In page 316 of Hartshorne's book Algebraic Geometry, exercise IV.3.12 asks:

For each value of $d=2,3,4,5$ and $r$ satisfying $0\le r\le\frac{1}{2}(d-1)(d-2)$, show that there exists an irreducible plane cure of degree $d$ with $r$ nodes and no other singularities.

I can find such curves for all cases, except the case $d=5,r=3$:

  1. For the case $r=0$: By Bertini's theorem, we can find an irreducible nonsingular plane curve of each degree $d$.
  2. For the case $\frac{1}{2}(d-1)(d-2)-d+3\le r\le\frac{1}{2}(d-1)(d-2)$, consider an irreducible nonsingular $X$ in $\mathbb{P}^3$ of degree $d$ and genus $0\le g\le d-3$, then a projection $\pi$ from a point $P\notin X$ (at general position) can give a plane curve $\pi(X)$ of degree $d$ with $r$ nodes. This includes the cases $(d,r)=(3,1),(4,3),(4,2),(5,6),(5,5),(5,4)$.
  3. For the case $r=1$: Inspired by Hartshorne exercise IV.3.7, we can find that the plane curve defined by equation $xyz^{d-2}+x^d+y^d=0$ is a curve with node $[0:0:1]$ as its unique singularity, assume $\mathrm{char}\,k\not\mid d$; if $\mathrm{char}\,k\mid d$, the curve $xyz^{d-2}+x^{d-1}z+y^d=0$ is an example. This includes the cases $(d,r)=(4,1),(5,1)$.
  4. For the case $(d,r)=(5,2)$: Inspired by Hartshorne exercise IV.5.4, consider a non-hyperellptic curve of genus $4$ in which has two $g_3^1$'s, then it can be realized as a $(3,3)$-type curve on Segre surface $\mathbb{P}^1\times\mathbb{P}^1\subset\mathbb{P}^3$. A projection $\pi$ from a point $P\in X$ (at general position) can give a plane quintic curve $\pi(X)$ with $2$ nodes, and no other singularities.

So it remains the case $(5,3)$. I first tried using the projection $\pi$ on a point in a curve $X$ of degree $6$ and genus $3$, then one can see that such curve cannot be contained in any quadratic surface in $\mathbb{P}^3$, so that it is not easy to show the projection can produce exactly $3$ nodes in $\pi(X)$. On the other hand, I also tried constructing an explicit plane quintic curve, such as $$x^3yz+y^3zx+z^3xy+x^3y^2+y^3z^2+z^3x^2=0,$$ and using Jacobi matrix to show it has $[1:0:0],[0:1:0],[0:0:1]$ as its singularities, but it is also not easy to solve such polynomial equations (I expect an equation that I can do by hand). Any help?


1 Answer 1


If you take a union of five distinct lines $V(h_1),\cdots,V(h_5)$, a conic $V(f)$ through exactly three of the points of intersection of the lines, and a sixth line $V(h_6)$ distinct from the other five, then for most $a,b\in k$ the curve cut out by $a\prod_{i=1}^5 h_i + bf^2h_6$ will work. This is because the curve has double points at the three points of intersections of the lines and nowhere else, a generic double point is a node, and a generic member of the pencil will be irreducible.

For instance, $y(y-z)(y-2z)(x+y+z)(x-y-z)+(yz-x^2+z^2)^2z$ works over any characteristic zero field: it is an irreducible curve with nodes at $(\pm1,0)$ and $(0,-1)$.

  • $\begingroup$ This is a nice construction! And it seems that we can use this to give a general construction for all $(d,r)$. $\endgroup$
    – ZCC
    Jul 26, 2023 at 3:57

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