Weierstrass semigroup at $m+1$ rational points in maximal curves which cannot be covered by the Hermitian curve

Abstract

We determine the Weierstrass semigroup $H(P_\infty ,P_1,\ldots ,P_m)$ at several rational points on the maximal curves which cannot be covered by the Hermitian curve introduced in Tafazolian et al. (J Pure Appl Algebra 220(3):1122–1132, 2016). Furthermore, we present some conditions to find pure gaps. We use this semigroup to obtain AG codes with better relative parameters than comparable one-point AG codes arising from these curves.

Appendices

Appendix A: genus of S

In this section we will determine the genus of the semigroup S generated by $\{q^3+1-iN-j:0\le i+j\le p^b\},$ with $q=(p^b)^r$, for a prime number p, some positives integers b, r, with $r\ge 2$, and $N=\frac{q^3+1}{q+1}$.

For this purpose we will first consider the numerical semigroup $S_{P,N,K}$ generated by $\{KN+aN-j:0\le j\le a\le P\}$, where P, N, K are positive integers with $P\mid N-1$, $P\mid K-1$, $K<N$. Notice that $S=S_{P,N,K}$ with $K=q+1-p^b$ and $P=p^b$, with a playing the role of $p^b-i$, and j, N playing their own role. The required conditions hold, indeed, $P\mid N-1$ since $N-1=\frac{q^3+1}{q+1}-1=(q^2-q+1)-1=q^2-q$ and, similarly, $P\mid K-1$.

1.1 A.1 Characterization

With the same notation as before, for an integer $M\ge 0$, let $S^M_{P,N,K}=\{MKN+aN-j:0\le j\le a\le MP\}$ and let $\widetilde{S}^M_{P,N,K}=\{MKN+aN-j: \max \{0,(M-1)P-K+1\}\le a \le MP, 0\le j\le \min \{a,N-1\}\}$. It is obvious that $S_{P,N,K}=\cup _{M\ge 0} S^M_{P,N,K}$ and that $\widetilde{S}^M_{P,N,K}\subseteq S^M_{P,N,K}$. Now, any element in $S^M_{P,N,K}$ is in at least one set $\widetilde{S}^{M'}_{P,N,K}$ for some $M'\le M$. This can be proved by induction on M. For $M=0$ and for $M=1$ it is straightforward. For $M>1$, suppose that an element of $S^M_{P,N,K}$ is $\ell =MKN+aN-j$ for some particular $0\le j\le a\le MP$. If $a\ge (M-1)P-K+1$ and $j\le N-1$ then $\ell \in \widetilde{S}^M_{P,N,K}$. Otherwise, if $0\le a<(M-1)P-K+1$, then $\ell =(M-1)KN+(K+a)N-j=(M-1)KN+a'-j$ with $a'=K+a\le (M-1)P$ and $0\le j\le a\le a'$, so $\ell \in S^{M-1}_{P,N,K}$ and the result follows by induction. If $a\ge (M-1)P-K+1$ but $j>N-1$, then suppose that Q is the quotient of the division of j by N. Then $\ell =MKN+(a-Q)N-(j-QN)=MKN+a'-j'$ with $a'=a-Q \ge a-j \ge 0$, and $a'\le a\le MP$. Furthermore, $j'=j-QN$, which is the remainder of the division of j by N, and which is between 0 and $N-1$. So, $\ell \in \widetilde{S}^M_{P,N,K}$. Consequently, we also have $S_{P,N,K}=\cup _{M\ge 0} \widetilde{S}^M_{P,N,K}$.

Now, $\widetilde{S}^M_{P,N,K}\subseteq [(M-1)(K+P)N+1,M(K+P)N]$. Indeed, the minimum of $\widetilde{S}^M_{P,N,K}$ is at least $(M-1)(KN+PN)+1$ since the elements in $\widetilde{S}^M_{P,N,K}$ satisfy $MKN+aN-j\ge MKN+((M-1)P-K+1)N-N+1=MKN+MPN-PN-KN+N-N+1=(M-1)(K+P)N+1$. On the other hand, the maximum of $\widetilde{S}^M_{P,N,K}$ is at most $M(K+P)N$ since the elements in $\widetilde{S}^M_{P,N,K}$ satisfy $MKN+aN-j\le MKN+MPN-0=M(K+P)N$.

In particular, the sets $\widetilde{S}^M_{P,N,K}$ are disjoint and, so, $S_{P,N,K}=\sqcup _{M\ge 0}\widetilde{S}^M_{P,N,K}.$

Let $M_0=\frac{K-1}{P}+1$, $M_1=\frac{N-1}{P}$, $M_2=\frac{K+N-2}{P}$.

1.2 A.2 Conductor of $S_{P,N,K}$

Now we are ready to determine the Frobenius number of $S_{P,N,K}$, that is, its largest gap. The conductor of $S_{P,N,K}$ is then the non-gap of $S_{P,N,K}$ right after its Frobenius number.

Observe that $\widetilde{S}_M=\cup _{a\ge \max \{0,(M-1)P-K+1\}}I_a$ with $I_a=\{MKN+aN-j: 0\le j\le \min \{a,N-1\}\}$. Since j ranges from 0 to $\min \{a,N-1\}$, there are gaps between the intervals $I_a$ and $I_{a-1}$ if and only if $\min \{a,N-1\}<N-1$, i.e., if and only if $a< N-1$. Since in $\widetilde{S}^M_{P,N,K}$a ranges from $\max \{0,(M-1)P-K+1\}$ to PN, the inequality $a<N-1$ occurs in $\widetilde{S}^M_{P,N,K}$ if and only if $(M-1)P-K+1<N-1$, that is, if and only if $M\le \frac{N+K-2}{P}$.

Let $M_F= \frac{N+K-2}{P}$. The last gap of $S_{P,N,K}$ will then be the gap previous to $I_a$ with $a=N-2$ in $\widetilde{S}^{M_F}_{P,N,K}$, that is, the Frobenius number will be $M_FKN+aN-a-1$ for $a=N-2$, i.e. $\frac{K+N-2}{P}KN+(N-2)N-N+1=\frac{K+N-2}{P}KN+N^2-3N+1$.

Simplifying by means of Sage we obtain that the conductor of $S_{P,N,K}$ is

$$\begin{aligned} c=\frac{N^2K + NK^2 + N^2P - 2NK - 3NP + 2P}{P}. \end{aligned}$$

1.3 A.3 Genus

Let $M_0=\frac{K-1}{P}+1$, $M_1=\frac{N-1}{P}$, $M_2=\frac{K+N-2}{P}=M_0+M_1-1$.

• If $0\le M\le M_0$ then $S_{P,N,K}^M=\widetilde{S}_{P,N,K}^M$ and $\#\widetilde{S}_{P,N,K}^M=\sum _{a=0}^{MP}(a+1)=\sum _{b=1}^{MP+1} b=\frac{(MP+1)(MP+2)}{2}$.

• If $M_0<M\le M_1$ then $\#\widetilde{S}_{P,N,K}^M=\sum _{a=(M-1)P-K+1}^{MP}(a+1)=\sum _{(M-1)P-K+2}^{MP+1}b=\frac{(MP+1)(MP+2)}{2}-\frac{(MP-P-K+1)(MP-P-K+2)}{2}$. This is because if $M\le M_1$, then $MP\le N-1$, so j will always range from 0 to a.

• If $M>M_1$ then $\#([(M-1)(K+P)N+1,M(K+P)N]{\setminus } \widetilde{S}_{P,N,K}^M)=$$\sum _{a=(M-1)P-K+1}^{N-2}(N-a-1)=\sum _{b=1}^{N-(M-1)P+K-2}b=$$\frac{(N-MP+P+K-2)(N-MP+P+K-1)}{2}$. This is because if $M>M_1$ then $MP>N-1$ and so at some point $a=N-1$. In this case, $[(M-1)(K+P)N+1,M(K+P)N] {\setminus }\widetilde{S}_{P,N,K}^M=\{MKN+aN-j: (M-1)P-K+1 \le a \le N-2, a+1\le j\le N-1\}$.

Now, using the formulas $\sum _{M=1}^n M=\frac{n(n+1)}{2}$ and $\sum _{M=1}^n M^2=\frac{n(n+1)(2n+1)}{6}$ we get to the final count of the genus:

$$\begin{aligned} g= & {} (M_1(P+K)N)\\&-\sum _{M=1}^{M_1}\frac{(MP+1)(MP+2)}{2}\\&+\sum _{M=M_0+1}^{M_1}\frac{(MP-P-K+1)(MP-P-K+2)}{2}\\&+\sum _{M=M_1+1}^{M_2}\frac{(N-MP+P+K-2)(N-MP+P+K-1)}{2}\\= & {} (M_1(P+K)N)\\&-\frac{1}{12}P^2M_1(M_1+1)(2M_1+1)-\frac{3}{4}PM_1(M_1+1)-M_1\\&+\frac{1}{12}P^2(M_1(M_1+1)(2M_1+1)-M_0(M_0+1)(2M_0+1))\\&+\frac{1}{4}P(3-2P-2K)(M_1(M_1+1)-M_0(M_0+1))\\&+\frac{1}{2}(P+K-1)(P+K-2)(M_1-M_0)\\&+\frac{1}{12}P^2(M_2(M_2+1)(2M_2+1)-M_1(M_1+1)(2M_1+1))\\&+\frac{1}{4}P(-2(P+K+N)+3)(M_2(M_2+1)-M_1(M_1+1))\\&+\frac{1}{2}(P+K+N-1)(P+K+N-2)(M_2-M_1)\\= & {} \frac{N^2K + NK^2 + N^2P + NKP - 3NK - 3NP + P + 1}{2P} \end{aligned}$$

1.4 A.4 Back to the originary problem

If we take $N=\frac{q^3+1}{q+1}$, $K=q+1-p^b$, $P=p^b$, then the genus is $g=\frac{q^5 - q^3p^b - q^3 + q^2}{2p^b},$ while the conductor is $c=\frac{q^5 - 2q^3p^b + q^2p^{2b} - q^2p^b - qp^{2b} + q^2 + qp^b + p^{2b} - p^b}{p^b}.$

Appendix B: genus of $S'$

Suppose we have $q=p^a$, $b\mid a$, $b\ne a$, n odd, $n\ge 3$. Let $M=\frac{q^n+1}{q+1}=q^{n-1}-q^{n-2}+q^{n-3}-q^{n-4}+\dots -q+1$. We want to prove that the genus of the semigroup $S'=\langle q^n+1-iM-j:0\le iM+jq^2\le q^{n-1}p^b \rangle$ is $\frac{q^{n+2}-p^bq^n-q^3+q^2}{2p^b}$.

1.1 B.1 Definition of $S'$ revisited

Lemma B.1

If $n\ge 3$,

$$\begin{aligned} S'=\left\langle k(q^{n-1}-q^{n-2})+\ell :q+1-p^b\le k\le q+1 \text{ and } q^{n-3}(q-p^b) \le \ell \le k\frac{q^{n-2}+1}{q+1} \right\rangle \end{aligned}$$

Equivalently, by setting $A=q^{n-1}-q^{n-2}$, $k_0=q+1-p^b$, $k_1=q+1$, $\ell _0=q^{n-3}(q-p^b)+1$, $\ell _1=\frac{q^{n-2}+1}{q+1}$, the semigroup $S'$ is $S'=\left\langle kA+\ell : k_0\le k\le k_1, \ell _0 \le \ell \le k\ell _1\right\rangle .$

Proof

We can rewrite $S'$ as $S'=\langle (q+1-i)M-j:0\le iM+jq^2\le q^{n-1}p^b \rangle .$ The integer i is then bounded as $0\le i\le \lfloor \frac{q^{n-1}p^b}{M}\rfloor$. The quotient and the remainder of the division of $q^{n-1}p^b$ by M are, respectively, $p^b$ and $p^b(q^{n-2}-q^{n-3}+\dots +q-1)$ (since this remainder is between 0 and $M-1$). Consequently, $0\le i\le p^b$. Now, setting $k=q+1-i$, the bounds for k are $q-p^b+1\le k\le q+1$. Hence, since $iM=(q+1-k)M=(q+1)M-kM=q^n+1-kM$,

$$\begin{aligned} S'=\langle kM-j:q-p^b+1\le k\le q+1 \text{ and } 0\le q^n+1-kM+jq^2\le q^{n-1}p^b \rangle \end{aligned}$$

Finally, we want to replace the bounds for $q^n+1-kM+jq^2$ by bounds on j. Reorganizing them, we obtain

$$\begin{aligned} kM-q^n-1\le jq^2\le kM-q^n+q^{n-1}p^b-1=kM-q^{n-1}(q-p^b)-1. \end{aligned}$$

Since $k\le q+1$, the lower bound is non-positive. So, $0\le jq^2$

As for the upper bound on j, $j\le \left\lfloor \frac{kM-q^{n-1}(q-p^b)-1}{q^2}\right\rfloor = \left\lfloor \frac{kq^{n-1}-kq^{n-2}+kq^{n-3}-\dots +kq^2-kq+k-q^{n-1}(q-p^b)-1}{q^2}\right\rfloor$$= \left\lfloor k\frac{q^{n-2}+1}{q+1}+\frac{-kq+k-1}{q^2}-q^{n-3}(q-p^b)\right\rfloor =\left\lfloor k\frac{q^{n-2}+1}{q+1} -q^{n-3}(q-p^b)-1 +\frac{q^2-k(q-1)-1}{q^2} \right\rfloor$. By the bounds on k we deduce that $0\le \frac{q^2-k(q-1)-1}{q^2}\le \frac{p^b(q-1)}{q^2}<1$. So,

$$\begin{aligned} S'=\langle kM-j:q+1-p^b\le k\le q+1 \text{ and } 0\le j\le k\frac{q^{n-2}+1}{q+1}-q^{n-3}(q-p^b)-1 \rangle \end{aligned}$$

Let now $\ell =k\frac{q^{n-2}+1}{q+1}-j$. Notice that $kM-j=k(M-\frac{q^{n-2}+1}{q+1})+\ell =k(q^{n-1}-q^{n-2})+\ell$. The bounds of $\ell$ are $q^{n-3}(q-p^b)+1 \le \ell \le k\frac{q^{n-2}+1}{q+1}$. $\square$

Let $G=\left\{ kA+\ell : k_0\le k\le k_1, \ell _0 \le \ell \le k\ell _1 \right\}$ and let $mG=\{a_1+\dots +a_m:a_i\in G\}$. Define $B_m=[(m-1)(q^n+1)+1,m(q^n+1)]\cap mG$.

Lemma B.2

The following statements hold.

• (1) $S'=\{0\}\cup \bigcup _{m\ge 1} mG$,

• (2) $S'=\{0\}\cup \sqcup _{m\ge 1} B_m$.

Proof

First of all, notice that $mG=\underbrace{[mk_0A+m\ell _0,mk_0(A+\ell _1)]}_{mk_0\ell _1 -m\ell _0 +1 } \cup \underbrace{[(mk_0+1)A+m\ell _0,(mk_0+1)(A+\ell _1)]}_{(mk_0+1)\ell _1 -m\ell _0 +1 } \cup \dots \cup \underbrace{[mk_1A+m\ell _0,mk_1(A+\ell _1)]}_{mk_1\ell _1-m\ell _0+1}$

• (1) The first part is obvious and follows from the definitions.

• (2) For the second part, it is obvious that the sets $B_m$ are disjoint and it is obvious the inclusion $\supseteq$. Let us prove for all m the inclusion $S'\cap [(m-1)(q^n+1)+1,m(q^n+1)]\subseteq B_m$ by induction on m.

  First of all we need to see that $S'\cap [1,q^n+1]=B_1$. The smallest element of 2G is $2(k_0A+\ell _0)= 2((q+1-p^b)(q^{n-1}-q^{n-2})+q^{n-3}(q-p^b)+1)= 2(q^n+1-p^b(q^{n-1}-q^{n-2}+q^{n-3}))$$= q^n+1+(q^n+1-2p^b(q^{n-1}-q^{n-2}+q^{n-3})) > q^n+1+(q^n+1-2p^b\frac{q^n+1}{q+1}) \ge q^n+1$ if $2p^b\le q+1$, which is a consequence of the fact that $p^b<q$.

  Now suppose $m>1$. Since the maximum of mG is $m(q^n+1)$, we have $mG\subseteq [0,m(q^n+1)]$. Now it will suffice to see that $mG\cap [0,(m-1)(q^n+1)]\subseteq (m-1)G$ and the result will follow by induction.

  Notice that mG is the union of the sets of the form $S_{m,\tilde{k}}=[\tilde{k} A+m\ell _0,\tilde{k}(A+\ell _1)]$ for some $\tilde{k}$ satisfying $mk_0\le \tilde{k}\le m k_1$, while $(m-1)G$ is the union of sets of the form $S_{(m-1),\tilde{\tilde{k}}}[\tilde{\tilde{k}} A+m\ell _0,\tilde{\tilde{k}}(A+\ell _1)]$ for some $\tilde{\tilde{k}}$ satisfying $(m-1)k_0\le \tilde{\tilde{k}}\le (m-1) k_1$. Suppose that $a\in mG\cap [0,(m-1)(q^n+1)]$. If $a\in S_{m\tilde{k}}$ with $mk_0\le \tilde{k}\le (m-1)k_1$, then, since $S_{m,\tilde{k}}\subseteq S_{m-1,\tilde{k}}$, we have $a\in S_{m-1,\tilde{k}}\subseteq (m-1)G$. On the other hand, if $a\in S_{m,\tilde{k}}\cap [0,(m-1)(q^n+1)]$ with $(m-1)k_1<\tilde{k} \le mk_1$, then $a\ge \tilde{k} A+ m\ell _0 >(m-1)k_1 A +(m-1)\ell _0$. So, $a\in S_{m-1,(m-1)k_1}\subseteq (m-1)G$. $\square$

1.2 B.2 Number of gaps by intervals

Let $C_m=[(m-1)(q^n+1)+1,m(q^n+1)]{\setminus } B_m$. In this section we wonder what are the elements in $C_m$. As before, we split the elements in mG into (not necessarily disjoint) blocks of the form $S_{m,\tilde{k}}=[\tilde{k} A+m\ell _0,\tilde{k}(A+\ell _1)]$ for some $\tilde{k}$ satisfying $mk_0\le \tilde{k}\le m k_1$.

Lemma B.3

• (1) Suppose that $mk_0< k\le mk_1$. Then the gaps between $S_{m,(k-1)}$ and $S_{m,k}$ are contained in $[(m-1)(q^n+1)+1,m(q^n+1)]$ if and only if $k\ge \max \left( mk_0+1,mq+m-q\right)$

• (2) $\max \left( mk_0+1,mq+m-q\right) =mk_0+1$ if and only if $m\le M_1:=p^{a-b}$.

Proof

• (1) Suppose that $mk_0< k\le mk_1$. Then the gaps between $S_{m,(k-1)}$ and $S_{m,k}$ are contained in $[(m-1)(q^n+1)+1,m(q^n+1)]$ if and only if $(k-1)\frac{q^n+1}{q+1}\ge (m-1)(q^n+1)$, that is, if and only if $k\ge (m-1)(q+1)+1=mq+m-q$.

• (2) $\max \left( mk_0+1,mq+m-q\right) =mk_0+1$ if and only if $mq+m-q\le m(q+1-p^b)+1$, that is, if and only if $-q\le -mp^b+1$, i.e., $mp^b\le q+1$. Now observe that the quotient of the Euclidean division of $q+1$ by $p^b$ is $p^{a-b}$ while the remainder is 1. So, the statement follows.$\square$

Lemma B.4

• (1) Suppose that $mk_0< k\le mk_1$. Then there are gaps between $S_{m,(k-1)}$ and $S_{m,k}$ if and only if $k\le \min \left( mk_1,\frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}\right) .$

• (2) If $n>3$, $\min \left( mk_1, \frac{q^{n}-q+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}\right) =mk_1$ if and only if $m\le M_2:=(q-1)p^{a-b}$.

• (3) If $n=3$, $\min \left( mk_1, \frac{q^{n}-q+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}\right) =mk_1$ if and only if $m\le \tilde{M_2}:=(q-1)p^{a-b}-1$.

Proof

• (1) Suppose that $mk_0< k\le mk_1$. Then there are gaps between $S_{m,( k-1)}$ and $S_{m, k}$ if and only if $( k-1)(A+\ell _1)\le k A+m\ell _0-2$, equivalently, $( k-1)\ell _1\le m\ell _0-2+A$, equivalently, $k\le \frac{m\ell _0-2+A}{\ell _1}+1= (q+1)\frac{m(q^{n-3}(q-p^b)+1)-2+q^{n-1}-q^{n-2}}{q^{n-2}+1}+1$$= \frac{m(q+1)(q^{n-3}(q-p^b)+1)-q-2+q^{n}-q^{n-2}+q^{n-2}+1}{q^{n-2}+1} = \frac{ q^{n} -q-1 +m((q+1)q^{n-3}(q-p^b)+1) }{q^{n-2}+1}$.

• (2) $\min \left( mk_1, \frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}\right) =mk_1$ if and only if $m(q+1)\le \frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}$, that is, if and only if $m(q^{n-1}+q^{n-2}+q+1)\le q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)$, i.e., $m(q^{n-1}+q^{n-2}+q+1)\le q^{n}-q-1+m(q^{n-2}(q-p^b)+q^{n-3}(q-p^b)+q+1)$, i.e., $0\le q^{n}-q-1-mp^b(q+1)q^{n-3}$, i.e., $m\le \lfloor \frac{q^n-q-1}{p^b(q+1)q^{n-3}}\rfloor$

  Here we notice that the Euclidean division of $q^n-q-1$ by $q^{n-2}p^b+q^{n-3}p^b$ has quotient $(q-1)p^{a-b}$ and remainder $q^n-q-1-(q-1)p^{a-b}(q^{n-2}p^b+q^{n-3}p^b)= q^n-q-1 -q^{n}-q^{n-1} +q^{n-1}+q^{n-2} = q^{n-2} -q-1$.

  So, the statement follows.

• (3) It can be proved as the previous item.$\square$

Lemma B.5

If $(q-1)p^{a-b}+1\le m\le qp^{a-b}-1$ then $\frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}$ is not an integer and $\lfloor \frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1} \rfloor =q^2-q+m(q-p^b+1)$.

Proof

$\frac{q^{n}-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}= \frac{q^{n}-q+1}{q^{n-2}+1}+ m\frac{(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}$ $=q^2-\frac{q^2+q+1}{q^{n-2}+1} +m(q-p^b+1)-mp^b\frac{q^{n-3}-1}{q^{n-2}+1}$ $= q^2-q+m(q-p^b+1)+q-\frac{q^2+q+1}{q^{n-2}+1}-mp^b\frac{q^{n-3}-1}{q^{n-2}+1}.$

Now, it is enough to see that $0\le q-\frac{q^2+q+1}{q^{n-2}+1}-mp^b\frac{q^{n-3}-1}{q^{n-2}+1}<1$.

On one hand, $q-\frac{q^2+q+1}{q^{n-2}+1} - mp^b( \frac{q^{n-3}-1}{q^{n-2}+1}) \ge q- \frac{q^2+q+1}{q^{n-2}+1} -(qp^{a-b}-1)(\frac{(q^{n-3}-1)p^b}{q^{n-2}+1})$$= q- \frac{q^2+q+1}{q^{n-2}+1} - \frac{(q^{n-3}-1)q^2-p^{b}(q^{n-3}-1)}{q^{n-2}+1}$$= q- \frac{q^2+q+1}{q^{n-2}+1} - \frac{q^{n-1}-q^2-q^{n-3}p^b+p^b}{q^{n-2}+1} = q+ \frac{-q^{n-1}+q^{n-3}p^b-p^b+q+1}{q^{n-2}+1}$$= \frac{q+q^{n-3}p^b-p^b+q+1}{q^{n-2}+1} = \frac{2q+p^b(q^{n-3}-1)+1}{q^{n-2}+1} >0$.

On the other hand, $q- \frac{q^2+q+1}{q^{n-2}+1} - mp^b\frac{q^{n-3}-1}{q^{n-2}+1} \le \frac{q^{n-1}+q-q^2-q-1-(qp^{a-b}-p^{a-b}+1)(q^{n-3}p^b-p^b) }{q^{n-2}+1}$$= \frac{q^{n-1}-q^2-1 -qp^{a-b}(q^{n-3}p^b-p^b) +p^{a-b}(q^{n-3}p^b-p^b) -(q^{n-3}p^b-p^b) }{q^{n-2}+1}$$= \frac{q^{n-1}-q^2-1 -q^{n-1}+q^2 +q^{n-2}-q -q^{n-3}p^b+p^b }{q^{n-2}+1} = \frac{ -1 +q^{n-2}-q -q^{n-3}p^b+p^b }{q^{n-2}+1}$$= \frac{ q^{n-2}+1-(2+q+(q^{n-3}-1)p^b) }{q^{n-2}+1} <1.$$\square$

Lemma B.6

Suppose that $mk_0< k\le mk_1$. Then the number of gaps between $S_{m,(k-1)}$ and $S_{m,k}$ is $m(q^{n-3}(q-p^b)+1)-{ k}\frac{q^{n-2}+1}{q+1}+\frac{q^n+1}{q+1}-1$.

Proof

The number of gaps between $S_{m,(k-1)}$ and $S_{m, k}$ is $kA+m\ell _0-(k-1)(A+\ell _1)-1= m\ell _0-k\ell _1 +A+\ell _1-1$, which yields the formula in the statement. $\square$

Lemma B.7

• (1) There are gaps that are at least $(m-1)(q^n+1)+1$ and which are smaller than the elements in $S_{m,mk_0}$ if and only if $m\le p^{a-b}=M_1$.

• (2) If $m\le p^{a-b}=M_1$, then the number of gaps between $(m-1)(q^n+1)+1$ and $S_{m,mk_0}$ is $q^n-mp^b(q^{n-1}-q^{n-2}+q^{n-3})$.

Proof

• (1) There are gaps that are at least $(m-1)(q^n+1)+1$ and which are smaller than the elements in $S_{m,mk_0}$ if and only if $mk_0A+m\ell _0\ge (m-1)(q^n+1)+2$. This is equivalent to $m(q+1-p^b)(q^{n-1}-q^{n-2})+m (q^{n-3}(q-p^b)+1)\ge (m-1)(q^n+1)+2$, that is, if and only if $mq^n-mq^{n-1}+mq^{n-1}-mq^{n-2}-mq^{n-1}p^b+mq^{n-2}p^b + mq^{n-2}-mq^{n-3}p^b+m \ge mq^n +m -q^n-1 +2$, which is equivalent to $q^n-1\ge mp^b(q^{n-1}-q^{n-2}+q^{n-3})$ i.e., $m\le \lfloor \frac{q^n-1}{p^b(q^{n-1}-q^{n-2}+q^{n-3})}\rfloor$.

  Here we remark that the Euclidean division of $q^n-1$ by $p^b(q^{n-1}-q^{n-2}+q^{n-3})$ has quotient $p^{a-b}$ and remainder $q^n-1-(q^{n}-q^{n-1}+q^{n-2}) =q^{n-1}-q^{n-2}-1$. So, the result follows.

• (2) It follows from the formula $mk_0A+m\ell _0-(m-1)(q^n+1)-1$ and a similar simplification as before. $\square$

Lemma B.8

Let $M_3=qp^{a-b}-1$. The set $C_m$ is not empty if and only if $m\le M_3$.

Proof

It is clear that for $m<M_2$, $C_m\ne \emptyset$. For $m\ge M_2$, $C_m\ne \emptyset$ if and only if

$$\begin{aligned} \frac{q^n-q-1+m(q+1)(q^{n-3}(q-p^b)+1)}{q^{n-2}+1}\ge m(q+1)-q. \end{aligned}$$

This is equivalent to $q^n-q-1+m(q+1)(q^{n-3}(q-p^b)+1-(q^{n-2}+1))\ge -q(q^{n-2}+1)$, which in turn is equivalent to $q^n-1-m(q+1)(p^bq^{n-3})\ge -q^{n-1}$, i.e., $m\le \lfloor \frac{(q+1)q^{n-1}-1}{(q+1)p^bq^{n-3}}\rfloor =qp^{a-b}-1$. $\square$

Corollary 1

$S'=\sqcup _{m=1}^{M_3}B_m$

Theorem 1

The genus of $S'$ is $\frac{q^{n+2}-p^bq^n-q^3+q^2}{2p^b}$.

Proof

By Lemmas B.1,  B.2,  B.3,  B.4,  B.5,  B.6,  B.7,  B.8, and Corollary 1, it easily follows that the genus of $S'$ is

$$\begin{aligned}&\sum _{m=1}^{M_1} \left( q^n-mp^b(q^{n-1}-q^{n-2}+q^{n-3}) \right) \\&\qquad + \sum _{m=1}^{M_1} \sum _{k=mk_0+1}^{mk_1} \left( mq^{n-3}(q-p^b)+m-{k}\frac{q^{n-2}+1}{q+1}+\frac{q^n+1}{q+1}-1\right) \\&\qquad + \sum _{m=M_1+1}^{M_2} \sum _{k=mq+m-q}^{mk_1} \left( mq^{n-3}(q-p^b)+m-{k}\frac{q^{n-2}+1}{q+1}+\frac{q^n+1}{q+1}-1\right) \\&\qquad + \sum _{m=M_2+1}^{M_3} \sum _{ k=mq+m-q}^{q^2-q+m(q-p^b+1)} \left( mq^{n-3}(q-p^b)+m-{k}\frac{q^{n-2}+1}{q+1}+\frac{q^n+1}{q+1}-1\right) \\&\quad =M_1q^n-p^b(q^{n-1}-q^{n-2}+q^{n-3})M_1(M_1+1)/2 \\&\qquad + \left( \frac{q^n+1}{q+1}-1\right) \\&\qquad \quad \left( \sum _{m=1}^{M_1} \sum _{k=mk_0+1}^{mk_1} 1 + \sum _{m=M_1+1}^{M_2} \sum _{k=mq+m-q}^{mk_1}1 + \sum _{m=M_2+1}^{M_3} \sum _{k=mq+m-q}^{q^2-q+m(q-p^b+1)}1 \right) \\&\qquad + (q^{n-3}(q-p^b)+1)\\&\qquad \quad \left( \sum _{m=1}^{M_1} m \sum _{k=mk_0+1}^{mk_1} 1 + \sum _{m=M_1+1}^{M_2} m \sum _{k=mq+m-q}^{mk_1}1 + \sum _{m=M_2+1}^{M_3} m \sum _{k=mq+m-q}^{q^2-q+m(q-p^b+1)}1 \right) \\&\qquad - \frac{q^{n-2}+1}{q+1} \left( \sum _{m=1}^{M_1} \sum _{k=mk_0+1}^{mk_1} k + \sum _{m=M_1+1}^{M_2} \sum _{k=mq+m-q}^{mk_1}k + \sum _{m=M_2+1}^{M_3} \sum _{k=mq+m-q}^{q^2-q+m(q-p^b+1)}k \right) \\&\quad =M_1q^n-p^b(q^{n-1}-q^{n-2}+q^{n-3})M_1(M_1+1)/2 \\&\qquad + \left( \frac{q^n+1}{q+1}-1\right) \left( A + B + C \right) \\&\qquad + (q^{n-3}(q-p^b) +1) \left( D + E + F \right) \\&\qquad - \frac{q^{n-2}+1}{q+1} \left( G + H + I \right) , \end{aligned}$$

where

$$\begin{aligned} A= & {} \sum _{m=1}^{M_1}\sum _{k=mk_0+1}^{mk_1}1= \sum _{m=1}^{M_1}(mk_1-mk_0)= (k_1-k_0)\frac{M_1(M_1+1)}{2}\\ B= & {} \sum _{m=M_1+1}^{M_2}\sum _{k=mq+m-q}^{mk_1}1= \sum _{m=M_1+1}^{M_2}(mk_1-mq-m+q+1)\\= & {} (q+1)(M_2-M_1)+(k_1-q-1)\sum _{m=M_1+1}^{M_2}m\\= & {} (q+1)(M_2-M_1)+(k_1-q-1)\left( \frac{M_2(M_2+1)-M_1(M_1+1)}{2}\right) \\ C= & {} \sum _{m=M_2+1}^{M_3}(q^2-mp^b+1)= (q^2+1)(M_3-M_2)-p^b\left( \frac{M_3(M_3+1)-M_2(M_2+1)}{2}\right) \\ D= & {} \sum _{m=1}^{M_1}m(mk_1-mk_0)= (k_1-k_0)\sum _{m=1}^{M_1}m^2= (k_1-k_0)\left( \frac{M_1(M_1+1)(2M_1+1)}{6}\right) \\ E= & {} \sum _{m=M_1+1}^{M_2}m(mk_1-mq-m+q+1)= (q+1)\sum _{m=M_1+1}^{M_2}m+(k_1-q-1)\sum _{m=M_1+1}^{M_2}m^2\\= & {} (q+1)\left( \frac{M_2(M_2+1)-M_1(M_1+1)}{2}\right) \\&+(k_1-q-1)\frac{M_2(M_2+1)(2M_2+1)-M_1(M_1+1)(2M_1+1)}{6}\\ F= & {} \sum _{m=M_2+1}^{M_3}m(q^2-mp^b+1)\\= & {} (q^2+1)\left( \frac{M_3(M_3+1)-M_2(M_2+1)}{2}\right) \\&-p^b\left( \frac{M_3(M_3+1)(2M_3+1)-M_2(M_2+1)(2M_2+1)}{6}\right) \\ G= & {} \sum _{m=1}^{M_1}\left( \frac{mk_1(mk_1+1)-mk_0(mk_0+1)}{2}\right) = \frac{k_1-k_0}{2}\sum _{m=1}^{M_1}m+\frac{k_1^2-k_0^2}{2}\sum _{m=1}^{M_1}m^2\\= & {} \frac{k_1-k_0}{2}\frac{M_1(M_1+1)}{2}+\frac{k_1^2-k_0^2}{2}\frac{M_1(M_1+1)(2M_1+1)}{6}\\= & {} \frac{(k_1-k_0)m(m+1)}{4}+\frac{(k_1^2-k_0^2)M_1(M_1+1)(2M_1+1)}{12}\\ H= & {} \sum _{m=M_1+1}^{M_2}\left( \frac{mk_1(mk_1+1)-(m(q+1)-q-1)(m(q+1)-q)}{2}\right) \\ {}= & {} -\frac{q(q+1)}{2}(M_2-M_1)+\left( \frac{k_1+(2q+1)(q+1)}{2}\right) \frac{M_2(M_2+1)-M_1(M_1+1)}{2}\\&+\frac{k_1^2-(q+1)^2}{2}\frac{M_2(M_2+1)(2M_2+1)-M_1(M_1+1)(2M_1+1)}{6}\\= & {} -\frac{q(q+1)}{2}(M_2-M_1)+\frac{(k_1+(2q+1)(q+1))(M_2(M_2+1)-M_1(M_1+1))}{4}\\&+\frac{(k_1^2-(q+1)^2)(M_2(M_2+1)(2M_2+1)-M_1(M_1+1)(2M_1+1))}{12} \\ I= & {} \sum _{m=M_2+1}^{M_3}\sum _{k=mq+m-q}^{q^2-q+m(q-p^b+1)}k\\= & {} \sum _{m=M_2+1}^{M_3}\left( \frac{(q^2-q+m(q-p^b+1))(q^2-q+1+m(q-p^b+1))-(m(q+1)-q-1)(m(q+1)-q)}{2}\right) \\= & {} \frac{(q^2-q)(q^2-q+1)-q(q+1)}{2}(M_3-M_2)\\&+ \frac{((q-p^b+1)(2q^2-2q+1)+(q+1)(2q+1)) (M_3(M_3+1)-M_2(M_2+1))}{4}\\&\quad \frac{((q-p^b+1)^2-(q+1)^2)(M_3(M_3+1)(2M_3+1)-M_2(M_2+1)(2M_2+1))}{12} \end{aligned}$$

A Sage simplification of this leads to

$$\begin{aligned}&= 1/2 q^{n+1}p^{a-b} - 1/2q^n - 1/2q^2p^{a-b} + 1/2qp^{a-b} \\&\quad = \frac{q^{n+1}p^{a-b} - q^n - q^2p^{a-b} + qp^{a-b}}{2} \end{aligned}$$

$\square$

We remark here that the result does not vary if we replace $M_2=(q-1)p^{a-b}$ by $M_2'=(q-1)p^{a-b}-1$.