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](https://math.stackexchange.com/questions/126559/noether-normalization-in-mathbbcx-1-x-n), 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!



Noether Normalization in $\mathbb{C}[[x_1,...,x_n]]$
Before proving the theorem I would, using the lemma, prove an intermediate result from which the theorem follows. Actually that intermediate result looks more like Noether normalization to me than the final theorem, but anyway. 



> 
> **Proposition:** If $A$ is a finite $C[[x\_1,...,x\_n]]$ algebra with structure map $i: C[[x\_1,...,x\_n]] \to A$ then there is a coordinate change $\varphi: x\_i \mapsto y\_i$ and $r \le n$ so that the composition
>  $$j:C[[y\_1,...,y\_r]] \hookrightarrow C[[y\_1,...,y\_n]] \xrightarrow[]{\varphi^{-1}}C[[x\_1,...,x\_n]] \xrightarrow[]{i}A$$
>  is injective and $A$ is finite over $C[[y\_1,...,y\_r]]$. 
> 
> 
> 


To prove your theorem, take $A=C[[x\_1,...,x\_n]]/I$ and compose $j$ with the isomorphism 
$$C[[x\_1,...,x\_n]]/I \xrightarrow[]{\bar{\varphi}}C[[y\_1,...,y\_n]]/\varphi(I).$$ 


The prove of the Proposition is by induction on $n$: If $i$ is injective, we are done. Otherwise let $f$ be in the kernel. As shown in your question, there is a coordinate change $\varphi\_1:x\_i \mapsto y\_i$ so that 
$$j\_1: C[[y\_1,...y\_{n-1}]] \hookrightarrow C[[y\_1,...y\_n]]/(\varphi\_1(f)) \cong C[[x\_1,...,x\_n]]/(f)$$ is injective and the latter is finite over the former. Hence $A$ is finite over $C[[y\_1,...,y\_{n-1}]]$ with structure map 
$$i\_1: C[[y\_1,...,y\_{n-1}]] \xrightarrow{j\_1} C[[x\_1,...,x\_n]]/(f) \xrightarrow[]{\bar{i}} A.$$
By induction hypothesis, there is another coordinate change $\varphi\_2: y\_i \mapsto z\_i$ (which we extend by $z\_n := y\_n)$ and $r \le n$ so that 
$$j\_2: C[[z\_1,...,z\_r]] \hookrightarrow C[[z\_1,...,z\_{n-1}] \xrightarrow[]{\varphi\_2^{-1}}C[[y\_1,...,y\_{n-1}]]\xrightarrow[]{i\_1}A$$
is injective and $A$ is finite over $C[[z\_1,...,z\_r]]$. Now $\varphi:= \varphi\_2 \circ \varphi\_1$ is the required coordinate change. 


