Minimal no. of generators of $mA_m$ , where $m$ is the maximal ideal $(\bar x -1 , \bar y -1)$ of $A=\mathbb C[x,y]/(x^3-y^2)$

Question

Let $A=\mathbb C[x,y]/(x^3-y^2)$ and consider the maximal ideal $m=(\bar x -1 , \bar y -1)$ of $A$ . Then how to compute the minimal no. of generators , $\mu(mA_m)$ , of $mA_m$ ?

Answer 1

Let us show that $\langle x-1,y-1\rangle$ is generated by just $y-1$ in $A_m$ (I'm going to be lazy and omit the bars over $x$ and $y$ and just directly use the relation $x^3=y^2$.)

So we want to find some rational function $r(x,y)\in \mathbb C(x,y)$ whose denominator is not in $\langle x-1,y-1 \rangle$ such that $$x-1=(y-1)r(x,y).$$

Okay, so $$r(x,y)=\frac{x-1}{y-1}=\frac{(x-1)(y+1)}{y^2-1}=\frac{(x-1)(y+1)}{x^3-1}=\frac{y+1}{x^2+x+1}.$$

Can we now check that $x^2+x+1$ is not contained in the ideal $\langle x-1,y-1\rangle$? Assume that it was and write $x^2+x+1$ as an $A$-combination of $x-1$ and $y-1$ and plug in $x=1$ and $y=1$ to get a contradiction. This will show that $r(x,y)\in A_m$ and finish our proof.

Extra: You might ask how I decided that this ideal is generated by just one element. And my answer is that the curve cut out by $y^2-x^3$ is singular (i.e. both partial derivatives vanish at a point on the curve) exactly at $(0,0)\in \mathbb C^2$. So at every other point (like (1,1) whose ideal is $m=\langle x-1,y-1\rangle$, the curve is smooth so the local ring at that point $A_m$ is a discrete valuation ring, which implies (among other things) that every ideal in that ring is principal.