So, I don’t really have a full length post in me at the moment. However, here’s a nice trick that I’ve learned recently, plus some motivation. With this, it’s easy to write down explicitly a curve of arbitrary degree in the plane, which has specified nodes. (So long as it’s possible to have that many nodes.)

First up, the motivation for why we want to write down lots of plane curves with nodes. First, up, we quote a theorem, which is proved thoroughly anywhere that curves are seriously investigated:

**Theorem**: Every algebraic curve can be embedded in .

A route to proving this is to show that hyperelliptic curves are plane curves, and that nonhyperelliptic curves are embedded by the canonical embedding, and that taking projections of curves gives isomorphisms down to .

However, not every curve is actually isomorphic to a plane curve. But the second theorem we’re going to quote gives us a way to represent every curve in the plane:

**Theorem: **Let be a curve in . Then there exists a point such that projection from the point maps surjectively onto a plane curve with only nodes as singularities.

So every curve can be represented as a curve of degree with nodes for some satisfying . So it’d be nice to have a good way to write down plane curves of a given degree and number of nodes, which then determines the genus.

The engine that makes this all work is Bertini’s Theorem. We’re going to construct a linear system of curves with some base points, and then a generic one will be the curve we want. In fact, it will be a one dimensional linear system, which we’ll refer to as a *pencil*.

So, fix a degree and the number of nodes that you want , so long as is small enough that such a curve can exist. In fact, we’re going to do this by constructing two curves with the maximal possible number of nodes.

Fix points in the plane, and take two configurations of lines such that the points are intersections of lines in each configuration, and no other points are. Also require that they have no whole lines in common. This is fairly easy to do by hand, because we’re just taking unions of lines. These give polynomials and .

Using and , we define a pencil of curves by looking at the zero locus of where are homogeneous coordinates on . Bertini then tells us that a general element of this pencil is smooth away from the points, which we made into base points (and which are the only ones). So the singular locus of the general curve here consists of points.

Finally, moving in the pencil will preserve the singularity types of these base points, which were originally nodes. Thus, we have constructed a whole family of genus , degree plane curves with nodes, so we can get concrete views of any curve in this way.

August 23, 2008 at 2:26 am

[...] Charles Siegel at Rigorous Trivialities with a post on algebraic geometry. He’s written about how to explicitly construct curves of arbitrary degree with specified [...]

April 27, 2009 at 12:14 am

Hi Charles,

Great blog! I’ve been reviewing it myself as I study for my own qualifying exam. You say here that hyperelliptic curves are plane curves. I don’t think this is quite right, (take genus 2, e.g.) but you can always get around it by just mapping the curve into P^n by using some big multiple of any point.

April 29, 2009 at 7:41 pm

Dear twistedcubic,

I’m not sure I get your comment about genus 2. If the genus is 2, then the canonical map is the hyperelliptic map to .

April 29, 2009 at 8:04 pm

I meant that as an example, genus 2 curves are hyperelliptic, yet not plane curves. I was refuting the fact that hyperelliptic curves are plane curves, not that genus 2 curves were hyperelliptic.

May 11, 2009 at 9:52 am

I apologize. I had been thinking about two different things that week, and got some wires crossed. What’s true is that any hyperelliptic curve can be mapped into such that there is only one singular point, and it lies on the line at infinity. I had been fiddling with the canonical embedding, but all you need for “all curves are space curves” is that every curve is projective, and there exists a point from which you can project down until you get to without creating any singularities.

July 24, 2011 at 12:43 pm

Very interesting. Do you have a reference for this construction? What in this construction ensures that sF+tG has nodes at those \delta points. Is it the requirement that they lie at the intersection of the lines? If that’s the case, why does one require a pencil when either F or G would have the prescribed nodes…or is the assertion that the pencil spans all plane curves with those \delta nodes?

July 24, 2011 at 7:49 pm

I don’t have a reference, and I’ve never seen it written down anywhere else. It was shown to me by my advisor while trying to solve a specific problem. The fact that the intersections of lines are nodes guarantees that things will be singular at those points, and nodes are general singularities, so a generic element of the pencil will be smooth away from those points, nodal at those points and, the reason we don’t use F or G, is because we’d like these curves to be irreducible.

July 24, 2011 at 8:14 pm

I think I see now how it works. Given two line configurations with \delta points lying in the intersection of intersection sets, the sum of products is guaranteed to have nodal singularities at the \delta points.

Can one always do this for any set of \delta points, 0 \leq \delta \leq (d-1)(d-2)/2? That seems like way too much freedom. I can see how this works for \delta small, but I can’t see how to make this work for g small. Take g=0, for example. In this case \delta=(d-1)(d-2)/2. d lines intersect at only d(d-1)/2 points, so it seems almost impossible to conceive of two *distinct* line configurations having those same \delta points.

It seems like a very clever construction, but is it general?

Cheers for the prompt reply!

December 5, 2011 at 6:18 am

Dear Charles,

I don’t understand exactly how you choose the lines. For example, if I want and and I choose , how do I choose 8 different lines such that their only points of intersection are these two points? Aren’t there necessarily going to be lines in each configuration that intersect outside these two points?

December 5, 2011 at 6:19 am

Sorry, the two lines are .

Thanks!

December 5, 2011 at 6:19 am

I mean points!