# Finding a plane quintic curve with 3 nodes. (Hartshorne exercise IV.3.12)

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?

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 $$(0,-1)$$.
• This is a nice construction! And it seems that we can use this to give a general construction for all $(d,r)$.