Remember that $R_{n}=\mathbb{C}\{x_{1}, \cdots, x_{n}\}$ is the ring of formal power series convergent on a neighborhood of $0$. Note that $R_n$ is a local ring with maximal ideal $m_{n}:=\langle x_{1}, x_{2}, \cdots, x_{n}\rangle$.
We say that $f\in R_{n}$ is regular of order $b$ in $x_{n}$ if $f(0,\cdots,0, x_n)=x^{b}_{n}\cdot h(x_{n})$ with $h\in \mathbb{C}\{x_n\}$, $h(0)\not=0$.
Let $f\in R_n$ regular of order $b$ in $x_{n}$, the quotient ring $M:=\frac{R_n}{\langle f\rangle}$ which is also an $R_{n-1}$-module. Also consider the following submodule $m_{n-1}M=\{\sum_{i=1}^{n-1}g_{i}\cdot x_{i}:g_{i}\in M\}$ of $M$.
My question is the following how to prove that $\frac{M}{m_{n-1}M}$ is isomorphic to $\frac{\mathbb{C}\{x_n\}}{\langle f(0,\cdots,0, x_n)\rangle}$.
I tried to prove this isomorfism using the following theorem
- Weierstrass Division Theorem: Let $h\in R_n$ be regular of order $b$ in $x_n$ and let $g\in R_n$. Then there is a unique $q\in R_n$ and a unique $r\in R_{n-1}[z_n]$ of degree $<b$ such that $g=q\cdot h + r.$
but i don't know how to define the morphism and then use the Noether's isomorphism theorem.
Thanks