My friend, who is currently taking an algebraic geometry course from an unnamed prolific poster on MO, told me about the following bonus question on one of his problem sets a few weeks ago.

Preliminary discussion.By Bézout's Theorem (which we will prove later in the course), two plane cubics in general position with regard to one another, and hence any two memebrs of the pencil that they span, intersect in $9$ points. The previous question shows that in fact this pencil is already determined by $8$ of the $9$ intersection points.From this you can deduce two things:

(a) $9$ points in general position

cannotarise as the intersection of two cubic projective curves.(b) $8$ points in general position determine a $9$th point (namely the $9$th common point of intersection of the pencil of cubics determined by the $9$ give points).

We can regard this process as defining a kind of $8$-ary operation$$(\mathbb{P}^2)^8\text{ }-\to \mathbb{P}^2.$$(We use a broken arrow because this operation is not really defined on

all$8$-tuples of points in $\mathbb{P}^2$, but only on those in general position. It is an example of what is called arational mapin algebraic geometry.)

Actual bonus question.Can you say anything about what this formula would look like (e.g. its degree)?

This begs the natural question, does there exist a natural moduli space $X$ of $9$ points so that our map $X \to$ $9$th point is actually a map and not just a rational thing? We could start with $\text{Hilb}^8 \,\mathbb{P}^2$ and figure out how we need to modify it.

At this point, we could remark the following. Presumably, it would be an open subscheme of $\text{Hilb}^8\,\mathbb{P}^2$? We can not get general position already as an open subscheme of $(\mathbb{P}^2)^8$ because we would have to account for infinitely many closed things to throw out. but perhaps we could get this to work on passing to the Hilbert scheme. Actually intuitively, probably not – we would probably have to add extra parameters or something to parameterize all the possible things we would need to throw out? But it definitely seems possible.

Or wait, linear dependence is definitely finitely encodable, so we can get finitely many things to throw out already in in the $(\mathbb{P}^2)^8$ case. Obviously, since it is a rational map, it is defined on an open subset. Basic degree questions like "if we fix $7$ points, and a target point, what is the locus of the possible $8$th point" can be addressed in that context geometrically too. So does the Hilbert scheme buy us anything? Obviously, the symmetry of the point arrangements and stuff – so we guess that is a goal in itself. And it would be useful, for example, if we wanted to consider "families" of points over something, the Hilbert scheme would allow us to see the possible behaviors more cleanly. We do not know if we could convince someone else that the Hilbert scheme adds something here, though.

Also, but wait, the weird blowup behavior of the Hilbert scheme of points only happens at diagonals, so in this case, the open subscheme of the Hilbert scheme of points would literally just be the symmetric quotient of $(\mathbb{P}^2)^8$, so in particular, any behavior of it is recoverable in an extremely simple way from the latter.

With regards to those remarks, the entire point of using the Hilbert scheme is we want to understand what happens at nongeneric points to do a lot of things (e.g. intersection theory) because we need properness, which we are going to get from modifying the Hilbert scheme in some, hopefully small, way.

At this point, we could remark that we do not see an obvious way to extend the rational map to all of $\text{Hilb}$. Conceivably, $\text{Hilb}$ might help, since the blowups give us extra information about how points approach the degenerate positions, but we do not know how to actually do it.

With regard to that remark, right, the rational map does not extend to $\text{Hilb}$, and that is why we need to modify $\text{Hilb}$ first.

At this point, we could remark yeah true, but we do not see why a priori modifying $\text{Hilb}$ to have this behavior is easier than just modifying $(\mathbb{P}^2)^8$.

With regard to that remark, we think $\text{Hilb}$ is the correct starting point because the intersection of $2$ different cubic curves is a length $9$ subscheme.

**Question.** Hopefully building on the blath I have above, is there anything more we can say about what this formula looks like?