System of Parameters

System of parameters

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

Corollary 1 Let \((A, \mathfrak{m})\) be a Noetherian local ring. Then \(\dim A\) is, among those for which the following condition

for sufficiently large \(n\), we always have \(\mathfrak{m}^n\subseteq (a_1,\ldots, a_d)\) for some \(d\) elements \(a_1,\ldots, a_d\in \mathfrak{m}\),

holds, the smallest such \(d\).

Proof

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

Conversely, suppose that \((A,\mathfrak{m})\) satisfies \(\dim A=d\). Then by definition, the supremum of lengths, \(d\), comes from a chain of prime ideals starting at \(\mathfrak{m}\), so it is precisely equal to \(\codim \mathfrak{m}\). Therefore, using §Krull Dimension, ⁋Corollary 8, we can ensure that \(\mathfrak{m}\) is a minimal prime containing the ideal \((a_1,\ldots, a_d)\). Then \(\mathfrak{m}\) is the unique prime ideal in \(A/(a_1,\ldots, a_d)\), so this must be precisely 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 \((A, \mathfrak{m})\),

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

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

Proof

Let \(n=\dim_K(\mathfrak{m}/\mathfrak{m}^2)\), choose a basis of \(\mathfrak{m}/\mathfrak{m}^2\), and lift its elements to \(y_1,\ldots, y_n\in \mathfrak{m}\). Since \(A\) is Noetherian, \(\mathfrak{m}\) is finitely generated, and hence by the second result of §Integral Extensions, ⁋Lemma 8 (Nakayama), \(y_1,\ldots, y_n\) generate \(\mathfrak{m}\). Then for any \(k\geq 1\), we have \(\mathfrak{m}^k\subseteq \mathfrak{m}=(y_1,\ldots, y_n)\), so by Corollary 1, \(\dim A\leq n\).

We now define the following.

Proposition–Definition 3 Let \((A,\mathfrak{m})\) be a local Noetherian ring of Krull dimension \(d\). Then for a family of elements of \(A\), \(a_1,\ldots, a_d\), 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. \(\mathfrak{m}=\sqrt{\mathfrak{a}}\) holds.
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 call \(\mathfrak{a}\) a parameter ideal.

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

Showing that these conditions are all equivalent is what we have already done in the previous post and the corollary above. Meanwhile, for a local ring \((A, \mathfrak{m})\), since

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

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

In the case of an \(A\)-module \(M\) and its parameter ideal \(\mathfrak{a}\), we know by the equivalence between the first and second conditions of §The Jordan-Hölder Theorem, ⁋Corollary 6 that \(M/\mathfrak{a}M\) having finite length is equivalent to sufficiently high powers of \(\mathfrak{m}\) always annihilating \(M/\mathfrak{a}M\). That is, \(\mathfrak{m}^k M \subseteq \mathfrak{a}M\) must hold, from which we see that the two definitions coincide when viewing \(A\) itself as an \(A\)-module. In a similar way, we can transfer our previous results on the relationship between ideals and dimension of a ring to results on parameter ideals of a module; for this, 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 equality

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

holds.

Proof

By §Properties of Localization, ⁋Corollary 8, it suffices to show that the set of prime ideals containing \(\ann(M/\mathfrak{a}M)\) and the set of prime ideals containing \(\mathfrak{a}+\ann(M)\) are precisely the same. Now, that a prime ideal \(\mathfrak{p}\) contains \(\ann(M/\mathfrak{a}M)\) is equivalent, by §Localization, ⁋Proposition 5, to \((M/\mathfrak{a}M)_\mathfrak{p}\neq 0\). 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 in turn is equivalent, by §Localization, ⁋Proposition 5, to \(\mathfrak{p}\supseteq \ann(M)\) and \(\mathfrak{p}\supseteq \mathfrak{a}\), that is, \(\mathfrak{p}\supseteq \mathfrak{a}+\ann(M)\), which yields 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 \(a\in\ann(M)\). For any \(x'\in M'\), we have \(u(ax')=au(x')=0\), and since \(u\) is injective, it follows that \(ax'=0\), so \(a\in\ann(M')\).

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

Now we can prove the following proposition.

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

1. The following are all equivalent:
   • \(M/\mathfrak{a}M\) has finite length.
   • For sufficiently large \(n\), \((\mathfrak{a}+\ann(M))\supseteq \mathfrak{m}^n\) always holds.
   • The \(A\)-module \(A/(\mathfrak{a}+\ann(M))\) has finite length.

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

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

   \(M/\mathfrak{a}M\) has finite length if and only if both \(M'/\mathfrak{a}M'\) and \(M''/\mathfrak{a}M''\) have finite length.

3. \(\dim M\) is the smallest natural number \(d\) such that \(M/\mathfrak{b}M\) has finite length for some ideal \(\mathfrak{b}\) generated by \(d\) elements.

Proof

1. First, suppose that \(M/\mathfrak{a}M\) has finite length. Then, by the argument examined immediately 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 have

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

   so we know that for sufficiently large \(n\), \(\mathfrak{m}^n\subseteq(\mathfrak{a}+\ann(M))\) must hold.
   Now suppose the second condition holds. Then, for sufficiently large \(n\), the product of maximal ideals \(\mathfrak{m}^n\) annihilates the \(A\)-module \(A/(\mathfrak{a}+\ann(M))\), so by the equivalence of the first and second conditions of §The Jordan-Hölder Theorem, ⁋Corollary 6, \(A/(\mathfrak{a}+\ann(M))\) has finite length.
   As for the last equivalence, it is clear from the following inclusion:

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

2. Suppose \(M/\mathfrak{a}M\) has finite length. Then by the first result, for sufficiently large \(n\), we have \(\mathfrak{m}^n\subseteq \mathfrak{a}+\ann(M)\), and since \(\ann(M)\subseteq \ann(M')\cap \ann(M'')\) by Lemma 5, the same inclusion holds for \(M'\) and \(M''\). Thus, again by the first result, \(M'/\mathfrak{a}M'\) and \(M''/\mathfrak{a}M''\) have finite length. Conversely, from the following 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 see that if \(M'/\mathfrak{a}M'\) and \(M''/\mathfrak{a}M''\) have finite length, then \(M/\mathfrak{a}M\) must also have finite length.

3. By definition, \(\dim M=\dim A/\ann(M)\), so this is clear from the first result and §Krull 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 dimension of the ring \(A/\ann(M/aM)\) is \(d\). 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, \((a,a_1,\ldots, a_d)\) is an ideal generated by \(d+1\) elements such that \(M/(a,a_1,\ldots, a_d)M\) has finite length. Therefore, by the third result of Proposition 6, \(\dim M\leq 1+d\).

Flat Morphisms and Dimension

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

Lemma 8 (Going down for flat extensions) Suppose we are given a ring homomorphism \(\phi: A \rightarrow B\) between Noetherian rings, through which \(B\) has a flat \(A\)-module structure. Then, for prime ideals \(\mathfrak{p}_2\subseteq\mathfrak{p}_1\subseteq A\), whenever \(\phi^{-1}\mathfrak{q}_1=\mathfrak{p}_1\) holds for a prime ideal of \(B\), \(\mathfrak{q}_1\), there exists in \(B\) a prime ideal \(\mathfrak{q}_2\subseteq \mathfrak{q}_1\) such that \(\phi^{-1}\mathfrak{q}_2=\mathfrak{p}_2\).

Proof

First, tensoring \(\phi: A \rightarrow B\) with \(A/\mathfrak{p}_2\otimes_A-\), we obtain the following 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. Hence it suffices to assume that \(\mathfrak{p}_2=0\) and that \(A\) is an integral domain. Then by §Flatness, ⁋Corollary 3, \(\phi\) must map non-zerodivisors of \(A\) to non-zerodivisors of \(B\).

Meanwhile, by [Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma), we know that contained in \(\mathfrak{q}_1\) there exists a minimal prime ideal \(\mathfrak{q}_2\). However, viewing \(B\) as a module over itself, we have \(\ann B=0\), 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 that theorem, \(\mathfrak{q}_2\) must consist entirely of zero-divisors. Therefore, by the property of \(\phi\) observed above, we see that \(\phi^{-1}(\mathfrak{q}_2)=0\).

If we keep \(B/\mathfrak{p}_2B\) as it is in the above proof, we see that when choosing \(\mathfrak{q}_2\), it suffices to choose one that is minimal among the prime ideals contained in \(\mathfrak{q}_1\) and containing \(\mathfrak{p}_2 B\).

Then the following holds.

Theorem 9 If there is a local homomorphism \(\phi:(A,\mathfrak{m}) \rightarrow (B, \mathfrak{n})\) between Noetherian local rings, then

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

holds, and if \(\phi\) makes \(B\) a flat \(A\)-module, then equality holds.

Proof

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

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

by §Krull Dimension, ⁋Theorem 7 (Principal Ideal Theorem), \(\dim B\leq d+e\) holds.

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

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

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

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

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

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

and from this we obtain the desired inequality.

The following corollaries are easily obtained from the theorem above.

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

Corollary 11 The following hold.

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

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References

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

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