# Proof verification of a weak version of Bezout's Theorem

I'd like to make sure here that my reasoning seems sound. I am working from Kirwan's book on algebraic curves. I was not totally happy with her proof of this theorem, so I wanted to see if I could rework it.

The simplicity of my proof suggests I've forgotten some important details. In particular, the original proof required that we change coordinates so that the point [0 : 0 : 1] was on neither curve, nor did it lie on any line defined by any two points of intersection. So please correct any mistakes, since I'm still learning this :)

Theorem: Given any two projective curves $P$ and $Q$ in $\mathbb{CP}^2$ of degrees $n$ and $m$ respectively which do not share a component, they meet at no more than $nm$ points.

By a projective change of coordinates, we can assume that all points of intersection are finite. (That is, they do not lie on the line at infinity, $z=0$).

I'm fairly sure this is easy enough to do by sending the line at infinity to any other line, as long as you're careful about what gets mapped to the point at infinity on the new line. Even if there is an error in the proof elsewhere, this seems like a productive thing to do since we can now work in affine space.

Momentarily considering $P$ and $Q$ as elements of the ring $\mathbb{C}[x,y][z]$ and define their resultant with respect to $z$, $R(x,y)$.

Because $P$ and $Q$ are homogeneous polynomials in $x$, $y$, and $z$, we can show that $R(x,y)$ is homogeneous and of degree $nm$. (This is a lemma Kirwan front-loads and proves at the end of the section). And so, we can factor $R(x,y)$ as $\kappa(\lambda_1 y - \mu_1 x) \dots (\lambda_{nm} y - \mu_{nm} x)$. Notably, it has exactly $nm$ factors.

Suppose $[\alpha : \beta : 1]$ is a common point of intersection. (We can assume the third compontent is $1$ since all such points are finite points). Then, as polynomials in $\mathbb{C}[\alpha, \beta][z]$, $P(\alpha, \beta, z)$ and $Q(\alpha,\beta, z)$ have a common zero (namely, when $z=1$) and so $R(\alpha, \beta)$ must be zero, and so consequently, $\alpha y - \beta x$ must be a factor of $R(x, y)$.

Finally, suppose that that we have at least $nm + 1$ distinct points of intersection: $[\alpha_1 : \beta_1 : 1] \dots [\alpha_{nm+1} : \beta_{nm+1} : 1]$. By the above argument, this makes $\alpha_i y - \beta_i x$ must be a factor of $R(x, y)$ for each $i$. But this gives us at least $nm + 1$ distinct factors, while we counted exactly $nm$ earlier. This is a contradiction and ends our proof.

• Oh,thank you for letting me know, @Hoot – Tac-Tics Jul 12 '15 at 16:31
• Also, I would suggest tagging this as algebraic geometry – Hoot Jul 12 '15 at 18:07

Your mistake : if $\alpha X - \beta Y$ divides $R(X,Y)$ it does not mean that it is one of the $\alpha_i X - \lambda_i Y$ (eg : $X-Y$ and $2X-2Y$ both divides $X-Y$). This is the reason why the author requires that the point $(0,0)$ does not lie on any line through two intersecting points.
Other mistake : did you proved that $R(X,Y)$ splits into factors of degree $1$ ?
Minor mistake : to deduce that $R$ is of degree $nm$ it requires $P$ and $Q$ to be coprime.