7
$\begingroup$

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

$\endgroup$
1

2 Answers 2

9
$\begingroup$

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).

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.