How to compute a minimal spanning set and the minimal spanning number of $\Bbb C[x,y]_{(x-1,y-1)}/(x^3-y^2)$?

Question

Let $A=\frac{\mathbb C[X,Y]}{(X^3-Y^2)}$. I am asked to show that $\mathbf m=(\overline X-1,\overline Y-1)$ is a maximal ideal of $A$ which I have shown successfully. Now I am asked to compute the $\mu(\mathbf mA_m)$ which is the unique cardinality of a minimal generating set of $\mathbf mA_m$ which I am unable to do. I know I have to calculate $\dim_{A_m/mA_m}(mA_m/(mA_m)^2)$. Can someone help me to calculate it? I know I have to use the formula $\mu(mA_m)=\dim_{A_m/mA_m}(mA_m/(mA_m)^2)$ but I got stuck while actually doing it. Some help is needed as I am just a beginner.

Answer 1

The first step is to simplify the algebraic expression $mA_m/(mA_m)^2$ a bit. As localization is exact, $mA_m/(mA_m)^2\cong (m/m^2)_m$, and since $A\setminus m$ acts invertibly on $m/m^2$, we get that $mA_m/(mA_m)^2\cong m/m^2$ as $A/m$-vector spaces.

Now we'll do even a bit more work to make our problem easier: there's a natural map from $(x-1,y-1)\subset\Bbb C[x,y]$ to $m$, and we can compose this with the map $m\to m/m^2$ to get a composite map $(x-1,y-1)\to m/m^2$, which has kernel generated by $(x-1,y-1)^2$ and $x^3-y^2$. So $m/m^2\cong (x-1,y-1)/((x-1,y-1)^2+(x^3-y^2))$, so $m/m^2$ has a surjective map from the two-dimensional vector space $(x-1,y-1)/(x-1,y-1)^2$ with kernel generated by $f=x^3-y^2$. Now all we have to do is figure out whether $f\in (x-1,y-1)^2$ or not: if $f\in(x-1,y-1)^2$, then the kernel is zero, and $m/m^2$ is two-dimensional, else the kernel is the span of $f$ and $m/m^2$ is one-dimensional.

I'll give you a shot at this computation on your own. You can check your work by looking under the spoiler:

$x^3-y^2\in (x-1,y-1)^2$ iff $(x+1)^3-(y+1)^2\in(x,y)^2$, and $(x+1)^3-(y+1)^2 = x^3+3x^2+3x-y^2-2y$ is not in $(x,y)^2=(x^2,xy,y^2)$. Therefore $m/m^2$ is one-dimensional.