# Exercise I 5.4 Hartshorne

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.

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.

• Thanks, all is clear except maybe a last point, I don't see why $\dim O_p = 2$ ...
– user171326
Commented Feb 28, 2015 at 7:41
• It is the local ring of a closed point in a two-dimensional variety.
– MooS
Commented Feb 28, 2015 at 10:10
• Oh, yes sure ! You're right thanks
– user171326
Commented Feb 28, 2015 at 16:46
• "catenarian" means "catenary"? Commented Mar 16, 2015 at 18:56
• Yes, I meant catenary. Commented Mar 17, 2015 at 17:21