I want to show that if $R=K[x_1,\dots,x_n]$ and $m$ a maximal ideal, then $R_m$ is regular where $K$ is algebraically closed. The definition I have comes from Introduction to commutative algebra Macdonald/Atiyah, that is, if $A$ is a noetherian local ring with maximal ideal $m$, then $m$ can be generated by $\dim A$ elements or $\dim_F(m/m^2)=\dim A$ where $F$ is the residue field. We also have another characterization with an isomorphism but I still don't know where it comes from.
I saw that if $R$ is a finitely generated $K$-algebra, $\dim R=\dim R_m$ for any maximal ideal $m\subset R$. Here we would have that $\dim R_m=n$ and so we want to find $n$ generators of $mR_m$ which is the maximal ideal of $R_m$.
My first idea was to consider the image of $x_1,\dots, x_n$ in $R_m$ but this way I don't know how to get elements of the form $\frac{k}{1}$ for $k\in K$. Note that by the weak Nullstenllensatz we have $m=m_{c_1,\dots,c_n}$ for some $c_1,\dots ,c_n \in K$ this might be useful, otherwise why would we suppose $K$ algebraically closed.
Can someone help me ?
Thank you for your attention
Best Answer
As you know, Nullstellensatz implies that $m= (x_1 -c_1, \dots, x_n-c_n)$. That means $m$ is generated by the $\lbrace x_i-c_i\rbrace$, and thus, $m$ is the ideal of all polynomials $p(x_1,\dots,x_n)$ such that $p(c_1,\dots,c_n)= 0$.
The residue field $R/m$ is isomorphic to $K$ (recall that the map $R\rightarrow K$ given by $p(x_1,\dots,x_n)\mapsto p(c_1,\dots,c_n)$ is surjective with kernel $m$). so you have to show that $m/m^2$ is $n$-dimensional as a $K$-vector space.
We are going to build a basis for this space, with $n$ elements.
Consider, for every $p(x_1,\dots,x_n)\in m$, the Taylor expansion at $(c_1, \dots, c_n)$:
$$p(x_1,\dots,x_n)= \sum_{i=1}^n \frac{\partial p}{\partial x_i}(c_1,\dots,c_n)(x_i-c_i) + R(x_1,\dots x_n) $$
With $R\in m^2$. Taking residues modulo $m^2$ shows that the class of $p$ in $m/m^2$ is a linear combination of the classes of $x_i-c_i$ (with coefficients equal to the partial derivatives), which are therefore a system of generators of $m/m^2$.
Nakayama's Lemma then implies that the classes of $x_i-c_i$ in $R_m$ are a system of generators of $mR_m$.
Still, to see that they are linearly independent, if there was a linear combination equal to $0$:
$$a_1(x_1-c_1) + \dots a_n(x_n-c_n) = 0\quad mod.\ m/m^2$$
Then $a_1(x_1-c_1) + \dots a_n(x_n-c_n)\in m^2$, so its first partial derivatives are zero at $(c_1, \dots, c_n)$, but they are equal to the $a_i$.