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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 $[a_i:b_i:c_i],[k:b_i:c_i]$?

jonathan
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  • A segment in polynom 1 may intersect several segments of polynom 2 – Ripi2 Nov 22 '19 at 17:48
  • @Ripi2 what do you mean by segment? Can you clarify what you mean? – jonathan Nov 22 '19 at 18:05
  • Imagine a polygon that approximates a curve.It may be (or not) easier to see polygons intersection than the intersection of two curves. – Ripi2 Nov 22 '19 at 19:05
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    @Ripi2 this is not applicable in this setting. In algebraic geometry, the usual topology on curves is the Zariski topology, which is turns out to be the cofinite topology on irreducible curves. So as soon as any two curves intersect in more than a finite number of points, they must have a common component, contrary to what the OP asked for. – KReiser Nov 22 '19 at 23:33

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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.

KReiser
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  • Thank you for your answer. What is wrong with the following argument: Change coordinates such that $[1:0:0] \notin C\cup D$ or any line going through two points of intersection. $R_{C,Q}(b_i,c_i)\equiv 0$ hence we can find an $a_i$ such that $P(a_i,b_i,c_i)=Q(a_i,b_i,c_i)=0$. If there was another distinct point of intersection $[\alpha:\beta:\gamma]\neq [a_i:b_i:c_i] \forall i$ with $b_i \gamma-c_i\beta =0$ then $[\alpha:\beta:\gamma], [a_j:b_j:c_j]$ and $[1:0:0]$ all lie on the line \begin{equation} b_jz=c_jy \end{equation}in contradiction to our conditions on $[1:0:0]$. – jonathan Nov 24 '19 at 15:23
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    The only thing wrong with that is that you didn't make these assumptions clear earlier - the argument is correct :) Let's look at what happens when we apply this to the first easier example. We make the change of variables $x=x+y$, $y=y-x$ so our equations are now $(y-x)z-(x+y)^2$ and $y-x-z$. Then it's clear that $[1:0:0]$ isn't on either of these curves nor the line between their intersection points $y-x-z$. Taking resultants, we get $4y(y-z)$ and the distinct linear factors correspond to distinct points of intersection. – KReiser Nov 24 '19 at 22:26
  • I've been working under the implicit assumption that this property was invariant under change of variables! Thank you very helping me realise that this is false :) – jonathan Nov 24 '19 at 22:40