Canonical trace ideal and residue for numerical semigroup rings

Abstract

For a numerical semigroup ring K[H] we study the trace of its canonical ideal. The colength of this ideal is called the residue of H. This invariant measures how far is H from being symmetric, i.e. how far is K[H] from being a Gorenstein ring. We remark that the canonical trace ideal contains the conductor ideal, and we study bounds for the residue. For 3-generated numerical semigroups we give explicit formulas for the canonical trace ideal and the residue of H. Thus, in this setting we can classify those whose residue is at most one (the nearly Gorenstein ones), and we show the eventual periodic behaviour of the residue in a shifted family.

Appendix A. Trace ideals as conductor ideals in local rings

Some statements in this paper can be formulated for extensions of 1-dimensional local domains where the conductor is defined. In this appendix we would like to understand the case when the trace of an ideal coincides with the conductor ideal, and some consequences this draws.

Hereafter, $(R,{\mathfrak m})$ is a 1-dimensional Cohen–Macaulay local ring with canonical module $\omega _R$. Assume now that $(R,{\mathfrak m})\subset ({\widetilde{R}},{\mathfrak n})$ is an extension of local rings, where ${\widetilde{R}}$ is a finite R-module and a discrete valuation ring such that ${\widetilde{R}}\subseteq Q(R)$. We also assume that the inclusion map $R\rightarrow {\widetilde{R}}$ induces an isomorphism $R/{\mathfrak m}\rightarrow {\widetilde{R}}/{\mathfrak n}$. The set

$$\begin{aligned} {\mathcal {C}}_{{\widetilde{R}}/R}= \{x \in R: x {\widetilde{R}} \subseteq R \} \end{aligned}$$

is called the conductor of this extension and it is an ideal of both R and ${\widetilde{R}}$. With the notation introduced we have

Proposition A.1

For any ideal $I\subset R$ one has ${\mathcal {C}}_{{\widetilde{R}}/R}\subseteq {\text {tr}}(I)$. In particular, ${\mathcal {C}}_{{\widetilde{R}}/R}\subseteq {\text {tr}}(\omega _R)$.

Proof

There exists $f\in I$ such that $I{\widetilde{R}}=(f){\widetilde{R}}$. Now let $g\in {\mathcal {C}}_{{\widetilde{R}}/R}$. Then $(g/f)I\subseteq (g/f)I{\widetilde{R}}=g{\widetilde{R}}\subseteq R$. Thus $g/f\in I^{-1}$. Since $g=(g/f)f$ with $f\in I$, it follows that $g\in I^{-1}\cdot I={\text {tr}}(I)$, where for the last equation we used [13, Lemma 1.1]. $\square$

Corollary A.2

R is nearly Gorenstein if ${\mathcal {C}}_{{\widetilde{R}}/R}={\mathfrak m}$.

For the rest of this section, the ideal quotients are computed in $Q(R)=Q({\widetilde{R}})$. The following lemma will be used in the proof of Proposition A.4.

Lemma A.3

${\widetilde{R}}=R:{\mathcal {C}}_{{\widetilde{R}}/R}$

Proof

It is clear from the definition of ${\mathcal {C}}_{{\widetilde{R}}/R}$ that ${\widetilde{R}} \subseteq R:{\mathcal {C}}_{{\widetilde{R}}/R}$.

For the reverse inclusion, let $f\in R:{\mathcal {C}}_{{\widetilde{R}}/R}$. If $f\notin {\widetilde{R}}$, then since $Q(R)=Q({\widetilde{R}})$ we may write $f=\epsilon t^{-a}$ with $\epsilon$ invertible in ${\widetilde{R}}$, a positive integer and t a generator of the maximal ideal of ${\widetilde{R}}$. Since ${\mathcal {C}}_{{\widetilde{R}}/R}$ is an ideal in ${\widetilde{R}}$, there exists $b>0$ such that ${\mathcal {C}}_{{\widetilde{R}}/R}=(t^b){\widetilde{R}}$.

The property $f\in R:{\mathcal {C}}_{{\widetilde{R}}/R}$ now reads as $\epsilon t^{-a} {\mathcal {C}}_{{\widetilde{R}}/R} \subseteq R$, or equivalently, that ${\mathcal {C}}_{{\widetilde{R}}/R} \subseteq t^a R$. This implies that $b\ge a$. Since we may write $t^{b-1}= f\cdot g$ where $g=\epsilon ^{-1}t^{a+b-1}$ is clearly in ${\mathcal {C}}_{{\widetilde{R}}/R}$, it follows that $t^{b-1}\in R$.

We claim that $t^{b-1}{\widetilde{R}} \subseteq R$. This will then lead to a contradiction, since $t^{b-1}\not \in {\mathcal {C}}_{{\widetilde{R}}/R}$. It is clear that $t^{b-1}{\mathfrak n}= {\mathcal {C}}_{{\widetilde{R}}/R}\subseteq R$. Thus is suffices to show that if $\nu \in {\widetilde{R}}$ is a unit, then $\nu t^{b-1}\in R$. To prove this, we use our assumption that $R/{\mathfrak m}\rightarrow {\widetilde{R}}/{\mathfrak n}$ is an isomorphism. In order to simplify notation we may assume that $R/{\mathfrak m}= {\widetilde{R}}/{\mathfrak n}$. Then this implies that there exists $h\in R$ such that $\nu -h\in {\mathfrak n}$. Thus, $\nu =h+h_1$ with $h\in R$ and $h_1\in {\mathfrak n}$, and therefore $\nu t^{b-1}=h t^{b-1}+h_1 t^{b-1}$. Since both summands on the right hand side of this equation belong to R, the claim follows.

This concludes the proof of the inclusion $R:{\mathcal {C}}_{{\widetilde{R}}/R} \subseteq {\widetilde{R}}$. $\square$

Our next result gives several characterizations of the situation when ${\text {tr}}(I)={\mathcal {C}}_{{\widetilde{R}}/R}$, in terms of the fractionary ideal $I^{-1}$. We first recall that for any fractionary ideal J of R, the ideal quotient J : J may be identified with the endomorphism ring ${\text {End}}(J)$ of J, see [19, Lemma 3.14].

Proposition A.4

Let $I\subset R$ be an ideal. The following conditions are equivalent:

(i):

${\text {tr}}(I)={\mathcal {C}}_{{\widetilde{R}}/R}$.

(ii):

${\text {End}}(I^{-1})={\widetilde{R}}$.

(iii):

$I^{-1}\cong {\widetilde{R}}$.

Proof

(i) ${}\Rightarrow {}$(ii): If ${\text {tr}}(I)={\mathcal {C}}_{{\widetilde{R}}/R}$, then using Lemma A.3 and the remark following it, we get

$$\begin{aligned} {\widetilde{R}}=R:{\mathcal {C}}_{{\widetilde{R}}/R} =R:(I \cdot I^{-1})=(R:I):I^{-1}={\text {End}}(I^{-1}). \end{aligned}$$

(ii)${}\Rightarrow {}$(i): Suppose that ${\text {tr}}(I)\ne {\mathcal {C}}_{{\widetilde{R}}/R}$. Then ${\mathcal {C}}_{{\widetilde{R}}/R}$ is properly contained in ${\text {tr}}(I)$. It follows that ${\mathcal {C}}_{{\widetilde{R}}/R}$ is properly contained in ${\text {tr}}(I){\widetilde{R}}$. Let t be generator of the maximal ideal of ${\widetilde{R}}$. Then there exist integers $a <b$ such that ${\text {tr}}(I){\widetilde{R}}=t^a{\widetilde{R}}$ and ${\mathcal {C}}_{{\widetilde{R}}/R}=t^b{\widetilde{R}}$. Clearly, $t^{b}\in {\mathcal {C}}_{{\widetilde{R}}/R}{\widetilde{R}}={\mathcal {C}}_{{\widetilde{R}}/R}$. Since ${\text {tr}}(I){\widetilde{R}}$ is a principal ideal, there exists $f\in {\text {tr}}(I)$ such that ${\text {tr}}(I){\widetilde{R}}= (f){\widetilde{R}}$. Therefore, there exists u invertible in ${\widetilde{R}}$ such that $t^a=u\cdot f$.

Since by assumption ${\widetilde{R}}={\text {End}}(I^{-1})$ (which is $R:{\text {tr}}(I)$), we obtain that $t^a\in R$. We may write $t^{b-1}=t^{b-a-1}\cdot t^a$, where $t^{b-a-1}\in {\widetilde{R}}$ and $t^a\in {\text {tr}}(I)$. Using again that ${\widetilde{R}}=R:{\text {tr}}(I)$, it follows that $t^{b-1}\in R$. Arguing as in the proof of Lemma A.3 we obtain $t^{b-1} \in {\mathcal {C}}_{{\widetilde{R}}/R}$, which is a contradiction.

(ii)${}\Rightarrow {}$(iii): If ${\text {End}}(I^{-1})={\widetilde{R}}$, then $I^{-1}{\widetilde{R}}=I^{-1}$. Since any nonzero ${\widetilde{R}}$-ideal is isomorphic to ${\widetilde{R}}$, the assertion follows.

(iii)${}\Rightarrow {}$(ii) is obvious. $\square$

For an R-module M we let e(M) denote its multiplicity.

Corollary A.5

Let R be as before, and assume in addition that R is of the form $R=S/J$ with $(S,{\mathfrak n})$ regular local ring of dimension 3 and $J\subseteq {\mathfrak n}^2$.

If ${\text {tr}}(\omega _R)= {\mathcal {C}}_{{\widetilde{R}}/R}$ then $e(R) \le \mu (J)$. Moreover, if R is an almost complete intersection, then R has minimal multiplicity.

Proof

Since R is a 1-dimensional domain, it it Cohen-Macaulay and ${\text {proj\,dim}}_S (R)=1$. Thus J has a minimal presentation $0 \rightarrow S^{g-1} \rightarrow S^{g} \rightarrow J$, where $g=\mu (J)$.

Now Proposition A.4(iii) together with ${\text {tr}}(\omega _R)= {\mathcal {C}}_{{\widetilde{R}}/R}$ imply that $\omega _R^{-1}$ and ${\widetilde{R}}$ are isomorphic R-modules, and they must have the same number of minimal generators over R. Hence $\dim _{R/{\mathfrak m}}({\widetilde{R}}/{\mathfrak m}{\widetilde{R}})=\mu (\omega _R^{-1}) \le g$, by [13, Corollary 3.4].

As ${\widetilde{R}}$ is a discrete valuation ring, there exists $f\in {\mathfrak m}$ such that ${\mathfrak m}{\widetilde{R}}=f{\widetilde{R}}$. Since ${\widetilde{R}}$ is a finitely generated R-module of rank 1, it follows that $e(R)=e({\widetilde{R}})$, see [3, Corollary 4.7.9]. Now $e({\widetilde{R}})=\dim _{R/{\mathfrak m}}{\mathfrak m}^k{\widetilde{R}}/{\mathfrak m}^{k+1}{\widetilde{R}}$ for $k\gg 0$. Since ${\mathfrak m}^k{\widetilde{R}}/{\mathfrak m}^{k+1}{\widetilde{R}}=f^k{\widetilde{R}}/f^{k+1}{\widetilde{R}}\cong {\widetilde{R}}/f{\widetilde{R}} \cong {\widetilde{R}}/{\mathfrak m}{\widetilde{R}}$, we conclude from the above considerations that $e(R)\le g$, as desired.

If R is an almost complete intersection, then $\mu (J)={\text {height}}(J)+1= \dim S$, hence $e(R) \le 3$. On the other hand, a celebrated inequality of Abhyankar gives $3={\text {emb\,dim}}R \le e(R)+\dim (R)-1=e(R)$. Thus $e(R)=3$, as desired. $\square$