Proof. We have \(2g(E)-2=(2g(E^G)-2)[E:E^G]+\deg\mathrm{Diff}(E/E^G)\) by the Hurwitz genus formula (see 6 ). Hence, \(\deg\mathrm{Diff}(E/E^G)>2g(E)-2\) if and only if \(g(E^G)=0\). This is equivalent to \(E^G\) being a rational function field by [33]. ◻

Proof. We have \(P_1+\dots+P_n\sim P_1'+\dots+P_n'\) if and only if \(P_1-O+\dots+P_n-O\sim P_1'-O+\dots+P_n'-O\), which is equivalent to \([P_1-O]+\dots+[P_n-O]=[P_1'-O]+\dots+[P_n'-O]\). This is equivalent to \(P_1\oplus \dots\oplus P_n=P_1'\oplus\dots\oplus P_n'\) by the group isomorphism \(\varphi\) in 8 . ◻

Proof. Let \(D_\infty:=gP_{\infty}\in \mathrm{Div}(E)\). Since \(D_0\sim D_0'\), we have \([D_0-D_{\infty}]=[D_0'-D_{\infty}]\in \mathrm{Cl}^0(E)\). By the uniqueness of the representative described in [41], we have \(D_0=D_0'\). ◻

Proof. Since \(\mathfrak{C}/\mathbb{F}_q\) is maximal, its \(L\)-polynomial is \(L(\mathfrak{C}/\mathbb{F}_q,T)=1+2g\sqrt{q} T+(\text{higher order terms of } T)=\prod_{i=1}^{2g}(1-\alpha_{i}\sqrt{q}T)\) with \(|\alpha_i|=1\) by [33]. Thus, \(\alpha_1=\dots=\alpha_{2g}=-1\) and the \(L\)-polynomial over \(\mathbb{F}_{q^s}\) is \(L(\mathfrak{C}/\mathbb{F}_{q^s},T)=(1-(-\sqrt{q})^sT)^{2g}\) by [33]. Since \(\mathfrak{C}\) is maximal (minimal, respectively) over \(\mathbb{F}_{q^s}\) if and only if its \(L\)-polynomial is \((1+q^{s/2}T)^{2g}\) (\((1-q^{s/2}T)^{2g}\), respectively), this lemma is proved. ◻

Proof. (i) By Lemma 1 (i), the set of all distinct places of \(E\) lying over \(O\cap F\) is \(\{\sigma(O):\; \sigma\in T_HA\}=H\). Since \(O\) is a rational place, we have \(f(O|O\cap F)=1\). Applying Lemma 1 (iii) yields \(|H|\cdot e(O|O\cap F)=|G|=|A||H|\), and hence, \(e(O|O\cap F)=|A|\geq 2\). By Dedekind’s different theorem (see 5 ), we have \(\deg\mathrm{Diff}(E/F)\geq \deg((e(O|O\cap F)-1)O)\geq 1\). By Lemma 2, \(F\) is a rational function field. Therefore, we have \(\dim_{\mathbb{F}_q}(\mathcal{L}_{F}(Q_{\infty}))=2\). Let \(z\in \mathcal{L}_F(Q_{\infty})\backslash \mathbb{F}_q\). We have \([F:\mathbb{F}_q(z)]=\deg((z)_{\infty}^F)=\deg(Q_{\infty})=1\) and \((z)^E_{\infty}=\mathrm{Con}_{E/F}((z)^F_{\infty})=\mathrm{Con}_{E/F}(Q_{\infty})=\sum_{j=1}^{r+\delta-1}P_j\).

(ii) Since the case \(r=1\) is trivial, we henceforth assume \(r\geq 2\). By the Riemann-Roch theorem (see 2 ), for any \(1\leq i\leq r-1\), we have \(\mathcal{L}_{E}(\sum_{j=1}^{i+1}P_{j})\backslash \mathcal{L}_{E}(\sum_{j=1}^{i}P_j)\neq \varnothing\). Let \(w_0:=1\) and \(w_i\in \mathcal{L}_{E}(\sum_{j=1}^{i+1}P_{j})\backslash \mathcal{L}_{E}(\sum_{j=1}^{i}P_{j})\) for each \(1\leq i\leq r-1\). We need to show that \(w_0,w_1,\dots,w_{r-1}\) are \(F\)-linearly independent. Assume towards a contradiction that there exist rational functions \({f}_{0}(z),\dots,{f}_{r-1}(z)\in F=\mathbb{F}_q(z)\), not all zero, such that \(\sum_{i=0}^{r-1}{f}_i(z)w_i=0\). By clearing denominators, we may assume that \({f}_{0}(z),\dots,{f}_{r-1}(z)\) are polynomials of \(z\). Let \(t:=\max\{0\leq i\leq r-1:\;\deg({f}_i(z))=\max\{\deg({f}_j(z)):\;0\leq j\leq r-1\}\}\), where we adopt the convention \(\deg(0):=-\infty\). It holds that \({f}_t(z)\neq 0\). Then we consider the following two possible cases of \(t\) towards deriving a contradiction. Before proceeding, we denote \(e:=e(P_1|Q_{\infty})=\dots=e(P_{r+\delta-1}|Q_{\infty})\), and observe that \(v_{P_1}(z)=\dots=v_{P_{r+\delta-1}}(z)=-e\).

• If \(t=0\), then we have \(\deg({f}_{0}(z))>\deg({f}_i(z))\) for any \(1\leq i\leq r-1\). Hence, for any \(1\leq i\leq r-1\) such that \({f}_i(z)\neq 0\), we have \[v_{P_{r+\delta-1}}({f}_{0}(z)w_0)=-e\deg({f}_{0}(z))<-e\deg({f}_i(z))-(e-1)\leq -e\deg({f}_i(z))+v_{P_{r+\delta-1}}(w_i)= v_{P_{r+\delta-1}}({f}_i(z)w_i),\] where the symbol “\(\leq\)” is due to \(w_i\in \mathcal{L}_{E}(P_1+\dots+P_{r+\delta-1}-P_{r+\delta-1})=\mathcal{L}_{E}(\sum_{P|Q_{\infty}}eP-P_{r+\delta-1})\). This implies \(v_{P_{r+\delta-1}}(\sum_{i=0}^{r-1}{f}_i(z)w_i)=v_{P_{r+\delta-1}}({f}_{0}(z)w_0)\neq \infty\) by the strict triangle inequality (see [33]).

• If \(1\leq t\leq r-1\), then we have \(\deg({f}_t(z))\geq \deg({f}_i(z))\) for any \(0\leq i\leq t-1\); and \(\deg({f}_t(z))>\deg({f}_i(z))\) for any \(t+1\leq i\leq r-1\). Hence, for any \(0\leq i\leq t-1\) such that \({f}_i(z)\neq 0\), we have \[\begin{align} \label{eq:202507131727} v_{P_{t+1}}({f}_t(z)w_t)=-e\deg({f}_t(z))+v_{P_{t+1}}(w_t)<-e\deg({f}_i(z))+v_{P_{t+1}}(w_i)=v_{P_{t+1}}({f}_i(z)w_i), \end{align}\tag{12}\] where “\(<\)” is due to \(\deg({f}_t(z))\geq \deg({f}_i(z))\) and \(v_{P_{t+1}}(w_t)<v_{P_{t+1}}(w_i)\) since \(v_{P_{t+1}}(w_t)\leq -v_{P_{t+1}}(\sum_{j=1}^{t+1}P_j)\),⁶ \(w_i\in \mathcal{L}_{E}(\sum_{j=1}^{t+1}P_j-P_{t+1})\); and for any \(t+1\leq i\leq r-1\) such that \({f}_i(z)\neq 0\), we have \[\begin{align} \label{eq:202507131728} v_{P_{t+1}}({f}_t(z)w_t)=-e\deg({f}_t(z))+v_{P_{t+1}}(w_t)<-e\deg({f}_i(z))+v_{P_{t+1}}(w_i)=v_{P_{t+1}}({f}_i(z)w_i), \end{align}\tag{13}\] where the symbol “\(<\)” is due to \(\deg({f}_t(z))> \deg({f}_i(z))\), \(v_{P_{t+1}}(w_t)\leq -1\) and \(v_{P_{t+1}}(w_i)\geq -e\) since \(v_{P_{t+1}}(w_t)\leq -v_{P_{t+1}}(\sum_{j=1}^{t+1}P_j)\) and \(w_i\in \mathcal{L}_{E}(\sum_{j=1}^{r+\delta-1}P_j)=\mathcal{L}_{E}(\sum_{P|Q_{\infty}}eP)\). Inequalities 12 and 13 imply \(v_{P_{t+1}}(\sum_{i=0}^{r-1}{f}_i(z)w_i) =v_{P_{t+1}}({f}_t(z)w_t)\neq \infty\).

The above two cases both lead to a contradiction with \(\sum_{i=0}^{r-1}{f}_i(z)w_i=0\). Therefore, \(w_0,w_1,\dots,w_{r-1}\) are \(F\)-linearly independent.

(iii) For any \(1\leq i\leq \ell\) and \(1\leq u_1<\cdots<u_{r} \leq r+\delta-1\), we will show that the submatrix consisting of the \(u_1,\dots,u_{r}\)-th columns of \(M_i\) in ?? is singular if and only if \(\oplus_{j=1}^{r}P_j= \oplus_{j=1}^{r}P_{i,u_j}\).

If the submatrix consisting of the \(u_1,\dots,u_{r}\)-th columns of \(M_i\) is singular, then there exist \(c_0,c_1,\dots,c_{r-1}\in \mathbb{F}_q\), not all zero, such that the vector \((c_0,c_1,\dots,c_{r-1})M_i\) vanishes at the positions \(u_1,\dots,u_{r}\). Thus, the function \(w:=\sum_{j=0}^{r-1}c_jw_j\) has \(r\) zeros \(P_{i,u_1},\dots,P_{i,u_r}\). Since \(w\in \mathcal{L}_{E}(P_1+\dots+P_{r})\), we have \(w \in \mathcal{L}_{E}(P_1+\dots+P_{r}-P_{i,u_1}-\dots-P_{i,u_r})\). This implies \((w)^E=-P_1-\dots-P_{r}+P_{i,u_1}+\dots+P_{i,u_r}\). And then it holds \(\oplus_{j=1}^{r}P_j= \oplus_{j=1}^{r}P_{i,u_j}\) by Lemma 3.

Conversely, \(\oplus_{j=1}^{r}P_j = \oplus_{j=1}^{r}P_{i,u_j}\) implies \(P_1+\dots+P_{r}\sim P_{i,u_1}+\dots+P_{i,u_r}\) by Lemma 3, i.e., there exists a nonzero function \(w\in E\) such that \((w)^E=-P_1-\dots-P_{r}+P_{i,u_1}+\dots+P_{i,u_r}\). Then we have \(w\in \mathcal{L}_{E}(\sum_{j=1}^{r}P_j)=\mathop{\mathrm{span}}_{\mathbb{F}_q}\{w_0,w_1,\dots,w_{r-1}\}\), and thus \(w=\sum_{j=0}^{r-1}c_jw_j\) for some \(c_0,c_1,\dots,c_{r-1}\in \mathbb{F}_q\) that are not all zero, which means that \((c_0,c_1,\dots,c_{r-1})M_i\) vanishes at the positions \(u_1,\dots,u_{r}\). And then the submatrix consisting of the \(u_1,\dots,u_{r}\)-th columns of \(M_i\) is singular.

The above statements imply the equivalences stated in Proposition 1 (iii), except for ?? . It remains to establish the equivalence between ?? and ?? .

Note that for any \(1\leq i\leq \ell\), \(\mathcal{L}_{F}(Q_{\infty}-Q_i)\neq \{0\}\) since \(F\) is a rational function field. Thus, there exists a nonzero function \(z'\in \mathcal{L}_{F}(Q_{\infty}-Q_i)\subseteq F\) such that \((z')^F=-Q_{\infty}+Q_i\), and then \((z')^E=\mathrm{Con}_{E/F}((z')^F)=\mathrm{Con}_{E/F}(-Q_{\infty}+Q_i)=(-P_1-\dots-P_{r+\delta-1})+(P_{i,1}+\dots+P_{i,r+\delta-1})\). This implies that \(\oplus_{j=1}^{r+\delta-1}P_j=\oplus_{j=1}^{r+\delta-1}P_{i,j}\) by Lemma 3. Therefore, \[\oplus_{j=1}^{r}P_{j}\notin\{\oplus_{j=1}^{r}P_{i,u_j}:\; 1\leq u_1< \dots < u_{r}\leq r+\delta-1\}\] is equivalent to \[\oplus_{j=r+1}^{r+\delta-1}P_{j}\notin\{\oplus_{j=1}^{\delta-1}P_{i,u_j}:\; 1\leq u_{1}< \dots < u_{\delta-1}\leq r+\delta-1\}.\] Consequently, ?? and ?? are equivalent. The proof is completed. ◻

Proof. By Proposition 1 (ii), \(w_0,w_1,\dots,w_{r-1}\) are \(F\)-linearly independent, which, along with the fact that \(1,z,\dots,z^t\in F\) are \(\mathbb{F}_q\)-linearly independent, implies that \(\dim_{\mathbb{F}_q}(V)=tr+1\). Note that \(V\subseteq \mathcal{L}_{E}(t(P_1+P_2+\dots+P_{r+\delta-1}))\). By Section 2.1, \(\mathcal{C}(\mathcal{P},V)\) is a linear code with parameters \([n=m(r+\delta-1),k=tr+1,d\geq (m-t)(r+\delta-1)]_q.\) In the following, we prove that \(\mathcal{C}(\mathcal{P},V)\) is an optimal \((r,\delta)\)-LRC.

Since \(z\in F\) and \(P_{i,1},\dots,P_{i,r+\delta-1}\) all lie over \(Q_{i}\in \mathbb{P}_F^1\) for each \(1\leq i\leq m\), we have \(z(P_{i,1})=\dots=z(P_{i,r+\delta-1})=z(Q_i)\), and thus \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}=\{(\phi(P_{i,1}),\dots,\phi(P_{i,r+\delta-1})):\; \phi\in \mathop{\mathrm{span}}_{\mathbb{F}_q}\{w_0,w_1,\dots,w_{r-1}\}\}\). The minimum distance of \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}\) is equal to \(\delta\) since the matrix \(M_i\) in ?? is a generator matrix of \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}\) and all \(r\times r\) submatrices of \(M_i\) are invertible by Proposition 1 (iii). Thus, \(\mathcal{C}(\mathcal{P},V)\) is an \((r,\delta)\)-LRC. By the Singleton-type bound 1 , we have \(d\leq n-k+1-(\lceil k/r\rceil-1)(\delta-1)=(m-t)(r+\delta-1).\) Therefore, the minimum distance \(d\) of \(\mathcal{C}(\mathcal{P},V)\) is determined to be \((m-t)(r+\delta-1)\) and \(\mathcal{C}(\mathcal{P},V)\) is an optimal \((r,\delta)\)-LRC. ◻

Proof. It suffices to prove the case \(r=2h-2\) and \(\delta=3\), from which the case \(r=2\) and \(\delta=2h-1\) can be deduced by Remark 4 (iii).

Let \(r=2h-2\) and \(\delta=3\). In the following, we select rational places \([P_1,\dots,P_{r+2}]\), \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) of \(E\) that satisfy the conditions of Proposition 1 (including the equivalent condition ?? ). To this end, we first consider the number of rational places of \(F\) that split completely in \(E/F\). We claim that this number is equal to \[\begin{align} \label{eq:estima951} \ell+1=\frac{N(E)-h}{2h}. \end{align}\tag{15}\] Indeed, the following two statements hold.

(1) for any \(P\in H\), \(P\cap F=O\cap F\) does not split completely in \(E/F\);

(2) for any \(P\in \mathbb{P}_E^1\backslash H\), \(P\cap F\) splits completely in \(E/F\).

To prove the statement (1) where \(P\in H\), it suffices to observe that \(\mathrm{Con}_{E/F}(O\cap F)=\sum_{R\in H}2R\) by Lemma 1. Consequently, \(e(P|P\cap F)=e(P|O\cap F)=2\), and thus the statement (1) holds.

To prove the statement (2) where \(P\in \mathbb{P}_E^1\backslash H\), we assume towards a contradiction that \(P\cap F\) does not split completely in \(E/F\). By Remark 1 (i), this implies that \(\sigma_1(P)=\sigma_2(P)\) for some \(\sigma_1\neq \sigma_2\in G=T_H\langle [-1] \rangle\), which further implies \[\begin{align} \label{eq:20250520951932} P=\sigma_1^{-1}\sigma_2(P). \end{align}\tag{16}\] Let \(\sigma_1^{-1}\sigma_2 = \tau_Q [-1]^i\) for some \(Q \in H\) and \(i \in \{0,1\}\). If \(i = 0\), then 16 implies that \(Q=O\), which would mean \(\sigma_1^{-1}\sigma_2= \mathrm{id}\), contradicting with \(\sigma_1\neq \sigma_2\). Thus, \(i=1\), and it follows that \[P=\sigma_1^{-1}\sigma_2(P)=\tau_Q [-1](P)=[-1]P\oplus Q.\] This implies \([2]P=Q\in H\), and then \(P=[N(E)+1]P=\ebrackets*{\frac{N(E)+1}{2}}([2]P)=\ebrackets*{\frac{N(E)+1}{2}}Q\in H\), which contradicts \(P\in \mathbb{P}_E^1\backslash H\). Therefore, the statement (2) holds, hence the number of rational places of \(F\) that split completely in \(E/F\) is exactly \(\ell+1=\frac{N(E)-h}{2h}.\) We denote the sets of all rational places of \(E\) that lie over these \(\ell+1\) rational places of \(F\) by \(\{P_1,\dots,P_{r+2}\}, \{P_{1,1},\dots,P_{1,r+2}\},\dots, \{P_{\ell,1},\dots,P_{\ell,r+2}\}\), respectively.

We claim that there exists a reorder of \(P_1,\dots,P_{r+2}\) such that \([P_1,\dots,P_{r+2}], \{P_{1,1},\dots,P_{1,r+2}\},\dots,\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) satisfy the condition ?? , i.e., \[\begin{align} \label{eq:tempcondition} P_{r+1}\oplus P_{r+2}\notin \bigcup_{i=1}^{\ell} \ebraces*{P_{i,u_1}\oplus P_{i,u_2}:\; 1\leq u_1< u_{2}\leq r+2}. \end{align}\tag{17}\] The proof of this claim is as follows. Since \(E/F=E/E^{G}\) is a Galois extension with \(\mathrm{Gal}(E/F)=G\), the group \(G=T_{H} \langle[-1]\rangle\) acts transitively on each of \(\{P_1,\dots,P_{r+2}\}, \{P_{1,1},\dots,P_{1,r+2}\}, \dots,\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) by Lemma 1 (i). Since the order of \(G\) is equal to \(r+2\), which is equal to the cardinality of these sets, these pairwise disjoint sets can be represented as \[\label{eq:202506120017} \begin{cases} \{P_1,\dots,P_{r+2}\}&=\{\sigma(P_1):\; \sigma\in G\}=\eparentheses*{P_1\oplus H}\sqcup \eparentheses*{[-1]P_1\oplus H},\\ \{P_{1,1},\dots,P_{1,r+2}\}&=\{\sigma(P_{1,1}):\; \sigma\in G\}=\eparentheses*{P_{1,1}\oplus H}\sqcup \eparentheses*{[-1]P_{1,1}\oplus H},\\ &\dots,\\ \{P_{\ell,1},\dots,P_{\ell,r+2}\}&=\{\sigma(P_{\ell,1}):\;\sigma\in G\}=\eparentheses*{P_{\ell,1}\oplus H}\sqcup \eparentheses*{[-1]P_{\ell,1}\oplus H},\\ \end{cases}\tag{18}\] where \(P_1\oplus H\) denotes the coset \(\{P_1\oplus Q:Q\in H\}\), and \(\sqcup\) denotes the union without intersection. We now prove that \[\eparentheses*{(P_1\oplus H) \oplus (P_1\oplus H)}\cap \eparentheses*{\bigcup_{i=1}^{\ell} \ebraces*{\oplus_{j=1}^{2}P_{i,u_j}: 1\leq u_1< u_{2}\leq r+2}}=\varnothing,\] i.e., by 18 , \[\eparentheses*{[2]P_1\oplus H}\cap \eparentheses*{\bigcup_{i=1}^{\ell}([2]P_{i,1}\oplus H)\cup H\cup ([-2]P_{i,1}\oplus H)}=\varnothing,\] and then we can choose two arbitrary distinct elements in \(P_1\oplus H\subseteq \{P_{1},P_2,\dots,P_{r+2}\}\) to serve as new \(P_{r+1}\), \(P_{r+2}\) such that 17 holds. Assume towards a contradiction that \[([2]P_1\oplus H)\cap \big( ([2]P_{i,1}\oplus H)\cup H\cup ([-2]P_{i,1}\oplus H)\big)\neq \varnothing\] for some \(1\leq i\leq \ell\). Then at least one of \([2]P_1\ominus[2]P_{i,1}, [2]P_1, [2]P_1\ominus[-2]P_{i,1}\) is in \(H\) by the property of cosets. We consider them separately and derive a contradiction in each case.

• \([2]P_1\ominus[2]P_{i,1}\in H\): In this case we have \([2](P_1\ominus P_{i,1})\in H\), which implies that \[P_1\ominus P_{i,1}=[N(E)+1](P_1\ominus P_{i,1})=\ebrackets*{\frac{N(E)+1}{2}}\big([2](P_1\ominus P_{i,1})\big)\in H.\] This leads to a contradiction with the fact that \((P_1\oplus H) \cap (P_{i,1} \oplus H)=\varnothing\) by 18 .

• \([2]P_1\in H\): In this case we have \[P_1=[N(E)+1](P_1)=\ebrackets*{\frac{N(E)+1}{2}}\big([2]P_1)\big)\in H.\] This leads to a contradiction with the fact that \(\{P_1,\dots,P_{r+2}\}=(P_1\oplus H)\sqcup ([-1]P_1\oplus H)\) is a set consisting of \(r+2=2h=2|H|\) distinct rational places by 18 .

• \([2]P_1\ominus[-2]P_{i,1}\in H\): In this case we have \([2](P_1\oplus P_{i,1})\in H\), which implies that \[P_1\oplus P_{i,1}=[N(E)+1](P_1\oplus P_{i,1})=\ebrackets*{\frac{N(E)+1}{2}}\big([2](P_1\oplus P_{i,1})\big)\in H.\] This leads to a contradiction with the fact that \((P_1\oplus H )\cap ([-1]P_{i,1} \oplus H)=\varnothing\) by 18 .

The claim is established, and therefore this theorem is proved. ◻

Proof. By [8], there exists an explicit maximal elliptic function field \(E/\mathbb{F}_q\) with \(N(E)=q+2\sqrt{q}+1\). Since \(2\nmid N(E)\) and there exists a subgroup \(H\leq \mathbb{P}_E^1\) of order \(h\) for any divisor \(h\geq 3\) of \(N(E)\), this corollary holds by Theorem 1. ◻

Proof. By Lemma 4 (i), there exists an elliptic function field \(E/\mathbb{F}_q\) with \(N(E)=q+2\sqrt{q}\). Since \(2\nmid N(E)\) and there exists a subgroup \(H\leq \mathbb{P}_E^1\) of order \(h\) for any divisor \(h\geq 3\) of \(N(E)\), this corollary follows from Theorem 1. ◻

Proof. It suffices to prove the case \(r=ah-2\) and \(\delta=3\), from which the case \(r=2\) and \(\delta=ah-1\) can be deduced by Remark 4 (iii).

Let \(r=ah-2\) and \(\delta=3\). By Lemma 1, we obtain \(\mathrm{Con}_{E/F}(O\cap F)=\sum_{P\in H}aP\). We directly define the list of rational places \([P_1,P_2,\dots,P_{r+2}]\) of \(E\) to be an arbitrary list such that \(\sum_{j=1}^{r+2}P_j=\mathrm{Con}_{E/F}(O\cap F)=\sum_{P\in H}aP\) (i.e., letting \(Q_{\infty}=O\cap F\) in Proposition 1) and \(P_{r+1}=P_{r+2}=O\), which is valid since \(a=|A|\geq 2\). In the following, we divide our discussion into two cases: \(2\nmid aN(E)\) and \(2\mid aN(E)\), and we separately select sets of rational places \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) of \(E\) satisfying the conditions in Proposition 1, including the condition ?? , which now takes the form: \[\begin{align} \label{eq:tempcondition2} P_{r+1}\oplus P_{r+2}=O\oplus O=O\notin \bigcup_{i=1}^{\ell} \ebraces*{P_{i,u_1}\oplus P_{i,u_2}:\; 1\leq u_1 < u_{2}\leq r+2}. \end{align}\tag{20}\]

(1) When \(2\nmid aN(E)\), we start to select sets of rational places \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) of \(E\) that satisfy the condition 20 . To this end, we first estimate the number of rational places of \(F\) that split completely in \(E/F\). We claim that this number is at least \[\begin{align} \label{eq:estima952} \ell=\eceil*{\frac{N(E)-(ah+3h)/2}{ah}}. \end{align}\tag{21}\] It can be proved as follows. Note that \(F\) is a rational function field by Proposition 1 (i). By the Hurwitz genus formula (see 6 ), we have \[\begin{align} \label{eq:20250522950002} 0=2g(E)-2&=(2g(F)-2)[E:F]+\deg\mathrm{Diff}(E/F)=-2ah+\deg\mathrm{Diff}(E/F). \end{align}\tag{22}\] By Lemma 1, we have \(\mathrm{Con}_{E/F}(O\cap F)=\sum_{P \in H}aP\). Let \(R_1,\dots,R_v\) denote all the distinct rational places of \(E\) outside \(H\) that are ramified in \(E/F\). Then we have \(e(R_i|R_i\cap F)\geq 3\) for each \(1\leq i\leq v\). This is because \(e(R_i|R_i\cap F)\mid [E:F]=|G|=ah\) by Lemma 1 (iii), and \(2\nmid ah\) by \(2\nmid aN(E)\). By Dedekind’s different theorem (see 5 ), we have \(\deg\mathrm{Diff}(E/F)\geq \deg\left(\sum_{P\in H}(a-1)P+\sum_{i=1}^{v}(e(R_i|R_i\cap F)-1)R_i \right)\geq (a-1)h+2v.\) Hence, it holds that \(2ah=\deg\mathrm{Diff}(E/F)\geq (a-1)h+2v\) by 22 , which implies \(v\leq \frac{ah+h}{2}\). Including those rational places in \(H\), there are at most \((ah+3h)/2\) rational places of \(E\) that are ramified in \(E/F\). Therefore, by Remark 1 (ii), there exist at least \(\ell=\eceil*{\frac{N(E)-(ah+3h)/2}{ah}}\) distinct rational places of \(F\) that split completely in \(E/F\) into \(\{P_{1,1},\dots,P_{1,r+2}\}, \dots, \{P_{\ell,1},\dots,P_{\ell,r+2}\}\subseteq \mathbb{P}_E^1\), respectively.

We claim that these \(\ell\) sets of rational places \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) satisfy 20 . To prove this claim, we only need to prove that for all \(1\leq i\leq \ell\) and \(1\leq u_1<u_2\leq r+2\), \[\begin{align} P_{i,u_1}\oplus P_{i,u_2}\neq O. \end{align}\] Assume towards a contradiction that \(P_{i,u_1}\oplus P_{i,u_2}=O\) for some \(1\leq i\leq \ell\) and \(1\leq u_1<u_2\leq r+2\). Then we have \(P_{i,u_2}=[-1]P_{i,u_1}\). Since \(|G|=ah=r+2\) and \(G\) acts transitively on \(\{P_{i,1},\dots,P_{i,r+2}\}\) by Lemma 1 (i), we have \[\begin{align} \label{eq:negativeeq} \{P_{i,1},\dots,P_{i,r+2}\}=\{\sigma([-1]P_{i,u_1}):\; \sigma\in G\}=\{[-1](\sigma(P_{i,u_1})):\; \sigma\in G\}=\{[-1]P_{i,1},\dots,[-1]P_{i,r+2}\}, \end{align}\tag{23}\] where the second “\(=\)” is due to \[\begin{align} \{\sigma[-1]:\; \sigma\in G\} \nonumber &=\{(\tau_P\alpha)[-1]:\; \tau_P \in T_{H}, \alpha \in A\}\\\nonumber &\xlongequal{\text{by Remark~\ref{rem:invo95of95EFF95and95isogeny95is95endo} {\rm (ii)}}}\{\tau_P([-1]\alpha):\; P\in H, \alpha \in A\}\\\nonumber &\xlongequal{\text{by Lemma~\ref{lem:groupstructure95of95Aut95of95EFF}}}\{[-1]\tau_{[-1]P}\alpha:\; P\in H, \alpha \in A\}\\\nonumber &=\{[-1]\tau_{P}\alpha:\; P\in H, \alpha \in A\}\\\nonumber &=\{[-1]\tau_{P}\alpha:\; \tau_P\in T_{H}, \alpha \in A\}\\ \label{eq:<[-1]>commutewithgroup} &=\{[-1]\sigma:\; \sigma\in G\}. \end{align}\tag{24}\] Note that \(2\nmid aN(E)\) implies \(2\nmid ah=r+2\). By 23 , there exists a rational place \(P_{i,j_0}\) for some \(1\leq j_0\leq r+2\) such that \(P_{i,j_0}=[-1]P_{i,j_0}\), i.e. \([2]P_{i,j_0}=O\), which implies \(P_{i,j_0}=[N(E)+1]P_{i,j_0}=[\frac{N(E)+1}{2}][2]P_{i,j_0}=[\frac{N(E)+1}{2}]O=O\). This leads to a contradiction since \(e(P_{i,j_0}|P_{i,j_0}\cap F)=1\) while \(e(O|O\cap F)>1\). Thus, these \(\ell\) sets of rational places \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) satisfy the condition 20 .

(2) When \(2\mid aN(E)\), we start to select sets of rational places \(\{P_{1,1},\dots,P_{1,r+2}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) of \(E\) that satisfy the condition 20 . We briefly outline our method before proceeding. We first define \(G':=G\langle[-1]\rangle\), a subgroup of order \(2ah=2(r+2)\) of \(\mathrm{Aut}(E/\mathbb{F}_q)\). We then consider the orbits of rational places of \(E\) under the action of \(G'\) that have length \(2(r+2)\). Every such orbit is divided into two orbits under the action of \(G\) that have length \(r+2\). These \(G\)-orbits of length \((r+2)\) are finally selected as \(\{P_{1,1},\dots,P_{1,r+2}\}\),…,\(\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) that satisfy the condition 20 .

Proceeding to the proof. Let \(G':=G\langle[-1]\rangle\). It is a subgroup of \(\mathrm{Aut}(E/\mathbb{F}_q)\) of order \(2(r+2)\) since \(\langle[-1]\rangle\cap G=\{\mathrm{id}\}\) and \(\langle[-1]\rangle G=G\langle[-1]\rangle\) by 24 . Let \(E^{G'}:=\{u \in E:\; \sigma (u)=u \text{ for all }\sigma \in G'\}\). The field \(E^{G'}\) is a subfield of the rational function field \(F=E^{G}\) (see Proposition 1 (i)), and thus is also a rational function field by Lüroth’s Theorem (see [33]). By the Hurwitz genus formula (see 6 ), we have \[\begin{align} \label{eq:20250522951016} 0=2g(E)-2=(2g(E^{G'})-2)[E:E^{G'}]+\deg\mathrm{Diff}(E/E^{G'})=-2(2ah)+\deg\mathrm{Diff}(E/E^{G'}). \end{align}\tag{25}\] By Lemma 1, we have \(\mathrm{Con}_{E/E^{G'}}(O\cap E^{G'})=\sum_{P \in H}2aP\). Let \(R_1,\dots,R_v\) denote all the distinct rational places of \(E\) outside \(H\) that are ramified in \(E/E^{G'}\). Then we have \(\deg\mathrm{Diff}(E/E^{G'})\geq \deg(\sum_{P\in H}(2a-1)P+\sum_{i=1}^{v}(2-1)R_i)\geq (2a-1)h+v\) by Dedekind’s different theorem (see 5 ). Thus, it holds \(4ah=\deg\mathrm{Diff}(E/E^{G'})\geq (2a-1)h+v\) by 25 , which implies \(v\leq 2ah+h\). Including those rational places in \(H\), there are at most \(2ah+2h\) rational places of \(E\) that are ramified in \(E/E^{G'}\). Hence, by Remark 1 (ii), there exist at least \(l=\lceil\frac{N(E)-2ah-2h}{2ah}\rceil\) distinct rational places of \(E^{G'}\) that split completely in \(E/E^{G'}\) into \(\{T_{1,1},\dots,T_{1,2(r+2)}\}, \dots, \{T_{l,1},\dots,T_{l,2(r+2)}\}\subseteq \mathbb{P}_E^1\), respectively. Note that \(|G'|=2(r+2)\) and \(G'=G\langle[-1]\rangle=\langle[-1]\rangle G.\) Since \(G'\) acts transitively on \(\{T_{i,1},\dots,T_{i,2(r+2)}\}\) for each \(1\leq i\leq l\) by Lemma 1 (i), we have \[\begin{align} \{T_{i,1},\dots,T_{i,2(r+2)}\}=\{\sigma(T_{i,1}):\; \sigma\in G'\} &=\{\sigma(T_{i,1}):\; \sigma\in G\}\sqcup \{[-1](\sigma(T_{i,1})):\; \sigma\in G\}\tag{26}\\ &=\{\sigma(T_{i,1}):\; \sigma\in G\}\sqcup \{\sigma([-1]T_{i,1}):\; \sigma\in G\}\tag{27}\\ &\xlongequal{\text{denote by}} \{P_{2i-1,1},\dots,P_{2i-1,r+2}\}\sqcup \{P_{2i,1},\dots,P_{2i,r+2}\} \tag{28}\\ &= \{P_{2i-1,1},\dots,P_{2i-1,r+2}\}\sqcup \{[-1]P_{2i-1,1},\dots,[-1]P_{2i-1,r+2}\},\tag{29} \end{align}\] where “\(\sqcup\)” refers to the union without intersection, and 29 follows from 26 . By 27 and 28 , we get \(\ell=2l=2\lceil\frac{N(E)-2ah-2h}{2ah}\rceil\) pairwise disjoint \(G\)-orbits: \(\{P_{1,1},\dots,P_{1,r+2}\},\dots,\{P_{\ell,1},\dots,P_{\ell,r+2}\}\), that lie over \(\ell\) distinct rational places of \(F\) other than \(Q_{\infty}=O\cap F\), respectively. Moreover, \(\{P_{1,1},\dots,P_{1,r+2}\},\dots,\{P_{\ell,1},\dots,P_{\ell,r+2}\}\) do satisfy the condition 20 ; otherwise, for some \(1\leq i\leq l\), the union in 28 would not be a disjoint union by 29 . The proof is complete. ◻

Proof. If \(E/\mathbb{F}_q\) is a maximal elliptic function field, then the group structure of \(\mathbb{P}_E^1\) is \[\label{eq:202507011500} \mathbb{P}_E^1\cong \mathbb{Z}/(\sqrt{q}+1)\mathbb{Z}\times \mathbb{Z}/(\sqrt{q}+1)\mathbb{Z}\tag{30}\] by Lemma 5. Therefore, for any positive divisor \(h_0\) of \(\sqrt{q}+1\), there exists a subgroup of \(\mathbb{P}_E^1\) of order \(h=h_0^2\) defined by \(H:=\ebraces*{P\in \mathbb{P}_E^1:\; \ebrackets*{h_0}P=O}\), which corresponds via the isomorphism in 30 to the subgroup \(<\overline{{(\sqrt{q}+1)}/{h_0}}>\times <\overline{{(\sqrt{q}+1)}/{h_0}}>\) of \(\mathbb{Z}/(\sqrt{q}+1)\mathbb{Z}\times \mathbb{Z}/(\sqrt{q}+1)\mathbb{Z}\). Let \(A\) be a subgroup of \(\mathrm{Aut}(E, O)\) of order \(a\). Then for any \(\sigma\in A\) and \(P\in H=\ebraces*{P\in \mathbb{P}_E^1:\; \ebrackets*{h_0}P=O}\), we have \([h_0](\sigma(P))=\sigma([h_0]P)=\sigma(O)=O\), where the first “\(=\)” follows from Remark 2 (ii). Hence \(\sigma(P)\in H\). By Lemma 9, \(G:=T_HA\) must be a subgroup of order \(ah\) of \(\mathrm{Aut}(E/\mathbb{F}_q)\). We now can prove (i), (ii), and (iii).

(i) Let \(q = 2^{2s}\) for a positive integer \(s\). There exists an explicit maximal elliptic function field \(E/\mathbb{F}_q\) with \(|\mathrm{Aut}(E, O)| = 24\), as shown in [8]. Let \(A\) be a subgroup of \(\mathrm{Aut}(E, O)\) of order \(a = 3\), which must not contain the elliptic involution \([-1]\) since \([-1]\) is of order \(2\). Then there exists a subgroup \(T_HA\) of \(\mathrm{Aut}(E/\mathbb{F}_q)\) of order \(3h\) by the above preceding discussion. Note that \(2 \nmid aN(E) = 3(2^{2s} + 2 \cdot 2^s + 1)\). By Theorem 2, the proof of (i) is completed.

(ii) Let \(q = 3^{2s}\) for a positive integer \(s\). There exists an explicit maximal elliptic function field \(E/\mathbb{F}_q\) with \(|\mathrm{Aut}(E, O)| = 12\), as shown in [8]. An argument analogous to the proof of item (i) yields item (ii). The only difference is that, in this case, \(2 \mid aN(E)=3(3^{2s}+2 \cdot 3^s+1)\), so we must employ the second estimate for \(\ell\) in 19 .

(iii) Let \(q = p^{2s}\) for an odd prime \(p \equiv 2 \pmod{3}\) and a positive integer \(s\). There exists an explicit maximal elliptic function field \(E/\mathbb{F}_q\) with \(|\mathrm{Aut}(E, O)| = 6\), as shown in [8]. The rest of the proof is the same as above. ◻

Proof. (i) Note that \(F\) is a rational function field. Let \(z\in \mathcal{L}_F(Q_{\infty})\backslash\mathbb{F}_q\). We have \((z)_{\infty}^F=Q_{\infty}\), \([F:\mathbb{F}_q(z)]=\deg((z)_{\infty}^F)=\deg(Q_{\infty})=1\) and \((z)^E_{\infty}=\mathrm{Con}_{E/F}((z)_\infty^{F})=\mathrm{Con}_{E/F}(Q_{\infty})=P_{1}+P_2+\dots+P_{r+2}.\)

(ii) Note that \(\mathcal{L}_{E}(P_1)\subseteq \mathcal{L}_{E}(P_1+P_2)\subseteq \dots \subseteq \mathcal{L}_{E}(P_1+\dots+P_{r+1})\), \(\dim_{\mathbb{F}_q}(\mathcal{L}_{E}(P_1))=1\), and \(\dim_{\mathbb{F}_q}(\mathcal{L}_{E}(P_1+\dots+P_{r+1}))=r+1+1-2=r\) by the Riemann-Roch theorem (see 2 ). Furthermore, \(\dim_{\mathbb{F}_q}(\mathcal{L}_E(\sum_{j=1}^{I+1}P_j))-\dim_{\mathbb{F}_q}(\mathcal{L}_E(\sum_{j=1}^{I}P_j))\leq 1\) for each \(1\leq I\leq r\) by [33]. Therefore, among the \(r\) sets \(V_I:=\mathcal{L}_{E}(\sum_{j=1}^{I+1}P_j)\backslash \mathcal{L}_{E}(\sum_{j=1}^{I}P_j)\) for \(1\leq I\leq r\), exactly \(r-1\) of them are non-empty. Let \(1\leq {I}_1<\dots<{I}_{r-1}\leq r\) be the \(r-1\) indexes such that \(V_{{I}_1},\dots,V_{{I}_{r-1}}\) are non-empty. Let \(w_0=1\), and let \(w_1,\dots,w_{r-1}\) be arbitrary elements in \(V_{{I}_1},\dots,V_{{I}_{r-1}}\), respectively.

Now we show that \(w_0,w_1,\dots,w_{r-1}\) are \(F\)-linearly independent. Assume towards a contradiction that there exist rational functions \({f}_{0}(z),\dots,{f}_{r-1}(z)\in F=\mathbb{F}_q(z)\), not all zero, such that \(\sum_{i=0}^{r-1}{f}_i(z)w_i=0\). By clearing denominators, we may assume that \({f}_{0}(z),\dots,{f}_{r-1}(z)\) are polynomials of \(z\). We define \(t:=\max\{0\leq i\leq r-1:\;\deg({f}_i(z))=\max\{\deg({f}_j(z)):\; 0\leq j\leq r-1\}\}\) (\(\deg(0):=-\infty\)). It holds that \({f}_t(z)\neq 0\). Then we consider the following two possible cases to derive a contradiction. Before proceeding, we denote \(e:=e(P_1|Q_{\infty})=\dots=e(P_{r+2}|Q_{\infty})\), and observe that \(v_{P_1}(z)=\dots=v_{P_{r+2}}(z)=-e\).

• If \(t=0\), then we have \(\deg({f}_{0}(z))>\deg({f}_i(z))\) for any \(1\leq i\leq r-1\). Therefore, we have \[v_{P_{r+2}}({f}_{0}(z)w_0)=-e\deg({f}_{0}(z))<-e\deg({f}_i(z))-(e-1)\leq -e\deg({f}_i(z))+v_{P_{r+2}}(w_i)=v_{P_{r+2}}({f}_i(z)w_i)\] for any \(1\leq i\leq r-1\) such that \({f}_i(z)\neq 0\), where the symbol “\(\leq\)” in the above inequality is due to \(w_i\in \mathcal{L}_{E}(P_1+\dots+P_{r+2}-P_{r+2})=\mathcal{L}_{E}(\sum_{P|Q_{\infty}}eP-P_{r+2})\). This implies \(v_{P_{r+2}}(\sum_{i=0}^{r-1}{f}_i(z)w_i)=v_{P_{r+2}}({f}_{0}(z)w_0)\neq \infty\).

• If \(1\leq t\leq r-1\), then we have \(\deg({f}_t(z))\geq \deg({f}_i(z))\) for any \(0\leq i\leq t-1\); and \(\deg({f}_t(z))>\deg({f}_i(z))\) for any \(t+1\leq i\leq r-1\). Therefore, for any \(0\leq i\leq t-1\) such that \({f}_i(z)\neq 0\), we have \[\begin{align} \label{eq:202507012157} v_{P_{{I}_t+1}}({f}_t(z)w_t)=-e\deg({f}_t(z))+v_{P_{{I}_t+1}}(w_t)<-e\deg({f}_i(z))+v_{P_{{I}_t+1}}(w_i)= v_{P_{{I}_t+1}}({f}_i(z)w_i) \end{align}\tag{31}\] where “\(<\)” is due to \(\deg({f}_t(z))\geq \deg({f}_i(z))\) and \(v_{P_{{I}_t+1}}(w_t)<v_{P_{{I}_t+1}}(w_i)\) since \(v_{P_{{I}_t+1}}(w_t)\leq -v_{P_{{I}_t+1}}(\sum_{j=1}^{{I}_t+1}P_j)\),⁸ \(w_i\in \mathcal{L}_{E}(\sum_{j=1}^{{I}_t+1}P_j-P_{{I}_t+1})\); and for any \(t+1\leq i\leq r-1\) such that \({f}_i(z)\neq 0\), we have \[\begin{align} \label{eq:202507012158} v_{P_{{I}_t+1}}({f}_t(z)w_t)=-e\deg({f}_t(z))+v_{P_{{I}_t+1}}(w_t)<-e\deg({f}_i(z))+v_{P_{{I}_t+1}}(w_i)=v_{P_{{I}_t+1}}({f}_i(z)w_i) \end{align}\tag{32}\] where the symbol “\(<\)” is due to \(\deg({f}_t(z))> \deg({f}_i(z))\), \(v_{P_{{I}_t+1}}(w_t)\leq -1\) and \(v_{P_{{I}_t+1}}(w_i)\geq -e\) since \(v_{P_{{I}_t+1}}(w_t)\leq -v_{P_{{I}_t+1}}(\sum_{j=1}^{{I}_t+1}P_j)\), \(w_i\in \mathcal{L}_{E}(\sum_{j=1}^{r+2}P_j)=\mathcal{L}_{E}(\sum_{P|Q_{\infty}}eP)\). Inequalities 31  and 32 imply \(v_{P_{{I}_t+1}}(\sum_{i=0}^{r-1}{f}_i(z)w_i)=v_{P_{{I}_t+1}}({f}_t(z)w_t)\neq \infty\).

In both cases, we arrive at a contradiction to the assumption \(\sum_{i=0}^{r-1}{f}_i(z)w_i=0\). Therefore, \(w_0,w_1,\dots,w_{r-1}\) are \(F\)-linearly independent.

(iii) Assume towards a contradiction that the submatrix consisting of the \(u_1,\dots,u_r\)-th columns of the matrix \(M_i\) in ?? is singular for some \(1\leq i\leq \ell\) and \(1\leq u_1<\dots<u_r\leq r+2\). Then there exist \(c_0,\dots,c_{r-1}\in\mathbb{F}_q\), not all zero, such that \((c_0,\dots,c_{r-1})M_{i}\) vanishes at the positions \(u_1,\dots,u_r\). That is, the function \(w:=c_0w_0+\dots+c_{r-1}w_{r-1}\) has zeros \(P_{i,u_1},\dots,P_{i,u_r}\). Note that \(w \in\mathcal{L}_E(\sum_{j=1}^{r+1}P_j)\). The principal divisor of \(w\) must be of the form \[\begin{align} \label{eq:temp2} \eparentheses*{w}^E=P-\eparentheses*{\sum_{j=1}^{r+1}P_{j}}+\sum_{j=1}^{r}P_{i,u_j}, \end{align}\tag{33}\] where \(P\in \mathbb{P}_E^1\) is an unknown rational place that will be discussed later. Note that \(\mathcal{L}_{F}(Q_{\infty}-Q_{i})\neq \{0\}\) since \(F\) is a rational function field. Let \(z'\) be a nonzero element in \(\mathcal{L}_F({Q_\infty}-Q_{i})\), then we have \[\begin{align} \label{eq:temp5} (z')^E=\mathrm{Con}_{E/F}((z')^F)=\mathrm{Con}_{E/F}(-Q_{\infty}+Q_{i})=(-P_1-\dots-P_{r+2})+(P_{i,1}+\dots+P_{i,r+2}). \end{align}\tag{34}\] By 33 and 34 , we have \[\begin{align} \label{eq:thetransformationofdivisor3} \eparentheses*{\frac{w}{z'}}^E=P+P_{r+2}-\sum_{u\in [r+2]\backslash\{u_1,\dots,u_r\}}P_{i,u}. \end{align}\tag{35}\] We now consider all possible \(P\in \mathbb{P}_E^1\), divided into two cases.

• \(P=P_{i,u'}\) for some \(u'\in [r+2]\backslash\{u_1,\dots,u_r\}\). By 35 , we have \[\eparentheses*{\frac{w}{z'}}^E=P_{r+2}-\sum_{u\in [r+2]\backslash\{u_1,\dots,u_r,u'\}}P_{i,u}.\] Then we have \(\ebrackets*{E:\mathbb{F}_q\eparentheses*{\frac{w}{z'}}}=\deg\eparentheses*{\eparentheses*{\frac{w}{z'}}^E_{0}}=\deg(P_{r+2})=1\). This contradicts the fact that \(E\) is a hyperelliptic function field rather than a rational function field.

• \(P\) is a rational place with \(P\neq P_{i,u'}\) for any \(u'\in [r+2]\backslash\{u_1,\dots,u_r\}\). By 35 , we have \[\eparentheses*{\frac{w}{z'}}^E=P+P_{r+2}-\sum_{u\in [r+2]\backslash\{u_1,\dots,u_r\}}P_{i,u}.\] The above equation leads to \(P+P_{r+2}\sim\sum_{u\in [r+2]\backslash\{u_1,\dots,u_r\}}P_{i,u}\), which is ridiculous by Lemma 10, along with conditions \((C1)\) and \((C2)\).

Both cases lead to a contradiction, thereby completing the proof of (iii). ◻

Proof. The proof is similar to that of Proposition 2. So we omit it. ◻

Proof. Since \(|G| \nmid N(E)\), there exists a rational place \(P'\) of \(E\) that is ramified in \(E/F\); otherwise, all rational places of \(E\) would be unramified in \(E/F\), and it would follow that \(|G|=[E:F] \mid N(E)\) by Lemma 1 (ii) and (iii). By Lemma 1 (ii), we have \(e(P|P'\cap F)-1>0\) for each \(P\in \mathbb{P}_E\) lying over \(P'\cap F\), and thus \[\deg\eparentheses*{\sum_{P|P'\cap F}\eparentheses*{e(P|P'\cap F)-1}P} \geq \frac{1}{2}\deg\eparentheses*{\sum_{P|P'\cap F}e(P|P'\cap F)P} =\frac{1}{2}\deg\eparentheses*{\mathrm{Con}_{E/F}(P'\cap F)} =\frac{1}{2}|G| \geq \frac{5}{2} >2,\] which implies \(\deg\mathrm{Diff}(E/F)>2\) by Dedekind’s different theorem (see 5 ). Thus, \(F\) is a rational function field by Lemma 2.

Let \([P_1,\dots,P_{r+2}]\) be a list of rational places of \(E\) such that \(P_{r+2}{=}P_{\infty}\) and \(\sum_{j=1}^{r+2}P_j{=}\mathrm{Con}_{E/F}(P_{\infty}\cap F)\) (i.e., letting \(Q_{\infty}=P_{\infty}\cap F\) in Proposition 3), then the condition \((C1)\) is satisfied. As for the selection of \(\{P_{1,1},\dots,P_{1,r+2}\},\dots,\{P_{\ell,1},\dots, P_{\ell,r+2}\}\) satifying \((C2)\), we use a similar method as that used in the proof of Theorem 2 for the case \(2\mid aN(E)\). Recall that the hyperelliptic involution \(\iota\) is of order \(2\) and commutes with all elements of \(\mathrm{Aut}(E/\mathbb{F}_q)\) (see Remark 3). Let \(G':=<\iota>G=\{\sigma:\sigma \in G\}\cup\{\iota\sigma:\sigma \in G\}\) be a new larger subgroup of order \(2|G|=2(r+2)\) of \(\mathrm{Aut}(E/\mathbb{F}_q)\). We now consider the function field extension \(E/E^{G'}\). Since \(F=E^{G}\) is a rational function field, its subfield \(E^{G'}\) is also a rational function field by Lüroth’s Theorem (see [33]). By the Hurwitz genus formula (see 6 ), we have \[\begin{align} 2=2g(E)-2=(2g(E^{G'})-2)[E:E^{G'}]+\deg\mathrm{Diff}(E/E^{G'})=-4|G|+\deg\mathrm{Diff}(E/E^{G'}). \end{align}\] Thus, by Dedekind’s different theorem (see 5 ), there are at most \(4|G|+2\) rational places of \(E\) that are ramified in \(E/E^{G'}\). Therefore, by Remark 1 (ii), there are at least \(l=\eceil*{\frac{N(E)-4|G|-2}{2|G|}}\) rational places of \(E^{G'}\) that split completely in \(E/E^{G'}\) into \(\{T_{1,1},\dots,T_{1,2(r+2)}\},\dots,\{T_{l,1},\dots, T_{l,2(r+2)}\} \subseteq \mathbb{P}_E^1\), respectively. Since \(|G'|=2(r+2)\) and \(G'\) acts transitively on each of these sets by Lemma 1 (i), for each \(1\leq i\leq l\) we have \[\begin{align} \{T_{i,1},\dots,T_{i,2(r+2)}\}=\{\sigma(T_{i,1}):\; \sigma\in G'\} &=\{\sigma(T_{i,1}):\; \sigma\in G\}\sqcup \{\iota(\sigma( T_{i,1})):\;\sigma\in G\} \tag{36} \\ \tag{37} &=\{\sigma(T_{i,1}):\;\sigma\in G\}\sqcup \{\sigma(\iota (T_{i,1})):\; \sigma\in G\} \\ \tag{38} &\xlongequal{\text{denote by}} \{P_{2i-1,1},\dots,P_{2i-1,r+2}\} \sqcup \{P_{2i,1},\dots,P_{2i,r+2}\} \\\tag{39} &= \{P_{2i-1,1},\dots,P_{2i-1,r+2}\}\sqcup \{\iota(P_{2i-1,1}),\dots,\iota(P_{2i-1,r+2})\}, \end{align}\] where “\(\sqcup\)” denotes the union without intersection, and 39 follows from 36 . By 37 and 38 , we get at least \(\ell{=}2l{-}1=2\eceil*{\frac{N(E)-4|G|-2}{2|G|}}-1\) pairwise disjoint \(G\)-orbits⁹: \(\{P_{1,1},\dots,P_{1,r+2}\},\dots,\{P_{\ell,1},\dots,P_{\ell,r+2}\}\), that lie over \(\ell\) distinct rational places of \(F\) other than \(Q_{\infty}=P_{\infty}\cap F\), respectively. Moreover, \(\{P_{1,1},\dots,P_{1,r+2}\},\) \(\dots,\) \(\{P_{\ell,1}, \dots, P_{\ell,r+2}\}\) do satisfy the condition \((C2)\) in Proposition 3; otherwise, for some \(1\leq i\leq l\), 38 would not be a disjoint union by 39 , leading to a contradiction. Based on the above selection of \(\{P_{1,1},\dots,P_{1,r+2}\}\), \(\{P_{2,1},\dots,P_{2,r+2}\}, \dots, \{P_{\ell,1},\dots,P_{\ell,r+2}\}\), this theorem is proved. ◻

Proof. (i) We consider the hyperelliptic curve defined by \(y^2=x^5+x\) over \(\mathbb{F}_q\), where \(q=5^{2s}\) for an odd positive integer \(s\). It is a maximal hyperelliptic curve with \(q+4\sqrt{q}+1\) rational points by Lemma 13 and Lemma 12. Let \(E/\mathbb{F}_q\) be its function field. It has an automorphism \(\sigma\in \mathrm{Aut}(E/\mathbb{F}_q)\) defined by the associated matrix \(\eparentheses*{\begin{array}{cc} \alpha & -1 \\ -1 & 0 \end{array}}\) (see 11 ), where \(\alpha\in \mathbb{F}_{25}\subseteq \mathbb{F}_q\) satisfies \(\alpha^2=2\). Let \(G:=<\sigma>\leq \mathrm{Aut}(E/\mathbb{F}_q)\). It is direct to verify that \(|G|=6\) and \(\iota \notin G\). Note that \(6\nmid N(E)=q+4\sqrt{q}+1=5^{2s}+4\cdot 5^s+1=(6-1)^{2s}+4\cdot(6-1)^s+1\) since \(2\nmid s\). Then by Theorem 3, item (i) holds.

(ii) We consider the hyperelliptic curve defined over \(\mathbb{F}_q\) by the equation \(y^2=x^5+x\), where \(q={\overline{q}}^{2s}\) for an arbitrary odd prime power \({\overline{q}}\neq 5\) with \({\overline{q}}\equiv 5 \text{ or } 7\pmod{8}\) and an odd positive integer \(s\). It is a maximal hyperelliptic curve by Lemma 13 and Lemma 12. Let \(E/\mathbb{F}_q\) be its function field. By Lemma 15 and some concrete computations, we obtain an automorphism \(\sigma_1\in \mathrm{Aut}(E/\mathbb{F}_q)\) of order \(2\) defined by the associated matrix \(\eparentheses*{\begin{array}{cc} 0& -1\\ -1& 0 \end{array}}\), and an automorphism \(\sigma_2 \in \mathrm{Aut}(E/\mathbb{F}_q)\) of order \(3\) defined by the associated matrix \(\eparentheses*{\begin{array}{cc} 2^{-1}(\alpha^2-1) & 2^{-1}(\alpha-\alpha^3) \\ 2^{-1}(\alpha^3-\alpha)& 2^{-1}(-\alpha^2-1) \end{array}}\) (see 11 ), where \(\alpha=u^{\frac{q-1}{8}}\), with \(u\) a primitive element of \(\mathbb{F}_{q}\). It is worth verifying that \(\sigma_2:(x\mapsto \frac{(\alpha^2-1)x+(\alpha-\alpha^3)}{(\alpha^3-\alpha)x+(-\alpha^2-1)},y\mapsto \frac{y}{2^{-3}((\alpha^3-\alpha)x+(-\alpha^2-1))^3})\) is indeed an automorphism in \(\mathrm{Aut}(E/\mathbb{F}_q)\). To this end, we first check that \((\sigma_2(y))^2=(\sigma_2(x))^5+\sigma_2(x)\), which is equivalent to verifying that \[2^6y^2=\big((\alpha^2-1)x+(\alpha-\alpha^3)\big)^5\big((\alpha^3-\alpha)x+(-\alpha^2-1)\big)+\big((\alpha^2-1)x+(\alpha-\alpha^3)\big)\big((\alpha^3-\alpha)x+(-\alpha^2-1)\big)^5.\] Note that \((1+\alpha^2)^2=2\alpha^2\) since \(\alpha^4=-1\). Multiplying both sides by \((1+\alpha^2)^6\), the above identity to be verified becomes \[(2\alpha^2)^3\cdot 2^6 y^2=(-2x+2\alpha)^5(-2\alpha x-2\alpha^2)+(-2x+2\alpha)(-2\alpha x-2\alpha^2)^5.\] The right hand side equals \((-2x+2\alpha)(-2\alpha x-2\alpha^2)((-2x+2\alpha)^4+\alpha^4(-2 x-2\alpha)^4)=-8\alpha^2\cdot 2^6(x^5+x)\), which is equal to the left hand side. Define \(\sigma_3\) as \(\sigma_3:(x\mapsto \frac{(-\alpha^2-1)x+(\alpha^3-\alpha)}{(\alpha-\alpha^3)x+(\alpha^2-1)},y\mapsto \frac{y}{2^{-3}((\alpha-\alpha^3)x+(\alpha^2-1))^3})\). We can similarly verify that \((\sigma_3(y))^2=(\sigma_3(x))^5+\sigma_3(x)\), and that \(\sigma_3\) is the inverse of \(\sigma_2\). Hence, \(\sigma_2\) is indeed an element of \(\mathrm{Aut}(E/\mathbb{F}_q)\).

Let \(G:=<\sigma_1,\sigma_2>\). It is a dihedral subgroup of \(\mathrm{Aut}(E/\mathbb{F}_q)\) of order \(6\), satisfying the relation \(\sigma_1\sigma_2\sigma_1=\sigma_2^{-1}\), and it does not contain the hyperelliptic involution \(\iota\). To apply Theorem 3, we examine under what conditions \(N(E)=q+4\sqrt{q}+1\) is not divisible by 6.

• \({\overline{q}}\equiv 5 \pmod 8\). In this case, we consider three subcases \({\overline{q}}\equiv 5,13,21 \pmod{ 24}\). In these three subcases, we have \(N(E)=q+4\sqrt{q}+1={\overline{q}}^{2s}+4{\overline{q}}^s+1\equiv -2,0,4 \pmod{6}\), respectively, where \(s\) is an odd positive integer. Thus, when the condition “\({\overline{q}}\equiv 5 \pmod 8\)” is strengthened to “\({\overline{q}}\equiv 5\text{ or }21 \pmod {24}\)”, we have \(6\nmid N(E)\).

• \({\overline{q}}\equiv 7 \pmod{8}\). In this case, we consider three subcases \({\overline{q}}\equiv 7,15,23 \pmod{24}\). In these three subcases, we have \(N(E)=q+4\sqrt{q}+1={\overline{q}}^{2s}+4{\overline{q}}^s+1\equiv 0,4,-2\pmod{6}\), respectively, where \(s\) is an odd positive integer. Thus, when the condition “\({\overline{q}}\equiv 7 \pmod 8\)” is strengthened to “\({\overline{q}}\equiv 15\text{ or } 23 \pmod{24}\)”, we have \(6\nmid N(E)\).

Based on the above discussion, the proof is complete by Theorem 3. ◻

Proof. For both (i) and (ii), we first consider the case where \(s\) is odd. The case where \(s\) is even will be handled using a slightly different twisted curve.

(i) Let \(q=(2g+1)^{2s}\) for an odd positive integer \(s\). We consider the hyperelliptic curve \(\mathfrak{C}\) defined over \(\mathbb{F}_q\) by the equation \(y^2=x^{2g+1}+x\). It is a maximal hyperelliptic curve of genus \(g\) with \(q+2g\sqrt{q}+1\) rational points by Lemma 13 and Lemma 12. Let \(E/\mathbb{F}_q\) be its function field. Since \(\mathbb{F}_{(2g+1)^2}\subseteq \mathbb{F}_q\), the equation \(\mathop{\mathrm{Tr}}_{(2g+1)^2/(2g+1)}(u)=u^{2g+1}+u=0\) has \(2g+1\) distinct roots \(\alpha_1,\dots,\alpha_{2g+1}\in \mathbb{F}_{q}\). Let \(G:=\{\sigma_i:\; 1\leq i\leq 2g+1\}\subseteq \mathrm{Aut}(E/\mathbb{F}_q)\), where \(\sigma_i\) is defined by \(\sigma_i: (x\mapsto x+\alpha_i,\; y\mapsto y)\). Then \(G\) is a subgroup of \(\mathrm{Aut}(E/\mathbb{F}_q)\) of order \(2g+1\). Let \(F:=E^{G}\). For the place at infinity \(P_{\infty}\) of \(E\), we have \(\sigma(P_{\infty})=P_{\infty}\) for any \(\sigma \in G\), and thus \(e(P_{\infty}|Q_{\infty})=2g+1\) by Lemma 1, where \(Q_{\infty}:=P_{\infty}\cap F\). By Dedekind’s different theorem (see 5 ), \(\deg\mathrm{Diff}(E/F)\geq 2g\), which, along with Lemma 2, implies that \(F\) is a rational function field. Take \(z\in \mathcal{L}_F(Q_{\infty})\backslash \mathbb{F}_q\). We define the vector space \(V\) of functions of \(E\) that will be used for evaluation by \[\begin{align} V:=\bigg\{a_{0,t}x^0z^t+\sum_{i=0}^{r-1}\sum_{j=0}^{t-1}a_{i,j}x^iz^j:a_{0,t}\in \mathbb{F}_q \text{ and }a_{i,j}\in \mathbb{F}_q \text{ for } 0\leq i\leq r-1, 0\leq j\leq t-1\bigg\}. \end{align}\]

We now prove that \(V\) is of dimension \(tr+1\). For this, it suffices to show that \(x^0,x^1,\dots,x^{r-1}\) are \(F\)-linearly independent. Assume towards a contradiction that there exist rational functions \({f}_{0}(z),\dots,{f}_{r-1}(z)\in F=\mathbb{F}_q(z)\), not all zero, such that \(\sum_{i=0}^{r-1}{f}_i(z)x^i=0\). By clearing denominators, we may assume that \({f}_{0}(z),\dots,{f}_{r-1}(z)\) are polynomials of \(z\). Note that \(v_{P_{\infty}}(x)=-2\) and \(v_{P_{\infty}}(z)=-(2g+1)\) since \((z)^{E}_{\infty}=\mathrm{Con}_{E/F}((z)^{F}_{\infty})=\mathrm{Con}_{E/F}(Q_{\infty})=(2g+1)P_{\infty}\). Let \(0\leq i_1,i_2\leq r-1\) be two (not necessarily distinct) integers such that \({f}_{i_1}(z)\neq 0,{f}_{i_2}(z)\neq 0\), and \(v_{P_{\infty}}({f}_{i_1}(z)x^{i_1})=v_{P_{\infty}}({f}_{i_2}(z)x^{i_2})\). Then we have \(-(2g+1)\deg({f}_{i_1}(z))-2i_1=v_{P_{\infty}}({f}_{i_1}(z)x^{i_1})=v_{P_{\infty}}({f}_{i_2}(z)x^{i_2})=-(2g+1)\deg({f}_{i_2}(z))-2i_2,\) which implies \(2i_1\equiv 2i_2 \pmod{2g+1}\), and therefore, \(i_1\equiv i_2 \pmod{2g+1}\). Since \(0\leq i_1,i_2\leq r-1=g-g'\leq 2g-1\), we have \(i_1=i_2\). Thus, the valuations \(v_{P_{\infty}}({f}_i(z)x^i)\) (\(0\leq i\leq r-1\)) with \({f}_i(z)\neq 0\) are pairwise distinct, which, together with the strict triangle inequality, leads to a contradiction with the assumption \(\sum_{i=0}^{r-1}{f}_i(z)x^i=0\). Therefore, \(x^0,x^1,\dots,x^{r-1}\) are \(F\)-linearly independent, and then \(\dim_{\mathbb{F}_q}(V)=tr+1\).

We then select the rational places that will be used for evaluation. For any affine rational place \(P_{(\alpha,\beta)}\) of \(E\), the orbit \(G\eparentheses*{P_{(\alpha,\beta)}}:=\{P_{(\alpha+\alpha_{i},\beta)}:\; 1\leq i\leq 2g+1\}\) is a \(G\)-orbit of length \(2g+1\). Since \(E\) has \(q+2g\sqrt{q}\) affine rational places, there are totally \(\ell=\frac{q+2g\sqrt{q}}{2g+1}\) \(G\)-orbits of the form \(\{P_{(\alpha+\alpha_i,\beta)}: 1\leq i\leq 2g+1\}\). We denote all these \(G\)-orbits by \(\{P_{1,1},\dots,P_{1,2g+1}\}\), …, \(\{P_{\ell,1},\dots,P_{\ell,2g+1}\}\). The rational places in these \(\ell\) orbits lie over \(\ell\) distinct rational places of \(F\) other than \(Q_{\infty}=P_{\infty}\cap F\), respectively. Let \(\mathcal{P}:=\{P_{1,1},\dots,P_{1,2g+1},\dots,P_{m,1},\dots,P_{m,2g+1}\}\) be the places for evaluation.

We define a linear code \(\mathcal{C}(\mathcal{P},V)\) by \[\begin{align} \mathcal{C}(\mathcal{P},V):=\{(\phi(P_{1,1}),\dots,\phi(P_{1,2g+1}),\dots,\phi(P_{m,1}),\dots,\phi(P_{m,2g+1})):\; \phi\in V\}. \end{align}\] Note that \(V\subseteq \mathcal{L}_E(t(2g+1)P_{\infty}+\max\{0,2(r-1)-(2g+1)\}P_{\infty})=\mathcal{L}_E(t(2g+1)P_{\infty}+\max\{0,-2g'-1\}P_{\infty})\) since \((x)_{\infty}^E=2P_{\infty}\) and \((z)_{\infty}^E=(2g+1)P_{\infty}\). By Section 2.1, the dimension of \(\mathcal{C}(\mathcal{P},V)\) is \(k=\dim_{\mathbb{F}_q}(V)=tr+1\) and the minimum distance \(d\) of \(\mathcal{C}(\mathcal{P},V)\) satisfies \(d\geq m(2g+1)-(t(2g+1)+\max\{0,-2g'-1\})=(m-t)(2g+1)+\min\{0,2g'+1\}\).

To prove that \(\mathcal{C}(\mathcal{P},V)\) is an \((r=g+1-g',\delta=g+1+g')\)-LRC, it suffices to note that for any \(1\leq i\leq m\), \[\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}=\{(\phi(P_{i,1}),\dots,\phi(P_{i,r+\delta-1})):\phi\in \mathop{\mathrm{span}}_{\mathbb{F}_q}\{1,x,\dots,x^{r-1}\}\},\] and that \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are pairwise distinct, which imply that \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}\) is a Reed-Solomon code with minimum distance \(((r+\delta-1)-(r-1))=\delta\). Invoking the Singleton-type bound 1 , we have \(d\leq (m-t)(2g+1)\). Thus, when \(0\leq g'\leq g-1\), the minimum distance \(d\) is determined to be \((m-t)(2g+1)\) and \(\mathcal{C}(\mathcal{P},V)\) is an optimal \((r=g+1-g',\delta=g+1+g')\)-LRC.

As for the case \(q=(2g+1)^{2s}\) for an even positive integer \(s\), we consider the (twisted) hyperelliptic curve \(\mathfrak{C}'/\mathbb{F}_q\) defined by \(\gamma y^2=x^{2g+1}+x\), where \(\gamma\in \mathbb{F}_q\backslash \{\beta^2:\beta\in \mathbb{F}_q\}\), i.e., a quadratic non-residue in \(\mathbb{F}_q\). Note that the number of distinct roots in \(\mathbb{F}_q\) of \(\gamma y^2=\alpha^{2g+1}+\alpha\) and the number of distinct roots in \(\mathbb{F}_q\) of \(y^2=\alpha^{2g+1}+\alpha\) sum to \(2\) for any \(\alpha\in \mathbb{F}_q\). Thus, the numbers of affine rational points of \(\mathfrak{C}'/\mathbb{F}_q\) and \(\mathfrak{C}/\mathbb{F}_q\) (defined by \(y^2=x^{2g+1}+x\)) sum to \(2q\). By Lemmas 13 and 12, the curve \(\mathfrak{C}/\mathbb{F}_q\) is a minimal curve of genus \(g\), and then \(\mathfrak{C}'/\mathbb{F}_q\) is a maximal hyperelliptic curve of genus \(g\). The rest of the proof is similar to the above argument where \(s\) is odd, using the curve \(\mathfrak{C}'/\mathbb{F}_q\).

(ii) Let \(q={\overline{q}}^{2s}\) for an odd prime power \({\overline{q}}\) satisfying \({\overline{q}}\equiv -1\pmod{2g+1}\) and an odd positive integer \(s\). We consider the hyperelliptic curve defined over \(\mathbb{F}_q\) by \(y^2=x^{2g+1}+1\). It is a maximal hyperelliptic curve of genus \(g\) by Lemma 14 and Lemma 12. Let \(E/\mathbb{F}_q\) be its function field. It has an automorphism \(\sigma\in \mathrm{Aut}(E/\mathbb{F}_q)\) defined as \(\sigma: (x\mapsto u^{\frac{q-1}{2g+1}}x,\; y\mapsto y)\), where \(u\) is a primitive element of \(\mathbb{F}_q\). Let \(G:=<\sigma>\), a cyclic group of order \(2g+1\), and let \(F:=E^{G}\). The rest of the proof is similar to the proof of (i). The only difference worth emphasizing is that the number \(\ell\) of \(G\)-orbits of length \(2g+1\) becomes \(\ell=\frac{q+2g\sqrt{q}-2}{2g+1}\), since there are two affine rational places \(P_{(\alpha,\beta)}\) of \(E\) satisfying \(\alpha=0\), \(P_{(0,1)}\) and \(P_{(0,-1)}\).

As for the case \(q={\overline{q}}^{2s}\) for an odd prime power \({\overline{q}}\) satisfying \({\overline{q}}\equiv -1\pmod{2g+1}\) and an even positive integer \(s\), we consider the hyperelliptic curve defined over \(\mathbb{F}_q\) by \(\gamma y^2=x^{2g+1}+1\), where \(\gamma\) is a quadratic non-residue in \(\mathbb{F}_q\). Arguing similarly to the end of the proof of (i), this is a maximal hyperelliptic curve of genus \(g\). The remainder of the proof is similar to the above (the definitions of \(E\), \(G\), and \(F\) are all the same, so we omit them). The only difference worth emphasizing is that the number \(\ell\) of \(G\)-orbits of length \(2g+1\) becomes \(\ell=\frac{q+2g\sqrt{q}}{2g+1}\), since any affine rational places \(P_{(\alpha,\beta)}\) of \(E\) satisfies \(\alpha\neq 0\).

Note that in all the above four subcases, it holds \(\ell=\efloor*{\frac{q+2g\sqrt{q}}{2g+1}}\). This theorem is proved. ◻

Proof. (i) Let \(\mathcal{P}:=\{P_{1,1},\dots,P_{1,r+\delta-1},\dots,P_{m,1},\dots,P_{m,r+\delta-1}\}\), and \(V:=\{a_{0,t}x^0y^t+\sum_{i=0}^{r-1}\sum_{j=0}^{t-1}a_{i,j}x^iy^j:\; a_{0,t}\in \mathbb{F}_{q}\text{ and }a_{i,j}\in \mathbb{F}_{q} \text{ for } 0\leq i\leq r-1,0\leq j\leq t-1\}\). Define a linear code \(\mathcal{C}(\mathcal{P},V)\) by \[\begin{align} \mathcal{C}(\mathcal{P},V):=\left\{(\phi(P_{1,1}),\dots,\phi(P_{1,r+\delta-1}),\dots,\phi(P_{m,1}),\dots,\phi(P_{m,r+\delta-1})):\; \phi\in V\right\}. \end{align}\]

First, we prove that \(\mathcal{C}(\mathcal{P},V)\) is an \((r=\efloor*{\frac{N-1}{M}}+1-b',\delta=N+1-r)\)-LRC. For this, it suffices to note that for any \(1\leq i\leq m\), \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}=\{(\phi(P_{i,1}),\dots,\phi(P_{i,r+\delta-1})):\; \phi\in \mathop{\mathrm{span}}_{\mathbb{F}_q}\{x^0,x^1,\dots,x^{r-1}\}\},\) and that \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are pairwise distinct, which imply that \(\mathcal{C}(\mathcal{P},V)|_{\{(i-1)(r+\delta-1)+1,\dots,i(r+\delta-1)\}}\) is a Reed-Solomon code with minimum distance \(((r+\delta-1)-(r-1))=\delta\).

Next, we prove that \(V\) has dimension \(tr+1\). To this end, it suffices to show that the \(tr+1\) functions \(x^0y^t,x^iy^j(0\leq i\leq r-1,0\leq j\leq t-1)\), which span \(V\), have pairwise distinct valuations at the place at infinity \(P_{\infty}\). Note that the functions \(x\) and \(y\) each have a unique pole \(P_{\infty}\). We have \(v_{P_{\infty}}(x)=-[E:\mathbb{F}_{q}(x)]=-[\mathbb{F}_q(x,y):\mathbb{F}_{q}(x)]=-M\) and \(v_{P_{\infty}}(y)=-[E:\mathbb{F}_{q}(y)]=-[\mathbb{F}_q(x,y):\mathbb{F}_{q}(y)]=-N\) by [33]. Thus, \(v_{P_{\infty}}(x^iy^j)=-Mi-Nj\). Since \(0< r-1=\efloor*{\frac{N-1}{M}}-b'<\frac{N}{M}\), the functions \(x^0y^t,x^iy^j(0\leq i\leq r-1,0\leq j\leq t-1)\) have pairwise distinct valuations at \(P_{\infty}\). Therefore, these \(tr+1\) functions are \(\mathbb{F}_{q}\)-linearly independent by the strict triangle inequality. Then we have \(\dim_{\mathbb{F}_{q}}(V)=tr+1\).

As a by-product of the above argument, we also obtain \(V\subseteq \mathcal{L}_{E}(tNP_{\infty})\). Hence, \(\mathcal{C}(\mathcal{P},V)\) has dimension \(tr+1\) and minimum distance at least \((m-t)N\) by Section 2.1. By the Singleton-type bound 1 , we conclude that the minimum distance of \(\mathcal{C}(\mathcal{P},V)\) is exactly \((m-t)N\), and \(\mathcal{C}(\mathcal{P},V)\) is an optimal \((r=\efloor*{\frac{N-1}{M}}+1-b',\delta=N+1-r)\)-LRC.

(ii) By interchanging \(x\) and \(y\), and swapping \(M\) with \(N\) in the above proof of (i), we can prove (ii). ◻

Proof. The genus of the superelliptic curve \(\mathfrak{C}\) is \(\frac{(M-1)(N-1)}{2}\) by Lemma 16. We now count the affine rational points of \(\mathfrak{C}/\mathbb{F}_{q}=\mathfrak{C}/\mathbb{F}_{{\overline{q}}^s}\). That is, the number of pairs \((\alpha,\beta)\in \mathbb{F}_{{\overline{q}}^s}^2\) satisfying \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}=\alpha^{{\overline{q}}^{s-c}}+\alpha^{{\overline{q}}^{s-2c}}+\dots+\alpha^{{\overline{q}}^{c}}+\alpha^{{\overline{q}}^0}=\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(\alpha)\).

By the property of the trace map \(\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}\) from \(\mathbb{F}_{{\overline{q}}^s}\) to \(\mathbb{F}_{{\overline{q}}^c}\), for any \(\beta\in \mathbb{F}_{{\overline{q}}^s}\) satisfying \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}\in \mathbb{F}_{{\overline{q}}^c}\), there are exactly \({\overline{q}}^{s-c}\) distinct \(\alpha\in \mathbb{F}_{{\overline{q}}^s}\), such that \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}=\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(\alpha)\). Also, for any \(\beta\in \mathbb{F}_{{\overline{q}}^s}\) satisfying \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}\notin \mathbb{F}_{{\overline{q}}^c}\), there exists no \(\alpha\in \mathbb{F}_{{\overline{q}}^s}\) such that \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}=\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(\alpha)\). Therefore, to determine the number of affine rational points of \(\mathfrak{C}/\mathbb{F}_{{\overline{q}}^s}\), we only need to count the elements \(\beta\in \mathbb{F}_{{\overline{q}}^s}\) satisfying \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}\in \mathbb{F}_{{\overline{q}}^c}\), and then multiply the result by \({\overline{q}}^{s-c}\). We consider the following two cases.

• \(\beta=0\). Clearly, \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}=0\in \mathbb{F}_{{\overline{q}}^c}\).

• \(\beta=u^i\) for a primitive element \(u\) of \(\mathbb{F}_{{\overline{q}}^s}\) and an integer \(0\leq i\leq {\overline{q}}^s-2\). Then \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}\in \mathbb{F}_{{\overline{q}}^c}\) if and only if \((u^{i\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}})^{{\overline{q}}^c-1}=1\), which is equivalent to \(\frac{b}{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})}\mid i\).

Thus, the number of elements \(\beta\in \mathbb{F}_{{\overline{q}}^s}\) satisfying \(\beta^{\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}}\in \mathbb{F}_{{\overline{q}}^c}\) is \(1+({\overline{q}}^s-1)/\frac{b}{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})}=\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})\cdot\frac{({\overline{q}}^s-1)}{b}+1\). Multiplying it by \({\overline{q}}^{s-c}\), together with the unique rational point at infinity (see Lemma 16), we complete the proof. ◻

Proof. (i) \(M<N.\) We use the superelliptic curve \(\mathfrak{C}\) given in Lemma 17, defined over \(\mathbb{F}_{q}=\mathbb{F}_{{\overline{q}}^s}\) by the equation \(y^{M}=x^{N}+x^{{\overline{q}}^{s-2c}}+\dots+x^{{\overline{q}}^c}+x^{{\overline{q}}^0}=\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(x)\). For each affine rational point \((\alpha,\beta)\in \mathfrak{C}(\mathbb{F}_{q})\), we have a set \(\{(\alpha+\alpha_i,\beta):1\leq i\leq N\}\subseteq \mathfrak{C}(\mathbb{F}_{q})\), where \(\alpha_1,\dots,\alpha_{N}\in \mathbb{F}_{q}\) are the \(N={\overline{q}}^{s-c}\) distinct roots of the equation \(\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(z)=0\). These sets are either disjoint or identical. There are totally \(\ell=\frac{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})\cdot \frac{(q-1)q}{b{\overline{q}}^c}+\frac{q}{{\overline{q}}^c}}{N}\) such sets since \(\mathfrak{C}/\mathbb{F}_{q}\) has \(\eparentheses*{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})\cdot \frac{(q-1)q}{b{\overline{q}}^c}+\frac{q}{{\overline{q}}^c}}\) affine rational points by Lemma 17. Let \(E/\mathbb{F}_{q}\) be the function field of \(\mathfrak{C}/\mathbb{F}_{q}\). We convert the affine rational points in each of these \(\ell\) sets into the corresponding affine rational places, and denote the \(\ell\) new sets of affine rational places by \(\{P_{i,1},\dots,P_{i,r+\delta-1}\}\) (\(1\leq i\leq \ell\)). For each \(1\leq i\leq \ell\), the values \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are pairwise distinct, while \(y(P_{i,1}),\dots,y(P_{i,r+\delta-1})\) are all the same. By Proposition 5 (i), the proof is complete.

(ii) \(M>N.\) We also use the superelliptic curve \(\mathfrak{C}\) in Lemma 17, defined over \(\mathbb{F}_{q}=\mathbb{F}_{{\overline{q}}^s}\) by the equation \(y^{M}=x^{N}+x^{{\overline{q}}^{s-2c}}+\dots+x^{{\overline{q}}^c}+x^{{\overline{q}}^0}=\mathop{\mathrm{Tr}}_{{\overline{q}}^s/{\overline{q}}^c}(x)\). For each affine rational point \((\alpha,\beta)\in \mathfrak{C}(\mathbb{F}_{q})\) with \(\beta\neq 0\), we have a set \(\{(\alpha,\beta_i\beta):1\leq i\leq M\}\subseteq \mathfrak{C}(\mathbb{F}_{q})\), where \(\beta_1,\dots,\beta_{M}\in \mathbb{F}_{q}\) are the \(M\) distinct roots of the equation \(z^{M}=1\) (note that \(M=\frac{{\overline{q}}^s-1}{b({\overline{q}}-1)}\mid (q-1)\)). These sets are either disjoint or identical. There are totally \(\ell=\frac{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})\cdot \frac{(q-1)q}{b{\overline{q}}^c}}{M}\) such sets since there are totally \(\eparentheses*{\gcd(b,\frac{{\overline{q}}^c-1}{{\overline{q}}-1})\cdot \frac{(q-1)q}{b{\overline{q}}^c}}\) affine rational points \((\alpha,\beta)\) of \(\mathfrak{C}/\mathbb{F}_{q}\) with \(\beta\neq 0\) by the proof of Lemma 17. Let \(E/\mathbb{F}_{q}\) be the function field of \(\mathfrak{C}/\mathbb{F}_{q}\). We convert the affine rational points in each of these \(\ell\) sets into the corresponding affine rational places, and denote the \(\ell\) new sets of affine rational places by \(\{P_{i,1},\dots,P_{i,r+\delta-1}\}\) (\(1\leq i\leq \ell\)). For each \(1\leq i\leq \ell\), the values \(y(P_{i,1}),\dots,y(P_{i,r+\delta-1})\) are pairwise distinct, while \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are all the same. By Proposition 5 (ii), the proof is complete. ◻

Proof. (i) By Lemma 17 (with the parameters in that lemma set to \(s=2\) and \(c=1\)), \(\mathfrak{C}/\mathbb{F}_{{\overline{q}}^2}\) is a superelliptic curve of genus \(\frac{(\frac{{\overline{q}}+1}{b}-1)({\overline{q}}-1)}{2}\), and has \(\frac{{\overline{q}}^3-{\overline{q}}}{b}+{\overline{q}}+1\) rational points. Thus, \(\mathfrak{C}/\mathbb{F}_{{\overline{q}}^2}\) is a maximal curve since \(\frac{{\overline{q}}^3-{\overline{q}}}{b}+{\overline{q}}+1={\overline{q}}^2+2\frac{(\frac{{\overline{q}}+1}{b}-1)({\overline{q}}-1)}{2}{\overline{q}}+1\). By Lemma 12, \(\mathfrak{C}/\mathbb{F}_{q}\) is also a maximal curve, where \(q={\overline{q}}^{2s}\) with \(2\nmid s\). The item (i) is proved.

(ii) Note that the number of distinct roots in \(\mathbb{F}_q\) of \(\gamma y^{2}=\alpha^{{\overline{q}}}+\alpha\) and the number of distinct roots in \(\mathbb{F}_q\) of \(y^{2}=\alpha^{{\overline{q}}}+\alpha\) sum to \(2\) for any \(\alpha\in \mathbb{F}_q\). The total number of affine rational points of \(\mathfrak{C}'/\mathbb{F}_q\) defined by \(\gamma y^{2}=x^{{\overline{q}}}+x\) and \(\mathfrak{C}/\mathbb{F}_q\) defined by \(y^{2}=x^{{\overline{q}}}+x\) must be \(2q\). Since \(\mathfrak{C}/\mathbb{F}_q\) is minimal by Lemma 12 and the proof of (i), it follows that \(\mathfrak{C}'/\mathbb{F}_q\) is maximal.

For the proof of (iii), we refer to the proof of [8], with the roles of \(x\) and \(y\) swapped. ◻

Proof. To apply Proposition 5, we consistently take \(M=\frac{{\overline{q}}+1}{b}\) and \(N={\overline{q}}\) below, although they do not explicitly appear in the proof. Note that \(\efloor*{\frac{N-1}{M}}=b-1\) when \(b>1\), and \(\efloor*{\frac{M-1}{N}}=1\) when \(b=1\).

(i) \(b>1\). Let \(q={\overline{q}}^{2s}\) for a prime power \({\overline{q}}\) and an odd positive integer \(s\). By Lemma 18 (i), the curve \(\mathfrak{C}/\mathbb{F}_{q}\) defined by \(y^{\frac{{\overline{q}}+1}{b}}=x^{{\overline{q}}}+x\) is a maximal superelliptic curve of genus \(\frac{(\frac{{\overline{q}}+1}{b}-1)({\overline{q}}-1)}{2}\). For each affine rational point \((\alpha,\beta)\in \mathfrak{C}(\mathbb{F}_{q})\), we have a set \(\{(\alpha+\alpha_i,\beta):1\leq i\leq {\overline{q}}\}\subseteq \mathfrak{C}(\mathbb{F}_{q})\), where \(\alpha_1,\dots,\alpha_{{\overline{q}}}\in \mathbb{F}_{q}\) are the \({\overline{q}}\) distinct roots of the equation \(z^{\overline{q}}+z=0\). These sets are either disjoint or identical. There are totally \(\ell=\frac{q+(\frac{{\overline{q}}+1}{b}-1)({\overline{q}}-1)\sqrt{q}}{{\overline{q}}}\) such sets since \(\mathfrak{C}/\mathbb{F}_{q}\) has \(q+(\frac{{\overline{q}}+1}{b}-1)({\overline{q}}-1)\sqrt{q}\) affine rational points. Let \(E/\mathbb{F}_{q}\) be the function field of \(\mathfrak{C}/\mathbb{F}_{q}\). We convert the affine rational points in each of these \(\ell\) sets into the corresponding affine rational places, and denote the \(\ell\) new sets of affine rational places by \(\{P_{i,1},\dots,P_{i,r+\delta-1}\}\) (\(1\leq i\leq \ell\)). For each \(1\leq i\leq \ell\), the values \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are pairwise distinct, while \(y(P_{i,1}),\dots,y(P_{i,r+\delta-1})\) are all the same. By Proposition 5 (i), the proof is complete.

As for the special case where \(q={\overline{q}}^{2s}\) with \(2\nmid {\overline{q}}\), \(b=\frac{{\overline{q}}+1}{2}\), and \(2\mid s\), we use the maximal curve \(\mathfrak{C}'/\mathbb{F}_{q}\) in Lemma 18 (ii) defined by \(\gamma y^{\frac{{\overline{q}}+1}{b}}=\gamma y^2=x^{{\overline{q}}}+x\), where \(\gamma\) is an arbitrary quadratic non-residue in \(\mathbb{F}_q\). The rest of the proof is the same as above.

(ii) \(b=1\). Let \(q={\overline{q}}^{2s}\) for a prime power \({\overline{q}}\) and an odd positive integer \(s\). By Lemma 18 (i), the curve \(\mathfrak{C}/\mathbb{F}_{q}\) defined by \(y^{\frac{{\overline{q}}+1}{b}}=y^{{\overline{q}}+1}=x^{{\overline{q}}}+x\) is a maximal superelliptic curve of genus \(\frac{{\overline{q}}({\overline{q}}-1)}{2}\). For each affine rational point \((\alpha,\beta)\in \mathfrak{C}(\mathbb{F}_{q})\) with \(\beta\neq 0\), we have a set \(\{(\alpha,\beta_i\beta):1\leq i\leq {\overline{q}}+1\}\subseteq \mathfrak{C}(\mathbb{F}_{q})\), where \(\beta_1,\dots,\beta_{{\overline{q}}+1}\in \mathbb{F}_{q}\) are the \({\overline{q}}+1\) distinct roots of the equation \(z^{{\overline{q}}+1}=1\) (note that \(({\overline{q}}+1)\mid (q-1)\)). These sets are either disjoint or identical. There are totally \(\ell=\frac{q+{\overline{q}}({\overline{q}}-1)\sqrt{q}-{\overline{q}}}{{\overline{q}}+1}\) such sets since there are totally \((q+{\overline{q}}({\overline{q}}-1)\sqrt{q}-{\overline{q}})\) affine rational points \((\alpha,\beta)\) of \(\mathfrak{C}/\mathbb{F}_{q}\) with \(\beta\neq 0\). Let \(E/\mathbb{F}_{q}\) be the function field of \(\mathfrak{C}/\mathbb{F}_{q}\). We convert the affine rational points in each of these \(\ell\) sets into the corresponding affine rational places, and denote the \(\ell\) new sets of affine rational places by \(\{P_{i,1},\dots,P_{i,r+\delta-1}\}\) \((1\leq i\leq \ell)\). For each \(1\leq i\leq \ell\), the values \(y(P_{i,1}),\dots,y(P_{i,r+\delta-1})\) are pairwise distinct, while \(x(P_{i,1}),\dots,x(P_{i,r+\delta-1})\) are all the same. Applying Proposition 5 (ii), the proof is complete.

As for the special case where \(q={\overline{q}}^{2s}\) with \({\overline{q}}=2\), \(b=1\), and \(2\mid s\), we use the maximal curve \(\mathfrak{C}''/\mathbb{F}_{q}\) in Lemma 18 (iii) defined by \(y^{\frac{{\overline{q}}+1}{b}}=y^3=x^{2}+x+\eta=x^{{\overline{q}}}+x+\eta\), where \(\eta\) is an arbitrary element in \(\mathbb{F}_q\backslash \{\alpha^2+\alpha:\alpha\in \mathbb{F}_q\}\). The rest of the proof is similar to the above. The only difference is that the number of the local repair groups becomes \(\ell=\frac{q+{\overline{q}}({\overline{q}}-1)\sqrt{q}}{{\overline{q}}+1}\) since there are totally \(q+{\overline{q}}({\overline{q}}-1)\sqrt{q}\) affine rational points \((\alpha,\beta)\) of \(\mathfrak{C}''/\mathbb{F}_{q}\) with \(\beta\neq 0\). In both cases, we have \(\ell=\lfloor\frac{q+{\overline{q}}({\overline{q}}-1)\sqrt{q}}{{\overline{q}}+1}\rfloor\), the proof of (ii) is complete. ◻

Proof. By [49], we have \(k\leq \min_{t\in \mathbb{Z}_{\geq 0}}\{tr+k_{\rm opt}^{(q)}(n-t(r+\delta-1),d)\},\) where \(k_{\rm opt}^{(q)}(n-t(r+\delta-1),d)\) denotes the maximal possible dimension of a linear code with length \(n-t(r+\delta-1)\) and minimum distance \(d\). Letting \(t=\big\lceil \frac{n-(1-\varepsilon)\frac{qd}{q-1}}{r+\delta-1}\big\rceil\), where \(\varepsilon=\frac{1}{q^2}\), we have \(t\geq 0\) since \(\frac{d}{n}\leq \frac{2}{3}\leq \frac{q}{q+1}\) for any prime power \(q\). By the plotkin bound, we have \(k_{\rm opt}^{(q)}(n-t(r+\delta-1),d)\leq k_{\rm opt}^{(q)}((1-\varepsilon)\frac{qd}{q-1},d)\leq \log_q(1/\varepsilon)=2\). Hence, we have \[\begin{align} \label{eq:202506061714} \eceil*{\frac{n-(1-\frac{1}{q^2})\frac{qd}{q-1}}{r+\delta-1}}r+2=tr+2\geq k. \end{align}\tag{41}\]

Moreover, we have a lower bound on \(k\). Let \(n_0=(r+\delta-1)\eceil*{\frac{n}{r+\delta-1}}-n\), \(n'=n+n_0\) and \(d'=d+n_0\). Since \(\mathcal{C}\) is optimal, it holds \(d=n-k+1-(\eceil*{\frac{k}{r}}-1)(\delta-1)\). Then we have \((r+\delta-1)\mid n'\) and \(d'=n'-k+1-(\eceil*{\frac{k}{r}}-1)(\delta-1)\). By [23], we have \[\begin{align} \tag{42} k=n'-d'+1-\frac{n'(\delta-1)}{r+\delta-1}+\eparentheses*{\efloor*{\frac{d'-\delta}{r+\delta-1}}+1}(\delta-1)&\geq n'-d'+1-\frac{\delta-1}{r+\delta-1}(n'-(d'-\delta))\\ \tag{43} &=n-d+1-\frac{\delta-1}{r+\delta-1}(n-d+\delta). \end{align}\] Combining 41 , 42 and 43 , we have \(\eparentheses*{\frac{n-(1-\frac{1}{q^2})\frac{qd}{q-1}}{r+\delta-1}+1}r+2\geq tr+2\geq k\geq n-d+1-\frac{\delta-1}{r+\delta-1}(n-d+\delta).\) Solving this inequality with respect to \(d\), we have \(d\leq \frac{(r+\delta-1)(r+1)+\delta(\delta-1)}{r}q.\) The proof is complete. ◻