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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!

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$\phi_1(f)$ and $p$ generate the same ideal. –  Martin Brandenburg Apr 19 '12 at 18:16
    
Hi Martin, and I have a very stupid typo in my question: this was clear to me, and I meant that then $\mathbb{C}[[x_1,...,x_{n-1}]]\to\mathbb{C}[[x]]/\varphi_1(I)$ is finite. This would be the step I don't understand. I'll edit my post, and sorry. –  InvisiblePanda Apr 20 '12 at 6:27
    
Just saw that in the stackexchange link the question is a bit different: "Why is $C[[x_1,...,x_{n-1}]] \to C[[x_1,...,x_n]]/\phi_1(I)$ finite ?" The answer is simple: You know that $C[[x_1,...,x_n]]/(p)$ is finite over $C[[x_1,...,x_{n-1}]]$. But since $(p)=(\phi_1(f)) \le \phi_1(I)$, the map $C[[x_1,...,x_n]]/(p) \to C[[x_1,...,x_n]]/\phi_1(I)$ is surjective and hence finite (generator is just $1$). Now apply the general (and easy to prove) result that if $R \to S$ and $S \to T$ are finite, then $R \to T$ is also finite. –  Ralph Apr 20 '12 at 6:30
    
Hello Ralph! How blind can one be? I somehow must have thought that I have 'to do more' or 'need more' for it to work since $\varphi_1(I)$ is bigger than $\langle p\rangle$, but we are looking at the quotient, so it actually gets easier and not harder... thanks for making this clear! –  InvisiblePanda Apr 20 '12 at 7:04
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1 Answer

up vote 4 down vote accepted

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.

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