Consider the ring $A=\mathbb{C}[x,y]/\left\langle x^2-y^3, 4y^2+x^4+5x^2y\right\rangle$. I proved that $A$ is a finite dimensional vector space over $\mathbb{C}$. More precisely $$A=\langle \overline{1},\overline{x},\overline{y},\overline{y}^2,\overline{y}^3,\overline{y}^4,\overline{y}^5,\overline{x}\overline{y},\overline{x}\overline{y}^2,\overline{x}\overline{y}^3,\overline{x}\overline{y}^4,\overline{x}\overline{y}^5\rangle$$ Thus, $A$ is Artinian and its dimension is $12$. Furthermore, there are nine maximal ideals in $A$, which are of the form $$\mathfrak{m}=(\overline{x-a},\overline{y-b})$$ for some $a,b\in \mathbb{C}$. Now, I am wondering for the dimension of $A_\mathfrak{m}$ for each maximal ideal $\mathfrak{m}$ as complex vector spaces, but I do not know how to find these.
Thank you in advance for your ideas.
Replace $x^2=y^3$ in $4y^2+x^4+5x^2y = y^2(y^2+1)(y^2+5)$, if $a\not=0\Leftrightarrow b\not=0$, it's clear that $b$ is not a repeated $0$ of $y$, and $x^2 - b^3=0$ has no repeated zero either. Therefore, $A_{\mathfrak m}\simeq \mathbb C[x,y]/\langle x-a, y-b\rangle\simeq\mathbb C$ has dimension $1$.
When $a=b=0$, then after the localization, the original ideal is the same as $\langle x^2, y^2\rangle$, and therefore the local ring has its maximal ideal generated by $\bar x,\bar y, \bar x \bar y$ over $\mathbb C$. Hence the dimension is $4$.