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) For a finitely generated \(d\)-dimensional \(\mathbb{K}\)-algebra \(A\), suppose we are given natural numbers satisfying

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

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

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

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 the identity

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

holds.

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

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

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

The proof of Theorem 1 (Noether normalization lemma) is then as follows.

Proof of Theorem 1

Since \(A\) is a finitely generated \(\mathbb{K}\)-algebra, we may write \(A=\mathbb{K}[\y_1,\ldots, \y_r]/\mathfrak{a}\). Given a chain of ideals satisfying the stated conditions, we consider the chain of their preimages in \(\mathbb{K}[\y_1,\ldots, \y_r]\)

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

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

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

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

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 conditions. For this, suppose we are given elements \(\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d\) satisfying the 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 \(i\), we have \(\mathfrak{a}_i\cap B_e\supset(\x_m,\ldots, \x_d)\), 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. Repeating this process, the fact that the final \(B=B_{d_m}\) obtained satisfies the desired conditions is obvious once we show that the inclusion in the second condition is actually an equality, and this follows immediately by considering the dimensions of the two ideals of \(B\) on either side.

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 let \(i\) be the smallest index such that \(e>d_i\). Then

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

Indeed, if this intersection were \(0\), then by the second condition we would have

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

but the ideal on the left has dimension \(d_i\) while the ideal on the right has dimension \(e\), a contradiction. Now choose \(\x_e\) to be any nonzero polynomial in this intersection; then by Lemma 2 we may replace the 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 \(\dim A=\trdeg_\mathbb{K}\Frac(A)\).

Theorem 3 generalizes naturally to the situation of taking a quotient, giving the following dimension formula.

Theorem 4 For a finitely generated \(\mathbb{K}\)-algebra domain \(A\) and its prime ideal \(\mathfrak{p}\), the following holds.

\[\dim A/\mathfrak{p}+\operatorname{ht}\mathfrak{p}=\dim A\]

Proof

The inequality \(\dim A/\mathfrak{p}+\operatorname{ht}\mathfrak{p}\leq\dim A\) holds for arbitrary rings (§Dimension, ⁋Definition 2), so it suffices to show the opposite inequality. Set \(n=\dim A\) and \(d=\dim A/\mathfrak{p}=\dim\mathfrak{p}\). Applying Theorem 1 (Noether normalization lemma) to the chain consisting of the single ideal \(\mathfrak{a}_1=\mathfrak{p}\) of \(A\), we obtain a subring \(B\cong\mathbb{K}[\x_1,\ldots, \x_n]\) of \(A\) such that \(A\) is a finitely generated \(B\)-module and \(\mathfrak{p}\cap B=(\x_{d+1},\ldots, \x_n)\). Since \(B\hookrightarrow A\) is an integral extension, by [Commutative Algebra] §Integral Extensions and Ideals, ⁋Proposition 1 and Integral Extensions and Ideals we have \(\operatorname{ht}_A\mathfrak{p}=\operatorname{ht}_B(\mathfrak{p}\cap B)\).

Now we compute the height of the ideal \((\x_{d+1},\ldots, \x_n)\) in the polynomial ring \(B=\mathbb{K}[\x_1,\ldots, \x_n]\). The chain

\[(0)\subseteq(\x_n)\subseteq(\x_{n-1},\x_n)\subseteq\cdots\subseteq(\x_{d+1},\ldots, \x_n)\]

shows that this ideal has height at least \(n-d\), and since the quotient ring \(B/(\x_{d+1},\ldots, \x_n)\cong\mathbb{K}[\x_1,\ldots, \x_d]\) has dimension \(d\) (§System of Parameters, ⁋Corollary 11), the inequality \(\dim+\operatorname{ht}\leq n\) shows that the height is at most \(n-d\). Therefore \(\operatorname{ht}_B(\x_{d+1},\ldots, \x_n)=n-d\), and hence \(\operatorname{ht}\mathfrak{p}=n-d=\dim A-\dim A/\mathfrak{p}\).

On the other hand, the polynomial subring provided by Theorem 1 (Noether normalization lemma) is compatible with enlarging the coefficients, which also yields that the dimension of a finitely generated \(\mathbb{K}\)-algebra does not change when the coefficient field is enlarged.

Proposition 5 For an extension \(\mathbb{K}\hookrightarrow \mathbb{L}\) of a field \(\mathbb{K}\) and a finitely generated \(\mathbb{K}\)-algebra \(A\), the following holds.

\[\dim(A\otimes_\mathbb{K}\mathbb{L})=\dim A\]

Proof

Set \(d=\dim A\) and apply Theorem 1 (Noether normalization lemma) with no chain of ideals to choose a subring \(B\cong\mathbb{K}[\x_1,\ldots, \x_d]\) of \(A\) such that \(A\) is a finitely generated \(B\)-module. Now \(\mathbb{L}\) is a \(\mathbb{K}\)-vector space and hence a free \(\mathbb{K}\)-module, so it is flat ([Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Definition 7), and therefore the ring homomorphism

\[B\otimes_\mathbb{K}\mathbb{L} \rightarrow A\otimes_\mathbb{K}\mathbb{L}\]

obtained by applying \(-\otimes_\mathbb{K}\mathbb{L}\) to the inclusion \(B\hookrightarrow A\) is again injective. Here \(B\otimes_\mathbb{K}\mathbb{L}\cong\mathbb{L}[\x_1,\ldots, \x_d]\), and if \(a_1,\ldots, a_r\) generate \(A\) as a \(B\)-module then the elements \(a_i\otimes 1\) generate \(A\otimes_\mathbb{K}\mathbb{L}\) as a \(B\otimes_\mathbb{K}\mathbb{L}\)-module, so this injection is finite and hence an integral extension. (§Integral Extensions, ⁋Lemma 4)

But integral homomorphisms preserve dimension (§Dimension, ⁋Proposition 4), so we obtain

\[\dim(A\otimes_\mathbb{K}\mathbb{L})=\dim\mathbb{L}[\x_1,\ldots, \x_d]=d\]

where the last equality is §System of Parameters, ⁋Corollary 11.

---

References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.