x⃗ is a R-sequence and a maximal N-sequence.

(x⃗) = (x_{1}, …, x_{d}) = 𝔪 is a minimal set of generators of 𝔪. By prop. 2.2.5., x⃗ is an R-sequence. Next, we prove that x⃗ is a system of parameters of N. This is easy, since N is a maximal Cohen-Macaulay module, dim N = dim R = d so we have the right number of elements, and

dim N/(x⃗)N = dim N/𝔪N
= dim R/ Ann(N/𝔪N)
= dim R/𝔪
= 0

From prop. 2.1.2. we get that x⃗ is an Nsequence. Now the length of any maximal N-sequence (in 𝔪) is equal to

depth N = dim N = dim R = d

so x⃗ is a maximal N-sequence. construct a minimal free resolution of k and 𝔪

The R-module k = R/𝔪 is generated by ¯1 = 1+𝔪, so this is a minimal set of generators of k. Denote F_{0} = Rα, the free R-module of rank 1. We construct a surjective R-morphism

φ_{0} : F_{0} ⟶ k : α ⟶ ¯1

Identify F_{0} to R, so the kernel becomes 𝔪. Take a minimal set of generators of 𝔪, say {y_{1}, …, y_{l}} and set F_{1} = ⨁ Rβ_{i} the free Rmodule of rank l. We have a R-morphism

φ_{1} : F_{1} ⟶ F_{0} : β_{i} → y_{i}

with Im φ_{1} = 𝔪 = ker φ_{0}, making exact

F_{1} ⟶ F_{0} ⟶ k → 0

By repeating this proces, one obtains a minimal free resolution of k:

F. : ⋯ ⟶ F_{1} ⟶ F_{0} ⟶ k → 0

Scince Im φ_{1} = 𝔪, we have a minimal free resolution of 𝔪

G. : ⋯ ⟶ F_{2} ⟶ F_{1} ⟶ 𝔪 → 0

Because the resolution F. has finite length, the resolution G. has finite lenght. 𝔪 contains a R-regular element x, not in 𝔪^{2}

First we prove that 𝔪 contains a R-regular element. The length of a maximal R-sequence in 𝔪 is

grade (𝔪, R) = depth R

If depth R ≠ 0, we are done. Should depth R = 0, then we use the Auslander - Buchsbaum formula (this is possible since projdim 𝔪 < ∞)

projdim 𝔪 + depth 𝔪 = depth R

to get that projdim 𝔪 = 0. But then 𝔪 is a free R-module. Taking the first element of any Rmodule basis, we have a R-regular element. Contradiction with the fact that depth R = 0.

Next we prove that our R-regular element x can be chosen outside 𝔪^{2}. If x ∈ 𝔪^{2}, then choose y ∈ 𝔪 \ 𝔪^{2}. Take N = Rx + Ry. Then N ⊄ 𝔭, ∀𝔭 ∈ Ass R. (remember that the union of all 𝔭 in Ass R is exactly the set of all zerodivisors of R). By using prime avoidance:

∃a ∈ R : y + ax ∉ 𝔭, ∀𝔭 ∈ Ass R

Therefore

y + ax ∉ 𝔪^{2}

and remains an R-regular element.

If 𝔪 = (z) then z is R-regular.

x ∈ 𝔪 so we may find an α ∈ R \ 𝔪 such that x = αz. Now suppose for any r ∈ R that

zr = 0
⇒ α^{−1}xr = 0 (α invertible because α ∈ R \ 𝔪)
⇒ α^{−1}r = 0 (x is R-regular )
⇒ r = 0

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

Obvious.

2. μ(𝔪/(x)) < μ(𝔪)

Notation: r̄ = r + (x) ∈ R/(x), ∀r ∈ R Let {y_{1}, …, y_{s}} be a minimal set of generators of 𝔪. Then {ȳ_{1}, …, ȳ_{s}} is a set of generators of 𝔪/(x). We prove that the last set cannot be minimal. Indeed, since x ∈ 𝔪, there are coefficients a_{1}, …, a_{s} in R such that

x = a_{1}y_{1} + … + a_{s}y_{s}

but x ∉ 𝔪^{2} so at least one of these coefficients is not in 𝔪, say a_{i} ∉ 𝔪. But then a_{i} is invertible in R, so

ȳ_{i} ∈ (ȳ_{1}, …, ȳ_{i−1}, ȳ_{i+1} , …, ȳ_{s})

3. projdim _{R/(x)} 𝔪/(x) < ∞

As a consequence of lemma 1.3.5

projdim _{R} 𝔪 = projdim _{R/(x)} 𝔪/(x)𝔪 < ∞

If we can show that 𝔪/(x) is a direct summand of 𝔪/(x)𝔪 then projdim _{R/(x)} 𝔪/(x) < ∞ and we are done.

Notation:

r̄ = r + (x) ∈ R/(x) ∀r ∈ R
ã = a + 𝔪^{2} ∈ R/𝔪^{2} ∀a ∈ 𝔪

Since x ∉ 𝔪^{2} we may find elements x_{2}, …, x_{l} ∈ 𝔪 such that

{x̃, x̃_{2}, …, x̃_{l}}

is a basis of the R/𝔪 -vectorspace 𝔪/𝔪^{2}. Let J = (x_{2}, …, x_{l}). One can check that

J ∩ (x) ⊂ (x)𝔪

Now build a composition of maps

J/J ∩ (x) ⟶ 𝔪/(x)𝔪 ⟶ 𝔪/(x)

by sending

x_{i} + J ∩ (x) ↦ x_{i} + (x)𝔪 ↦ x_{i} + (x)

But

𝔪/(x) = (J + (x))/(x) ≅ J/J ∩ (x)

so in this composition, x̄_{i} is sent to itself, so the composition is the identity. We conclude that 𝔪/(x) is a direct summand of 𝔪/(x)𝔪.

4. 𝔪̄/𝔪̄^{2} is free as R/𝔪-module

Take the basis of the R/𝔪 -vectorspace 𝔪/𝔪^{2} from part 3.

{x̃, x̃_{2}, …, x̃_{l}}

Then one easily checks that the set

{x̄_{2} + 𝔪̄^{2}, …, x̄_{l} + 𝔪̄^{2}}

generates 𝔪̄/𝔪̄^{2} as an R̄/𝔪̄-module. In fact these generators are linear independent.