On the set of trace ideals of a Noetherian ring

Abstract

We consider trace ideals in Noetherian rings and focus our attention to one-dimensional analytically irreducible local rings. For such rings we classify those Gorenstein rings which admit only a finite number of trace ideals.

1 Introduction

Let R be a Noetherian ring, M a finitely generated R-module, and $\varphi : M\rightarrow R$ an R-module homomorphism. Then $\varphi (M)$ is an ideal in R. The sum of all such ideals is called the trace of M and is denoted by ${\text {tr}}(M)$. We call an ideal $I\subseteq R$ a trace ideal if there exists a finitely generated R-module M such that $I={\text {tr}}(M)$. The concept of trace is a classical notion which appears in various contexts. The trace of a module, which we consider here, is not related to the trace map of an endomorphism which plays an important role in algebraic geometry, see Kunz (2005, Appendix H). Very recently, the trace of the canonical module has led to the notion of nearly Gorenstein rings, see Herzog et al. (2021) and Herzog et al. (2019). Of course in general not all ideals are trace ideals. Lindo and Pande (2018) proved that a local ring is an artinian Gorenstein ring if and only if every ideal is a trace ideal, and Kobayashi and Takahashi (2019) classified all Noetherian local rings for which every ideal is isomorphic to a trace ideal.

In this paper, we focus on the problem to identify trace ideals in R. In the first section we recall a few basic properties of trace ideals which are mostly borrowed from the paper of Lindo (2017). The following important facts help us to identify the trace ideals. Namely, an ideal I is a trace ideal if and only if $I={\text {tr}}(I)$. One also has ${\text {tr}}(I)=II^{-1}$ for any ideal $I\subsetneq R$ with ${\text {grade}}(I)>0$. Here $I^{-1}$ is the set of elements f in the full ring of quotients Q(R) of R with the property that $fI\subsetneq R$. We use the fact that I is a trace ideal if and only if ${\text {tr}}(I)=I$ to show in Proposition 2.7 that a generically Gorenstein ring with canonical ideal $\omega _R$ is Gorenstein if and only if $\omega _R\cong {\text {tr}}(\omega _R)$.

Another interesting fact is that the sum of trace ideals is again a trace ideal. This has the consequence that for any given ideal I there exists a unique, with respect to inclusion, maximal trace ideal contained in I which we denote by ${\text {subtr}}(I)$, and we have ${\text {subtr}}(I)\subseteq I\subseteq {\text {tr}}(I)$. In general there is no smallest trace ideal containing I. In particular, ${\text {tr}}(I)$ may not be the smallest trace ideal containing I. A smallest trace ideal containing I would exist if the intersection of trace ideal is again a trace ideal, which also fails to be true in general. We close Sect. 1 by showing in Theorem 2.10 that in a local Cohen–Macaulay domain R the set of trace ideals is precisely the set of ideal of height $>1$, if and only if R is normal. In Theorem 2.10 the hypothesis that R is a domain is required. Indeed, we give an example of a trace ideal $I\subsetneq R$ of height 1, where R is Cohen–Macaulay and normal, but not a domain. These results leave the question open for which general local rings an ideal I is a trace ideal if and only if ${\text {height}}(I)>1$.

In Sect. 2, we study the trace ideals of one-dimensional analytically irreducible local rings. Among them are the semigroup rings attached to numerical semigroups. If $(R,{{\mathfrak {m}}})$ is a one-dimensional analytically irreducible local ring, then its integral closure $({\overline{R}}, {{\mathfrak {n}}})$ is a discrete valuation ring and a finite R-module. We assume in addition that R and ${\overline{R}}$ have the same residue class field. The ideal $C=R:{\overline{R}}$ is called the conductor of R. If ${{\mathfrak {n}}}=(t)$, then $C=t^c{\overline{R}}$ for some integer $c\ge 0$. Trace ideals are rigid in the sense that if I and J are trace ideals and $I\cong J$, then $I=J$. We denote by ${{\mathcal {T}}}_R$ the set of trace ideals of R and by ${{\mathcal {C}}}_R$ the set of ideals I with $C\subseteq I\subseteq R$. It can be seen that ${{\mathcal {T}}}_R$ is a subset of ${{\mathcal {C}}}_R$, which in most cases is a proper subset. There is an explicit set of trace ideals. Indeed for each $f\in R{\setminus } {{\mathfrak {n}}}C$, the ideal $f{\overline{R}}\cap R$ is a trace ideal. These are exactly the integrally closed ideals which contain the conductor. Let v denote the valuation induced by ${\overline{R}}$, and $H=v(R)$ the value semigroup of R. There exist exactly $n(H)+1$ inegrally ideals containing the conductor, where n(H) denotes the number of elements $h\in H$ with $h<c$.

We denote by ${{\mathcal {O}}}_R$ the set of overrings $R'$ of R which are contained in ${\overline{R}}$. For the rest of the paper Proposition 3.10 is crucial. This result is due to Goto et al. (2020, Corollary 2.8), and it says that if R is Gorenstein, then the map $\alpha : {{\mathcal {T}}}_R\rightarrow {{\mathcal {O}}}_R$, $I\mapsto I^{-1}$ is bijective. By using this fact and assuming that R is Gorenstein, the maximal trace ideal contained in I can be computed for an ideal $I\in {{\mathcal {C}}}_R$ as follows: there is a smallest integer $k_0$ such that $(I^{-1})^{k_{0}}=(I^{-1})^{k}$ for all $k\ge k_0$. Then ${\text {subtr}}(I)=J^{-1}$ where $J=(I^{-1})^{k_{0}}$. Another application is Corollary 3.16 which asserts that if R is Gorenstein and $|R/{{\mathfrak {m}}}|=\infty$ and H is the value semigroup of R, then the following conditions are equivalent: (1) R has only finitely many trace ideals, (2) $H=\langle 2,a\rangle$ where $a>2$ is an odd number, or $H=\langle 3,4\rangle$ or $H=\langle 3,5\rangle$, (3) all trace ideals of R are integrally closed.

In the last section, we consider certain binary operations on the set of trace ideals in one-dimensional analytically irreducible local domains. As mentioned before, intersections of trace ideals are in general not trace ideals. The same is the case for products of trace ideals. If we replace the usual product of two ideals I and J by the product $I*J=(I^{-1}J^{-1})^{-1}$, then it turns out that $I*J$ is a trace ideal if R is Gorenstein and if I and J are trace ideals. Moreover, if R is Gorenstein, this product is associative and coincides with the largest trace ideal contained in $I\cap J$. In general one has $I*(J+L)\subseteq I*J+I*L$. Equality holds if R is the semigroup ring of a numerical semigroup and the ideals I, J and L are monomial ideals. Hence in this case ${{\mathcal {T}}}_R$ has the structure of a semiring.

We gratefully acknowledge the use of the numericalsgps package (Delgado et al. 2016) in GAP for our computations.

2 The trace of a module and trace ideals

Let R be a Noetherian commutative ring and ${\text {Mod}}_R$ be the category of finitely generated R-modules. Let $M\in {\text {Mod}}_R$. The trace of M, denoted ${\text {tr}}_R(M)$, is defined to be the sum of the ideals $\varphi (M)$ with $\varphi \in {\text {Hom}}_R(M,R)$. Thus

$$\begin{aligned} {\text {tr}}_R(M)=\sum _{\varphi \in {\text {Hom}}_R(M,R)}\varphi (M). \end{aligned}$$

For simplicity we write ${\text {tr}}(M)$ instead of ${\text {tr}}_R(M)$, if no confusion is possible.

It follows immediately from the definition of the trace that ${\text {tr}}(M)={\text {tr}}(N)$ if $M\cong N$.

We call an ideal $I\subsetneq R$ a trace ideal, if there exists $M\in {\text {Mod}}_R$ with $I={\text {tr}}(M)$.

Lindo (2017, Proposition 2.8(iv)) observed that trace ideals are characterized as follows.

Proposition 2.1

Let $I\subsetneq R$ be an ideal. Then $I\subseteq {\text {tr}}(I)$, and $I={\text {tr}}(I)$ if and only if I is a trace ideal.

Corollary 2.2

Let $I, J\subsetneq R$ be non-zero ideals. Then the following holds:

• (a) Let I and J be trace ideals. Then I is isomorphic to J if and only if $I=J$.

• (b) Suppose $I\ne (0)$ and $I=fJ$, where f is a non zero-divisor which is contained in all maximal ideals of R. Then I is not a trace ideal.

Proof

 

• (a) Suppose $I\cong J$. Then $I={\text {tr}}(I)={\text {tr}}(J)=J.$

• (b) Since $I\cong J$ it follows that ${\text {tr}}(I)={\text {tr}}(J)$. Suppose I is a trace ideal. Then $fJ=I={\text {tr}}(I)={\text {tr}}(J)\supseteq J$. Therefore, $fJ=J$, which implies that $f^2J=fJ=J$. Thus, by induction, $f^kJ=J$ for all k, and hence $J=\bigcap _{k\ge 0}f^kJ=(0)$, by Krull’s intersection theorem, see Kunz (1985, Chapter V, Corollary 5.7). This is a contradiction since $J\ne (0)$.

$\square$

In the next section, we give an example which shows that the converse of Corollary 2.2(b) is not true in general, see Example 3.5.

Products of trace ideals need not be trace ideals. But as shown in Lindo (2017, Proposition 2.8(ii)) one has

Proposition 2.3

The sum of trace ideals is again a trace ideal. Indeed,

$$\begin{aligned}{\text {tr}}(M\oplus N) ={\text {tr}}(M)+{\text {tr}}(N) \quad \text {for} \quad M,N\in {\text {Mod}}_R.\end{aligned}$$

Corollary 2.4

Let $I\subseteq R$ be an ideal. Then there exists a unique trace ideal contained in I which is maximal with respect to inclusion, which we denoted by ${\text {subtr}}(I)$. Hence

$$\begin{aligned} {\text {subtr}}(I)\subseteq I\subseteq {\text {tr}}(I), \end{aligned}$$

and ${\text {subtr}}(I)\subsetneq I$ if and only if $I\subsetneq {\text {tr}}(I)$.

Proof

The ideal ${\text {subtr}}(I)$ is just the sum of all trace ideals which are contained in I. This sum is not empty, because $(0)\subseteq I$ is a trace ideal. Moreover, ${\text {subtr}}(I)\subseteq I$. Since R is Noetherian, this sum is a finite sum. Therefore, by Proposition 2.3, ${\text {subtr}}(I)$ is a trace ideal. By its construction, ${\text {subtr}}(I)$ is the largest trace ideal contained in I. $\square$

Example 2.5

Let K be a field, $R=K[|t^3,t^4|]$ and let $I=(t^3,t^8)$. Then ${\text {tr}}(I)=(t^3,t^4)$ which is the maximal ideal of R. Since $I\ne {\text {tr}}(I)$, I is not a trace ideal. On the other hand, the ideal $J=(t^6,t^7,t^8)$ is a trace ideal. Note that $I/J\cong R/{{\mathfrak {m}}}$. Therefore, any ideal $L\subseteq I$ with $J\subsetneq L$ is equal to I. This shows that J is the biggest ideal contained in I which is different from I. Hence, $J={\text {subtr}}(I)$. So, in this case we have

$$\begin{aligned} {\text {subtr}}(I)=(t^6,t^7,t^8)\subsetneq (t^3,t^8)\subsetneq (t^3,t^4)={\text {tr}}(I). \end{aligned}$$

Similarly to Lindo (2017, Proposition 2.8(iii)), we have

Proposition 2.6

Let R be a local ring or a positively graded K-algebra where K is a field, and let $M\in {\text {Mod}}_R$. When R is graded we assume that M is also graded. Under these assumptions the following conditions are equivalent:

• (i) ${\text {tr}}(M)=R$.

• (ii) There exists a module N such that $M\cong N\oplus R$.

Proof

(i) ${}\Rightarrow {}$(ii): Since ${\text {tr}}_R(M)=\sum _{\varphi \in {\text {Hom}}_R(M,R)}\varphi (M)=R$, our assumption on the ring R implies that there exists $\varphi : M\rightarrow R$ with $\varphi (M)=R$. Therefore $\varphi : M\rightarrow R$ is surjective. Since R is a free R-module, the map $\varphi$ splits. This implies that $M\cong {\text {Ker}}(\varphi )\oplus R$.

(ii) ${}\Rightarrow {}$(i): Since $M\cong N\oplus R$, it follows from Proposition 2.3 that ${\text {tr}}(M)={\text {tr}}(N\oplus R)={\text {tr}}(N)+{\text {tr}}(R)=R$. This shows that ${\text {tr}}(M)=R$. $\square$

Let Q(R) be a full ring of fractions of R. Then for any ideal I with ${\text {grade}}(I)\ge 1$, one has the well-known fact that

$$\begin{aligned} {\text {tr}}(I)=II^{-1}, \end{aligned}$$

(1)

where $I^{-1}= \{f \in Q(R):\; fI\in R\}$, see Lindo (2017, Proposition 2.8(iii)). Therefore, if ${\text {grade}}(I)\ge 1$, Proposition 2.1 implies that I is a trace ideal if and only if $I=II^{-1}$.

Observe that $I^{-1}$ is a fractionary ideal. Recall that any finitely generated R-submodule of Q(R) is called a fractionary ideal. In other words, if I is a fractionary ideal, then there exist $f_1,\ldots ,f_r\in Q(R)$ such that $I=(f_1,\ldots ,f_r)$. We have $f_i=h_i/g_i$ with $g_i,h_i\in R$, where $g_i$ is a non zero-divisor. It follows that gI is an ideal in R for $g=g_1g_2\ldots g_r$ and shows that, as an R-module, any fractionary ideal is isomorphic to an ideal of R.

We use the identity (1) to give the following characterization of Gorenstein rings.

Proposition 2.7

Let R be a local Cohen–Macaulay ring and suppose that R admits a canonical module $\omega _R$. Then R is Gorenstein if and only if $\omega _R\cong {\text {tr}}(\omega _R)$.

Proof

If R is Gorenstein, then $\omega _R\cong R$, and hence ${\text {tr}}(\omega _R)={\text {tr}}(R)=R=\omega _R$.

Conversely, suppose that $\omega _R\cong {\text {tr}}(\omega _R)$. Let P be a minimial prime ideal of R. Then $\omega _{R_P}\cong (\omega _R)_P\cong {\text {tr}}(\omega _R)_P\cong {\text {tr}}(\omega _{R_P})$. Therefore, considering the length of the modules, we get $\ell ({\text {tr}}(\omega _{R_P}))=\ell (\omega _{R_P})=\ell (R_P)$, where the last equality follows from Matlis duality. This implies that ${\text {tr}}(\omega _{R_P})=R_P$. Thus $R_P$ is Gorenstein. Thus, our hypothesis that $\omega _R\cong {\text {tr}}(\omega _R)$ implies that R is generically Gorenstein.

If $\dim R=0$, then R is Gorenstein and hence $\omega _R\cong R$ and there is nothing to prove. Suppose now that $\dim R>0$. Since R is generically Gorenstein, it follows that $\omega _R$ is isomorphic to a fractionary ideal, see Herzog and Kunz (1971, Korollar 6.7) or Bruns and Herzog (1998, Proposition 3.3.18). Thus, our assumption implies that $\omega _R\cong \omega _R\omega _R^{-1}$. Then

$$\begin{aligned} R=\omega _R:\omega _R\cong \omega _R: (\omega _R\omega _R^{-1})=(\omega _R:\omega _R):\omega _R^{-1}=R:\omega _R^{-1}=(\omega _R^{-1})^{-1}. \end{aligned}$$

The first equality follows from Herzog and Kunz (1971, Satz 6.1) or Bruns and Herzog (1998, Theorem 3.3.10). Taking once again the inverse on both sides of this equation, we find that $R=((\omega _R^{-1})^{-1})^{-1}=\omega _R^{-1}$. For the last equation we use the fact that $I^{-1}$ is reflexive for all ideals I containing a non zero-divisor. Now since $\omega _R^{-1}$ is a principal ideal, Herzog and Kunz (1971, Korollar 7.33) implies that $\omega _R$ itself is a principal ideal, which in turn implies that R is Gorenstein. $\square$

Proposition 2.8

Let K be a field, and let R be a local ring or a positively graded K-algebra. Furthermore, let ${{\mathfrak {m}}}$ be the maximal (graded) ideal of R. If ${{\mathfrak {m}}}$ is not a principal ideal, then ${{\mathfrak {m}}}$ is a trace ideal.

Proof

Suppose ${{\mathfrak {m}}}$ is not a trace ideal. Then ${\text {tr}}({{\mathfrak {m}}})\ne {{\mathfrak {m}}}$. Since ${{\mathfrak {m}}}\subsetneq {\text {tr}}({{\mathfrak {m}}})$, it follows that ${\text {tr}}({{\mathfrak {m}}})=R$. Then, by Proposition 2.6, we have ${{\mathfrak {m}}}\cong N\oplus R$ for some module N. This means that there exists an ideal $I\subsetneq {{\mathfrak {m}}}$ and a non zero-divisor $f\in {{\mathfrak {m}}}$ such that ${{\mathfrak {m}}}= I+(f)$ and $I\cap (f)=(0)$. Suppose $I\ne (0)$. Then there exists $0\ne g\in I$. Since f is a non zero-divisor $fg\ne 0$. Since $fg\in I\cap (f)$ it follows that $I\cap (f)\ne (0)$, a contradiction. Hence ${{\mathfrak {m}}}=(f)$ and this means ${{\mathfrak {m}}}$ is a principal ideal. $\square$

Let $R=K[x]/(x^2)$. Then the maximal ideal of R is a principle ideal and also a trace ideal. This shows that the converse of Proposition 2.8 is in general not correct.

Finally we recall the following result (Lindo 2017, Example 2.4).

Proposition 2.9

Suppose $I\subsetneq R$ is an ideal with ${\text {grade}}(I)>1$. Then I is a trace ideal. In particular, if R is Cohen–Macaulay and ${\text {height}}(I)>1$, then I is a trace ideal.

Now we show

Theorem 2.10

Let R be a Cohen–Macaulay domain. Then the following conditions are equivalent:

• (i) I is a trace ideal of R if and only if ${\text {grade}}(I)>1$.

• (ii) R is normal.

Proof

(i)${}\Rightarrow {}$(ii): By our assumption, if I is an ideal with ${\text {grade}}(I)=1$, then I is not a trace ideal. Hence, if P is a prime ideal of height one, it follows that $P\ne {\text {tr}}(P)$. Therefore, $P\subsetneq {\text {tr}}(P)$. Hence, after localization $R_P={\text {tr}}(P)R_P={\text {tr}}(PR_P)$. Here we use that trace localizes. Since $PR_P\ne R_P$, this shows that $PR_P$ is not a trace ideal. By Proposition 2.8, $PR_P$ is a principal ideal. Therefore $R_P$ is a discrete valuation ring. It follows that R satisfies the Serre condition $R_1$. By Serre’s normality criterion (Matsumura 1971, Sect. 23) this implies that R is normal.

(ii)${}\Rightarrow {}$(i): Suppose there exists a trace ideal I with ${\text {grade}}(I)=1$. Then, since R is Cohen–Macaulay, ${\text {height}}(I)=1$, and hence there exists a prime ideal P of height one with $I\subseteq P$. Now since I is a trace ideal we have $I={\text {tr}}(I)$ and hence $IR_P={\text {tr}}(IR_P)$. Therefore, $IR_P$ is a trace ideal of $R_P$ which is different from $R_P$. By assumption, R is normal. Therefore $R_P$ is a discrete valuation. This implies that $IR_P$ is a principal ideal generated by non a zero-divisor, contradicting Corollary 2.2(b). $\square$

Example 2.11

Let K be a field, $S=K[x,y]$ the polynomial ring over K with the variables x and y and $R=S/(xy)$. Then R is a one dimensional Cohen–Macaulay ring. Also R is normal because it is the direct product of normal domains, see Matsumura (1971, Sect. 23). Indeed, $R\cong K[x]\times K[y]$.

Set ${{\mathfrak {m}}}=(x,y)$. Next we claim that ${\text {tr}}({{\mathfrak {m}}})={{\mathfrak {m}}}.$ Since ${\text {grade}}({{\mathfrak {m}}})=1$ and R is normal, this example shows that in Theorem 2.10 we cannot drop the hypothesis that R is a domain.

In order to prove the claim, we assume to the contrary that ${\text {tr}}({{\mathfrak {m}}})\ne {{\mathfrak {m}}}$. Then ${\text {tr}}({{\mathfrak {m}}})=R$ and hence there exists a surjective R-module homomorphism $\varepsilon :{{\mathfrak {m}}}\rightarrow R$. Let Z be the kernel of this epimorphism and let $t=x+y$. Then $t\in {{\mathfrak {m}}}$ is a non zero-divisor and hence ${{\mathfrak {m}}}R_t=R_t$. Here $R_t$ is the ring which is obtained from R by localization with respect to powers of t. It follows that $\varepsilon$ after localization becomes an isomorphism. Therefore, $ZR_t=0$. This implies that $t^kZ=0$ for some k. Hence Z is a $R/(t^k)$-module. Since $\dim R=1$ and since t is a non zero-divisor it follows that $\dim R/(t^k)=0$, and so, $\dim Z=0$. In particular, ${\text {depth}}Z=0$. Since Z is a submodule of ${{\mathfrak {m}}}$, ${\text {depth}}{{\mathfrak {m}}}=0$. This is a contradiction, because ${\text {depth}}{{\mathfrak {m}}}=1$.

Question 2.12

Which are the rings with the property that an ideal I is a trace ideal if and only if ${\text {height}}(I)>1$?

3 Trace ideals of one-dimensional analytically irreducible local rings

In this section, we assume that $(R,{{\mathfrak {m}}})$ is a 1-dimensional domain with quotient field Q(R) whose integral closure ${\overline{R}}\subsetneq Q(R)$ is a discrete valuation ring and a finite R-module. This is equivalent to saying that R is analytically irreducible. Let $v: Q(R)\rightarrow {\mathbb Z}$ be the valuation attached to R. The set $v(R)=\{v(r): r\in R{\setminus }\{0\} \}$ is called the value semigroup of R. For details we refer to Herzog and Kunz (1971).

For a non-empty subset $A\subseteq {\mathbb Z}_{\ge 0}$ we denote by $\langle A\rangle$ the smallest additively closed subset containing A. Let $H=\langle A\rangle$. One has $\gcd (A)=1$ if and only if there exists an integer $c\ge 0$ such that $H =c+{\mathbb Z}_{\ge 0}$. In this case, H is called a numerical semigroup. The elements of A are called the generators H. Any numerical semigroup is finitely generated. The value semigroups v(R) of R is a numerical semigroup. The smallest element $h\in H$ with $h\ne 0$ is called the multiplicity of H which is denoted by e(H).

Note that ${\overline{R}}$ is a finitely generated R-module of rank one, and hence a fractionary ideal containing R. This implies that $C=R:{\overline{R}}$ is an ideal in R, which is called the conductor of R. We first recall a few well-known facts.

Lemma 3.1

The conductor C is the largest ideal in R such that $C{\overline{R}}=C$.

Proof

We first show that $C{\overline{R}}=C$. Indeed, let $f\in C$ and $g\in {\overline{R}}$. Then

$$\begin{aligned} (fg){\overline{R}}= f(g{\overline{R}})\subseteq f{\overline{R}}\subseteq R. \end{aligned}$$

This shows that $C{\overline{R}}\subseteq C$. The converse direction is trivial. Now let $D\subseteq R$ be an ideal such that $D{\overline{R}}=D$. Let $f\in D$. Then $f{\overline{R}}\subseteq D{\overline{R}}=D\subseteq R$. Therefore, $f\in R: {\overline{R}}=C$. This shows that $D\subseteq C$. $\square$

Proposition 3.2

Let $I\ne 0$ be a trace ideal of R. Then $C\subseteq I\subseteq R$.

Proof

It is clear that trace ideals are ideals in R. Next we show that $C\subseteq I$. We have $I=II^{-1}$ because I is a trace ideal. There exists $f\in I$ such that $I{\overline{R}}=f{\overline{R}}$. Now let $g\in C$. Then

$$\begin{aligned} (g/f)I\subseteq (g/f)I{\overline{R}}=(g/f)f{\overline{R}}=g{\overline{R}} \subseteq C\subseteq R. \end{aligned}$$

This shows that $g/f\in I^{-1}$. Hence $g=f(g/f)\in II^{-1}=I$, and so $C\subseteq I$. $\square$

We denote by ${{\mathfrak {n}}}$ the maximal ideal of ${\overline{R}}$. Then ${{\mathfrak {n}}}=(t)$ for some $t\in {\overline{R}}$. Note that ${{\mathfrak {m}}}=R\cap {{\mathfrak {n}}}$ because $R\subseteq {\overline{R}}$ is an integral extension. Since the kernel of the map $R\rightarrow {\overline{R}}/{{\mathfrak {n}}}$ is $R\cap {{\mathfrak {n}}}$ it follows that the induced map $R/{{\mathfrak {m}}}\rightarrow R/{{\mathfrak {n}}}$ is injective. For the rest of this paper we assume that $R/{{\mathfrak {m}}}\rightarrow R/{{\mathfrak {n}}}$ is an isomorphism. This is equivalent to saying that for each $g\in {\overline{R}}$, there exists $f\in R$ such that $f-g\in {{\mathfrak {n}}}$.

Lemma 3.3

Let $C=t^c{\overline{R}}$. Then $t^{c-1}\not \in R$.

Proof

Suppose $t^{c-1}\in R$. Then $D=t^{c-1}R+C$ is an ideal in R. We claim that $D{\overline{R}} =D$. Indeed, let $h\in D$ and $g\in {\overline{R}}$. Since $gC\subsetneq C$, it suffices to show that $gt^{c-1}\in D$. There exists $f\in R$ such that $g=f+s$ with $s\in {{\mathfrak {n}}}$. Then $t^{c-1}g =t^{c-1}f+t^{c-1}s$. Since $t^{c-1}f\in t^{c-1}R$ and $t^{c-1}s\in t^c{\overline{R}}=C$, it follows that $hg\in D$. This proves the claim. Since $C\subsetneq D$ and $D{\overline{R}} =D$, this contradicts Lemma 3.1. $\square$

As in Lemma 3.3, let $C=t^c{\overline{R}}$ be the conductor of R and let $H=v(R)$. Then $a\in H$ for all $a\ge c$ and $c-1\not \in H$. The elements in ${\mathbb Z}_{\ge 0}{\setminus } H$ are called the gaps of H. The largest gap of H is $c-1$ and it is called the Frobenius number of H. The number of gaps of H is denoted by g(H) and called the genus of H. The elements $a\in H$ with $a<c-1$ are called the non-gaps of H. The number of non-gaps is denoted by n(H). It is easy to see that $n(H)\le g(H)$. If equality holds, then H is called symmetric. Kunz (1970) showed that R is a Gorenstein ring if and only if $H=v(R)$ is symmetric.

Proposition 3.4

Let $I=fJ$ with $f\in {{\mathfrak {m}}}$ and $J\subseteq R$. Then $C\not \subseteq I$.

Proof

We have $C=t^c{\overline{R}}$. Since $fJ\subseteq (f)$, it is enough to show that $C\not \subseteq (f)$. Indeed, $ft^{c-1}\notin (f)$ because of Lemma 3.3. But we have $ft^{c-1}\in C$ because $f\in {{\mathfrak {m}}}$. $\square$

In Proposition 3.2 we have seen that $C\subseteq I\subseteq R$ if I is a trace ideal. The following example shows that the converse of this statement is not true in general.

Example 3.5

Let $R=K[| t^5,t^6,t^7|]$ and let $I= (t^6, t^{10}, t^{14})$. Then

$$\begin{aligned} (t^{10},t^{11}, t^{12}, t^{13}, t^{14})=C\subseteq I\subseteq R. \end{aligned}$$

But I is not a trace ideal because $(t^6, t^{7}, t^{10})={\text {tr}}(I)\ne I$.

We also find that I is not of the form fJ, see Proposition 3.4.

Let $I\subsetneq R$ be an ideal. Then there exists $f\in I$ such that $I{\overline{R}}= f{\overline{R}}$. The ideal I is integrally closed if and only if $I=f{\overline{R}}\cap R$.

Proposition 3.6

Let I be an integrally closed ideal of R with $C\subseteq I$. Then I is a trace ideal.

Proof

Suppose $I\subsetneq R$ is an integrally closed ideal containg the conductor, but I is not a trace ideal. Then $I^{-1}I\ne I$. Hence there exists $g\in I^{-1}$ such that $gI\subsetneq I$. Therefore there exists $h\in I$ such that $gh\not \in I$. Since $I=f{\overline{R}}\cap R$ with $f\in I$, and $gh\in R$, it follows that $gh\not \in f{\overline{R}}$. This implies that $gh/f\not \in {\overline{R}}$. Since ${\overline{R}}$ is discrete valuation ring, we see that $f/gh\in {{\mathfrak {n}}}$. Now we use that $h\in I$ and therefore $h=fl$ with $l\in {\overline{R}}$. Then we conclude that $1/gl\in {{\mathfrak {n}}}$ which implies that $1/g\in {{\mathfrak {n}}}$, too. Hence $t^{c-1} (1/g)\in C\subseteq I$. It follows that $t^{c-1}= gt^{c-1} (1/g)\in I^{-1}I\subseteq R$, a contradiction to Lemma 3.3. $\square$

Corollary 3.7

Let H be the value semigroup of R. Then the number of trace ideals is at least $n(H)+1$.

Proof

Let a be a non-gap of H. Then there exists an element $f\in R$ with $v(f)=a$. It follows that $f{\overline{R}}=t^a{\overline{R}}$. The integrally closed ideal $I_a=f{\overline{R}}\cap R$ is a trace ideal. If $a'$ is a non-gap with $a'\ne a$, then $I_a\ne I_{a'}$. This shows that together with C, which is also a trace ideal, we have at least $n(H)+1$ trace ideals. $\square$

Let $I\subseteq R$ be a non-zero ideal. Then $I\subseteq (I^{-1})^{-1}$. I is called reflexive if $I= (I^{-1})^{-1}$.

For the proof of next results we need two lemmata. The first lemma can be found in Berger (1962).

Lemma 3.8

(a) Let $I\subseteq R$ be a non-zero ideal. Then $I^{-1}$ is reflexive.

(b) R is Gorenstein if and only if all fractionary ideals of R are reflexive.

It follows from Lemma 3.8(a) that $(I^{-1})^{-1}$ is reflexive. In fact, $(I^{-1})^{-1}$ is a smallest reflexive ideal containing I. Therefore, $(I^{-1})^{-1}$ is called the reflexive hull of I. Furthermore, Lemma 3.8(b) implies that if I and J are fractionary ideals in a Gorenstein ring, then $I=J$ if and only if $I^{-1}=J^{-1}$.

Let I be a fractionary ideal of R. Then ${\text {Hom}}_R(I,I)$ is the endomorphism ring of I which we denote by ${\text {End}}_R(I)$. It follows that ${\text {End}}_R(I)=I:I$, because ${\text {Hom}}_R(I,J)=J:I$ for any two fractionary ideals I and J of R. We recall the following well-known facts.

Lemma 3.9

Let I be a fractionary ideal. Then

• (a) $R\subseteq {\text {End}}_R (I)\subseteq {\overline{R}}$.

• (b) If $R'$ is a ring with $R\subseteq R' \subseteq {\overline{R}}$, then there exists a fractionary ideal I such that $R'={\text {End}}_R(I)$.

• (c) If $R'$ is a ring with $R\subseteq R' \subseteq {\overline{R}}$, then $R'$ is a fractionary ideal and $R'={\text {End}}_R(R')$.

Proof

(a) Since I is a fractionary ideal, it follows that $RI\subseteq I$. This shows that $R\subseteq {\text {End}}_R(I)$. Now let $f\in {\text {End}}_R(I)$, then $fI{\overline{R}}\subseteq I{\overline{R}}$. Since $I{\overline{R}}= (t^a){\overline{R}}$ for some a, it follows that $ft^a{\overline{R}}\subseteq t^a{\overline{R}}$, which implies that $f{\overline{R}}\subseteq {\overline{R}}$. Since $1\in {\overline{R}}$, we see that $f=f1\in {\overline{R}}$.

(c) ${}\Rightarrow {}$(b) is clear. For the proof of (c) note that $R'$ is a fractionary ideal, because $RR'\subseteq R'$. Since $R'$ is a ring, we have $R'R'=R'$, and hence $R'\subseteq {\text {End}}_R(R')$. Conversely, let $f\in {\text {End}}_R(R')$. Then $f=f1\in R'$. This shows that ${\text {End}}_R(R')\subseteq R'$. $\square$

Let ${\mathcal {F}}_R$ be the set of non-zero fractionary ideals of R. We define subsets of ${\mathcal {F}}_R$ as follows: ${{\mathcal {C}}}_R=\{ I\in {\mathcal {F}}_R: \; C\subseteq I\subseteq R\}$, ${{\mathcal {T}}}_R=\{I\in {\mathcal {F}}_R: \; I\text { is a trace ideal of } R\}$, and ${{\mathcal {O}}}_R=\{ I\in {\mathcal {F}}_R: \; I\text { is an overring of }R\}$.

For the proof of the main result of this section we need the following fact, see Goto et al. (2020, Corollary 2.8).

Proposition 3.10

Let R be Gorenstein, and let $\alpha :{{\mathcal {T}}}_R\rightarrow {\mathcal {F}}_R$ be the map with $I\mapsto I^{-1}$. Then ${\text {Im}}\alpha \subseteq {{\mathcal {O}}}_R$ and $\alpha : {{\mathcal {T}}}_R\rightarrow {{\mathcal {O}}}_R$ is bijective.

Proof

Since $I\in {{\mathcal {T}}}_R$, we have $I=II^{-1}$. Therefore, $\alpha (I)=I^{-1}=(II^{-1})^{-1}=I^{-1}:I^{-1}$. By Lemma 3.9(a), $I^{-1}:I^{-1}\in {{\mathcal {O}}}_R$.

$\alpha$ is injective: Suppose $I^{-1}=J^{-1}$. Since R is Gorenstein, $I=(I^{-1})^{-1}=(J^{-1})^{-1}=J$, see Lemma 3.8(b).

$\alpha$ is surjective: Let $R'\in {{\mathcal {O}}}_R$. Then $R'=L$, where L is a fractionary ideal and $R'=L:L$, see Lemma 3.9(b). Let $I=L^{-1}$. Then $I^{-1}=(L^{-1})^{-1}=L$ and $R'=I^{-1}:I^{-1}$. Since $(II^{-1})^{-1}=I^{-1}:I^{-1}=R'$ and since $II^{-1}\in {{\mathcal {T}}}_R$, the assertion follows. $\square$

In general, not all trace ideals are reflexive.

Examples 3.11

 

• (a) Let $R=K[| t^{14},t^{15},t^{20},t^{21},t^{25} |]\subsetneq {\overline{R}}=K[|t|]$. Then $I=(t^{21},t^{28},t^{29},t^{30},t^{34})$ is a trace ideal which is not reflexive. Indeed,

  $$\begin{aligned} (I^{-1})^{-1}=(t^{21},t^{28},t^{29},t^{30},t^{34},t^{39},t^{40}). \end{aligned}$$

• (b) Let $R=K[| t^3,t^4,t^5|]\subsetneq {\overline{R}}=K[|t|]$. In Goto et al. (2020, Example 3.4(2)), it is shown that $\alpha :{{\mathcal {T}}}_R\rightarrow {{\mathcal {O}}}_R$ is not bijective. Goto et al. (2020, Proposition 5.1(3),(4)) shows that $\alpha$ may be bijective, even when R is not Gorenstein.

For Gorenstein rings the next result provides an efficient method to compute the largest trace ideal contained in a given ideal.

Corollary 3.12

Let R be Gorenstein and $I\in {{\mathcal {C}}}_R$. Then there exists an integer $k_0$ such that $(I^{-1})^{k_0} =(I^{-1})^{k}$ for all $k\ge k_0$. Let $J=(I^{-1})^{k_0}$. Then ${\text {subtr}}(I)=J^{-1}$.

Proof

If follows from Proposition 3.10 that ${\text {subtr}}(I)^{-1}$ is the smallest overring in ${{\mathcal {O}}}_R$ which contains $I^{-1}$. It is clear that any overring containing $I^{-1}$ also contains $(I^{-1})^k$ for all $k\ge 0$. Since $I^{-1} \subseteq (I^{-1})^2\subseteq \cdots$ and since R is Noetherian, this chain of inclusions stabilizes and the desired result follows. $\square$

Example 3.13

Let $R=K[|t^7,t^{10}|]$. Then R is Gorenstein. The ideal $I=(t^{14},t^{50})$ belongs to ${{\mathcal {C}}}_R$, but I is not a trace ideal. Indeed, ${\text {tr}}(I)=(t^{14}, t^{20})$. We have $I^{-1}=(1,t^6)$, $(I^{-1})^2 =(1,t^6,t^{12})$ and $(I^{-1})^k = (I^{-1})^3=(1,t^6,t^{12}, t^{18})$ for all $k\ge 3$. Therefore, ${\text {subtr}}(I)=((I^{-1})^3)^{-1}= (t^{38},t^{42}, t^{44}, t^{50})$.

In general R may have infinitely many trace ideals. The following example is given in Goto et al. (2020, Example 3.4(1)): let $R=K[| t^4,t^5|]\subsetneq {\overline{R}}=K[|t|]$, and for $a\in K$ let $R_a=K[| t^4,t^5,t^6+at^7|]$. Note that R is Gorenstein and $R\subsetneq R_a\subsetneq {\overline{R}}=K[|t|]$. It is shown in Goto et al. (2020) that $R_a\ne R_{a'}$ for $a\ne a'$. Therefore, Proposition 3.10 implies that $|{{\mathcal {T}}}_R|=\infty$ if $|K|=\infty$.

Our next goal is to characterize all analytically irreducible Gorenstein rings R for which ${{\mathcal {T}}}_R$ is finite. First we observe

Lemma 3.14

Let $R_1,R_2\in {{\mathcal {O}}}_R$, and suppose that $R_1\subseteq R_2$. Then $R_1=R_2$ if and only if $v(R_1)=v(R_2)$.

Proof

It is enough to show that $R_1=R_2$ if $v(R_1)=v(R_2)$. Let $t^c{\overline{R}}$ be the conductor of R. If $f\in R_2$ and $v(f)\ge c$, then $f\in {\overline{R}}$ and hence $f\in R_1$. Suppose $R_1\subsetneq R_2$. Then there exists $f\in R_2{\setminus } R_1$. It follows that $v(f)< c$. We may choose $f\in R_2{\setminus } R_1$ with v(f) maximal. Since $v(R_1)=v(R_2)$, we find $g\in R_1$ such that $v(f)=v(g)$. Then there exists a unit $\varepsilon \in {\overline{R}}$ such that $f=\varepsilon g$. Since $R/{{\mathfrak {m}}}={\overline{R}}/{{\mathfrak {n}}}$, we have $\eta +{{\mathfrak {m}}}=\varepsilon +{{\mathfrak {n}}}$ for some $\eta \in R$. This implies that $\varepsilon -\eta \in {{\mathfrak {n}}}$, and we get

$$\begin{aligned} {}f-\eta g=f-\varepsilon g+(\varepsilon -\eta )g=(\varepsilon -\eta )g. \end{aligned}$$

Since $\varepsilon -\eta \in {{\mathfrak {n}}}$, it follows that $v(f-\eta g)=v((\varepsilon -\eta )g)>v(g)=v(f)$. Then $f-\eta g\in R_1$, by the choice of f. Since $\eta \in R\subsetneq R_1$ and $g\in R_1$, this implies that $f\in R_1$, a contradiction. $\square$

Theorem 3.15

Assume that $|R/{{\mathfrak {m}}}|=\infty$. Let H be the value semigroup of R. Then the following conditions are equivalent:

• (i) $|{{\mathcal {O}}}_R|<\infty$.

• (ii) $H=\langle 1\rangle$ or $H=\langle 2,a\rangle$ where $a>2$ is an odd number, or H is one of the semigroups: $\langle 3,4\rangle$, $\langle 3,4, 5\rangle$, $\langle 3,5\rangle$ or $\langle 3,5,7\rangle$.

Proof

(i)${}\Rightarrow {}$(ii): Suppose H is not one of the semigroups in (ii). Then two cases are possible:

Case 1: $e(H)\ge 4$

Case 2: $e(H)=3$ and if a is the smallest integer in H with $a>3$, then $a>4,5$.

For $r\in R$ we set ${\overline{r}}=r+{{\mathfrak {m}}}$. Then r is a unit in R if and only if ${\overline{r}}\ne 0$. Since $| R/{{\mathfrak {m}}}|=\infty$, for each $i\ge 1$ there exists $r_i\in R$ such that $\overline{r_i} \ne \overline{r_j}$ for $i\ne j$. Then $\overline{r_i-r_j}=\overline{r_i}-\overline{r_j}\ne 0$. Therefore $r_i-r_j$ is a unit in R for all $i\ne j$.

Now we discuss Case 1: Since $e(H)\ge 4$, it follows that $2,3\not \in H$. Therefore by Lemma 3.14, $t^2,t^3\not \in R$. Now for each i, let $f_i=t^2+r_it^3$ and $R_i=R[f_i]$. Then $R\subsetneq R_i\subsetneq {\overline{R}}$.

We claim that $R_i\ne R_j$ for $i\ne j$. This will be a contradiction to (i). Assume that $R_i= R_j$ for some $i\ne j$. Then $f_j\in R_i$. Therefore $f_i-f_j=(r_i-r_j)t^3$. Since $r_i-r_j$ is a unit, it follows that $t^3\in R_i$. Hence, there exists an integer $m\ge 1$ such that $t^3=g_0+g_1f_i+g_2f_i^2+\cdots +g_mf_i^m$ with $g_i\in R_i$. So we may write $t^3=g_0+g_1f_i+g$ with $g\in R_i$ and $v(g)\ge 4$. Suppose that $g_0\not \in {{\mathfrak {m}}}$ then $g_0\not \in {{\mathfrak {n}}}$. Therefore $g_0\not \in R_i\cap {{\mathfrak {n}}}={{\mathfrak {m}}}_i$ which is the maximal ideal of $R_i$, and this is implies that $g_0$ is a unit in $R_i$. Then $g_1f_i+g=t^3-g_0$ and $t^3-g_0$ is a unit. This is a contradiction, because $v(g_1f_i+g)\ge \min \{v(g_1f_i), v(g)\}\ge 2$ and $v(t^3-g_0)=0$. Hence $g_0\in {{\mathfrak {m}}}$, and then $v(g_0)\ge 4$. Let $t^3=g_1f_i+h$ where $h=g-g_0$. Then $v(h)\ge 4$. It follows that, $v(g_1)+2=v(g_1f_i)= v(t^3-h)=3$. Therefore $v(g_1)=1$, a contradiction.

For the proof of Case 2, we set $f_i=t^4+r_it^5$. Again we claim that $R_i\ne R_j$ for $i\ne j$. Suppose that $R_i= R_j$ for some $i\ne j$, then $f_j\in R_i$. Therefore, as in the Case 1, $t^5\in R_i$ and $t^5=g_0+g_1f_i+g$ where $g_0, g_1\in R$ and $g\in R_i$ with $v(g)\ge 8$. As in case 1, $g_0$ cannot be a unit. Therefore, $v(g_0)\ge 3$. If $v(g_0)>3$ then $v(g_0)\ge 6$ and $5=v(g_1f_i)=v(g_1)+4$. Hence $v(g_1)=1$, a contradiction. If $v(g_0)=3$, then it follows that $5=v(t^5)=v(g_0+g_1f_i+g)=3$, again a contradiction.

(ii)${}\Rightarrow {}$(i): If $H=\langle 1\rangle$, then $R={\overline{R}}$ and the assertion is trivial. Next we consider the case $H=\langle 2,a\rangle$ where $a>2$ is an odd number. Let $R'\in {{\mathcal {O}}}_R$ with $R'\ne R$. Then $R\subsetneq R'\subseteq {\overline{R}}$. By Lemma 3.14 we have $v(R)\subsetneq v(R')$. Therefore, we conclude that $v(R')=\langle 2,b\rangle$ where b is odd and $b<a$. We claim that $R'=R[t^b]$. Suppose $t^b\not \in R'$. Then $R'\subsetneq R'[t^b]$. Again by using Lemma 3.14, $v(R')\subsetneq v(R'[t^b])$. On the other hand, each element in $R'[t^b]{\setminus } R'$ has a value bigger than b and hence $v(R'[t^b])=\langle 2,b\rangle =v(R')$, a contradiction. Thus $t^b\in R'$ which is implies that $R[t^b]\subsetneq R'$. Since $v(R[t^b])=v(R')$. Then by Lemma 3.14, $R'=R[t^b]$. Because $0<b<a$, the numbers of b is finite and $|{{\mathcal {O}}}_R|<\infty$.

For the remaining cases, it is enough to show that $|{{\mathcal {O}}}_R|<\infty$, if $v(R)=\langle 3,4\rangle$ or $v(R)=\langle 3,5\rangle$. Because if $v(R)=\langle 3,4,5\rangle$ or $v(R)=\langle 3,5,7\rangle$, then they are overrings of rings whose value semigroup is $\langle 3,4\rangle$ or $\langle 3,5\rangle$, and so $|{{\mathcal {O}}}_R|<\infty$, as well. We indicate the arguments for the case $H=\langle 3,4\rangle$. Let $R'\in {{\mathcal {O}}}_R$. We may assume that $R\subsetneq R'\subsetneq {\overline{R}}$. Then $v(R)\subsetneq v(R')$. Therefore, $v(R')=\langle 2,3\rangle$ or $v(R')=\langle 3,4,5\rangle$. In the first case suppose that $t^2\not \in R'$. Then $R'\subsetneq R'[t^2]$, and so $v(R')\subsetneq v(R'[t^2])$. On the other hand, $\langle 2,3\rangle \subsetneq v(R'[t^2])$, and since $1\not \in v(R'[t^2])$, it follows that $v(R')= v(R'[t^2])$, a contradiction. Thus $t^2\in R'$ and hence $R[t^2]\subseteq R'$. Since $v(R[t^2])=v(R')$ we conclude that $R'=R[t^2]$. The same argument shows that if $v(R')=\langle 3,4,5\rangle$, then $R'=R[t^5]$. $\square$

We call R small if R has only finitely many trace ideals. In the next result the small Gorenstein rings are classified.

Corollary 3.16

Assume that $|R/{{\mathfrak {m}}}|=\infty$, R is Gorenstein and $H=v(R)$. Then the following conditions are equivalent:

• (i) R is small.

• (ii) $H=\langle 1\rangle$ or $H=\langle 2,a\rangle$ where $a>2$ is an odd number, or $H=\langle 3,4\rangle$ or $H=\langle 3,5\rangle$.

• (iii) All trace ideals are integrally closed.

Proof

(i)${}\Longleftrightarrow {}$(ii): Since R is Gorenstein, Proposition 3.10 implies that $|{{\mathcal {O}}}_R |<\infty$, if and only if R is small. Hence the desired conclusion follows from Theorem 3.15, if we observe that v(R) cannot be $\langle 3,4,5\rangle$ or $\langle 3,5,7\rangle$, since R is Gorenstein.

(i), (ii)${}\Rightarrow {}$(iii): We have seen in the proof of Theorem 3.15 that for each gap of H there exists exactly one proper overring of R. Therefore, Proposition 3.10 implies that, together with R, we have precisely $g(H)+1$ trace ideals. Since H is symmetric, $n(H)=g(H)$, and therefore, the number of trace ideals is equal $n(H)+1$, which is the number of integrally closed ideals.

(iii)${}\Rightarrow {}$(i): Since the number of integrally closed ideals is finite, (iii) implies that the number of trace ideals is finite, that is, R is small. $\square$

4 Binary operations on the set of trace ideals

In this section we assume that R is a one-dimensional analytically irreducible local domain. By Proposition 2.3, the sum of trace ideals is a trace ideal. However this is not the case for products of trace ideals. For example, C is a trace ideal, but $C^2$ can not be a trace ideal if $R\ne {\overline{R}}$ because, in that case $C^2\subsetneq C$, which by Proposition 3.2 is not possible if $C^2$ is a trace ideal.

Therefore, we define a binary operation $*$ on ${{\mathcal {C}}}_R$ which replaces the product. In the case that R is Gorenstein it has the property that the $*$-product of trace ideals is again a trace ideal.

Let $I, J\in {{\mathcal {C}}}_R$. We set $I*J=(I^{-1}J^{-1})^{-1}.$

Proposition 4.1

Let $I,J\in {{\mathcal {C}}}_R$. Then

• (a) $I*J\in {{\mathcal {C}}}_R$.

• (b) $I*J\subseteq (I^{-1})^{-1}\cap (J^{-1})^{-1}$.

Proof

 

• (a) Since $C\subseteq I, J\subseteq R$, it follows that $R\subseteq I^{-1}, J^{-1}\subseteq {\overline{R}}$. Therefore, $R\subseteq I^{-1} J^{-1}\subseteq {\overline{R}}$, and hence $C\subseteq (I^{-1} J^{-1})^{-1}=I*J \subseteq R$.

• (b) Since $I^{-1}J^{-1}\supseteq I^{-1}+J^{-1}$, it follows that

  $$\begin{aligned} {}I*J\subseteq (I^{-1}+J^{-1})^{-1}=(I^{-1})^{-1}\cap (J^{-1})^{-1}. \end{aligned}$$

$\square$

In general, the intersection of trace ideals is not a trace ideal. But when R is Gorenstein we will see that the $*$-product is the best approximation of a trace ideal to the intersection. In the proof of the next results we use the fact that if R is Gorenstein, then the non-zero fractionary ideals are reflexive.

Proposition 4.2

Let R be Gorenstein and let $I,J\in {{\mathcal {T}}}_R$. Then

• (a) $I*J\in {{\mathcal {T}}}_R$.

• (b) $I*J=J*I$, $R*I=I*I=I$ and $I*C=C$.

• (c) $I*J={\text {subtr}}(I\cap J)\subseteq I\cap J$. In particular, $I\cap J$ is a trace ideal if and only if $I\cap J=I*J$.

Proof

 

• (a) By Proposition 3.10, $I^{-1}, J^{-1}\in {{\mathcal {O}}}_R$. Therefore, $I^{-1} J^{-1}\in {{\mathcal {O}}}_R$. Since $\alpha (I*J) =I^{-1}J^{-1}$, Proposition 3.10 implies that $I*J\in {{\mathcal {T}}}_R$.

• (b) It is clear that $I*J=J*I$ and that $R*I=I*I=I$. By Proposition 3.10, $I^{-1}$ is a ring. Therefore, $I^{-1}I^{-1}=I^{-1}$. This shows that $I*I=I$. Finally, $I*C=(I^{-1}C^{-1})^{-1}=(I^{-1}{\overline{R}})^{-1}=({\overline{R}})^{-1}=C$.

• (c) First note that $I^{-1}+J^{-1}=(I^{-1}+J^{-1})^{-1})^{-1}= ((I^{-1})^{-1}\cap (J^{-1})^{-1})^{-1}= (I\cap J)^{-1}$. Now let $L\in {{\mathcal {T}}}_R$ with $I*J\subseteq L\subseteq I\cap J$, then $I^{-1}+J^{-1}=( I\cap J)^{-1}\subseteq L^{-1}\subseteq I^{-1}J^{-1}$. Since $L^{-1}$ is a ring, and since $I^{-1}\subseteq L^{-1}$ and $J^{-1}\subseteq L^{-1}$, it follows that $I^{-1}J^{-1}\subseteq L^{-1}$. Therefore, $I^{-1}J^{-1}=L^{-1}$ and hence $L=I*J$.

$\square$

The zero ideal (0) is also a trace ideal. We set $(0)*I=I*(0)$. The following result together with Proposition 4.2 implies that if $R=K[|H|]$ is a semigroup ring with H a symmetric semigroup, then the set of monomial trace ideals of R with the operations $*$ and $+$ has the structure of a finite semiring.

Proposition 4.3

Let R be Gorenstein and let $I,J, L\in {{\mathcal {T}}}_R$.

• (a) $(I*J)*L=I*(J*L)$.

• (b) $J*I+L*I\subseteq (J+L)*I$. Equality holds, if R is a semigroup ring and the ideals I, J, L are monomial ideals in R.

Proof

 

• (a) $(I*J)*L= (I^{-1} J^{-1})^{-1}*L=(((I^{-1} J^{-1})^{-1})^{-1}L^{-1})^{-1}=(I^{-1} J^{-1}L^{-1})^{-1}$. For the last equation we used that R is Gorenstein, which implies that $(A^{-1})^{-1}=A$ for any fractionary ideal A of R. In the same we get $I*(J*L)=(I^{-1} J^{-1}L^{-1})^{-1}$.

• (b) Let $I_1,I_2 \in {{\mathcal {C}}}_R$. Then

  $$\begin{aligned}&I_1*I_2=(I_1^{-1}I_2^{-1})^{-1}= (I_1^{-1})^{-1}:I_2^{-1}=I_1:I_2^{-1}.\\&(J+L)*I=(J+L):I^{-1}\supseteq J:I^{-1}+L: I^{-1}= I*J+ I*L. \end{aligned}$$

  Equality holds, if R is a semigroup ring and the ideals I, J, L are monomial ideals in R.

$\square$