# Projective roots of a homogeneous polynomial

Suppose that $f(X,Y)\in\mathbb C[X,Y]$ is a homogeneous polynomial of degree $n$, then we can consider it as a function on $\mathbb P^1_\mathbb C$. It has at most $n+1$ projective roots (points of $\mathbb P^1_\mathbb C$), namely the "affine" roots (that are $n$) plus possibly the point at infinity.

Now consider the following two homogeneous polynomials in three variables (over $\mathbb C$):

$$F(X,Y,Z)=Z^m+Z^{m-1}F_1(X,Y)+\ldots+F_m$$ $$G(X,Y,Z)=Z^n+Z^{n-1}G_1(X,Y)+\ldots+G_n$$

where $F_i$ and $G_i$ are homogeneous of degree $i$. The resultant $R_{F,G}$ respect to the variable $Z$ is a homogeneous polynomial of degree $mn$, so it has at most $mn+1$ projective roots. I don't understand why, in some proofs of the Bezout theorem, authors say this fact leads to an absurd. It seems that $R_{F,G}$ can only have $mn$ projective roots but i don't understand the reason.

• your inital statement is wrong. a homo poly of degree $n$ has at most $n$ roots including infinity. Commented Jul 18, 2014 at 19:26
• Ok, but I don't understand why. Consider $F(X,Y)=F_0(Y)X^n+F_1(Y)X^{n-1}+\ldots+F_n(Y)$; I have at most $n$ solution of the type $(1:y)$ plus possibly $(0:1)$. Where is the mistake? Commented Jul 18, 2014 at 19:31
• $(0 : 1)$ is a zero if and only if $F_n (Y) = 0$; but in that case, $F (X, Y)$ is divisible by $Y$, etc. Commented Jul 18, 2014 at 19:35
• Yeah, I dont get the notation, but write, $a_nx^n+a_{n-1}x^{n-1}y +\cdots$ then infinity is a solution iff $a_n=0$ so this takes away from the number of finite solutions. Commented Jul 18, 2014 at 19:37
• Your first sentence is false for every $n\gt0$ Commented Jul 18, 2014 at 20:25

Let me answer by showing how a homogenous polynomial of degree $n$ in the projective line has $n$ solutions.
$$a_nx^n+a_{n-1}x^{n-1}y+\cdots +a_0y^n$$
let us assume that $a_{m}\neq 0$ is the first nonzero coefficient then the equation factors
$$a_m y^{n-m}(x-\beta_1 y) \cdots (x-\beta_m y)$$ now for $y=1$ we have the finite roots $[\beta_1:1], \ldots , [\beta_m:1]$
Now $y=0$ is a root only if $n-m\neq 0$ in which case this root occurs $n-m$ times so the total number of roots counting multiplicity is always $$m+(n-m)=n$$