Intersection multiplicities Let $X \subset \Bbb P^n$ be a curve defined by $f_1 = \dots = f_r = 0$ and $H$ an hyperplane. How to compute the multiplicity $m$ at a point $p \in X \cap H$ ?
Specific example :
In my case, $X \subset \Bbb P^3$ is the twisted cubic, given by $xw = yz, xz = y^2, yw = z^2$. It can be also parametrised as $[s^3:s^2t:st^2:t^3]$, and $H$ is the hyperplane given by $y=0$.
What I tried :
Set theoretically, $X \cap H = \{ [1:0:0:0]\} \cup \{ [0:0:0:1]\}$. However, since $X$ is a cubic curve one should have multiplicity $2$. Around the first point, we can set $x=1$ and take local coordinates $y,z,w$. The curve is then given by $(t,t^2,t^3)$ and $y=0$ b the equation is $t=0$, multiplicity one. Around the second point, we can take $w=1$ and then local coordinates $x,y,z$.  The curve is given by $(s^3,s^2,s)$ and $y=0$ becomes $s^2 = 0$, hence the multiplicity should be 2.
Specific questions :

*

*Is my argument correct ?

*How to get the same answer without using the parametrisation, i.e only using the equations defining $X$ ?

 A: Your argument is correct. To do this directly from the defining equations, just apply the same procedure (you're basically just intersecting things in a different order). For the first point, the equations on the first chart are $w=yz,z=y^2,yw=z^2$. Plugging in $y=0$ we get $w=0,z=0,z^2=0$. While we have an equation of multiplicity $2$, it's redundant because we have the simpler $z=0$ as well, hence the multiplicity is $1$. Geometrically, in the affine plane $y=0$ we have the lines $w=0$ and $z=0$, as well as the double line $z^2=0$. The intersection between a reduced curve and a "fat" version is just the reduced curve, so we are ultimately looking just at $(w,z)$, a transverse intersection of lines hence a single reduced point.
For the second point, the equations on the second chart are $x=yz,xz=y^2,y=z^2$, which reduce to $x=0,xz=0,z^2 = 0$. Here we have an equation of multiplicity $2$ that cannot be simplified out of existence. Geometrically, the intersection of $x=0$ and $xz=0$ is just $x=0$, so introducing $z^2=0$ we see that the total intersection is topologically transverse, but one of the lines has a double structure. In other words, the scheme-theoretic intersection is $(x,z^2)$ which has colength $2$ in $k[x,z]$.
