Auslander-Reiten Sequences for Gorenstein Rings of Dimension One

Abstract

Let R be a complete local Gorenstein ring of dimension one, with maximal ideal $\mathfrak {m}$. We show that if M is a Cohen-Macaulay R-module which begins an AR-sequence, then this sequence is produced by a particular endomorphism of $\mathfrak {m}$ corresponding to a minimal prime ideal of R. We apply this result to determining the shape of some components of stable Auslander-Reiten quivers, which in the considered examples are shown to be tubes.

Appendix

In this Appendix we record some lemmas for reduced connected graded rings of dimension one. The following theorem is well-known.

Theorem A.1

LetB be one-dimensional, noetherian, local domain with integral closureB¯ and$\mathfrak {m}$_B-adiccompletion $\hat {B}$. Then the following are equivalent.

• (1) $\hat {B}$isa domain. (“B is analytically irreducible.”)

• (2) $\overline {B}$islocal and $\hat {B}$isreduced. (“B is unibranched and analytically unramified.”)

• (3) $\overline {B}$islocal and finitely generated as aB-module.

Notation A.2

If R is a connected graded ring, let $\hat {R}$ denote the completion of R with respect to its graded maximal ideal, $\mathfrak {m}$.

Lemma A.3

Let R be a reduced connected graded ring. Then:

• 1. The integral closure ofR inR[nonzerodivisors]⁻¹coincideswith the integral closure ofR inQ = R[graded nonzerodivisors]⁻¹, our definition of$\overline {R}$. Moreover, $\overline {R} =\bigoplus _{i \ge 0} \overline {R}_{i}$isanℕ-gradedsubring ofQ.

• 2. We have $\hat {R}= {\prod }_{i \ge 0} R_{i}$.

• 3. The completion,R̂, is also reduced. IfR is a domain, thenR̂ isa domain.

• 4. The integral closure,$\overline {R}$, is finitely generated as anR-module.

• 5. The integral closure of the completion, $\overline {\hat {R}}$, is finitely generated as anR̂-module.

Proof

Statement (1) is [11, Corollary 2.3.6]. Statement (2) can be checked by noting that $\{\mathfrak {m}^{i}\}_{i}$ is cofinal with $\{\bigoplus _{j \ge i} R_{j}\}_{i}$, and checking that the completion of R with respect to the latter filtration is isomorphic to ${\prod }_{i \ge 0} R_{i}$. From (2) we see that $\hat {R}$ is reduced, resp. a domain, if R is such. As R is a finitely generated algebra over the field R₀, (4) is a consequence of [14, Theorem 72]. The last assertion is a consequence of Theorem A.1 (alternatively, it follows from (4)). □

Lemma A.4

LetD be a connected graded domain of dimension one, andlet $\mathfrak {q}=\bigoplus _{i \ge 1} D_{i}$, and $\mathfrak {n}=\bigoplus _{i \ge 1} \overline {D}_{i}$. Then

• (a) ${\overline {D}}_{0}$isa field, and

• (b) ${\prod }_{i \ge 0} {\overline {D}}_{i} =\hat {\overline {D}}^{\mathfrak {n}}=\hat {\overline {D}}^{\mathfrak {q}}=\overline {\hat {D}}$.

Proof

The notation ${\overline {D}}_{i}$ means $({\overline {D}})_{i}$, and makes sense due to Lemma A.3, as does $\overline {\hat {D}}$. Since ${\overline {D}}$ is an ℕ-graded domain, $\mathfrak {n}$ is a prime ideal, and is thus maximal since $\dim {\overline {D}}=\dim D = 1$. So ${\overline {D}}_{0}$ is a field. Now ${\prod }_{i \ge 0} {\overline {D}}_{i}=\hat {\overline {D}}^{\mathfrak {n}}$ by Lemma A.3. Note that $X_{\mathfrak {n}} \neq 0$ for all graded ${\overline {D}}$-modules X≠ 0. Now ${\overline {D}}_{\mathfrak {n}}/(\mathfrak {q} {\overline {D}}_{\mathfrak {n}})$ is an artinian local ring, so there exists i ≥ 1 such that $(({\mathfrak {n}} ^{i} +\mathfrak {q}{\overline {D}})/\mathfrak {q}{\overline {D}} )_{\mathfrak {n}} = 0$, hence $({\mathfrak {n}} ^{i} +\mathfrak {q} {\overline {D}})/\mathfrak {q} {\overline {D}} = 0$. Thus $\{{\mathfrak {n}} ^{i}\}_{i}$ and $\{\mathfrak {q}^{i} {\overline {D}}\}_{i}$ are cofinal, so $\hat {\overline {D}}^{\mathfrak {n}}=\hat {\overline {D}}^{\mathfrak {q}}$. Lastly we show $\hat {\overline {D}}^{\mathfrak {q}}=\overline {\hat {D}}$. Note that $\overline {D} \hookrightarrow \overline {\hat {D}}$, and since $\overline {\hat {D}}$ is complete by Lemma A.3 Rbar and Rhat are gdd, we have $\overline {\hat {D}} \supseteq \hat {\overline {D}}^{q} \supseteq \hat {D}$. It remains to observe that $\hat {\overline {D}}^{q}$ is normal. But any I-adic completion of an excellent, normal ring, such as ${\overline {D}}$, is normal ([14, Theorem 79]). □

Lemma A.5

LetD be a connected graded domain of dimension one , andlet $l= \min \{ i > 0 | \overline {D}_{i} \neq 0 \}$. Lett be any nonzero element of ${\overline {D}}_{l}$. Then ${\overline {D}}=\bigoplus _{i\ge 0} {\overline {D}}_{0} t^{i}$isthe polynomial ring over the field ${\overline {D}}_{0}$inthe variable t; and $\hat {\overline {D}}={\prod }_{i\ge 0} {\overline {D}}_{i} t^{i}$isthe power series ring.

Proof

By the previous lemmas, ${\overline {D}}$ is connected graded, so we may assume $D={\overline {D}}$ to improve notation. Then the previous lemma also shows that $\hat {D}={\prod }_{i \ge 0} D_{i}$ is a normal domain. Thus it is a DVR; let $\pi \in \hat {D}$ be a uniformizing parameter. So $\pi \hat {D}={\prod }_{i \ge 1} D_{i}$. Then t = uπⁱ for some unit $u \in {\prod }_{i\ge 0} D_{i}$, and it follows that i = 1, hence t is a uniformizing parameter for $\hat {D}$. It follows that D_i = 0 for i∉ℕl, and D_i = D₀t^i/l for i ∈ℕl. The lemma follows. □

Remark A.6

If D₀ is algebraically closed, Lemmas A.4 and A.5 show that D has the form $k[t^{i_{1}},...,t^{i_{n}}]$.