I have started to read 3264 and all that from Harris and Eisenbud. In particular I was eager to understand the concrete examples in it (Example 2.21 p75).
I'm at a loss to understand something that must be very basic about the cone over a smooth conic, so a quadric cone in $\mathbb{P}^3$. They give an argument for why the ruling of the cone cannot be Cartier, I know how to prove this, it is easy to prove so by looking at the Zariski tangent space at the vertex of the cone, but they give a more geometrically flavored argument along the following lines.
If $C$ is an odd degree curve contained in the cone, and it meets a general ruling at $d$ points away from the vertex and at a point of multiplicity $m$ at the vertex then the degree of the curve is $2d+m$ forcing $m$ to be odd. Why do each point away from the vertex count as 2 in the degree?
I'm guessing this has to do with the curve being tangent to the tangent plane along the ruling away from the vertex, and the fact that the intersection being proper, the intersection cycle is given by intersection multiplicities, and away from the vertex the cone is smooth, so this tangency forces the intersection multiplicity to be greater than 1? But why 2? Is there an easy way to see this? I feel like this is very "hand wavy".
You measure the degree of $C$ by intersecting with a hyperplane $H \subset \mathbb P^3$. If you choose $H$ such that it intersects the cone in the ruling, then $H$ is tangent to the cone. Note that $H$ intersects the cone in twice the ruling; this is where the factor $2$ comes from. So any point where $C$ intersects the ruling is a point of multiplicity $2$ in $C \cap H$.