Does each factor of the resultant correspond to exactly one intersection point? Consider the homogeneous polynomials $P,Q\in k[x,y,z]$ that both define a projective curve. Assume their GCD is 1 so they have no common component. We can identify them as polynomials in $x$ by writing $P=\sum a_i(y,z)x^i,Q=\sum b_i(y,z)x^i $ and calculating the resultant. The resultant then is a homogenous polynomial in $k[y,z]$ and it splits over an algebraically closed field as a product of linear factors $b_i z-c_iy$. The resultant is zero if and only if there is a point of intersection of the curves. Note that we could of done the same construction by writing the polynomials in $y$ or $z$.
My question is as follows: for each tuple $(b_i,c_i)$ arising from this factorisation, does each correspond to exactly one point of intersection $[a_i:b_i:c_i]$? Or could there be two distinct points of intersection $[a_i:b_i:c_i],[k:b_i:c_i]$?
 A: Let's try some examples. On the affine patch $z=1$, we take the curves given by $y=x^2$ and $y=1$. These homogenize to $yz-x^2$ and $y-z$, respectively, and their resultant is $(y-z)^2$. So we see that the two intersection points $[\pm 1:1:1]$ on the line $y=z$ correspond to the two factors of this polynomial. Similarly, for $y=x^2$ and $y=-\frac23x^2+x^2+\frac23x$ which homogenize to, we have the resultant is (up to scaling) $yz^3(y-z)^2$, and we have a triple intersection at $[0:1:0]$, then single intersections at $[0:0:1]$ as well as $[\pm1:1:0]$.
The natural generalization is as follows: there's a 1-1 correspondence between distinct factors of the resultant and lines where $V(f)$ and $V(g)$ have intersections, and the degree of these distinct factors correspond to how many intersections are on this line. We can see this from the fact that the resultant of $f$ and $g$ belongs to the ideal $(f,g)$ - by localizing at the homogeneous ideal determining a line $L$, we see that in order to be in this ideal, the resultant must vanish to at least the order of the number of intersections on that line. But the resultant is of total degree equal to the number of intersections, so in fact it must vanish to order exactly the number of intersections along each line.
