Hello everyone,

though I am quite sure my question is not on a research level, I am a bit helpless as of where to ask it. I posted it on math.stackexchange here, but there has been no answer helping me with my actual problem, so I decided to try this site. If it totally does not fit here, feel free to close, though I'd be very thankful for any help!

**Theorem.** Let $I\subseteq\mathbb{C}[[x]]=\mathbb{C}[[x_1,...,x_n]]$ be an ideal, then there exists an $r\in\mathbb{N}$ and a linear coordinate change $\varphi:\mathbb{C}[[x]]\to\mathbb{C}[[x]]$ such that $\mathbb{C}[[x_1,...,x_r]]\subseteq\mathbb{C}[[x]]/\varphi(I)$, and $\mathbb{C}[[x]]/\varphi(I)$ is finite as $\mathbb{C}[[x_1,...,x_r]]$-module.

To show this, we used the following:

**Lemma.** Let $f\in\mathbb{C}[[x]]$. Then there exists a linear coordinate change $\varphi:\mathbb{C}[[x]]\to\mathbb{C}[[x]]$ s.t. $\varphi(f)$ is $x_n$-regular of some order.
*Proof.* Let $f=\sum_{\mu\geq m}f_\mu$ be the homogeneous decomposition of $f$, $f_m\neq 0$. Take $(a_1,...,a_{n-1})\in\mathbb{C}^{n-1}$ with $f(a_1,...,a_{n-1},1)\neq 0$, and define $\varphi(x_i):=x_i+a_i x_n$ for $i\leq n-1$, $\varphi(x_n):=x_n$. Then $$\varphi(f_m)=f_m(x_1+a_1x_n,...,x_{n-1}+a_{n-1}x_n,x_n)$$ $$=f_m(a_1,...,a_{n-1},1)\cdot x_n^m+\mbox{lower degree in }x_n.$$ Then $\varphi(f)$ is $x_n$-regular of order $m$.

As for the proof of the theorem, assume $I\neq 0$. Let $0\neq f\in I$, then we find a linear coordinate change $\varphi_1$ s.t. $\varphi_1(f)$ is $x_n$-regular. By Weierstraß Preparation, there exists a unit $u\in\mathbb{C}[[x]]$ and a Weierstraß polynomial $p$ w.r.t. $x_n$ such that $u\varphi_1(f)=p$. In particular, $\mathbb{C}[[x_1,...,x_{n-1}]]\hookrightarrow\mathbb{C}[[x]]/p$ is finite, hence also $\mathbb{C}[[x_1,...,x_{n-1}]]\to\mathbb{C}[[x]]/\varphi_1(I)$ is finite. (Then the proof proceeds by induction)

I don't understand the last step here. Wouldn't it be necessary for $\varphi_1$ to work for every nonzero element in $I$ to make this possible?

Thank you very much in advance!