[续帖]不可约n次曲线最多有$(n-1)(n-2)\over2$个二重点

hbghlyj

续前年这帖 昨天我在这篇文章的第18页好像找到了一个具体的构造(Severi’s construction of irreducible nodal curves) we take the union of $d$ straight lines in general position, which is a curve with $\frac{d(d-1)}{2}$ nodes (the maximum by (27)). Then choose some line and deform the curve by smoothing out all $d−1$ intersection points of this line with the other lines, obtaining an irreducible, rational curve with $\frac{(d-1)(d-2)}{2}$ nodes (see Figure 1). Finally, we take another deformation by smoothing out $\frac{(d-1)(d-2)}{2}-n$ nodes and obtain an irreducible curve of degree $d$ with $n$ nodes as desired. In the reducible case, we take irreducible curves of degrees $d_1, ..., d_s$ with $n_1, ..., n_s$ nodes respectively and place them in general position in the plane.

Czhang271828

回复 1# hbghlyj

可以学学 Algebratic Geometry, 会大受启发的.

非常推荐阅读 В. И́. Арно́льд 为中学生写的科普书Real Algebratic Geometry (不得不说, 中国的数学科普真的拉跨). 随后可以看看 Ried Miles 为本科生简化的 Algebratic Geometry 教材 UAG, 大概学完半本近世代数即可上手. 学完交换代数后可以看看 R. Hartshorne 的经典教材 GTM 52. 然后当然是著名的 Éléments de géométrie algébrique 了 (读完第二节放弃...).

当然, A. Gathmann 的 Plane Algebratic Curve 不错, 零基础即可通读.

hbghlyj

回复 2# Czhang271828
好像找到了构造方法:对于整数n≥3,考虑参数曲线族\begin{cases}x=\cos((n-1) t)\\y =\cos(nt)\end{cases}

n=3:	n=4:	n=5:	n=6:	n=7:

根据mathcurve.com上的资料(截图),
⑴ 这个曲线是n次代数曲线
⑵ 这个曲线恰有$\frac{(n-1)(n-2)}2$个实的二重点.

尝试证明一下:
⑴ $T_n$是切比雪夫多项式，要证明从$x=T_{n-1}(\cos t),y=T_n(\cos t)$消去$t$后得到的x,y的多项式是$n$次的
见forum.php?mod=viewthread&tid=12110
⑵ 见forum.php?mod=viewthread&tid=12115