Let $\mathbb{K}$ be a field, $A=\mathbb{K}[x_1,\dots,x_n]$ and let $\mathcal{M}$ a maximal ideal in $A.$ I want to prove that the localisation $A_\mathcal{M}$ is a regular ring. I don’t know many things about regularity, only the definition, so I want to prove that the Krull dimension of $A_\mathcal{M},$ that is, the height of the ideal $\mathcal{M}$ is equal to $\dim_{A/\mathcal{M}}{\mathcal{M}/\mathcal{M}^2}$ (the dimension of $\mathcal{M}/\mathcal{M}^2$ as $A/\mathcal{M}-$vector space, or else the minimum number of generators of the ideal, by Nakayama). I know the following results:
1) If $A$ is a finitely generated $\mathbb{K}-$algebra, and if $\mathcal{P}\subset\mathcal{Q}$ are two prime ideals in $A,$ then each maximal chain of primes between $\mathcal{P},\mathcal{Q}$ has length $$\dim{\frac{A}{\mathcal{P}}}-\dim{\frac{A}{\mathcal{Q}}}$$
2) With the same notations of 1), if $A$ is a domain then $\dim{A}=\dim{A_\mathcal{P}}+\dim{A/\mathcal{P}}.$
3) If $A$ is a Noetherian ring, $a_1,\dots,a_n\in A$ and $\mathcal{P}$ is a minimal prime containing these elements, then $\dim{A_\mathcal{P}}\leq n.$
4) If $(A,\mathcal{M})$ is a Noetherian local ring, then
$\hspace{0.5cm}$ A) $\dim{A}<\infty$
$\hspace{0.5cm}$ B) If $a\in\mathcal{M}$ then $\dim{A/(a)}\geq \dim{A}-1$, with equality if $a$ is not a zero-divisor
$\hspace{0.5cm}$ C) $\dim{A}=\min\{n\in\mathbb{N}:\exists I\subset A \text{ an ideal, } $I$ \text{ is } \mathcal{M}-\text{primary and generated by }n\text{ elements}\}$
My attempt: $A$ is a finitely generated $\mathbb{K}-$algebra which is also a domain. By 2) each maximal chain of primes bewtween $(0)$ and $\mathcal{M}$ has length $$\dim{A/(0)}-\dim{A/\mathcal{M}}=\dim{A/(0)}=\dim{A},$$ since $A/\mathcal{M}$ is a field. On the other side, using 2) we get $$\dim{A/(0)}-\dim{A/\mathcal{M}}=\dim{A_\mathcal{M}}$$ Hence each maximal ideal in $A$ has the same height $\dim{A}.$ It seems to me that what I have to do is to prove that each maximal ideal is generated by $\dim{A}$ elements, but I’m stuck at this point. Surely, if the field $\mathbb{K}$ is algebraically closed I can conclude, since by Nullstellensatz each maximal ideal is of the form $(x_1-a_1,\dots,x_n-a_n),$ but I don’t know what to do if $\mathbb{K}$ is not algebraically closed (we didn’t do anything about how to construct systems of generators for a maximal ideal). How can I conclude following this path? Or is there a better way to prove this fact with these tools?