# Noether Normalization in $\mathbb{C}[[x_1,…,x_n]]$

I have a problem in understanding the proof of the following theorem:

Let $I\subseteq\mathbb{C}[[x_1,...,x_n]]$ be an ideal. Then there exists a $k\in\mathbb{N}$ and a linear coordinate change $\phi:\mathbb{C}[[x_1,...,x_n]]\to\mathbb{C}[[x_1,...,x_n]]$ such that $\mathbb{C}[[x_1,...,x_k]]\subseteq\mathbb{C}[[x_1,...,x_n]]/\phi(I)$ and $\mathbb{C}[[x_1,...,x_n]]/\phi(I)$ is finite as $\mathbb{C}[[x_1,...,x_k]]$-module.

As for the proof, assume $I\neq0$. Let $0\neq f\in I$, then one finds a linear coordinate change $\phi_1:\mathbb{C}[[x_1,...,x_n]]\to\mathbb{C}[[x_1,...,x_n]]$ such that $\phi_1(f)$ is $x_n$-regular. By the Weierstraß Preparation Theorem, there is a unit $u$ and a Weierstraß polynomial $p$ w.r.t. $x_n$ such that $u\phi_1(f)=p$. In particular, $\mathbb{C}[[x_1,...,x_{n-1}]]\hookrightarrow\mathbb{C}[[x_1,...,x_n]]/p$ is finite. Hence $\mathbb{C}[[x_1,...,x_{n-1}]]\to\mathbb{C}[[x_1,...,x_n]]/\phi_1(I)$ is finite.

The bold part is where I am still stuck. First, finite means the right hand side is a finitely generated module over the left hand side, is this correct (or is it over the image of the lhs)? Assuming this, the first part is due to $p$ being a Weierstraß polynomial w.r.t. $x_n$, i.e. $p$ has terms containing $x_n$ only up to a certain order.

Edit: If $f=\sum_{\mu\geq m}f_\mu$ is the homogeneous decomposition of $f$ with $f_m\neq 0$, take any $(a_1,...,a_{n-1})\in\mathbb{C}^{n-1}$ with $f_m(a_1,...,a_{n-1},1)\neq 0$, and define $\phi_1(x_i):=x_i+a_i x_n$ for $i<n$, $\phi_1(x_n):=x_n$. Then $\phi_1$ is what we wanted. But still, I don't get why this $\phi_1$ seems to work for all $f\neq 0$ in $I$ simultaneously. Or is this not what we need for the proof to work? How can I 'patch' these to get the result about $\mathbb{C}[[x_1,...,x_n]]/\phi_1(I)$?

And does it have anything to say that once the arrow $\hookrightarrow$ is used (which indicates injectivity to me), and in the next sentence it's just a 'normal' map?

Thank you in advance!

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The second map is finite because it is the composition of the first one with the canonical surjection $C[[x_1,...,x_n]]/(p) \to C[[x_1, ..., x_n]]/(\phi_1(I)$. As both maps are finite, the composition is finite. –  user18119 Aug 26 '12 at 21:25

## 1 Answer

About the first question: A ring $B$ is finite over a ring $A$ iff $B$ is a finite $A$-module. Now, let $A = \mathbb C[[x_1,\ldots,x_{n-1}]]$. Then $$C[[x_1,\ldots,x_n]]/\langle p \rangle \cong \bigoplus_{i=0}^{\deg p - 1} A x_n^i$$ as $A$-module, so it is obviously finite over $A$.

For the second part, I'm not 100% sure, but I think you simply apply the $\phi_1$ that was used for the construction of $p$ above and apply it to all elements of the ideal. Since it is linear, the result is again an ideal.

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Hello @Johannes and thanks so far! I think I have to check up on how such a $\phi_1$ can be constructed to see if for any element of the ideal, the same $\phi_1$ makes all of them regular w.r.t. $x_n$. Why / how do you use linearity there to say that $\phi_1(I)$ is again an ideal? Wouldn't surjectivity alone be enough? –  InvisiblePanda Apr 1 '12 at 8:43
I'll have to look up the details for that. Indeed, $\phi_1$ surjective implies $\phi_1(I)$ ideal, but for the construction, it was somehow useful to assume $\phi_1$ linear. Actually, in the commutative algebra class I took, we proved that $\phi_1$ was given by a block matrix of the form $$\left(\begin{array}{c|ccc} 1 & & * & \\\hline & 1 & & * \\ 0 & & \ddots & \\\ & 0 & & 1 \end{array} \right)$$ –  Johannes Kloos Apr 1 '12 at 8:53
Just a short idea: you can probably build $\phi_1$ inductively in the given form, by extending the matrix by one row and one column for each $x_k$. I guess there should be some argument for this in the Greuel/Pfister book somewhere... –  Johannes Kloos Apr 1 '12 at 9:01
Hello @Johannes, sorry I didn't reply by now. I don't know if $\phi_1$ should be built inductively, since it is only meant to work for $x_n$. Indeed, the actual $\phi$ in the claim is built inductively. I will soon edit the above post to include the construction of $\phi_1$. Hopefully - as it often is - the actual idea how it works comes to me while writing ;) if not, we'll see. Thanks so far! The Greuel / Pfister book was a nice reference, shame on me I didn't think of it, since the class I took was held by Prof. Pfister... –  InvisiblePanda Apr 16 '12 at 5:31
@Randal'Thor: If you want, I can send you a copy of my notes from Thomas Markwig's lecture - since he's in the same group as Pfister, I suppose the proof (or a variation of it) should be in there. –  Johannes Kloos Apr 16 '12 at 9:53