Noether Normalization

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

Theorem 1 (Noether normalization lemma) For a finitely generated $d$-dimensional $\mathbb{K}$-algebra $A$, suppose that non-negative integers satisfying the following inequality

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

and, satisfying $\dim \mathfrak{a}_i=d_i$, a descending chain of ideals of $A$

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

are given. Then there exists in $A$ a suitable subring $B\cong \mathbb{K}[\x_1,\ldots, \x_d]$ such that $A$ is finitely generated as a $B$-module and the following equation

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

holds.

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

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

For a sufficiently large integer $e$, we can choose $\x_i'=\x_i-\x_r^{e^i}$.
If $\mathbb{K}$ is an infinite field, then for suitable $a_i\in \mathbb{K}$ we can choose $\x_i'=\x_i-a_i\x_r$.

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 $A=\mathbb{K}[\y_1,\ldots, \y_r]/\mathfrak{a}$. Given a chain of ideals satisfying the given 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 $\mathbb{K}[\y_1,\ldots, \y_r]$

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

so that it suffices to prove the assertion only for the polynomial ring $A=\mathbb{K}[\y_1,\ldots, \y_r]$. In this setting, the dimension of the newly inserted $\mathfrak{a}_0$ is $d_0=\dim \mathfrak{a}_0=\dim A$, and what takes the place of $d$ in the theorem in this reduced setting is the dimension of the new ambient ring, namely $\dim \mathbb{K}[\y_1,\ldots, \y_r]=r$ given by the first result of §System of Parameters, ⁋Corollary 11. Therefore, below we set $d=r$ and write the original $\dim A$ as $d_0$. If the claim holds for polynomial rings, the general result is obtained as follows. Suppose that in $\mathbb{K}[\y_1,\ldots, \y_r]$ there exists a subring $B\cong \mathbb{K}[\x_1,\ldots, \x_r]$ such that $\mathbb{K}[\y_1,\ldots, \y_r]$ is a finitely generated $B$-module and $\mathfrak{a}_0\cap B=(\x_{d_0+1},\ldots, \x_r)$, $\tilde{\mathfrak{a}}_i\cap B=(\x_{d_i+1},\ldots, \x_r)$ hold. Let $\pi:\mathbb{K}[\y_1,\ldots, \y_r] \rightarrow A$ be the quotient map and let $\bar{B}=\pi(B)$. Since the kernel of the restriction of $\pi$ to $B$ is $\mathfrak{a}_0\cap B$, we have $\bar{B}\cong B/(\mathfrak{a}_0\cap B)\cong \mathbb{K}[\x_1,\ldots, \x_{d_0}]$, and since $\mathbb{K}[\y_1,\ldots, \y_r]$ is finite over $B$, its quotient $A$ is finite over $\bar{B}$. Moreover, for $b\in B$, $\pi(b)\in \mathfrak{a}_i$ is equivalent to $b\in \tilde{\mathfrak{a}}_i$, so $\mathfrak{a}_i\cap \bar{B}=\pi(\tilde{\mathfrak{a}}_i\cap B)$, and since $\x_{d_0+1},\ldots, \x_r$ all belong to the kernel of $\pi$, this is equal to $(\pi(\x_{d_i+1}),\ldots, \pi(\x_{d_0}))$. That is, $\bar{B}$ is the subring required by the theorem for the original $A$.

Now, to construct the elements $\x_i$ of the theorem, we first set $\x_i'=\y_i$ and successively modify them to find $\x_i$ satisfying the given conditions. To this end, consider the following two conditions:

$A$ is a finitely generated $B_e=\mathbb{K}[\x_1',\ldots, \x_e',\x_{e+1},\ldots, \x_d]$-module.
For each $i$, the inclusion $\mathfrak{a}_i\cap B_e\supseteq(\x_k,\ldots, \x_d)$ holds, where $k=\max(d_i+1, e+1)$.

Suppose we are given elements $\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d$ satisfying these conditions; we show that from these we can find new elements $\x_1',\ldots, \x_{e-1}'$ and $\x_e$ such that the above conditions continue to hold. Repeating this process, to show that the final subring $B=B_{d_m}$ obtained satisfies the desired conditions, it suffices to show that the inclusion in the second condition is actually an equality, which is verified as follows. First, since $A$ is a finitely generated $B$-module, $B\hookrightarrow A$ is an integral extension by §Integral Extensions, ⁋Lemma 4, and therefore $\dim(\mathfrak{a}_i\cap B)=\dim \mathfrak{a}_i=d_i$ by §Krull Dimension, ⁋Proposition 4. On the other hand, the second condition for $B=B_{d_m}$ is $\mathfrak{a}_i\cap B\supseteq(\x_{d_i+1},\ldots, \x_d)$, so if we let the image in $R=B/(\x_{d_i+1},\ldots, \x_d)\cong \mathbb{K}[\x_1,\ldots, \x_{d_i}]$ of $\mathfrak{a}_i\cap B$ be $J$, then $B/(\mathfrak{a}_i\cap B)\cong R/J$, and in particular $\dim R/J=d_i$. However, since $\dim \mathfrak{a}_i=d_i$, $\mathfrak{a}_i$ is a proper ideal, and hence $J$ is also a proper ideal of $R$. If $J\neq 0$, any element belonging to $J$ that is not $0$, say $g$, is not a unit, that is, a non-constant polynomial. Now, any prime ideal containing $J$, say $\mathfrak{p}$, contains $g$, so among the minimal prime ideals containing $(g)$, there is one contained in $\mathfrak{p}$, say $\mathfrak{q}$; by §Krull Dimension, ⁋Theorem 6 (Codimension one Principal Ideal Theorem), $\codim \mathfrak{q}\leq 1$, but since $R$ is a domain and $g\neq 0$, we have $\mathfrak{q}\neq(0)$, so $\codim \mathfrak{q}=1$. Then from the inequality $\dim \mathfrak{q}+\codim \mathfrak{q}\leq \dim R=d_i$ (§Krull Dimension, ⁋Definition 2, §System of Parameters, ⁋Corollary 11), we obtain $\dim R/\mathfrak{p}\leq \dim \mathfrak{q}\leq d_i-1$, and taking the supremum over all such $\mathfrak{p}$ yields $\dim R/J\leq d_i-1$, which is a contradiction. Therefore $J=0$, which means $\mathfrak{a}_i\cap B=(\x_{d_i+1},\ldots, \x_d)$.

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

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

holds. Indeed, if we assume this intersection is $0$, when writing an arbitrary element $g\in B_e$ as the sum $g=g_0+h$ of $g_0\in \mathbb{K}[\x_1',\ldots, \x_e']$ and $h\in (\x_{e+1},\ldots, \x_d)$, since $h\in \mathfrak{a}_i$ by the second condition, if $g\in \mathfrak{a}_i$, then $g_0\in \mathfrak{a}_i\cap \mathbb{K}[\x_1',\ldots, \x_e']=0$. Therefore,

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

holds, but the ideal on the left-hand side is $d_i$-dimensional, while the dimension of the ideal on the right-hand side is $e$, which is a contradiction. Now choose $\x_e$ to be any non-zero polynomial belonging to the intersection above. Since $\mathfrak{a}_i$ is a proper ideal, $\x_e$ is not a constant. Here, since $A$ is finite over $B_e$, we have $\dim B_e=\dim A=d$ by §Krull Dimension, ⁋Proposition 4, and since $B_e$ is generated by $d$ elements as a $\mathbb{K}$-algebra, it is a quotient of a polynomial ring in $d$ variables; by the previous argument that taking the quotient by a proper ideal other than $0$ strictly decreases the dimension, the kernel of this quotient map must be $0$. That is, $\x_1',\ldots, \x_e',\x_{e+1},\ldots, \x_d$ are algebraically independent, and in particular $\mathbb{K}[\x_1',\ldots, \x_e']$ is a polynomial ring, so setting $f=\x_e$, we can apply Lemma 2. Then we obtain new elements $\x_1',\ldots, \x_{e-1}'$ such that $\mathbb{K}[\x_1',\ldots, \x_e']$ is finite over $\mathbb{K}[\x_1',\ldots, \x_{e-1}', \x_e]$.

That the subring $B_{e-1}=\mathbb{K}[\x_1',\ldots, \x_{e-1}',\x_e,\x_{e+1},\ldots, \x_d]$ obtained in this way still satisfies the two conditions above is verified as follows. For the first condition, since $B_e$ is finite over $B_{e-1}$ and $A$ is finite over $B_e$, the ring $A$ is also finite over $B_{e-1}$. For the second condition, what is required for $B_{e-1}$ is $\mathfrak{a}_j\cap B_{e-1}\supseteq(\x_k,\ldots, \x_d)$, $k=\max(d_j+1, e)$, so we divide into the cases $j\geq i$ and $j<i$. First, if $j\geq i$, then $d_j\leq d_i<e$, so $k=e$; moreover, $\x_e\in \mathfrak{a}_i\subseteq \mathfrak{a}_j$, and $\x_{e+1},\ldots, \x_d$ already belong, by the second condition for $B_e$, to $\mathfrak{a}_j$, which gives the desired inclusion. On the other hand, if $j<i$, then from the fact that $i$ is the smallest index satisfying $e>d_i$, we have $d_j\geq e$, so $k=d_j+1$; the required generators $\x_{d_j+1},\ldots, \x_d$ in this case are all among the variables $\x_{e+1},\ldots, \x_d$ untouched in this step and hence belong to $B_{e-1}$, and their belonging to $\mathfrak{a}_j$ is precisely the second condition for $B_e$.

Consequences

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

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

Theorem 3 naturally generalizes to the quotient setting, 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:

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

Proof

Since the inequality $\dim A/\mathfrak{p}+\codim\mathfrak{p}\leq\dim A$ holds for any ring (§Krull Dimension, ⁋Definition 2), it suffices to show the reverse inequality. Set $n=\dim A$ and $d=\dim A/\mathfrak{p}=\dim\mathfrak{p}$. Applying Theorem 1 (Noether normalization lemma) to a chain of ideals in $A$ consisting of the single ideal $\mathfrak{a}_1=\mathfrak{p}$, there exists in $A$ a subring $B\cong\mathbb{K}[\x_1,\ldots, \x_n]$ such that $A$ is a finitely generated $B$-module and $\mathfrak{p}\cap B=(\x_{d+1},\ldots, \x_n)$ holds. Then $B\hookrightarrow A$ is an integral extension. Here, if we contract a chain of prime ideals starting at $\mathfrak{p}$ in $A$ to $B$, the inclusions remain strict by §Integral Extensions and Ideals, ⁋Corollary 4, so we obtain $\codim_A\mathfrak{p}\leq\codim_B(\mathfrak{p}\cap B)$.

The opposite direction we need corresponds to going-down, lifting a chain starting at $\mathfrak{p}\cap B$ in $B$ to below $\mathfrak{p}$; since $B$ is a polynomial ring, it is a UFD ([Ring Theory] §Polynomial Rings, ⁋Theorem 16) and thus a normal domain (§Integral Extensions, ⁋Proposition 9), and since $A$ is a domain by assumption, repeatedly applying §Integral Extensions and Ideals, ⁋Theorem 6 (Going down) yields, whenever a chain below $\mathfrak{p}\cap B$ is given, a chain below $\mathfrak{p}$ lying over it. The elements of this chain contract to distinct prime ideals, so the chain is again strict, and therefore $\codim_A\mathfrak{p}=\codim_B(\mathfrak{p}\cap B)$.

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

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

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

Meanwhile, since the polynomial subring given by Theorem 1 (Noether normalization lemma) behaves well under extension of scalars, it also follows that the dimension of a finitely generated $\mathbb{K}$-algebra does not change even when extending the base field.

Proposition 5 For a field $\mathbb{K}$, an extension $\mathbb{K}\hookrightarrow \mathbb{L}$, and a finitely generated $\mathbb{K}$-algebra $A$, the equality

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

holds.

Proof

Let $d=\dim A$, and applying Theorem 1 (Noether normalization lemma) without a chain of ideals, choose a subring of $A$, $B\cong\mathbb{K}[\x_1,\ldots, \x_d]$, such that $A$ is a finitely generated $B$-module. Now, since $\mathbb{L}$ is a $\mathbb{K}$-vector space, it is a free $\mathbb{K}$-module and hence flat ([Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Definition 7), so the ring homomorphism obtained from the inclusion $B\hookrightarrow A$ by applying $-\otimes_\mathbb{K}\mathbb{L}$

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

is also injective. Here, $B\otimes_\mathbb{K}\mathbb{L}\cong\mathbb{L}[\x_1,\ldots, \x_d]$, and with $A$ generated as a $B$-module by elements $a_1,\ldots, a_r$, 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 thus an integral extension (§Integral Extensions, ⁋Lemma 4).

However, since an integral homomorphism preserves dimension (§Krull Dimension, ⁋Proposition 4), 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.

Generic Freeness

So far, we have treated the situation where the base is a field. If we extend this to a Noetherian integral domain $A$, there is no reason for a finite type $A$-algebra $B$ and a finitely generated module $M$ over it to be free as an $A$-module. However, the following result shows that it can always be made free merely by inverting a single element of $A$ other than $0$; since the localization $A_a$ has the same field of fractions as $A$, this means that freeness holds generically.

Theorem 6 (Grothendieck’s generic freeness) Let a Noetherian integral domain $A$, a finite type $A$-algebra $B$, and a finitely generated $B$-module $M$ be given. Then there exists a non-$0$ $a\in A$ such that $M_a$ is a free $A_a$-module.

Proof

Since $B$ is a finite type algebra over the Noetherian integral domain $A$, it is a Noetherian ring by §Basic Notions, ⁋Theorem 12 (Hilbert basis theorem).

First, we reduce the problem by dévissage applying §Associated Primes, ⁋Lemma 6 to $B$ and $M$. Applying this gives a filtration

\[0=M_0\subseteq M_1\subseteq\cdots\subseteq M_n=M,\qquad M_i/M_{i-1}\cong B/\mathfrak{q}_i\]

If we assume that the conclusion holds for each $B/\mathfrak{q}_i$ and let $a$ be the product of those elements, then since the right-hand term of the exact sequence

\[0 \longrightarrow (M_{i-1})_a \longrightarrow (M_i)_a \longrightarrow (B/\mathfrak{q}_i)_a \longrightarrow 0\]

is a free module, this exact sequence splits, and by induction on $i$, all $(M_i)_a$ are free $A_a$-modules. Therefore, it suffices to show the case where $M=B/\mathfrak{q}$, that is, replacing $B$ by $B/\mathfrak{q}$, the case where $B$ is an integral domain and $M=B$.

The case where the kernel of $\varphi: A \rightarrow B$ is not $0$ is handled immediately. In this case, choosing $0\neq a\in \ker \varphi$, in $B_a=B\otimes_AA_a$ we have $\varphi(a)=0$ while $a$ is a unit, so $B_a=0$, which is a free module of rank $0$. Therefore, we may assume $A\subseteq B$.

Now setting $K=\Frac(A)$, since $B\otimes_AK$ is the localization at $A\setminus\{0\}$ of $B$, it is a non-$0$ integral domain and a finite type algebra over $K$. We show the remaining assertion by induction on $d=\dim (B\otimes_AK)$.

Applying Theorem 1 (Noether normalization lemma) without a chain of ideals, there exist algebraically independent elements $y_1,\ldots, y_d\in B\otimes_AK$ such that $B\otimes_AK$ is a finite module over $K[y_1,\ldots,y_d]$. Since algebraic independence and finiteness are preserved even if we multiply each $y_i$ by an element of $A$ other than $0$, we may assume from the start that $y_i\in B$.

If we let $B$ be generated as an $A$-algebra by elements $b_1,\ldots, b_m$, then each $b_j$ is integral over $K[y_1,\ldots,y_d]$. Since there are only finitely many coefficients appearing in these integral equations, we can choose $0\neq a_0\in A$ clearing all their denominators. Then each $b_j$ is integral over

\[C=A_{a_0}[y_1,\ldots, y_d]\]

and therefore $B_{a_0}$ is a finite $C$-module. Here, since the $y_i$ are algebraically independent over $K$, $C$ is a polynomial ring over $A_{a_0}$, and in particular a free $A_{a_0}$-module.

Now, applying dévissage again to the Noetherian ring $C$ and the finite $C$-module $B_{a_0}$, we obtain a finite filtration whose quotients are of the form $C/\mathfrak{p}$. Therefore, it suffices to show the conclusion for each $C/\mathfrak{p}$.

The base of induction is the case $d=0$. In this case, $C=A_{a_0}$, so the quotients given by the dévissage above are of the form $A_{a_0}/\mathfrak{p}$; if $\mathfrak{p}=0$, it is a free module, and if $\mathfrak{p}\neq 0$, by inverting an element of $\mathfrak{p}$ other than $0$, we can make it $0$.

Now assume $d>0$. If $\mathfrak{p}=0$, then $C$ itself is a free $A_{a_0}$-module, so there is nothing to show. Assume $\mathfrak{p}\neq 0$. If $\mathfrak{p}\cap A_{a_0}\neq 0$, then choosing a non-$0$ element $a_1$ in it, we have $(C/\mathfrak{p})_{a_1}=0$, so the conclusion holds. Otherwise, for $S=A_{a_0}\setminus\{0\}$, we have $S^{-1}C=K[y_1,\ldots,y_d]$ and $\mathfrak{p}\cap S=\emptyset$, so $\mathfrak{p}K[y_1,\ldots,y_d]$ is a prime ideal of $K[y_1,\ldots,y_d]$ other than $0$. Now,

\[(C/\mathfrak{p})\otimes_{A_{a_0}}K=K[y_1,\ldots,y_d]/\mathfrak{p}K[y_1,\ldots,y_d]\]

and if we choose $0\neq g\in \mathfrak{p}K[y_1,\ldots,y_d]$, the images of $y_1,\ldots,y_d$ in this quotient ring satisfy the algebraic relation $g=0$, so they cannot be algebraically independent. Since this quotient ring is a $K$-algebra generated by these images, the transcendence degree of its field of fractions is less than $d$, and by Theorem 3,

\[\dim\big((C/\mathfrak{p})\otimes_{A_{a_0}}K\big)<d\]

holds. Since $A_{a_0}$ is also a Noetherian integral domain having $K$ as its field of fractions and $C/\mathfrak{p}$ is a finite type integral domain over it, applying the induction hypothesis shows that there exists $0\neq a_\mathfrak{p}\in A_{a_0}$ such that $(C/\mathfrak{p})_{a_\mathfrak{p}}$ is a free $(A_{a_0})_{a_\mathfrak{p}}$-module. If necessary, multiplying by a power of $a_0$, we may regard $a_\mathfrak{p}\in A$.

For the finitely many $\mathfrak{p}$ appearing in the filtration, if we let the product of the $a_\mathfrak{p}$ and $a_0$ be $a$, then $B_a=(B_{a_0})_a$ is a free $A_a$-module.

Two points are decisive in the proof: one is using dévissage to decompose $M$ into quotients of the form $B/\mathfrak{q}$ to transfer the problem to the case of a domain, and the other is using Theorem 1 (Noether normalization lemma) to make $B$ finite over a polynomial ring over $A_{a_0}$, and then running the induction by decreasing the transcendence degree using the dimension calculation of Theorem 3. Since free modules are flat, Theorem 6 (Grothendieck’s generic freeness) can also be read as saying that flatness holds generically, and in algebraic geometry this form yields the fact that a morphism of finite type is flat over a dense open subset of the base, as well as Chevalley’s theorem that the image of a morphism of finite type is constructible.

References

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

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Noether Normalization