Abstract. Let X denote the rational curve with n+1 nodes obtained from the Riemann sphere by identifying 0 with ∞ and ζ j with −ζ j for j = 0, 1, . . . , n−1, where ζ is a primitive (2n)th root of unity. We show that if n is even, then X has no smooth Weierstrass points, while if n is odd, then X has 2n smooth Weierstrass points.C. Widland [14] showed that the rational curve with three nodes obtained from P 1 C by identifying 0 with ∞, 1 with −1, and i with −i has no smooth Weierstrass points. This curve may be realized as the projective plane curve x 2 y 2 + y 2 z 2 = x 2 z 2 , which has "biflecnodes" at the points (1, 0, 0), (0, 1, 0), and (0, 0, 1), with each biflecnode being a Weierstrass point of weight 8. In this note, we show that this curve is one member of an infinite family of rational nodal curves with no smooth Weierstrass points. We remark that a general rational nodal curve with n nodes has n(n − 1) smooth Weierstrass points [9].Let n be a positive integer. Let ζ denote a primitive (2n)th root of unity. Let X denote the rational curve with n + 1 nodes obtained from P 1 C by identifying 0 with ∞, and ζ j with −ζ j for j = 0, 1, . . . , n − 1. Our main result is:(0.1) Theorem.(1) If n is even, then X has no smooth Weierstrass points.(2) If n is odd, then X has 2n smooth Weierstrass points at the (2n)th roots of −1.In the first section, we prove a theorem that can be used to show that fixed points of an automorphism are frequently Weierstrass points, which extends known results for smooth curves. This is then used in the second section to prove the main result.We thank Pierre Conner for very helpful discussions.
1In this section, X and Y will denote integral projective curves over the complex numbers. If C is a curve, we let p a (C) denote the arithmetic genus of C and p g (C) denote the geometric genus of C. If P ∈ C, then we letÕ P denote the normalization