I came across this question in Algebraic Geometry: A Problem Solving Approach: Let P(x,y,z) be an irreducible homogeneous second degree polynomial. Show that the intersection multiplicity of V(P) with any line l is at most 2.
I tried writing down the general form of a second degree polynomial and taking second order partials and showed that no point can have intersection multiplicity more than 2 this way. However, I did not use the fact that P was irreducible at any point and presume there was some error I made. Any advice into the error and how to solve it would be greatly appreciated.
Also, I'm just starting to teach myself algebraic geometry so I don't yet have much familiarity with the more abstract concepts.
I guess this is an example for Bezout's theorem: https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem.
For a more concrete approach, parametrize $\mathbb{P}^2$ by homogeneous coordinates $(x : y : z)$. Then we can apply a change of coordinates (acting on the coordinates by $\operatorname{GL}_3$) so that the line is $L = V(z)$. Then intersection $L \cap V(P)$ is then $(x : y : 0)$ where $P(x, y, 0) = 0$, which is multiplicity 2 for algebraically closed fields since $f(x, y) = P(x, y, 0)$ is a quadratic.