Computing intersection multiplicity using primary ideals In $\mathbb{P^2}$, $f = x^2-yz, g = (x+z)^2-yz$, Compute the intersection multiplicity of the curves $V(f),V(G)$ at $p = [-2:1:4]$, using the fact that intersection multiplicity of two curves at point $p$ is the hilbert polynomial of the $I(P)$-primary component of $(f)+(g)$.
This is an example that I cooked up to practice computing intersection multiplicity, but I'm having trouble calculating the $I(p)$-primary component of $(f)+(g)$, I'm thinking it might be messy? I wanted an example where your point of intersection isn't something basic.

I think you can save yourself some pain by working in the local ring $\left(\frac{k[X,Y,Z]}{(X^2 - YZ, (X+Z)^2 - YZ)}\right)_{I(p)}$, where $I(p) = (X+2, Y-1, Z-4)$.  We can find a more convenient set of generators for the ideal $I = (f,g)_{I(p)}$ as follows.
\begin{align*}
I &= (X^2 - YZ, (X+Z)^2 - YZ) = (X^2 - YZ, (X+Z)^2 - X^2) = (X^2 - YZ, (2X+Z)Z)
\end{align*}
Since $Z \notin I(p)$, then $Z$ is a (local) unit so $(X^2 - YZ, (2X+Z)Z) = (X^2 - YZ, 2X+Z)$.  Then
\begin{align*}
I &= (X^2 - YZ, 2X+Z) = (X(X+2Y), 2X+Z) = (X+2Y,2X+Z)
\end{align*}
since, as above, $X$ is a unit.  For more on how to use this local computation, see Lemma 12.22 of Andreas Gathmann's notes.
