Some basic facts

Proposition Let (R, 𝔪) be a Noetherian local ring. Then

𝔪 = (x_{1}, …, x_{n}) ⇔ x̄_{1}, …, x̄_{n} generates 𝔪/𝔪^{2}

as an R/𝔪 -vectorspace

We often use the following tool:

Lemma (prime avoidance) Let I = (x_{1}, …, x_{n}) be an ideal of a ring R. Let 𝔭_{1}, …, 𝔭_{l} ∈ Spec R. If

I ⊄ 𝔭_{j} for any j

then

∃ a_{2}, …, a_{l} ∈ R : x_{1} + n ; ∑ ; i=2 a_{i}x_{i} ∉ 𝔭_{j} for any j

dim R

Let R be a ring and 𝔭 ∈ Spec R. The height of 𝔭 is the supremum of the lengths t of (strict) chains

𝔭_{t} ⊂ 𝔭_{t−1} ⊂ … 𝔭_{1} ⊂ 𝔭_{0} = 𝔭 (𝔭_{i} ∈ Spec R)

A fundamental theorem on height is

Theorem (Krull’s Principal Ideal Theorem) Let R be a Noetherian ring. Let I = (x_{1}, …, x_{n}) be a proper ideal. Let

𝔭 minimal in {𝔮 ∈ Spec R | I ⊂ 𝔮}

Then

height 𝔭 ≤ n

The (Krull) dimension of R is dim R = sup {height 𝔭 | 𝔭 ∈ Spec R}

Note that for any Noetherian ring R dim R = 0 ⇔ R is Artinian

For (R, 𝔪) a Noetherian local ring we have

dim R = n ⇔ n is the infimum of all l for which

∃x_{1}, …, x_{l} : dim R/(x_{1}, …, x_{l}) = 0

We call x_{1}, …, x_{n} a system of parameters of R if

dim R = n and dim R/(x_{1}, …, x_{n}) = 0 dim M

Let R be a ring and M an R-module. The dimension of M is the supremum of the lengths t of (strict) chains

𝔭_{t} ⊂ 𝔭_{t−1} ⊂ … 𝔭_{1} ⊂ 𝔭_{0} = 𝔭 (𝔭_{i} ∈ Supp M)

Note that dim M ≤ dim R. In case M is f.g. then

Supp M = {𝔭 ∈ Spec R | Ann M ⊂ 𝔭}

and therefore

dim M = dim (R/ Ann M)

In case (R, 𝔪) is local and M ≠ 0, we call x_{1}, …, x_{n} ∈ 𝔪 a system of parameters of M if

dim M = n and dim M/(x_{1}, …, x_{n})M = 0 The following inequality is often usefull:

Proposition A.4 Let (R, 𝔪) be a Noetherian local ring. Let M be a f.g. R-module. Let x_{1}, …, x_{l} ∈ 𝔪. Then

dim M/(x_{1}, …, x_{l})M ≥ dim M − l

and equality ⇔ x_{1}, …, x_{l} is part of a system of parameters of M

PROOF: First replace M by R/ Ann M, so we may assume that M is a ring. Setting dim R/(x_{1}, …, x_{l}) = r, we choose y_{1}, …, y_{r} ∈ 𝔪 such that

dim (R/(x_{1}, …, x_{l})) /(ȳ_{1}, …, ȳ_{r}) = 0 But then

dim (R/(x_{1}, …, x_{l}, y_{1}, …, y_{r})) = 0

so l + r ≥ dim R, and we have equality exactly when x_{1}, …, x_{l}, y_{1}, …, y_{r} is a system of parameters of R.

R-sequence and M-sequence

Let R be a ring and M an R-module. We call x ∈ R an M-regular element if x is not a zerodivisor on M, thus

∀z ∈ M : xz = 0 ⇒ z = 0

A sequence x⃗ = x_{1}, …, x_{n} ∈ R is an M-(regular) sequence if

(i) x_{i} is an M/(x_{1}, …, x_{i−1})M regular element
i = 1, …, n
(ii) M/(x̄)M ≠ 0

An M-sequence x_{1}, …, x_{n} is maximal if x_{1}, …, x_{n}, x_{n+1} is not an M-sequence for any x_{n+1} ∈ R. grade (I, M) and depth M

Let R be a Noetherian ring. Let M be a f.g. R-module. Let I an ideal of R with M ≠ IM. We define the grade of I on M as the lenght of any (!) maximal M-sequence in I. Rees proved that

grade (I, M) = min {i | Ext _{R}^{i}(R/I, M) ≠ 0}

Now let R be a Noetherian local ring. Let M be a f.g. R-module. We define the depth of M as

depth M = grade (m, M) recall the Auslander-Buchsbaum formula:

Theorem 1.3.3 Let (R, 𝔪) be a Noetherian local ring. Let 0 ≠ M be a f.g. R-module. If projdim M < ∞ then

projdim M + depth M = depth R

We proved this with help of the following

Lemma 1.3.5 Let (R, 𝔪) be a Noetherian local ring. Let M be a f.g. R-module. If x ∈ 𝔪 is R-regular and M-regular then

projdim _{R} M = projdim _{R/(x)} M/xM Let (R, 𝔪) be a Noetherian local ring. Let M een f.g. R-module, and x ∈ 𝔪. In general, we have (prop. A.4)

dim M/xM ≥ dim M − 1

Now let x be an M-regular element. Then

x ∉ 𝔭 ∀𝔭 ∈ Ass M
⇒ x ∉ 𝔭 ∀𝔭 minimal in Supp M
⇒ dim M/xM ≤ dim M − 1

So

dim M/xM = dim M − 1

And by prop. A.4, x is part of a system of parameters of M. Now use induction. If x⃗ = x_{1}, …, x_{n} is an M-sequence in 𝔪 then

dim M/(x⃗)M = dim M − n

Thus n ≤ dim M , so the length of any (maximal) M-sequence in 𝔪 is smaller than dim M, proving Proposition 1.2.12 Let (R, 𝔪) be a Noetherian local ring. Let 0 ≠ M be a f.g. R-module. Then

x⃗ is an M-sequence ⇒ x⃗ is part of a system

of parameters of M.

In particular,

depth M ≤ dim M

With the same assumptions of the proposition, we call M a Cohen-Macaulay module if

depth M = dim M

We say that R is a Cohen-Macaulay ring if R (as R-module) is a Cohen-Macaulay module. Also, we have in general

dim M ≤ dim R

We say that a Cohen-Macaulay module M is maximal if we have equality.

Now assuming that M is Cohen-Macaulay in prop. 1.2.12, we can get the other implication:

Proposition 2.1.2 Let (R, 𝔪) be a Noetherian local ring. Let 0 ≠ M be a Cohen-Macaulay R-module. Let x⃗ = x_{1}, …, x_{n} any sequence in R. Then

x⃗ is an M-sequence ⇔ x⃗ is part of a system

of parameters of M.

This can easily be proved using

Proposition 1.6.19 Let (R, 𝔪) be a Noetherian local ring. Let 0 ≠ M be a f.g. R-module. Let I = (x_{1}, …, x_{n}) ⊂ 𝔪. Then

grade (I, M) = n ⇔ x_{1}, …, x_{n} is an M-sequence

(the proof involves Koszul complexes)

Let (R, 𝔪) be any Noetherian local ring. As a consequence of Krull’s principal ideal theorem we have

𝔪 = (x_{1}, …, x_{n}) ⇒ dim R ≤ n.

Denote the minimal number of generators of 𝔪 by μ(𝔪), so we have

dim R ≤ μ(𝔪).

We say that R is regular if we reach equality. Some (trivial) basic facts are

1. Let R be regular. Then any minimal set of generators of R has dim R elements and forms a system of parameters of R. We call such a sequence a regular system of parameters of R.

2. R is regular ⇔ R has a system of parameters generating 𝔪 Proposition 2.2.4 Let (R, 𝔪) be regular. Let I ≠ 0 be an ideal of R. Then

R/I is regular ⇔ I is generated by a subset

of a regular system of parameters of R

PROOF ⇐ Suppose I = (x_{1}, …, x_{m}) with x_{1}, …, x_{m}, …, x_{d} a regular system of parameters of R (d = dim R). Then 𝔪̄ = 𝔪/I = (x̄_{m+1} , …, x̄_{d}) is the maximal ideal of R/I. It is easy to show that

1. dim (R/I)/(x̄_{m+1} , …, x̄_{d}) = 0

2. dim (R/I)/(ȳ_{1}, …, ȳ_{s}) ≠ 0 for any s < d − m and any ȳ_{i} ∈ 𝔪̄

3. dim R/I = d − m So 𝔪̄ can be generated by dim R/I elements.

Proposition 2.2.3 Let (R, 𝔪) be regular. Then R is an integral domain.

PROOF: If dim R = 0 then 𝔪 = 0 and we are done. Now assume dim R > 0 and set 𝔪 = (x_{1}, …, x_{d}) with d = dim R. Construct an element

x = x_{1} + d ; ∑ ; i=2 a_{i}x_{i}

in 𝔪 but not in any minimal prime of R. Then x ∉ 𝔪^{2}. Furthermore, x is a part of a regular sequence of parameters of R

𝔪 = (x_{1}, …, x_{d}) = (x, x_{2}, …, x_{d})

and from prop.2.2.4

R/(x) is regular and dim (R/(x)) = d − 1

using induction, R/(x) is an integral domain. So (x) is a prime ideal, thus contains a minimal prime of R, say 𝔭. But then 𝔭 = (x)𝔭 and from Nakayama’s lemma we get 𝔭 = 0, proving that R is an integral domain.

Proposition 2.2.5 Let (R, 𝔪) be a Noetherian local ring. Let {x⃗} = {x_{1}, …, x_{n}} be a minimal set of generators of 𝔪. Then

R is regular ⇔ x⃗ is an R-sequence

PROOF: If R is regular, then x⃗ is a regular system of parameters of R. Choose any i between 1 and n−1. Now (x_{1}, …, x_{i}) is generated by a subset of a regular system of parameters of R and by prop.2.2.4 we get

R/(x_{1}, …, x_{i}) is regular.

In fact, it is a domain, and therefore

x_{i+1} is an R/(x_{1}, …, x_{i}) regular element.

So x⃗ is an R-sequence. If x⃗ is an R-sequence, then it is part of a system of parameters of R (prop. 1.2.12) so n ≤ dim R.

Corollary 2.2.6 Let (R, 𝔪) be a Noetherian local ring. Then

R is regular ⇒ R is Cohen-Macaulay.

PROOF: In general, depth R ≤ dim R. But if R is regular, then

𝔪 = (x_{1}, …, x_{d}) with d = dim R

so x_{1}, …, x_{d} is a minimal set of generators of 𝔪 and by prop. 2.2.5, it is an R-sequence, so

depth R ≥ d = dim R Theorem (Auslander-Buchsbaum-Serre)

Let (R, 𝔪, k) be a Noetherian local ring. Then the following are equivalent:

(a) R is regular

(b) projdim M < ∞, ∀ f.g. R-module M

(c) projdim k < ∞

(a) ⇒ (b): We may choose a regular system of parameters x⃗ = x_{1}, …, x_{d} of R (d = dim R). Take a free resolution of M and let N be the d-th syzygy. Then N = 0 or N is a maximal Cohen-Macaulay module. (use ex. 2) Now x⃗ is an R-sequence and a maximal Nsequence. (Indeed, x⃗ is a system of parameters of N and use prop. 2.1.2) So one can use induction on lemma 1.3.5 to get

projdim _{R} N = projdim _{R/(x⃗)} N/(x⃗)N

Now dim N/mN = 0 as R-module, but also as k-module, so

N/mN ≅ 0 as k-module.

So N is a free R-module and projdim M < ∞.

(b) ⇒ (c): Simply because 𝔪 is a f.g. R-module.

(c) ⇒ (a): Use theorem 2.2.8. with I = 𝔪 In order to check projdim 𝔪 < ∞, construct a minimal free resolution of k

F. : ⋯ ⟶ F_{1} ⟶ F_{0} ⟶ k → 0

with F_{0} = R and ker φ_{0} = 𝔪 = Im φ_{1}. One obtains a free resolution of finite lenght of m. By th. 2.2.8 we have 𝔪 = (x_{1}, …, x_{n}) for some R-sequence. So we have

dim R ≤ n ≤ depth R

which yields dim R = depth R, and thus R is regular.

Theorem 2.2.8 (Ferrand-Vasconcelos)

Let (R, 𝔪) Noetherian local ring. Let I ≠ 0 an ideal with projdim I < ∞ Let I/I^{2} be free as R/I-module. Then I can be generated by an R-sequence.

We prove the theorem for I = 𝔪. 𝔪 contains an R-regular element x, not in 𝔪^{2}. If μ(𝔪) = 1 then 𝔪 = (z) for some z ∈ R. But then z is R-regular, and we are done. Suppose μ(𝔪) > 1. Use induction, passing from R to R/𝔪 and from 𝔪 to 𝔪/(x). Check

1. 𝔪̄ = 𝔪/(x) is the maximal ideal in R/(x)

2. μ(𝔪/(x)) < μ(𝔪)

3. projdim _{R/(x)} 𝔪/(x) < ∞

4. 𝔪̄/𝔪̄^{2} is free as an R/𝔪-module