May two smooth quadratic curves have an intersection of degree 3?

Question

The underlying field is assumed to be algebraically complete. By Bézout's theorem, the number of intersections of two distinct smooth projective quadratic curves, counted with multiplicity, is $4$. But is it possible that they have exactly two distinct intersections, of degree $1$ resp. $3$ ?

I guess that the answer is no (at least in characteristic $0$), but I do not know how to approach this problem.

Furthermore, is there a similar statement about smooth curves of arbitrary degree ? That is, is there some restriction or symmetry about the number of distinct intersections and their multiplicities, depending upon the respective degrees of the curves in question ?

Answer 1

$F$ and $G$, defined below, have an intersection of degree $3$ at $(x, y) = (0, 0)$, and an intersection of degree $1$ at $(x, y) = (0, 1)$. $$ F = x^2 + y^2 - y \\ G = x^2 - x \cdot y + y^2 - y $$

Also, both $F$ and $G$ are smooth.

Answer 2

$$ xy = 0 \\ x^2 + (y-1)^2 = 1 $$ have two intersections, at $(x, y) = (0,0)$ and at $(x, y) = (0, 2)$. The second has degree 1. The first has degree $3$ (in a sort of peculiar way: a tangent intersection with the $x$-axis, and a crossing-intersection with the $y$-axis).

I suspect that to get this kind of $(1, 3)$ pair of indices, at least one of the curves must be degenerate, like my curve $xy = 0$.

I want to be clear that I'm incompetent in algebraic geometry --- I've got more a topologist's mindset. Given that understanding, here's a rationale: at an intersection point, pick some local coordinates, where the curves can each be parameterized, so we have $x_1(s), y_1(s), x_2(t), y_2(t)$. Shift the parameters so thatthe curves intersect when $s = t = 0$. And we might as well change coordinates so that they intersect at the origin $(x,y)= (0,0)$.

By rotating coordinates, we can arrange that $y_1'(0) = 0$; by scaling the $x$-coordinate, we can make $x_1'(0) = 1$.

If the curves intersect to third order, let's see what that says.

$0$th order, and $xy$-coordinate translation, parameter translation: $$ x_1(0) = x_2(0) = y_1(0) = y_2(0) = 0. $$ $1$st order, $xy$-coordinate rotation and scale: $$ x_1'(0) = 1, y_1'(0) = 0, x_2'(0) = ?, y_2'(0) = 0. $$ By scaling the parameter $t$ (replacing it with $ct$ for some nonzero $c$), we can arrange that $x_2'(0) = 1$ as well, so we have $1$st order, $xy$-coordinate rotation and scale: $$ x_1'(0) = 1, y_1'(0) = 0, x_2'(0) = 1, y_2'(0) = 0. $$

Agree to second order? That tells you something about $x_i''(0), y_i''(0)$ ... does it actually make $x_0''(0) = x_1''(0)$, and similarly for $y$?

I suspect that if you write out the details, you'll find that agreement to third order entails the two quadratics being proportional, with division by $x'(0)$ being a key step in showing that...so that you can get 3rd-order agreement without equality only if one of the quadratics has a zero derivative at the intersection point. But I don't have time to work through all the details here.

I suspect that you can probably do something clever by looking at the pencil of quadratic curves described by $$ (1-t)F(x, y) + tG(x, y) = 0 $$ and observing that all such curves end up meeting the originals to 1st and 3rd order at the two special points but I'm not witty enough to actually carry that out. I think that Pierre Samuel's introductory book might do so.