I'm stuck with an exercise which implies the strong version of the Bezout's theorem.

$\textbf{Statement}:$ Let $C_1[x,y,z], C_2[x,y,z]$ be the two algebraic curves on $CP^2$ of degrees $m,n$ respectively. Let $A$ be the common point of the two curves such that the degree of $A$ for $C_1$ is $r$ and for $C_2$ it is $s$. Show that the corresponding to $A$ solution of $$ \mathrm{Res}(C_1,C_2)[y,z] = 0 $$ has the degree $s\cdot r$.

This exercise proves the Bezout's theorem almost directly. Since the resultant equation is of order $mn$ then the number of common points is exactly $mn$ counting the multiplicities (by the homogeneity of $\mathrm{Res}$ and fundamental thm. of algebra) . And the aforementioned exercise shows that the multiplicity of the root in the resultant is actually the product of multiplicities of the root $A$ for $C_1,C_2$.

The problem is that the multiplicity of a point for an algebraic curve is defined through reduction of the equation of the curve on the lines and looking for the order of the root for this one-dimensional polynom.So reducting the resultant on some lines gives big formulas in which I cannot see what is the order of the corresponding root.

  • $\begingroup$ What definition of the resultant are you using? In the matrix version, the degree is almost immediate by computation. $\endgroup$ – Chris C Oct 24 '17 at 15:38
  • $\begingroup$ Just as a usual determinant of a matrix. But it is not clear for me how it follows almost immediate. $\endgroup$ – Fedor Goncharov Oct 24 '17 at 17:49

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