I have the following problem;
Let $C = \{Q:=x_0x_2^2 -x_1(x_1-x_0)(x_1+x_0)=0\}$ and $L = \{ax_0 + bx_1 = 0\}$ be two projective curves with $(a,b) \ne (0,0)$. Let $p=[0,0,1]$, then I am asked to calculate the intersection multiplicity $I(p, C, L)$.
Is this a correct argument? (I feel it isn't!)
Observe that $p$ is a smooth point of $Q$ since, $\partial_{x_0}Q(p) = 1$, hence the intersection multiplicity has to be equal to 0 or 1. Since $p \in L \cap C$ it follows that the intersection multiplicity cannot equal 0. Therefore $I(p, C, L) =1$.
I don't really want to have to start calculating the resultant of $C$ and $L$ and computing determinants since this wont give me any intuition, I want to use properties of the intersection multiplicities instead.
Thanks for your help!
Your argument is almost correct.
Since $p$ is a smooth point of $C$, a line through $p$ has multiplicity intersection $1$ with $C$ unless the line is the tangent to $C$ at $p$.
But the tangent to $C$ at $p$ is the line $x_0=0$ which is different from $L$ (since $b\neq0$) .
Hence the required intersection number is indeed $1$.