Showing that $K[[x_1...x_n]] / (f)$ is a free $K[[x_1,....x_{n-1}]]$ module

Question

Let $A = K[[x_1,...,x_n]]$.

For any nonzero $f \in A$, define the order of $f$ with respect to $x_n$ as the least integer $r$ such that $f$ has a term $a_rx_n^r$ in its expression, where $a_r \neq 0 \in K$.

Let $f \in A$ have order $d$ with respect to $x_n$.

Show that $A/(f)$ is a free $K[[x_1,...x_{n-1}]]$ module of rank $d$ generated by {1,$\bar x_n$ , $\bar x_n^2$ , ... , $\bar x_n^{d-1}$} (where $\bar x_n^i = $ image of $x_n^i$ in $A/(f)$).

I've been stuck on this question for a while and I'm not really sure how to proceed with it. It seems like if we mod out by a degree $d$ term in $x_n$ then it should work intuitively, though that's my intuition of the case in the polynomial ring. I'm not very good working with the power series case and translating that intuition and so I'm trying to understand a proof of this.

I'd appreciate any help!

Answer 1

Let $K[[\pmb{x}]]=K[[x_{1},\cdots,x_{n}]]$ and $K[[\pmb{x}']]=K[[x_{1},\cdots,x_{n-1}]]$. Say that $f\in K[[\pmb{x}]]$ is $x_{n}$-general of order $d$ if $$f(0,\cdots,0,x_{n})=\alpha\cdot x_{n}^{d} + \text{higher order terms in }x_{n}$$ where $\alpha\in K\setminus\{0\}$. This is the same as the order you describe above.

Theorem (Weierstrass division theorem)

Let $f\in K[[\pmb{x}]]$ be $x_{n}$-general of order $d$ and $g\in K[[\pmb{x}]]$. Then there is a unique $h\in K[[\pmb{x}]]$, $r\in K[[\pmb{x}']][x_{n}]$ such that $$ g = f\cdot h + r$$ where $\text{deg}_{x_{n}}(r)\leq d-1$.

It is clear that $K[[\pmb{x}]]$ is a $K[[\pmb{x}']]$-module, and the above theorem describes this module structure. You can then quotient out by $(f)$ and get the result that you want.

The above can be found in more generality in the first pages of the book Introduction to singularities and deformations by Greuel, G. M., Lossen, C., & Shustin, E. I. (2007)