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My objective is to show that $\mathcal{O}_{P}/(f,g)$ is of finite length as a $\mathcal{O}_{P}$-module.

$\mathcal{O}_{P}$ is the local ring of $P = (0, 0)$. In other words it's $k[x, y]_{(x,y)}$. $f$ and $g$ are irreducible polynomials that correspond to distinct curves in $\mathbb{A}^{2}$.

I think I have the proof, but I would like to see if it's correct. I've found a couple of solutions on this site, but they use heavier machinery, and I'm wondering if my simpler solution is somehow flawed.

First of all, the $\mathcal{O}_{P}$-submodules of $\mathcal{O}_{P}/(f,g)$ are the same as the $\mathcal{O}_{P}/(f,g)$-submodules since $\mathcal{O}_{P}/(f,g)$ is annihilated by $(f,g)$. It is sufficient to show that $\mathcal{O}_{P}/(f,g)$ is Artinian.

Since $k[x,y]$ is Noetherian, so is $\mathcal{O}_{P}/(f,g)$.

Now this is the interesting part. I want to show that $\mathcal{O}_{P}/(f,g)$ is of dimension $0$. $\dim k[x,y] = 2$. If $(f,g)$ generates $k[x,y]$ the result is clear. Suppose not. $(f,g)$ contains the prime ideal $(f)$ and it is contained in a maximal ideal that is distinct from $(f)$ since $g \not\in (f)$. This forces $\operatorname{height}(f,g)=2$. Thus $\dim k[x,y]/(f,g)=0$. I conclude that the localisation $\mathcal{O}_{P}/(f,g)$ also has dimension $0$.

Any flaws in this argument?

Thank you.

user26857
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PeterM
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  • possible duplicate of Quotient of a local ring at a point is a finite dimensional vector space – user26857 Aug 17 '14 at 18:25
  • @user26857 This question asks "are there any flaws in this argument", not "how to prove the statement?" –  Aug 17 '14 at 21:22
  • @user26857 I checked those answers. Thanks for that. This answer is similar to mine, but uses a result related to nonzero divisors in Noetherian rings. I'm not familiar with that. My argument doesn't use such a result. My question is whether my argument is correct or not. – PeterM Aug 17 '14 at 23:33
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    You probably want to ask yourself what you mean by "this forces". I'd argue that since $(f)$ is a prime ideal of height $1$ and $g \notin (f)$, height $(f,g)$ is at least two. – Youngsu Aug 18 '14 at 01:07
  • @PeterM: In addition to Youngsu's point, let me just explicitly state the inequality you're using, which is $\dim R/I \le \dim R - \text{ht}(I)$ which holds for any ideal $I$ in any ring $R$. Your argument is correct – zcn Aug 18 '14 at 06:41
  • Youngsu, zcn Thank you both. Yes this is what I had in mind. Feel free to post an answer and I'll accept. – PeterM Aug 18 '14 at 23:31

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