# Computing the intersection divisors of a smooth projective cubic curve.

Let $X$ be the smooth projective plane cubic curve defined by $y^2z=x^3-xz^2.$ Compute the intersection divisors of the lines defined by $x=0,y=0,$ and $z=0$ with $X$.

Here is an idea:

Any point $(x_0,y_0,z_0)$ of intersection between the line $x=0$ and the curve must satisfy $x_0=0$ and $y_0^2z_0=0.$ This means the point $(x_0,y_0,z_0)$ must satisfy $x_0=0$ and one of the following : $y_0=0$ and $z_0=0$.

Is this the right way to go about it? Any suggestions/hints will be helpful. Thanks. Edit: I removed some statements that weren't refined.

For the intersection of $L:x=0$ with your curve, you have correctly shown that $[0:0:1]$ and $[0:1:0]$ lie in the set-theoretic intersection. However, these intersections can have non-trivial multiplicities. Consider the intersection at $[0:0:1]$ for example. The intersection multiplicity is given by the length of the local ring $$\left(k[x,y]/(x,y^2-x^3+x)\right)_{(x,y)},$$ which we get by dehomogenizing with respect to $z.$ This ring is $$\left(k[x,y]/(x,y^2-x^3+x)\right)_{(x,y)}\cong \left(k[y]/(y^2)\right)_{(y)}\cong k[y]/y^2,$$ which shows it has length two. This implies that the summand in the intersection divisor coming from the intersection point $[0:0:1]$ on $L\cap X$ is actually $2[0:0:1].$
Let's try the point $[0:1:0]$ on $L\cap X.$ In this case we want to find the length of the local ring $$\left(k[x,z]/(x,z-x^3+xz^2)\right)_{(x,z)},$$ which comes from dehomogenizing with respect to $y.$ This ring is $$\left(k[x,z]/(x,z-x^3+xz^2)\right)_{(x,z)}\cong \left(k[z]/(z)\right)_{(z)}=k,$$ which has length one.
Thus, we have computed the intersection divisor of $L\cap X$ to be $2[0:0:1]+[0:1:0].$ (Note that Bezout's theorem tells us that we're done, since the total intersection multiplicity of a line with a cubic is three, and we have a divisor of degree $2+1=3.$) The other intersections can be computed similarly.
• It's essentially an algebraic way of picking out the exponents on the condition $y^2z = 0$ that we get by setting $x$ to be zero. The idea is that because of the $y^2$ term, the function $y^2z$ vanishes to order two at the intersection point $[0:0:1]$ (notice that near this point $z\neq 0$ since $x=0$) while it vanishes to order one at $[0:1:0]$ (near which we cannot have $y^2=0$). So "length" here essentially means "order of vanishing." – Andrew Mar 21 '13 at 20:44
• Hi @Elliot, yes, the answer already does it this way. E.g. consider the line $x=0$ intersecting our curve inside the chart $z \neq 0$. This is equivalent to $x=0$ intersecting the curve $y^2 = x^3 - x$ in $\mathbb{A}^2$, from dehomogenizing $z$. Hence, the intersection is given by the equation $y^2 = 0$, which is the line $y=0$ with multiplicity $2$. You can play a similar game and dehomogenize $y$ to find that the intersection inside the $y \neq 0$ chart has multiplicity $1$. Cheers – Andrew Feb 22 '17 at 19:59