System of Parameters

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.

System of parameters

Combining §Dimension, ⁋Theorem 7 (Principal Ideal Theorem) and §Dimension, ⁋Corollary 8 from the previous post, we obtain the following.

Corollary 1 Let \((A, \mathfrak{m})\) be a Noetherian local ring. Then \(\dim A\) is the smallest \(d\) for which there exist \(d\) elements \(a_1,\ldots, a_d\in \mathfrak{m}\) such that \(\mathfrak{m}^n\subseteq (a_1,\ldots, a_d)\) for all sufficiently large \(n\).

Proof

First, assume that \(\mathfrak{m}^n\subseteq (a_1,\ldots, a_d)\). Then by §The Jordan-Hölder Theorem, ⁋Corollary 8, \(\mathfrak{m}\) is a minimal prime ideal containing \((a_1,\ldots, a_d)\). Therefore, by §Dimension, ⁋Theorem 7 (Principal Ideal Theorem), we have \(\codim \mathfrak{m}\leq d\).

Conversely, suppose \((A,\mathfrak{m})\) satisfies \(\dim A=d\). Then by definition, the supremum of lengths \(d\) arises from chains of prime ideals starting at \(\mathfrak{m}\), so it is exactly \(\codim \mathfrak{m}\). Thus, using §Dimension, ⁋Corollary 8, we can arrange that \(\mathfrak{m}\) is a minimal prime containing the ideal \((a_1,\ldots, a_d)\). Then \(\mathfrak{m}\) becomes the unique prime ideal in \(A/(a_1,\ldots, a_d)\), so it must be exactly the nilradical of \(A/(a_1,\ldots, a_d)\) (§Properties of Localization, ⁋Corollary 8), and therefore we obtain the desired result.

Proposition 2 For a Noetherian local ring \((A, \mathfrak{m})\), we have

\[\dim_K(\mathfrak{m}/\mathfrak{m}^2) \ge \dim A\]

where \(K = A/\mathfrak{m}\).

Proof

Let \(\dim A = d\). Then there exists a system of parameters \(x_1, \ldots, x_d \in \mathfrak{m}\). By Nakayama’s lemma, their images are linearly independent in \(\mathfrak{m}/\mathfrak{m}^2\). Therefore \(\dim_K(\mathfrak{m}/\mathfrak{m}^2) \ge d\).

We now make the following definition.

Proposition–Definition 3 Let \((A,\mathfrak{m})\) be a local Noetherian ring of Krull dimension \(d\). Then for a family of elements \(a_1,\ldots, a_d\) of \(A\) and \(\mathfrak{a}=(a_1,\ldots, a_d)\), the following are all equivalent.

1. \(\mathfrak{m}\) is minimal among prime ideals containing \(\mathfrak{a}\).
2. \(\mathfrak{m}=\sqrt{\mathfrak{a}}\).
3. Some power of \(\mathfrak{m}\) is contained in \(\mathfrak{a}\).
4. \(\mathfrak{a}\) is an \(\mathfrak{m}\)-primary ideal.

If these conditions hold, we call \(a_1,\ldots, a_d\) a system of parameters of \(A\), and \(\mathfrak{a}\) a parameter ideal.

More generally, for a finitely generated \(A\)-module \(M\) of rank \(d\), we call elements \(a_1,\ldots, a_d\) a system of parameters if the \(A\)-module \(M/\mathfrak{a}M\) has finite length, and we call \(\mathfrak{a}\) a parameter ideal of \(M\).

The equivalence of these conditions was already established in the previous post and in Corollary 1 above. Meanwhile, for a local ring \((A, \mathfrak{m})\) we have

\[d=\dim A=\codim \mathfrak{m}\]

so by §Dimension, ⁋Theorem 7 (Principal Ideal Theorem), any parameter ideal \(\mathfrak{a}\) of \(A\) must be generated by at least \(d\) elements.

In the case of a parameter ideal \(\mathfrak{a}\) of an \(A\)-module \(M\), we know from the equivalence between the first and second conditions of §The Jordan-Hölder Theorem, ⁋Corollary 6 that \(M/\mathfrak{a}M\) has finite length if and only if some sufficiently large power of \(\mathfrak{m}\) annihilates \(M/\mathfrak{a}M\). That is, we must have \(\mathfrak{m}^k M \subseteq \mathfrak{a}M\), and from this we see that the two definitions agree when we regard \(A\) itself as an \(A\)-module. In a similar manner, we can translate the results we examined earlier on ideals and dimension for rings into results about parameter ideals for modules; to do so, we first need the following simple lemmas.

Lemma 4 For a Noetherian ring \(A\), a finitely generated \(A\)-module \(M\), and an ideal \(\mathfrak{a}\) of \(A\), the following equality holds:

\[\sqrt{\ann(M/\mathfrak{a}M)}=\sqrt{\mathfrak{a}+\ann(M)}\]

Proof

By §Properties of Localization, ⁋Corollary 8, it suffices to show that the set of prime ideals containing \(\ann(M/\mathfrak{a}M)\) is exactly the same as the set of prime ideals containing \(\mathfrak{a}+\ann(M)\).

Now, a prime ideal \(\mathfrak{p}\) contains \(\ann(M/\mathfrak{a}M)\) if and only if \((M/\mathfrak{a}M)_\mathfrak{p}\neq 0\) by §Localization, ⁋Proposition 5. Then \((M/\mathfrak{a}M)_\mathfrak{p}=M_\mathfrak{p}/\mathfrak{a}M_\mathfrak{p}\neq 0\) is equivalent, by §Integral Extensions, ⁋Lemma 8 (Nakayama), to \(M_\mathfrak{p}\neq 0\) and \(\mathfrak{a}A_\mathfrak{p}\subseteq \mathfrak{p}A_\mathfrak{p}\). This is again equivalent, by §Properties of Localization, ⁋Corollary 8, to \(\mathfrak{p}\supseteq \ann(M)\) and \(\mathfrak{p}\supseteq \mathfrak{a}\), that is, to \(\mathfrak{p}\supseteq \mathfrak{a}+\ann(M)\), so we obtain the desired result.

Also, the following holds.

Lemma 5 For a short exact sequence of \(A\)-modules

\[0 \rightarrow M' \overset{u}{\longrightarrow} M \overset{v}{\longrightarrow} M'' \rightarrow 0\]

we have \(\ann(M)\subseteq \ann(M')\cap \ann(M'')\).

Proof

Let \(a\in\ann(M)\). Then for any \(x'\in M'\), we have \(u(ax')=au(x')=0\), and since \(u\) is injective, we get \(ax'=0\), so \(a\in\ann(M')\).

Similarly, for any \(x''\in M''\), since \(v\) is surjective there exists \(x\in M\) with \(v(x)=x''\), and then \(ax''=av(x)=v(ax)=0\), so \(a\in\ann(M'')\).

We can then prove the following proposition.

Proposition 6 For a Noetherian local ring \((A,\mathfrak{m})\), an ideal \(\mathfrak{a}\) of \(A\), and a finitely generated \(A\)-module \(M\), the following hold.

1. The following are equivalent.
   • \(\mathfrak{a}\) is a parameter ideal of \(M\).
   • \((\mathfrak{a}+\ann(M))\supseteq \mathfrak{m}^n\) for all sufficiently large \(n\).
   • \(\mathfrak{a}\) is a parameter ideal of \(A/\ann(M)\).

2. For a short exact sequence of \(A\)-modules

   \[0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0\]

   \(\mathfrak{a}\) is a parameter ideal of \(M\) if and only if it is a parameter ideal of both \(M'\) and \(M''\).

3. \(\dim M\) is the smallest natural number \(d\) for which there exists a parameter ideal of \(M\) generated by \(d\) elements.

Proof

1. First, assume that \(\mathfrak{a}\) is a parameter ideal of \(M\). Then by the argument examined right after Proposition–Definition 3, we know that a sufficiently large power of \(\mathfrak{m}\) always annihilates \(M/\mathfrak{a}M\), and combining this with Lemma 4 we obtain

   \[\mathfrak{m}\subseteq \sqrt{\ann(M/\mathfrak{a}M)}=\sqrt{\mathfrak{a}+\ann(M)}\]

   so we see that \(\mathfrak{m}^n\in(\mathfrak{a}+\ann(M))\) must hold for sufficiently large \(n\).

   Now assume the second condition. Then in the ring \(A'=A/\ann(M)\), \(\mathfrak{m}+\ann(M)\) is the unique maximal ideal, and from the assumption we know that \((\mathfrak{m}+\ann(M))^n\) belongs to \(\mathfrak{a}+\ann(M)\) for sufficiently large \(n\), so \(\mathfrak{a}+\ann(M)\) is a parameter ideal of \(A/\ann(M)\) (as a ring), and regarding \(A/\ann(M)\) as an \(A\)-module, we obtain the desired result.

   The last equivalence is obvious from the inclusion

   \[\mathfrak{m}\subseteq \sqrt{\mathfrak{a}+\ann(M)}=\sqrt{\ann(M/\mathfrak{a}M)}\]

2. Suppose \(\mathfrak{a}\) is a parameter ideal of \(M\). Then by Lemma 5, \(\ann(M)\subseteq \ann(M')\cap \ann(M'')\), so it is obvious that \(\mathfrak{a}\) is a parameter ideal of both \(M'\) and \(M''\).

   Conversely, from the exact sequence obtained by applying \(A/\mathfrak{a}\otimes-\)

   \[M'/\mathfrak{a}M' \rightarrow M/ \mathfrak{a}M \rightarrow M''/\mathfrak{a}M'' \rightarrow 0\]

   we know that if \(M'/\mathfrak{a}M'\) and \(M''/\mathfrak{a}M''\) have finite length, then so must \(M/\mathfrak{a}M\).

3. By definition \(\dim M=\dim A/\ann(M)\), so this is obvious from the first result and §Dimension, ⁋Corollary 8.

Corollary 7 Let \((A, \mathfrak{m})\) be a Noetherian local ring and \(M\) a finitely generated \(A\)-module. Then for any \(a\in \mathfrak{m}\),

\[\dim M/ aM \geq \dim M-1\]

holds.

Proof

By definition, \(\dim M/aM=d\) means that the ring \(A/\ann(M/aM)\) has dimension \(d\). Then by Corollary 1, \(A/\ann(M/aM)\) has a parameter ideal \(\mathfrak{a}=(a_1,\ldots, a_d)\) generated by \(d\) elements, and by the first result of Proposition 6, this is also a parameter ideal of \(M/aM\). Then since

\[\frac{M/aM}{\mathfrak{a}(M/aM)}\cong \frac{M}{((a)+\mathfrak{a})M}=\frac{M}{(a,a_1,\ldots, a_d)M}\]

has finite length, \((a,a_1,\ldots, a_d)\) becomes a parameter ideal of \(M\). Therefore, by the third condition of Proposition 6, we have \(\dim M\leq 1+d\).

Flat Morphisms and Dimension

By the definition of dimension, to compare the dimensions of \(A\) and \(B\) via a ring homomorphism \(\phi: A \rightarrow B\), §Integral Extensions and Ideals, ⁋Proposition 1 is essential. The following lemma is also used for a similar purpose, but produces prime ideals in the opposite direction.

Lemma 8 (Going down for flat extensions) Let \(\phi: A \rightarrow B\) be a ring homomorphism between Noetherian rings, and suppose that \(B\) has a flat \(A\)-module structure via this map. Then for prime ideals \(\mathfrak{p}_2\subseteq\mathfrak{p}_1\subseteq A\) and a prime ideal \(\mathfrak{q}_1\) of \(B\) satisfying \(\phi^{-1}\mathfrak{q}_1=\mathfrak{p}_1\), there exists a prime ideal \(\mathfrak{q}_2\) of \(B\) such that \(\phi^{-1}\mathfrak{q}_2=\mathfrak{p}_2\).

Proof

First, applying \(A/\mathfrak{p}_2\otimes_A-\) to \(\phi: A \rightarrow B\), we obtain the ring homomorphism

\[\phi\otimes_A\id_{A/\mathfrak{p}_2}: A/\mathfrak{p}_2\cong A\otimes_A A/\mathfrak{p}_2 \rightarrow B\otimes_A A/\mathfrak{p}_2\cong B/\mathfrak{p}_2B\]

and from the assumption that \(\phi\) is flat, we know that this is also flat. Thus we may assume that \(\mathfrak{p}_2=0\) and that \(A\) is an integral domain. Then by §Flatness, ⁋Corollary 3, \(\phi\) must send non-zerodivisors of \(A\) to non-zerodivisors of \(B\).

Meanwhile, by [Set Theory] §Axiom of Choice, ⁋Theorem 4, we know that there exists a minimal prime ideal \(\mathfrak{q}_2\) contained in \(\mathfrak{q}_1\). However, if we regard \(B\) as a module over itself, then \(\ann B=0\), so by the first result of §Associated Primes, ⁋Theorem 7, we have \(\mathfrak{q}_2\in \Ass B\), and again by the second result of the same theorem, \(\mathfrak{q}_2\) must consist only of zero-divisors. Therefore, by the property of \(\phi\) examined above, we know that \(\phi^{-1}(\mathfrak{q}_2)=0\) must hold.

Keeping \(B/\mathfrak{p}_2B\) as in the above proof, we see that when choosing \(\mathfrak{q}_2\), it suffices to take a minimal prime ideal among those contained in \(\mathfrak{q}_1\) and containing \(\mathfrak{p}_2 B\).

Then the following holds.

Theorem 9 If there exists a map \((A,\mathfrak{m}) \rightarrow (B, \mathfrak{n})\) between Noetherian local rings, then

\[\dim B\leq \dim A +\dim A/\mathfrak{m}\]

holds, and if \(\phi:A \rightarrow B\) is flat, then equality holds.

Proof

For convenience, write \(\dim A=d\) and \(e=\dim B/\mathfrak{m}B\). First, by Corollary 1, there exist \(a_1,\ldots, a_d\) such that \(\mathfrak{m}^s\subseteq (a_1,\ldots, a_d)\) for all sufficiently large \(s\), and similarly there exist \(b_1,\ldots, b_e\in B\) such that \(\mathfrak{n}^t\subseteq \phi(\mathfrak{m})B+(b_1,\ldots, b_e)\) for all sufficiently large \(t\). Then

\[\mathfrak{n}^{st}=(\mathfrak{n}^t)^s\subseteq (\phi(\mathfrak{m})B+(b_1,\ldots, b_e))^s\subseteq \phi(\mathfrak{m}^s)B+(b_1,\ldots, b_e)\subseteq (\phi(a_1),\ldots, \phi(a_d), b_1,\ldots, b_e)\]

so by §Dimension, ⁋Theorem 7 (Principal Ideal Theorem), we have \(\dim B\leq d+e\).

Now assume that \(\phi:A \rightarrow B\) makes \(B\) a flat \(A\)-module, and let us show the reverse inequality. To do so, consider a chain of prime ideals giving the dimension of \(B/\phi(\mathfrak{m})B\); then there exists a prime ideal \(\mathfrak{q}\) of \(B\) such that \(\dim \mathfrak{q}=\dim B/\phi(\mathfrak{m})B\), and in particular \(\mathfrak{q}\) is minimal among prime ideals containing \(\phi(\mathfrak{m})B\). Then from the inequality

\[\dim B\geq\dim \mathfrak{q}+\codim \mathfrak{q}=\dim B/\phi(\mathfrak{m})B+\codim \mathfrak{q}\]

we see that we need to show \(\codim \mathfrak{q}\geq\dim A\). But by definition \(\phi^{-1}(\mathfrak{q})=\mathfrak{m}\), so by Lemma 8 (Going down for flat extensions), whenever a chain of prime ideals of \(A\) starting at \(\mathfrak{m}\)

\[\mathfrak{m}\supseteq \mathfrak{p}_1\supseteq \mathfrak{p}_2\supseteq\cdots\]

is given, there exists a chain of prime ideals of \(B\) starting from \(\mathfrak{q}\)

\[\mathfrak{q}\supseteq \mathfrak{q}_1\supseteq \mathfrak{q}_2\supseteq\cdots\]

and from this we obtain the desired inequality.

The following corollaries are obtained without difficulty from the above theorem.

Corollary 10 For a Noetherian local ring \((A, \mathfrak{m})\) and the \(\mathfrak{m}\)-adic completion \(\widehat{A}\) of \(A\), we have \(\dim A=\dim \widehat{A}\).

Corollary 11 The following hold.

1. For a field \(\mathbb{K}\), we have \(\dim \mathbb{K}[\x_1,\ldots, \x_r]=r\).
2. For any ring \(A\), we have \(\dim A[\x]=1+\dim A\).
3. For any prime ideal \(\mathfrak{p}\) of \(A\), there exists a prime ideal \(\mathfrak{q}\) of \(A[\x]\) satisfying \(\mathfrak{q}\cap A=\mathfrak{p}\), and for a maximal such \(\mathfrak{q}\) among those satisfying this property, the formula \(\dim A[\x]_\mathfrak{q}=1+\dim A_\mathfrak{p}\) holds.

---

References

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

---