $\deg \Gamma = dd_1d_2 - m(d+d_1+d_2)+2m+ \sum _{i \ne j} L_i\cdot L_j$ , which gives

$$ \begin{align*}\deg\Gamma= 15\bigg(\frac{p-1}{2}\bigg)^3-3\bigg(\frac{p-1}{2}\bigg)^2\bigg(9\bigg(\frac{p-1}{2}\bigg)\bigg)+6\bigg(\frac{p-1}{2}\bigg)^2+6\bigg(\frac{p-1}{2}\bigg)^2(p-2)=0.\end{align*} $$

Thus, $S\cap S_1\cap S_2$ does not contain isolated rational points.

Proof. Let $\zeta _5$ be a primitive 5th root of unity in $\mathbb C$ and $\mu _5({\mathbb C})$ the group of 5th roots of unity in ${\mathbb C}$ . In the cyclotomic number field ${\mathbb Q}(\zeta _5)$ , we consider the sums $x_1+x_2+x_3+x_4+x_5$ , where $x_i\in \mu _5({\mathbb C})\cup \{0\}$ . We are looking for the primes $p\equiv 1\pmod 5$ dividing the norm of such sums.

Consider, for example, the algebraic integer $\alpha :=1+1+\zeta _5+\zeta _5+\zeta _5\in {\mathbb Q}(\zeta _5)$ . The norm of $\alpha $ is equal to

$$ \begin{align*}N_{{\mathbb Q}(\zeta_5)/{\mathbb Q}}(\alpha)=\prod_{(k,5)=1}\sigma_k(\alpha),\end{align*} $$

where $\sigma _k(\zeta _5)=\zeta _5^k$ . Hence,

$$ \begin{align*}N_{{\mathbb Q}(\zeta_5)/{\mathbb Q}}(\alpha)=3^4\prod_{k=1}^4\bigg(-\frac{2}{3}-\zeta_5^k\bigg)=3^4\Phi_5\bigg(-\frac{2}{3}\bigg),\end{align*} $$

where $\Phi _5(X)=\prod _{k=1}^4(X-\zeta _5^k)=X^4+X^3+X^2+X+1$ is the 5th cyclotomic polynomial. Thus, $N_{{\mathbb Q}(\zeta _5)/{\mathbb Q}}(\alpha )=5\times 11$ .

Since $11\equiv 1\pmod 5$ , the prime $p=11$ totally splits in the extension ${\mathbb Q}(\zeta _5)/{\mathbb Q}$ and the quotient of the ring of integers ${\mathbb Z}[\zeta _5]$ by the ideal generated by $11$ is a product of fields

$$ \begin{align*}{\mathbb Z}[\zeta_5]/11{\mathbb Z}[\zeta_5]\simeq {\mathbb F}_{11}^4.\end{align*} $$

Then, we look at the image of the algebraic integer $2+3\zeta _5$ by the reduction morphisms which send $\zeta _5$ to one of the primitive 5th roots of unity in ${\mathbb F}_{11}$ which are $\{3,9,5,4\}$ . We find that $2+3\times 3\equiv 0\pmod {11}$ , that is, $1+1+3+3+3$ is a sum of five nondistinct 5th roots of unity in ${\mathbb F}_{11}$ which is equal to zero.

In the same way, we show that the algebraic integer $1+4\zeta _5\in {\mathbb Z}[\zeta _5]$ has norm equal to $5\times 41$ , and the reduction morphism which sends $\zeta _5$ to $10$ , which is a primitive 5th root of unity in ${\mathbb F}_{41}$ , gives $1+10+10+10+10=0$ , a sum of five nondistinct 5th root of unity in ${\mathbb F}_{41}$ which is equal to zero.

Again, the algebraic integers $1+3\zeta _5$ and $3+\zeta _5$ have norms equal to 61 and provide the following sums of four 5th roots of unity in ${\mathbb F}_{61}$ which equal zero: $20+20+20+1=0$ and $1+1+1+58=0$ .

Finally, if $\alpha =x_1+x_2+x_3+x_4+x_5$ , where $x_i\in \mu _5({\mathbb C})\cup \{0\}$ , then $N_{{\mathbb Q}(\zeta _5)/{\mathbb Q}}(\alpha )$ is a product of at most five complex numbers of modulus at most 5 and, thus, the norm of $\alpha $ is bounded above by $5^4=625$ . One can show that, aside from $11$ , $41$ and $61$ , no other prime number $\equiv 1\pmod 5$ and less than $625$ appears as a factor of the norm of such an element $\alpha $ . Thus, if $p\not = 11, 41, 61$ , then the only way to have a sum of at most five 5th roots of unity in ${\mathbb F}_p$ to be zero, up to permutations, is $1+\zeta +\zeta ^2+\zeta ^3+\zeta ^4=0$ , where $\zeta $ is a primitive 5th root of unity in ${\mathbb F}_p$ .

Proof. Let us prove the first assertion. Let H be the hyperplane given by the equation $x=0$ . The intersection $S\cap H$ does not contain any rational point since this would provide a sum of at most three 5th roots of unity equal to zero and this cannot happen by Lemma 4.1. We deduce that S does not contain a line defined over ${\mathbb F}_p$ (because such a line would intersect the plane H at a rational point).

We now have to count the number of points $(x:y:z:w)\in {\mathbb P}^3({\mathbb F}_p)$ such that $x^{2(p-1)/5} + y^{2(p-1)/5} + x^{(p-1)/5} y^{(p-1)/5}+z^{2(p-1)/5} + w^{2(p-1)/5}=0$ . If $p\not =11,41, 61$ , by Lemma 4.1, this is necessarily a sum of five distinct 5th roots of unity. Consider the morphism

$$ \begin{align*} \begin{matrix} \psi:&{\mathbb F}_p^{\ast} & \longrightarrow &{\mathbb F}_p^{\ast}\cr &x & \longmapsto & x^{2(p-1)/5}.\cr \end{matrix} \end{align*} $$

The kernel has order ${(p-1)}/{5}$ and the image is the group of 5th roots of unity $\mu _5({\mathbb F}_p)$ in ${\mathbb F}_p^{\ast }$ . For the choice of x, any element of ${\mathbb F}_p^{\ast }$ works, so we have $p-1$ choices. Then, for y, one can choose $p-1-{(p-1)}/{5}={4(p-1)}/{5}$ different elements and, finally, $x^{(p-1)/5} y^{(p-1)/5}$ is completely determined. There remain ${2(p-1)}/{5}$ choices for z and, then, ${(p-1)}/{5}$ choices for w. Dividing $8(p-1)({(p-1)}/{5})^3$ by $p-1$ to find projective points, we get the result.

For $p=11, 41$ and $61$ , a direct computation gives the result. Extra points come from sums of less than five 5th roots of unity which are equal to zero, or from sums of five nondistinct 5th roots of unity which are also equal to zero.

For example, $1+1+3+3+3=1+5+5=1+1+1+4+4=1+1+9=1+9\hspace{0.2pt}+\hspace{0.2pt}4\hspace{0.2pt}+\hspace{0.2pt}4\hspace{0.2pt}+\hspace{0.2pt}4=\hspace{0.2pt}1\hspace{0.2pt}+\hspace{0.2pt}3\hspace{0.2pt}+\hspace{0.2pt}9\hspace{0.2pt}+\hspace{0.2pt}9\hspace{0.2pt}=\hspace{0.2pt}1\hspace{0.2pt}+\hspace{0.2pt}1\hspace{0.2pt}+\hspace{0.2pt}4\hspace{0.2pt}+\hspace{0.2pt}5\hspace{0.2pt}=\hspace{0.2pt}3\hspace{0.2pt}+\hspace{0.2pt}3\hspace{0.2pt}+\hspace{0.2pt}5\hspace{0.2pt}=\hspace{0.2pt}1\hspace{0.2pt}+\hspace{0.2pt}3\hspace{0.2pt}+\hspace{0.2pt}3\hspace{0.2pt}+\hspace{0.2pt}4\hspace{0.2pt}=\hspace{0.2pt}1\hspace{0.2pt}+\hspace{0.2pt}5\, +$ $9+9+9=0$ in ${\mathbb F}_{11}$ .