Happy new year to all readers! This year I hope will be full of mathematical reflexions here, alas certainly again in English, in order to reach new audience (I’ve just noticed I’m referenced on Terry Tao’s weblog for instance).

I have already discussed what this same Tao calls the dichotomy principle (« a massive lot or very few ») for polynomials. It was essentially the case of polynomials in two variables, as in the traditional Bezout’s theorem: two coprime polynomials $P,Q\in \mathbb{R}[X,Y]$ of degrees $m,n$ have at most $mn$ common zeros,
where $\mathbb{R}$ can in fact be replaced by any field, and
with the inequality being an equality in the algebraic closure, if one counts points at infinity and multiplicities of intersection. This can be proven either with the resultant, or by some argument of dimensions.

Lines in curves, planes in surfaces

For instance it implies that if a polynomial of degree $m$ vanishes on $m+1$ distinct points of the line $ax+by+c=0$, then $ax+by+c$ is a factor of this polynomial. However it’s not relevant at all in this case to invoke Bezout’s theorem, because the statement is quite immediate in the case where the line has equation $x=0$ – expand according to the powers of $X$, so latex $P=\sum_{k=0}^{m}c_k(Y)X^k$ and notice that this line is in fact fairly general (by change of variables).
What about a polynomial in three variables now, so an algebraic surface of the space instead of an algebraic curve in the plane? An algebraic surface can very well include a full line and still be irreducible, as show ruled surfaces like the hyperbolic paraboloid $z=x^2-y^2$. Lines are not the zero set of a truly irreducible polynomial anymore; indeed a line is defined by two equations of degree 1 $f=0$ and $g=0$. But this system of equations is in $\mathbb{R}$ synonymous with $f^2+g^2=0$, which involves a polynomial of degree 2 which is irreducible over $\mathbb{R}$ but not over $\mathbb{C}$. But an algebraic surface cannot contain a plane, without the degree 1 equation polynomial defining the plane to factor out in it.

Intersection points, intersection lines
In dimension $n$, a reasonable generalization of Bezout’s theorem must involve enough hypersurfaces (of codimension 1) in order for the intersection to be expected to be points, so of dimension 0, hence you need $n$ hypersurfaces. If their degrees are $d_1,\cdots,d_n$, it is reasonable to guess $d_1\cdots d_n$ intersection points. But the example of the hyperbolic paraboloid (degree 2) with a line included in it, defined by two equations of degree 1, yields infinitely many points of intersections, instead of the expected 2… Although all the equations are coprime. But the correct statement is that if the number of intersections is finite, it’s at most the product of the degrees.

However, a variant of Bezout’s theorem involving two surfaces in $\mathbb{R}^3$ is still possible. Surely, you won’t bound the number of points of intersection, but you can bound the number of lines of intersection: the intersection of two coprime surfaces of degree $n$ and $m$ contains at most $nm$ points. A proof of this can be seen on Larry Guth’s lecture notes on polynomial method lecture 13.

Monge-Cayley-Salmon: flecnodal points, ruled surfaces
Salmon’s theorem, hinted at by Monge earlier and also stated by Cayley, says that if an algebraic surface $S=\{(x,y,z)\mid P(x,y,z)=0\}$ is such that at every point $x$ there exists a line $l_x\ni x$ tangent up to order 3 to $S$ (flecnodal point), then for all points $x$, the line $l_x$ is in fact fully included and so the surface is ruled. As a line tangent up to order 3, on a surface of degree 2 has to be included, we suppose that $p$ is squarefree and of degree at most 3. Then flecnodal points of $p$ are determined by a polynomial $Fl(p)$ of degree $11d-24$. Together with Bezout’s theorem, it yields the following dichotomy for a surface of degree $d$ :
1) Either ruled, so one line in $S$ through each point
2) Or only $O(d^2)$ on $S$.

This is one of the basic ingredients for Guth-Katz solution of the Erdös distinct distance. problem.