advanced34 min studyLesson 528 in Mathematics

Regular sequences and depth

Measure successive nonzerodivisor cuts in a finite module over a Noetherian local ring and distinguish maximal sequence length from an arbitrary first choice.

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What you’ll understand

  • Check regularity on successive quotients and retain a nonzero final quotient.
  • Define depth with Noetherian local and finite nonzero module hypotheses.
  • Compute a Cohen–Macaulay example and an embedded-prime depth-zero example.

A useful equation cuts a module without killing a nonzero element through multiplication. A regular sequence repeats that requirement after each earlier equation has been imposed. Depth is the number of such cuts available inside a local ring’s maximal ideal. It measures a homological form of thickness rather than simply the size of the support. The theorem that all maximal regular sequences have the same length requires a finite nonzero module over a Noetherian local ring; without those settings, the familiar depth picture cannot be assumed unchanged.

Test each equation after the preceding cuts

For an R-module M, a sequence a₁,…,a_r is M-regular when multiplication by a_i is injective on M/(a₁,…,a_{i−1})M for each i, and the final quotient is nonzero. This last condition rules out counting a unit as a meaningful cut: a unit acts injectively but makes the quotient zero. The order is part of the definition. Global rings can have regular sequences whose permutations fail regularity, so one should not reorder equations without a theorem supporting that step.

We now specialize to a Noetherian local ring (R,m) and a finite nonzero module M, with all chosen elements in m. In this setting regular sequences can be permuted, and a regular sequence can be extended until it is maximal. The nonzero quotient requirement follows along such sequences from the local finite-module setting and Nakayama’s lemma, discussed next. For finding a first regular element, avoid the union of Ass(M), which is the zero-divisor set. Avoiding only minimal support primes can fail because an embedded associated prime may still contain the chosen element.

ai:M/(a1,…,ai−1)M⟶M/(a1,…,ai−1)M injective,M/(a1,…,ar)M≠0a_i:M/(a_1,\ldots,a_{i-1})M\longrightarrow M/(a_1,\ldots,a_{i-1})M\text{ injective},\qquad M/(a_1,\ldots,a_r)M\ne0
Regularity is a successive-quotient condition. The nonzero quotient excludes a unit, even though multiplication by a unit is injective.↔ Scroll wide equations horizontally.

Depth is a maximal length under local finiteness hypotheses

For the stated Noetherian local R and finite nonzero M, depth M is the supremum of lengths of M-regular sequences in m. It is finite, and every maximal such sequence has that same length. Here maximal means no further admissible element can be appended, not simply that a particular list seems long. A theorem using associated primes and regular-element reduction shows that different maximal choices agree. We state this theorem rather than substituting one successful sequence as proof that no longer sequence exists.

The fundamental inequality is depth M≤dim Supp M. If a∈m is M-regular, then depth(M/aM)=depth M−1. A module is Cohen–Macaulay when its depth equals its dimension. This equality does not require the module to be a domain or reduced: a quotient can have nilpotents and still have the maximum possible regular-sequence length. The zero module is conventionally assigned infinite depth in many texts; our computations use nonzero M so that this convention cannot distort a finite-dimensional example.

depth⁡RM=max⁡{r:∃ M-regular sequence of length r in m}≤dim⁡Supp⁡M\operatorname{depth}_RM=\max\{r:\exists\ M\text{-regular sequence of length }r\text{ in }\mathfrak m\}\le\dim\operatorname{Supp}M
The maximum and equality of maximal sequence lengths use a finite nonzero module over a Noetherian local ring. Depth is not the length of an arbitrary chosen sequence.↔ Scroll wide equations horizontally.

Depth zero can reveal an embedded obstruction

Depth is zero exactly when every element of m is a zero divisor on M. In the finite Noetherian local setting, this is equivalent to m being an associated prime: there is a nonzero element killed by the whole maximal ideal. If the support has positive dimension, this can be an embedded obstruction. The module may still have a large-looking geometric component, but no first regular cut through the chosen local point. This explains why support dimension alone cannot determine depth.

One should also separate regular sequences from regular local rings. A regular local ring has embedding dimension dim_k m/m² equal to its Krull dimension, which is a stronger structural property than merely displaying some nonzerodivisors. Cohen–Macaulay modules and rings need not be regular. When claiming a depth value, exhibit a regular sequence for a lower bound and use dimension or a zero-divisor obstruction for the upper bound. The worked examples follow that two-sided method rather than assuming the first plausible candidate list is automatically maximal.

A reducible hypersurface with depth two

Let R=k[x,y,z]_(x,y,z) and M=R/(xy). Prove z,x+y is M-regular and determine depth M.

  1. Multiplication by z is injective on k[x,y,z]/(xy), since polynomial coefficients in z can be compared independently; localization preserves that injection.

  2. After quotienting by z, the module is k[x,y]_(x,y)/(xy). Its associated primes are (x) and (y), and x+y belongs to neither, so it is a nonzerodivisor.

  3. The final quotient is k[x]_(x)/(x²), nonzero. Thus depth M≥2. Its support has dimension two, so depth M≤2 and equality follows.

The result

M is Cohen–Macaulay of dimension two even though the hypersurface is reducible. Regularity of the sequence was checked after each earlier cut.

Explore the model

Interactive lab

Interactive lab · modular

Injective multiplication and admissible regular cuts differ

Pause and Reset, then set modulus m=12 and additive step=4. Predict a three-element orbit, showing the nonzero scalar 4 has an annihilator. For a fresh comparison Reset and set m=12,step=5; predict all twelve residues, so 5 is a unit modulo 12 and multiplication by it is injective. A unit still cannot count as a regular cut under the nonzero-quotient convention. This residue model does not calculate local-module depth.

Injective multiplication and admissible regular cuts differAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA nonzero scalar can be a zero divisor, while a unit can be injective but still be excluded from a meaningful regular sequence.

Check your understanding

For a finite nonzero module over a Noetherian local ring, how is depth related to maximal regular sequences in the maximal ideal?

Try it yourself

Let R=k[x,y]_(x,y) and M=R/(x²,xy). Determine dim M and depth M, exhibiting the obstruction to even one regular element in m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the nonzero class of x and compare its annihilator with the maximal ideal.

Compare with the solution

The radical of the annihilator ideal is (x), so the support is the local y-axis and dim M=1. The class of x is nonzero and is killed by both x and y; hence its annihilator is m=(x,y)R. Every element of m kills that same nonzero class, so every possible first regular element is a zero divisor. Therefore depth M=0. This embedded associated prime makes M fail to be Cohen–Macaulay despite its one-dimensional support.

Take it with you

  • A regular sequence tests injectivity on successive quotients and leaves a nonzero quotient.
  • Depth is the common maximal regular-sequence length in the finite Noetherian local setting.
  • Embedded associated primes can force depth zero while support dimension remains positive.

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