Exercise I 5.4 Hartshorne

Question

I have no idea how to begin this exercise from the Hartshorne:

If $Y$, $Z$ are two varieties of $\mathbb A^2$ given by the equation $f=0$ and $g=0$, the intersection multiplicity at $P$ is the lengh of the $\mathcal O_p$-module $M = \mathcal O_p/(f,g)$.

The first question is about show that the intersection number is finite. I was thinking to use this characterization : $M$ has finite lenght $\Leftrightarrow$ $M$ is Artinian and Noetherian. I know that $M$ is Noetherian (localization of a Noetherian ring) but I have no idea on how to show that $M$ is Artinian.

Thanks in advance.

Answer 1

The ring $\mathcal O_P$ is a two-dimensional, catenarian, local, integral, noetherian ring. Therefore $f$ is a non-zero-divisor (and not a unit). Therefore every minimal prime $\mathfrak q$ over $(f)$ in $\mathcal O_P$ is of height $1$. (Krull's Theorem). Because $\mathcal O_p$ is catenarian, it follows that $\dim \mathcal O_P/(f) = 1$.

As $V(f)$ is a variety the ring $A=\mathcal O_P/(f)$ is integral. Therefore $g$ is also a non-zero-divisor in $A$ (and not a unit). Therefore every minimal prime $\mathfrak q'$ in $A$ over $g$ is of height $1$. As $\mathcal O_P$ is catenarian of dimension two it follows, that $\dim A/(g) = \dim \mathcal O_P/(f,g) = 0$. So $\mathcal O_P/(f,g)$ is a zero-dimensional noetherian ring, therefore artinian and of finite length.