Let $R=\mathbb{Z}[x,y]$, $A=R/(y^3-x^3-4)$ and $m=(x,y,2)$. Now I want to find out what the dimension is of $A_m$ and say whether it is regular or not. My prefered definition of the ring $A_m$ being regular is:
"Let $A_m$ be a Noetherian local ring of dimension $d$ with $m$ its maximal ideal and $k=A_m/m$ its residue field. Then $A_m$ is regular if $\text{dim}_k(m/m^2)=d$ (or $m$ can be generated by $d$ elements)."
To determine the dimension of $A_m$, first we look at $R$ and $R_m$ I guess, because for determining dimensions we can interchange localisations and taking quotients. The dimension of $R$ is $3$ because $(0)\subset(p)\subset(p,x)\subset(p,x,y)$ with $p\in\mathbb{Z}$ prime (is that how I write it down?). Then we know dim$R_m\leqslant3$. We have $(0)\subset(x)\subset(x,y)\subset(x,y,2)$ in $R_m$, so dim$R_m=3$. Now what is dim$R_m/(y^3-x^3-4)=$dim$A_m$? We know that $y^3-x^3-4$ is an irreducible polynomial by the Eisenstein criterion. So $(y^3-x^3-4)$ is prime in $R_m$. Maybe there's a formula we can use about some chain of ideals containing this ideal $(y^3-x^3-4)$ and the dimension of $R_m$ using the upperbound we have?
Now, say $A_m$ has dimension $2$, then $A_m$ can't be regular since $m=(x,y,2)$ is generated by $3$ elements. How to determine $\text{dim}_km/m^2$? Can we compute it in $R$ also, to generalise?
Is there a step by step plan or a list of things to consider in these kind of computations in order to solve them?