This question is very related to this other question. I have an alternative solution to the ones proposed in the answers, and I'd like to know if it is correct.

I want to find the dimension of $A=\mathbb{C}[x,y]/(x^3-y^2)$ using the dimension theory for Noetherian local rings that is extensively developed in chapter $11$ of the book Introduction to Commutative Algebra by Atiyah and MacDonald. If $R$ is any Noetherian ring we have: $$\dim R = \sup_{m\subset R\mathrm{\ max.\ id.}}\dim R_m$$ (since every chain of primes is contained in a maximal ideal). Let $m\subset A$ be a maximal ideal. If $x^3-y^2\notin m$, then $A_m=0$ and thus $\dim A_m=0$, else if $x^3-y^2\in m$ we use exactness of the localization functor to see that: $$A_m=\mathbb{C}[x,y]_m/(x^3-y^2)_m=\mathbb{C}[x,y]_m/(x^3-y^2)$$ where we denote again by $x^3-y^2$ the element $\frac{x^3-y^2}{1}\in\mathbb{C}[x,y]_m$. Since $x^3-y^2$ is not a zero-divisor in $\mathbb{C}[x,y]_m$, we have (by corollary $11.18$ in Atiyah/MacDonald): $$\dim A_m=\dim \mathbb{C}[x,y]_m - 1=1$$ Thus $\dim A=1$.

Is this a valid way to proceed or am I doing something illicit? Also, would this work also for quotients by ideals generated by more than one element? For example, say I'd like to compute $\dim \mathbb{C}[x,y,z]/(x^2+y,x^3y^2)$. I would use the fact that: $$A/(f,g)\cong[A/(f)]/[(f,g)/(f)]$$ Then I would compute the dimension of $A/(f)$ as above. $(f,g)/(f)=(\bar{g})$ (generated by the image of $g$ in $A/(f)$), and repeat.

Daniel Robert-Nicoud
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  • This is known as the *cusp curve*, so the dimension is probably ... – Martin Brandenburg Jan 26 '14 at 01:55
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    @MartinBrandenburg The question here is not "what is the dimension", but rather "is this method to compute the dimension correct? If not, where does it fail?" – Daniel Robert-Nicoud Jan 26 '14 at 02:01
  • This is a valid way to proceed. It's probably the way I would solve the problem. However, your $\in$ and $\notin$ symbols in the above derivation should be switched. Of course, when you have more than one element generating your ideal, you have to worry about whether the image of $g$ is a zero-divisor on $A/(f)$. – neilme Jan 26 '14 at 02:04
  • This is probably trivial but I don't see why $\frac{x^3-y^2}{1}$ is not a zero divisor in $\mathbb{C}[x,y]_m$ and why $\frac{x^3-y^2}{1}$ is not a unit in $\mathbb{C}[x,y]_m$. Could you help me, please? Thanks. – user8463524 Jun 16 '15 at 19:32
  • @jeffrey Because $x^3-y^2$ is not a zero divisor in $\mathbb{C}[x,y]$, and localizing we're just inverting the elements not in $m$. – Daniel Robert-Nicoud Jun 16 '15 at 23:30
  • @DanielRobert-Nicoud, sorry but I don't understand. Ok, $f=x^3-y^2$ is no zero divisor in $\mathbb{C}[x,y]$. If $\frac{f}{1}$ was a zero divisor in the localization, this is $\frac{f}{1}\cdot \frac{g}{h}=0\Leftrightarrow \exists t\in \mathbb{C}[x,y]\setminus m: t(fh-g)=0$ which is equivalent to $fh=g$ since $\mathbb{C}[x,y]$ is a domain. Here, $g\in \mathbb{C}[x,y]$ and $h\in \mathbb{C}[x,y]\setminus m$. Why can't such elements exist? – user8463524 Jun 17 '15 at 06:31
  • @jeffrey Your condition is wrong, it should be (...) such that $t(fg-h)=0$. Then it's pretty obvious... – Daniel Robert-Nicoud Jun 17 '15 at 18:25
  • @DanielRobert-Nicoud, ah, I am sorry, you are right, I see. Thank you. – user8463524 Jun 17 '15 at 18:56

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