# Quotient of a local ring at a point is a finite dimensional vector space

$f,g\in \mathbb{C}[x,y]$ are irreducible polynomials, and the varieties $V_1=V(f)$ and $V_2=V(g)$ are not equal. Is the ring $\mathcal{O}_p/(f,g)$ a finite dimensional vector space over $\mathbb{C}$? (Here $\mathcal{O}_p$ denotes the local ring of $\mathbb{A}^2$ at the point $p\in \mathbb{A}^2$.)

How can i prove this? Any hints/suggestions would be highly appreciated.

Is there anyway that i could relate the ideal $(f,g)$ to the maximal ideal $\mathfrak{m}_p=(x-p_1,y-p_2)$ where $p=(p_1,p_2)\in \mathbb{A}^2$? If so how can i proceed further?

• Aug 17, 2014 at 22:10

Saying what Youngsu said, slightly more geometrically:

You know that $$V(f)$$ is a dimension $$1$$ topological space, and since $$V(g)\cap V(f)\subsetneq V(f)$$ (why?) this implies that $$V(f)\cap V(g)$$ is a proper closed subset of $$V(f)$$. But, since $$V(f)$$ is Noetherian, you know that $$V(f)\cap V(g)$$ can be decomposed into a finite union of irreducible closed subsets of $$V(f)$$ which, by dimension considerations, must be dimension $$0$$.

Thus, we see that $$\text{Spec }(\mathbb{C}[x,y]/(f,g))$$ is a finite $$\mathbb{C}$$-variety, and thus must be Artinian, and so finite as a $$\mathbb{C}$$-space.

In less fancy words, the intersection of the two curves must be finite, purely by dimension considerations, and since the only varieties supported on finitely many points are finite dimensional $$\mathbb{C}$$-spaces, this implies your desired result.

Implicitly, I am using this very nice, and very useful theorem:

Theorem: Let $$k$$ be a field, and let $$A$$ be a finite type $$k$$-algebra. Then, the following are equivalent:

1. $$A$$ is a finitely generated $$\mathbb{C}$$-module.
2. $$\text{Spec}(A)$$ is finite.
3. $$\text{Spec}(A)$$ is discrete.
4. $$\text{MaxSpec}(A)$$ is finite.

Geometrically, this is saying something about the fibers of quasifinite morphisms. Namely, that they are finite type morphisms whose fibers, over any point $$p$$, satisfy any of the above equivalent properties over the residue field at $$p$$ (usually they are defined to be finite type morphisms whose fibers are finite).

Since $V_i$ are varieties and they are not equal, $f,g$ are relatively prime elements. Therefore, $O_p/(f)$ is a $1$-dimensional integral domain and the image of $g$ is a non zerodivisor. Therefore, $O_p/(f,g)$ is of dimension zero; hence it is of finite length.

• I don't understand why $\mathcal{O}_p$ and the ideal $(f,g)$ are equal so that the quotient would be of dimension zero? Feb 28, 2014 at 4:07
• @User101: I did not say $O_p = (f,g)$. Feb 28, 2014 at 6:50
• @User101: Dear User, Youngsu means "of dimension zero" as a ring (not a zero-dimensional vector space, which seems to have been your interpretation). Regards, May 13, 2014 at 11:22