$f,g\in \mathbb{C}[x,y]$ 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?


2 Answers 2


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.

  • $\begingroup$ I don't understand why $\mathcal{O}_p$ and the ideal $(f,g)$ are equal so that the quotient would be of dimension zero? $\endgroup$
    – User101
    Feb 28, 2014 at 4:07
  • $\begingroup$ @User101: I did not say $O_p = (f,g)$. $\endgroup$
    – Youngsu
    Feb 28, 2014 at 6:50
  • $\begingroup$ @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, $\endgroup$
    – Matt E
    May 13, 2014 at 11:22

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