Noether Normalization

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Noether Normalization

The goal of this post is to prove the following theorem and examine its consequences.

Theorem 1 (Noether normalization lemma) Let \(A\) be a finitely generated \(d\)-dimensional \(\mathbb{K}\)-algebra. Suppose we are given natural numbers satisfying

\[d_1>d_2>\cdots>d_m>0\]

and a descending chain of ideals of \(A\)

\[\mathfrak{a}_1\subset \mathfrak{a}_2\subset\cdots\subset \mathfrak{a}_m\]

with \(\dim \mathfrak{a}_i=d_i\). Then there exists a suitable subring \(B\cong \mathbb{K}[\x_1,\ldots, \x_d]\) of \(A\) such that \(A\) is finitely generated as a \(B\)-module and

\[\mathfrak{a}_i\cap B=(\x_{d_i+1},\ldots, \x_d)\qquad\text{for $i=1,\ldots, m$}\]

holds.

This can be proved using the following lemma, whose proof we omit.

Lemma 2 Let \(\mathbb{K}\) be a field and let \(f\in B=\mathbb{K}[\x_1,\ldots, \x_r]\) be a non-constant polynomial. Then there exist elements \(\x_1',\ldots, \x_{r-1}'\in B\) such that, if we let \(B'\) be the \(\mathbb{K}\)-subalgebra of \(B\) generated by \(\x_1',\ldots, \x_{r-1}', f\), then \(B\) is a finitely generated \(B'\)-module. Moreover, these elements can be chosen as follows.

1. For a sufficiently large integer \(e\), we can take \(\x_i'=\x_i-\x_r^{e}\).
2. If \(\mathbb{K}\) is an infinite field, then we can take \(\x_i'=\x_i-a_i\x_r\) for suitable \(a_i\in \mathbb{K}\).

Then the proof of Theorem 1 (Noether normalization lemma) is as follows.

Proof of Theorem 1

First, since \(A\) is a finitely generated \(\mathbb{K}\)-algebra, we can write \(A=\mathbb{K}[\y_1,\ldots, \y_r]/\mathfrak{a}\). Then, given a chain of ideals satisfying the given condition, we consider the chain of their preimages in \(\mathbb{K}[\y_1,\ldots, \y_r]\)

\[\tilde{\mathfrak{a}}_1\subset \tilde{\mathfrak{a}}_2\subset\cdots\subset \tilde{\mathfrak{a}}_m\]

and insert \(\mathfrak{a}_0=\mathfrak{a}\) to view it as a descending chain of ideals in \(\mathbb{K}[\y_1,\ldots, \y_r]\)

\[\mathfrak{a}\subset \tilde{\mathfrak{a}}_1\subset \tilde{\mathfrak{a}}_2\subset\cdots\subset \tilde{\mathfrak{a}}_m\]

so it suffices to prove the claim for the polynomial ring \(A=\mathbb{K}[\y_1,\ldots, \y_r]\). In this case, by §System of Parameters, ⁋Corollary 11, we must have \(r=d\).

Now, to construct the elements \(\x_i\) of the theorem, we first set \(\x_i'=\y_i\) and then modify them to find \(\x_d\) satisfying the given condition. To this end, suppose we are given elements \(\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d\) satisfying the following two conditions:

1. \(A\) is a finitely generated \(B_e=\mathbb{K}[\x_1',\ldots, \x_e',\x_{e+1},\ldots, \x_d]\)-module.
2. For each \(i\), \(\mathfrak{a}_i\cap B_e\supset(\x_m,\ldots, \x_d)\) holds, where \(m=\max(d_i+1, e+1)\).

We show that from these we can find new elements \(\x_1',\ldots, \x_{e-1}'\) and \(\x_e\) such that the above conditions are preserved. Then, repeating this process, it is obvious that the final \(B=B_{d_m}\) obtained satisfies the desired condition once we show that the inclusion in the second condition is in fact an equality, which is clear by considering the dimensions of the two ideals of \(B\) on both sides.

Now, to complete this induction, let \(e\) satisfy \(d\geq e>d_m\), suppose we are given \(\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d\) satisfying the two conditions above, and assume that \(i\) is the smallest index satisfying \(e>d_i\). Then

\[\mathfrak{a}_i\cap \mathbb{K}[\x_1',\ldots, \x_e']\neq 0\]

If this intersection were \(0\), then by the second condition

\[\mathfrak{a}_i\cap B_e\supseteq (\x_{e+1},\ldots, \x_d)\]

would hold, but the ideal on the left-hand side has dimension \(d_i\) while the ideal on the right-hand side has dimension \(e\), a contradiction. Now take \(\x_e\) to be any nonzero polynomial in the above intersection, and then use Lemma 2 to replace the new elements \(\x_1',\ldots, \x_{e-1}'\) with new ones as well.

Consequences

Theorem 1 (Noether normalization lemma) yields the following result.

Theorem 3 Let \(A\) be an integral domain that is a finitely generated \(\mathbb{K}\)-algebra. Then \(\dim A=\trdeg_\mathbb{K}\Frac(A)\).