In a projective plane over a finite field, complete $(k,n)$-arcs with few characters are rare but interesting objects with several applications to finite geometry and coding theory. Since almost all known examples are large, the construction of small ones, with k close to the order of the plane, is considered a hard problem. A natural candidate to be a small $(k,n)$-arc with few characters is the set $\mathrm{\Omega }(\mathcal{C})$ of the points of a plane curve $\mathcal{C}$ of degree n (containing no linear components) such that some line meets $\mathcal{C}$ transversally in the plane, i.e. in n pairwise distinct points. Let $\mathcal{C}$ be either the Hermitian curve of degree $q+1$ in $\mathrm{PG}(2,q^{2r})$ with $r\ge 1$, or the rational BKS curve of degree $q+1$ in $\mathrm{PG}(2,q^r)$ with q odd and $r\ge 1$. Then $\mathrm{\Omega }(\mathcal{C})$ has four and seven characters, respectively. Furthermore, $\mathrm{\Omega }(\mathcal{C})$ is small as both curves are either maximal or minimal. The completeness problem is investigated by an algebraic approach based on Galois theory and on the Hasse-Weil lower bound. Our main result for the Hermitian case is that $\mathrm{\Omega }(\mathcal{C})$ is complete for $r\ge 4$. For the rational BKS curve, $\mathrm{\Omega }(\mathcal{C})$ is complete if and only if r is even. If r is odd then the uncovered points by the $(q+1)$-secants to $\mathrm{\Omega }(\mathcal{C})$ are exactly the points in $\mathrm{PG}(2,q)$ not lying in $\mathrm{\Omega }(\mathcal{C})$. Adding those points to $\mathrm{\Omega }(\mathcal{C})$ produces a complete $(k,q+1)$-arc in $\mathrm{PG}(2,q^r)$, with $k=q^r+q$. The above results do not hold true for $r=2$ and there remain open the case $r=3$ for the Hermitian curve, and the cases $r=3,4$ for the rational BKS curve. As a by product we also obtain two results of interest in the study of the Galois inverse problem for $\mathrm{\text{PGL}}(2,q)$.

在有限域上的射影平面中，具有很少字符的完整 $(k,n)$ 弧是罕见但有趣的对象，在有限几何和编码理论中具有多种应用。由于几乎所有已知的例子都很大，因此构建 k 接近平面阶的小例子被认为是一个难题。具有很少字符的小 $(k,n)$ 弧的自然候选者是 n 阶平面曲线 $\mathcal{C}$ 的点的集合 $\mathrm{\Omega }(\mathcal{C})$ （不包含线性分量），使得某些线在平面上横向相交 $\mathcal{C}$ ，即在 n 两对不同的点上相交。令 $\mathcal{C}$ 为 $\mathrm{PG}(2,q^{2r})$ 中 $r\ge 1$ 的 $q+1$ 次 Hermitian 曲线，或 $q+1$ 中，q 为奇数且 $r\ge 1$ 。那么 $\mathrm{\Omega }(\mathcal{C})$ 分别有四个和七个字符。此外， $\mathrm{\Omega }(\mathcal{C})$ 很小，因为两条曲线要么是最大值，要么是最小值。通过基于伽罗瓦理论和 Hasse-Weil 下界的代数方法研究完整性问题。我们对 Hermitian 情况的主要结果是 $\mathrm{\Omega }(\mathcal{C})$ 对于 $r\ge 4$ 来说是完整的。对于有理 BKS 曲线，当且仅当 r 为偶数时， $\mathrm{\Omega }(\mathcal{C})$ 才是完整的。如果 r 是奇数，则 $(q+1)$ 割线到 $\mathrm{\Omega }(\mathcal{C})$ 的未覆盖点正是 $\mathrm{PG}(2,q)$ 中不位于 $\mathrm{\Omega }(\mathcal{C})$ 中的点。将这些点添加到 $\mathrm{\Omega }(\mathcal{C})$ 中会在 $\mathrm{PG}(2,q^r)$ 中生成完整的 $(k,q+1)$ 弧，并带有 $k=q^r+q$ 。上述结果对于 $r=2$ 并不成立，对于 Hermitian 曲线的情况 $r=3$ 以及有理 BKS 曲线的情况 $r=3,4$ 仍然存在。作为副产品，我们还获得了研究 $\mathrm{\text{PGL}}(2,q)$ 伽罗瓦反问题的两个有趣结果。