STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS

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1 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS ANDREAS REINHART Astract. Let H e a monoid (resp. an integral domain) and r an ideal system on H. In this paper we investigate the r-ideal semigroup of H. One goal is to specify monoids such that their r-ideal semigroup possesses semigroup-theoretical properties, like almost completeness, π-regularity and completeness. Moreover if H is an integral domain and a star operation on H, then we provide conditions on H such that the idempotents of the -ideal semigroup are trivial or such that H is π -stale. 0. Introduction In 1961 E. C. Dade, O. Taussky and H. Zassenhaus [9] investigated the semigroup structure of the ideal class semigroup of one-dimensional noetherian domains with a focus on non-principal orders in algeraic numer fields. In the sequel, this paper seems to have fallen into olivion. It was reconsidered and generalized y F. Halter-Koch [17], who put the results into the context of the structure theory of semigroups as presented in [14]. In recent times, starting with a paper y P. Zanardo and U. Zannier [23], the structure of the ideal class semigroup of an integral domain attracted the interest of several authors. In particular, the Clifford and the Boolean property of ideal class semigroups was investigated. First, this was done for valuation domains in [7]. In the sequel, S. Bazzoni provided a general theory, focused on Prüfer domains [2], [3], [4], [5]. Among others, she proved that a Prüfer domain has a Clifford semigroup if and only if it has finite character, and she highlighted the connection with stale domains. Following S. Bazzoni, S. Kaaj and A. Mimouni, in [20], [21] and [22], investigated the Clifford and Boolean property of the t-ideal class semigroup. Among others, they characterized Prüfer v-multiplication domains with Clifford t-class semigroup and determined the structure of their constituent groups. A more general approach, ased on the theory of ideal systems on commutative cancellative monoids, was recently presented y F. Halter-Koch [18]. He oserved that the semigroup structure of the ideal class semigroup is essentially determined y the semigroup structure of the ideal semigroup itself. Among others, he proved that the semigroup of fractional r-ideals of an r-prüfer monoid is a Clifford semigroup if and only if it has the local invertiility property (a generalization of Bazzoni s conjecture for Prüfer domains), and he characterized v-domains with a Clifford v-ideal semigroup. A valuale overview of Clifford and Boolean properties of (t-)ideal class semigroups of integral domains may e found in the survey article y S. Bazzoni and S. Kaaj [6]. A collection of important results concerning the ideal class semigroups of Prüfer domains can e found in [10]. In this paper, we reconsider the more general semigroup theoretical properties of ideal semigroups as this was done in [9] and [17], ut now we depart from the special cases of one-dimensional or Prüfer domains. Following [18], we present our theory as far as possile in the language of ideal systems on commutative monoids [19]. We provide conditions on a monoid or domain which entail nice semigrouptheoretic properties (as almost completeness, π-regularity and completeness) of the ideal class semigroup in question Mathematics Suject Classification. 13A15, 20M12, 13F05. Key words and phrases. semigroup, star operation, ideal system, regular, stale. 1

2 2 ANDREAS REINHART Our main suject of interest is the v-ideal (class) semigroup of a Mori domain. The main results are in Section 4, and there the case of = v is of particular interest. The strength of our results in the ring-theoretical case is illustrated y a series of examples and counter-examples in Section 5. In Section 1 we recall the relevant definitions and facts from the structure theory of commutative semigroups, essentially following [14]. In Section 2 we introduce ideal systems and r-ideal semigroups, summarize some of their elementary properties and provide first results concerning staility and completeness. In Section 3 we discuss preparatory ring-theoretical results concerning Mori domains and a general prime avoidance lemma (Lemma 3.2) which seems to e new. As already mentioned, Section 4 contains the main results, in particular, criteria for the idempotents of the ideal semigroup to e trivial and criteria for πr-staility. 1. Commutative Semigroups Throughout this section, let S e a multiplicative commutative semigroup. Our main reference for the theory of commutative semigroups is [14]. We denote y E(S) the set of all idempotents of S, endowed with the Rees order, defined y e f if ef = e. Note that ef e, for all e, f E(S). For a S, let E(a) = {e E(S) ea = a} e the set of all idempotents elonging to a and set E (a) = k N E(ak ). Then E(a) E (a) E(S) S are susemigroups. If I E(S) is a susemigroup, then every minimal element of I (with respect to the Rees-order) is a least element. In particular, if I is finite, then I has a minimum. An element a S is called regular if there exist elements e E(a) and S such that a = e (equivalently: There exists some S such that a 2 = a). π-regular if a n is regular for some n N. The semigroup S is called regular or a Clifford semigroup if every a S is regular. π-regular if every a S is π-regular. almost complete if for every a S, the set E(a) possesses a minimum (in the Rees-order). complete if it is π-regular and almost complete. For a E(S), let Pa = as\ E(S),< S a S e the partial Ponizovski factor of a. Note that the partial Ponizovski factors are essential invariants for the structure of S (see [14] and [17, Theorems 2.4 and 2.5]). Lemma 1.1. Let a S. 1. If n N such that a n is regular, then a k is regular for all k N n. 2. If there exist k, n N such that k n and a k = a n, then a is π-regular. 3. If a is π-regular, then E (a) has a smallest element in the Rees order. Proof. 1. Let n N e such that a n is regular and let k N n. There exists some l N such that nl > k. Since a n is regular, there exist e E(a n ) and S such that a n = e. Let = a nl k l S. Then a k = a nl l = e l = e. Since e E(a k ), it follows that a k is regular. 2. Let k, n N e such that k < n and a k = a n. It is straightforward to show that a k = a k+i(n k) for all i N. There exists some j N such that j(n k) > k, hence a j(n k) = a k a j(n k) k = a k+j(n k) a j(n k) k = a 2j(n k). It follows that a j(n k) is regular, and thus a is π-regular. 3. Let a e π-regular. Then there exist n N, e E(a n ) and S such that a n = e. This implies that e E (a). We assert that e = min E (a). Let f E (a). There is some m N such that fa m = a m. If l N is such that nl > m, then ef = e l f = a nl l f = a nl m a m f l = a nl m a m l = a nl l = e l = e.

3 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 3 2. Ideal systems and r-ideal semigroups Next we recall the most important facts aout ideal systems. For a asic introduction into ideal systems see [15]. We riefly gather the required terminology. In the following a monoid is always a multiplicative commutative semigroup that possesses an identity element and a zero element and where every non-zero element is cancellative. A sumonoid of a monoid H always contains the zero element and the identity element of H. A monoid K is called quotient monoid of a sumonoid H if K\{0} is a quotient group of H\{0}. Quotient monoids of monoids are essentially unique. A sumonoid T of a quotient monoid of a monoid H is called overmonoid of H if H T. For a monoid H let H = H\{0}. Throughout this section, let H e a monoid and K a quotient monoid of H. A suset X K is called fractional if there exists some c H such that cx H. Let F(H) denote the set of all fractional susets of K. We set F (H) = {I F(H) I\{0} = }. For aritrary X, Y K let (X : Y ) = {z K zy X}, X 1 = (H : X) and R(X) = (X : X). Let Ĥ denote the complete integral closure of H. Note that for all X F (H), R(X) is a sumonoid of Ĥ. A suset X H is called multiplicatively closed if X contains the identity element and XX X. A suset P H is called prime ideal of H if HP = P and H\P is multiplicatively closed. For a multiplicatively closed suset T H let T 1 H = {t 1 x t T, x H} and for a prime ideal P of H let H P = (H\P ) 1 H. If T and S are sumonoids of K such that T S, then let F S/T = (T : S). Especially if T is an overmonoid of H, then F T/H = T 1. A map r : F(H) F(H), I I r is called ideal system on H, if for all X, Y F(H) and all c K it follows that: X {0} X r. If X Y r, then X r Y r. (cx) r = cx r. H r = H. Throughout this section let r e an ideal system on H. Oserve that for all X, Y F (H) it follows that (XY ) r = (X r Y ) r = (X r Y r ) r and (X r : Y ) r = (X r : Y ) = (X r : Y r ). (For a proof see [15, Propositions 2.3 and 11.7]). Furthermore it is straightforward to show that if FĤ/H {0}, then Ĥ, F Ĥ/H F r (H). Next we introduce the most important ideal systems. Let s : F(H) F(H) e defined y s( ) = {0} and s(i) = IH for all I F(H)\{ }. Let v : F(H) F(H) e defined y I v = (I 1 ) 1 for all I F(H). If H is a domain, then let d : F(H) F(H) e defined y I d = (I) R for all I F(H). Oserve that s, v and d are ideal systems on H. Let F r (H) = {I F (H) I r = I}. If X F(H), then let R r (X) = R(X r ). If r is another ideal system on H, then we set r r if F r (H) F r (H) (equivalently: For all I F (H) it follows that I r I r ). r is called finitary if for all X F(H) it follows that X r = F X, F < F r. Let r : F r (H) F r (H) F r (H) e defined y I r J = (IJ) r, for all I, J F r (H). Then (F r (H), r) is a commutative semigroup, called the r-ideal semigroup. We say that H is r-noetherian, if H satisfies one of the following equivalent conditions: Every ascending sequence of elements of {I F r (H) I H} ecomes stationary. Every suset M {I F r (H) I H} has a maximal element. If I F (H), then there exists some finite E I such that I r = E r. If r = v, then these conditions are equivalent to each the following conditions: Every descending sequence of elements of F r (H) with non-zero intersection ecomes stationary. Every suset M F r (H) with I M I {0} has a minimal element. For I, J F r (H), we set I r J if there exists some c K such that I = cj. It is easily checked that r is an equivalence relation on F r (H). Let S r (H) = F r (H)/ r. For I F r (H) we denote y [I] r

4 4 ANDREAS REINHART the equivalence class of I. Let r : S r (H) S r (H) S r (H) e defined y [I] r r [J] r = [(IJ) r ] r, for all I, J F r (H). Then (S r (H), r ) is a commutative semigroup, called the r-ideal class semigroup. From now on we assume that s r. Let r-spec(h) = {P F r (H) P is a prime ideal of H} and let r-max(h) e the set of all maximal elements of {I F r (H) I H, IH = I}. If r is finitary and T H is mutliplicatively closed, then let T 1 r : F(T 1 H) F(T 1 H) e defined y T 1 r(t 1 X) = T 1 X r, for all X F(H). Then T 1 r is an ideal system on T 1 H. If P is a prime ideal of H, then let r P = (H\P ) 1 r. If H F r (H) is an overmonoid of H, then let r[h ] : F(H) F(H) e defined y r[h ](X) = (XH ) r, for all X F(H). Oserve that r[h ] is an ideal system on H. Note that this construction is investigated in [16] in the more general context of module systems. Next we introduce the concept of r-staility and πr-staility. Note that the r-staility of H is investigated in [11] for integral domains and in [18] for aritrary monoids. An element I F r (H) is called r-invertile if (II 1 ) r = H. r-regular if I is regular as an element of F r (H). πr-regular if I is π-regular as an element of F r (H). r-stale if there exists some J F r (H) such that (IJ) r = R(I) (equivalently: (I(R(I) : I)) r = R(I)). πr-stale if (I n ) r is r-stale for some n N (equivalently: (I n (R r (I n ) : I n )) r = R r (I n ), for some n N). Note that every r-invertile element of F r (H) is r-stale and every r-stale element of F r (H) is r-regular. The monoid H is called r-stale if every I F r (H) is r-stale. πr-stale if every I F r (H) is πr-stale. Lemma 2.1. Let H e r-noetherian, T a sumonoid of H and H F r (H) an overmonoid of H. 1. H is r[h ]-noetherian. 2. T 1 H is T 1 r-noetherian. 3. For all I, J F r (H), P r-max(h) I P = I and T 1 (I :J) = (T 1 I :T 1 J). Proof. 1. Let I F (H ) = F (H). Then there exists some finite E I such that E r = I r. This implies that E r[h ] = (EH ) r = (E r H ) r = (I r H ) r = (IH ) r = I r[h ]. Therefore it follows that H is r[h ]-noetherian. 2. This follows y [15, Theorem 4.4]. 3. Let I, J F r (H). It follows y [15, Theorem 11.3] that P r-max(h) I P = I and it follows y [15, Proposition 11.7] that T 1 (I :J) = (T 1 I :T 1 J). In the following we investigate the r-ideal semigroup of H with respect to the properties defined in Section 1. Note that R(I) E(I) E(F r (H)) for all I F r (H). Therefore {I F r (H) I = R(I)} E(F r (H)) and it is natural to ask, when equality holds. In this context it is worth remarking, that the structure of the idempotents of the t-ideal class semigroup is studied in [22]. Lemma 2.2. Let I F r (H). 1. If J F r (H) is such that (IJ) r = I, then (II 1 ) r (I(J : I)) r J R(I) and F R(I)/H J 1 R(I). 2. If (I 2 I 1 ) r = I, then I is r-regular. 3. If E E(I), then (I(E :I)) r {J F r (H) (II 1 ) r J R(I), E E(J) and J 2 J}. 4. If E (I) possesses a least element, then there exists some n N such that R r (I k ) = R r (I n ) for all k N n.

5 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 5 5. Assume that n N, E E((I n ) r ), and let J F r (H) e such that (I n J) r = E. Then it follows that (I k (R r (I k ):I k )) r = E and R((R r (I k ):I k )) = R r (I k ) = R(E) for all k N n. Proof. 1. Let J F r (H) e such that (IJ) r = I. Then IJ (IJ) r = I, hence J R(I). Of course, J 1 = (H : J) (IH : IJ) ((IH) r : (IJ) r ) = (I : I) = R(I). Clearly I 1 = (H : I) (HJ : IJ) ((HJ) r : (IJ) r ) = (J : I) and hence (II 1 ) r (I(J : I)) r J. Since J R(I), it follows that F R(I)/H = (H :R(I)) (H :J) = J Let (I 2 I 1 ) r = I. Since I 1 F r (H), it follows that I is r-regular. 3. Let E E(I). Of course, (I(E : I)) r F r (H) and y 1. it follows that (II 1 ) r (I(E : I)) r R(I). Since E E(F r (H)) and ((I(E : I)) r E) r = (IE(E : I)) r = ((IE) r (E : I)) r = (I(E : I)) r, it follows that E E((I(E : I)) r ). Finally, it follows that ((I(E : I)) r ) 2 (((I(E : I)) r ) 2 ) r = (I 2 (E : I) 2 ) r (I(E : I)E) r = (I(E :I)) r. 4. Let E e a least element of E (I). There exists some n N such that (EI n ) r = (I n ) r. Let k N n. Then there exists some l N such that ln k. Since R r (I ln ) E((I ln ) r ) E (I), it follows that E R r (I ln ). This implies that (ER r (I ln )) r = E, hence R r (I ln ) R(E). Therefore it follows that R(E) R r (EI n ) = R r (I n ) R r (I k ) R r (I ln ) R(E), hence R r (I k ) = R r (I n ). 5. Let n N, k N n, E E((I n ) r ), and J F r (H) e such that (I n J) r = E. There exists some l N such that ln k. It follows that R r (I n ) R r (I k ) R r (I ln ) R r (I ln J l ) = R r (E l ) = R(E) R r (EI n ) = R r (I n ), hence R r (I ln ) = R r (I k ) = R r (I n ) = R(E). It follows that E = (I n J) r (I n (E : I n )) r E, hence (I n (E : I n )) r = E. Since (E : I n ) = (E : (I n ) r ) = (E : (EI n ) r ) = (E : EI n ) = (R(E) : I n ) = (R r (I n ) : I n ), it follows that (I n (R r (I n ) : I n )) r = E. This implies that E = (E l ) r = (I ln (R r (I n ) : I n ) l ) r (I ln (R r (I n ) : I ln )) r (I k (R r (I n ) : I k )) r (I n (R r (I n ) : I n )) r = E, hence (I k (R r (I k ) : I k )) r = (I k (R r (I n ) : I k )) r = E. Finally, this implies that R((R r (I k ) : I k )) = (R r (I k ) : I k (R r (I k ):I k )) = (R(E):E) = R(E). Theorem The following conditions are equivalent: a. If I E(F r (H)), then I = R(I) (equivalently: E(F r (H)) = {I F r (H) I = R(I)}).. For all E, F E(F r (H)), E F implies F E. c. For all E, F E(F r (H)), F E implies E F. d. For all I F r (H) we have min(e(i)) = R(I). If these equivalent conditions are satisfied, then F r (H) is almost complete. 2. Let I, E F r (H). Then the following conditions are equivalent: a. E is a least element of E(I) with respect to the Rees-order.. E is a least element of {F E(I) F R(I)} with respect to inclusion. c. E is a minimal element of {F E(I) F R(I)} with respect to inclusion. 3. The following conditions are equivalent: a. H is πr-stale.. F r (H) is π-regular and E(F r (H)) = {E F r (H) E = R(E)}. 4. Suppose that E(F r (H)) = {I F r (H) I = R(I)} and let I E(F r (H)). Then I r F r (H) = {J F r (H) I R(J)} and P I = {J F r (H) I = R(J)}. Proof. 1. a.. If E, F E(F r (H)) and E F, then (EF ) r = F. Hence F (F R(E)) r = (F E) r = E. a. c. If E, F E(F r (H)) and E F, then (EF ) r (E 2 ) r = E and E (ER(F )) r = (EF ) r. This implies that (EF ) r = E, hence E F.. d. Let I F r (H) and F E(I). Then we have to show that (F R(I)) r = R(I). It follows that (F R(I)) r E(I) and (F R(I)) r R(I), hence R(I) (F R(I)) r. Since F E(I), it follows y Lemma that F R(I), hence (F R(I)) r (R(I) 2 ) r = R(I). Finally, this implies that (F R(I)) r = R(I). c. a. Let E E(F r (H)). Then E R(E) y Lemma , hence R(E) E. Since (ER(E)) r = E,

6 6 ANDREAS REINHART it follows that E R(E), hence E = R(E). d. a. Let E E(F r (H)). Then E E(E), hence R(E) E. Since (ER(E)) r = E, it follows that E R(E), hence E = R(E). 2. a.. Of course, E E(I). Since R(I) E(I), it follows that E R(I). Let F E(I) e such that F R(I). It follows that (EF ) r E(I) and (EF ) r E, hence E = (EF ) r. By Lemma it follows that E R(I), hence E = (EF ) r (R(I)F ) r = F.. c. Trivial. c. a. Let F E(I). Then it is sufficient to show that (EF ) r = E. It follows that (EF ) r E R(I) and (EF ) r E(I), hence (EF ) r {G E(I) G R(I)}. By Lemma F R(I), hence (EF ) r (ER(I)) r = E. This implies that E = (EF ) r. 3. a.. Let I F r (H). Then there is some n N such that (I n (R r (I n ) : I n )) r = R r (I n ). Since R r (I n ) E((I n ) r ), it follows that I is πr-regular. This implies that F r (H) is π-regular. Let F E(F r (H)). There is some n N such that (F n (R r (F n ) : F n )) r = R r (F n ), hence F = (F R(F )) r = (F (R(F ):F )) r = (F n (R r (F n ):F n )) r = R r (F n ) = R(F ).. a. Let I F r (H). Since I is πr-regular, there exist some n N, E E((I n ) r ) and J F r (H) such that (I n J) r = E. Consequently, Lemma implies that (I n (R r (I n ):I n )) r = E = R(E) = R r (I n ). 4. Let us show the first equality. : Let J I r F r (H). Then there exists some A F r (H) such that J = (IA) r. It is clear that IJ (IJ) r = (I 2 A) r = ((I 2 ) r A) r = (IA) r = J, hence I R(J). : Let J F r (H) e such that I R(J). It follows that J JI JR(J) = J, hence J = J r = (IJ) r = I r J I r F r (H). Now we show the second equality. It follows y 1. that A E(F r (H)),A<I A r F r (H) = A E(F r (H)),I A{J F r (H) A R(J)} = {J F r (H) there exists some A E(F r (H)) such that I A R(J)} = {J F r (H) I R(J)}. This implies that PI = {J F r (H) I R(J)}\{J F r (H) I R(J)} = {J F r (H) I = R(J)}. It is not difficult to see that algeraic properties of the r-ideal semigroup F r (H) induce corresponding properties of the r-ideal class semigroup S r (H). Oviously E(S r (H)) = {[I] r I E(F r (H))} and if I, J E(F r (H)), then [I] r [J] r holds if and only if I J. Moreover if I F r (H), then P[I] r = {[J] r J PI }. I is (π)r-regular if and only if [I] r is a (π-)regular element of S r (H). Consequently, S r (H) is regular[π-regular, almost complete, complete] if and only if F r (H) is regular[π-regular, almost complete, complete]. Moreover let r e an ideal system on H satisfying r r. If F r (H) is regular[π-regular], then F r (H) is regular[π-regular]. If H is r-stale[πr-stale], then H is r -stale[πr -stale]. For an astract semigroup-theoretic formalism ehind these statements, see [18, Lemma 2.2]. Lemma If I F v (H), then (II 1 ) v = F R(I)/H, and if I E(F v (H)), then FĤ/H I Ĥ and FĤ/H I 1 Ĥ. 2. If I F v (H) and (IF R(I)/H ) v = I, then I is v-regular. 3. Let Ĥ e factorial and I F v (H). Then there exists some c K such that ((FĤ/H ) 2 ) v ci Ĥ. Proof. 1. If I F v (H), then (II 1 ) v = (H : (H : II 1 )) = (H : ((H : I 1 ) : I)) = (H : (I : I)) = (H : R(I)) = F R(I)/H. Let I E(F v (H)). Then it follows y Lemma that FĤ/H F R(I)/H = (II 1 ) v I R(I) Ĥ and F Ĥ/H F R(I)/H I 1 R(I) Ĥ.

7 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 7 2. Let I F v (H) e such that (IF R(I)/H ) v = I. Then it follows y 1. that (I 2 I 1 ) v = (I(II 1 ) v ) v = (IF R(I)/H ) v = I. Therefore Lemma implies that I is v-regular. 3. Since Ĥ is factorial, it follows that every J F (Ĥ) is principal. Since (Ĥ :I) F vĥ vĥ (Ĥ), there exists some c K such that (Ĥ : I) = cĥ. By 1. it follows that F Ĥ/H F R(I)/H = (II 1 ) v (I(Ĥ : I)) v = (IcĤ) v. This implies that ((FĤ/H ) 2 ) v ((IcĤ) vfĥ/h ) v = c(ifĥ/h ) v ci ciĥ = I(Ĥ :I) Ĥ. Lemma 2.4. shows that every idempotent of the v-ideal semigroup contains the conductor. In Section 5 we present several examples of integral domains, where the conductor itself is an idempotent of the v- ideal semigroup. In the following we summarize some properties of the v-ideal semigroup of v-noetherian monoids and consider the extremal case that the conductor is non-trivial and v-idempotent. Lemma 2.5. Let H e v-noetherian and FĤ/H E(F v (H)). 1. Let I F v (H) e such that FĤ/H I Ĥ. If there exist some k, l N such that k l and (I k ) v (I l ) v, then I is πv-regular. 2. Let Ĥ e factorial and I F v (H) such that R(I) = Ĥ. Then I is πv-regular. Proof. 1. Let k, l N e such that k l and (I k ) v (I l ) v. Case1: k < l: Since FĤ/H {0}, we have Ĥ = (F Ĥ/H ) 1 F v (H). If 0 c FĤ/H and r N, then c(i k+r(l k) ) v c(i k+(r+1)(l k) ) v c(ĥk+(r+1)(l k) ) v = cĥ H. Since H is v-noetherian, there is some n N such that c(i k+n(l k) ) v = c(i k+(n+1)(l k) ) v. This implies that (I k+n(l k) ) v = (I k+(n+1)(l k) ) v and consequently Lemma implies that I is πv-regular. Case2: l < k: If r N, then (I l+(r+1)(k l) ) v (I l+r(k l) ) v. This implies that {0} FĤ/H = ((FĤ/H ) l+(r+1)(k l) ) v (I l+(r+1)(k l) ) v (I l+r(k l) ) v. Since H is v-noetherian it follows that there exists some n N such that (I l+(n+1)(k l) ) v = (I l+n(k l) ) v. Therefore Lemma implies that I is πv-regular. 2. Of course, (IĤ) v = (IR(I)) v = I v = I. It follows y Lemma that there exists some c K such that FĤ/H ci Ĥ. Since (ci)2 ciĥ c(iĥ) v = ci, it follows y 1. that ci is πv-regular. This implies that I is πv-regular. Theorem 2.6. Let H e v-noetherian. 1. F v (H) is almost complete. 2. If I F v (H) and if there exists some n N such that R v (I n ) = R v (I k ) for all k N n, then E (I) has a least element. 3. If Ĥ is factorial, FĤ/H E(F v (H)) and {J F v (H) J = R(J)} = {H, Ĥ}, then F v (H) is complete. Proof. 1. Let I F v (H) and M = {F E(I) F R(I)}. Since R(I) M, it follows that M. It follows y Lemma and Lemma that B M B F R(I)/H {0}. Since H is v-noetherian, there is some E M that is minimal with respect to inclusion. Finally, it follows y Theorem that E is a least element of E(I) with respect to the Rees-order. 2. Let I F v (H) and n N such that R v (I k ) = R v (I n ) for all k N n. Let M = {F E (I) F R v (I n )}. Since R v (I n ) E((I n ) v ) E (I), it follows that M. Let us show that B M B F Rv(I n )/H. Let B M. Then there is some k N n such that B E((I k ) v ). It follows y Lemma and Lemma that B F Rv(I k )/H = F Rv(I n )/H. This implies that B M B {0}, hence (since H is v-noetherian) there exists some E M that is minimal with respect to inclusion. Let F E (I). Then we have to show that (EF ) v = E. There exists some k N n such that F E((I k ) v ), hence Lemma implies that (EF ) v (ER v (I k )) v = (ER v (I n )) v = E. It follows that (EF ) v E (I)

8 8 ANDREAS REINHART and (EF ) v E R v (I n ), hence (EF ) v M and (EF ) v = E. 3. Let Ĥ e factorial, F Ĥ/H E(F v (H)) and {J F v (H) J = R(J)} = {H, Ĥ}. By 1. it follows that F v (H) is almost complete. Let I F v (H). Then R(I) {H, Ĥ}. Case1: R(I) = H: Since F R(I)/H = H E(I), it follows y Lemma that I is v-regular. Case2: R(I) = Ĥ: It follows y Lemma that I is πv-regular. Note that Theorem proves the converse of Lemma , if r = v and H is v-noetherian. We were not ale to decide whether the r-ideal semigroup of a r-noetherian monoid is almost complete. Theorem gives a hint how to construct examples of monoids which are not πv-stale ut whose v-ideal semigroup is π-regular. 3. Preparations for the main results Let R e an integral domain and K a field of quotients of R. By F (R) we denote the set of all nonzero fractional ideals of R and y R we denote the integral closure of R in K. R is called G-domain if P spec(r),p {0} P {0}. An ideal system r on R is called (extended) star operation on R if d r. Note that an ideal system r on R is a star operation if and only if r F (R) is a star operation on R in the sense of [13]. Oserve that for all X F (R), R(X) is an intermediate ring of R and R. Lemma 3.1. Let R e a Mori domain, x R and S R a multiplicatively closed set. 1. spec 1 (R) v-spec(r), and {P v-spec(r) x P } is finite. 2. S 1 R = Ŝ 1 R. 3. If R is not a field, then R\R = M v-max(r) M. 4. If F R/R {0}, then R is a Krull domain and S 1 F R/R = FŜ 1 R/S 1 R. 5. If R is local, dim(r) = 1 and R is a Krull domain, then R is a semilocal principal ideal domain. Proof. By spec 1 (R) we denote the set of height-one prime ideals of R. Oserve that R is a Mori domain if and only if R is a v-noetherian monoid in the sense of [12]. Note that if R denotes the complete integral closure of R in its quotient group, then R = R. 1. The first assertion follows y [12, Proposition ], and the second assertion follows y [12, Theorem ]. 2. See [12, Theorem ]. 3. Let R e not a field. The claim is an immediate consequence of [12, Proposition ]. 4. Let F R/R {0}. The first assertion follows y [12, Theorem ], and the second assertion follows y 2. and [12, Proposition ]. 5. Let R e local, dim(r) = 1 and R a Krull domain and let M e the maximal ideal of R. It is an immediate consequence of [12, Proposition (c)] that dim( R) = 1. Therefore R is a Dedekind domain. Of course, M R = M for all M max( R), hence max( R) = P(M R). Since R is noetherian this implies that max( R) is finite. Therefore R is a semilocal Dedekind domain, hence R is a semilocal principal ideal domain. Lemma 3.2 (Prime avoidance lemma). Let S e a commutative ring with identity and I S an additively closed suset. For k N, let I k = { k i=1 z i (z i ) k i=1 I[1,k] } and I k = { m i=1 z i m N, (z i ) m i=1 (I k ) [1,m] }. Let n N and (P i ) n i=1 spec(s)[1,n] e such that I P i for all i [1, n]. Then it follows that I (n 1)! n i=1 P i.

9 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 9 Proof. We use induction on n. The assertion is ovious for n = 1. n n + 1: Let n N and (P i ) n+1 i=1 spec(s) [1,n+1] e such that I P i for all i [1, n + 1]. If i [1, n + 1], then I (n 1)! n+1 j=1,j i P j (y the induction hypothesis), and thus there exists some (y i ) n+1 i=1 S[1,n+1] such that y i I (n 1)! \ n+1 j=1,j i P j for all i [1, n + 1]. Now we set y = yn+1 n + n i=1 y i I n! and assume (contrary to our assertion) that I n! n+1 i=1 P i. For every i [1, n + 1] we have yi n I n! n+1 j=1 P j and thus y i n+1 j=1 P j, hence y i P i. Since y I n! n+1 i=1 P i, there exists some m [1, n + 1] such that y P m. Case1: m [1, n]: Since y m P m it follows that n i=1 y i P m. Since y P m, we have yn+1 n P m, hence y n+1 P m, a contradiction. Case2: m = n + 1: Since y n+1 P n+1 and y P n+1, it follows that n i=1 y i P n+1, hence there exists some l [1, n] such that y l P n+1, a contradiction. 4. Main results In [9] it was shown that orders in quadratic numer fields are (d-)stale. This result is ased on two other results. The first one states that the ideal semigroup of a noetherian 1-dimensional integral domain is πd-stale. The second result gives a common upper ound on the index of staility of all elements I (i.e. the smallest n N for which I n is stale) of the ideal semigroup. The main goal of this paper is to extend the first result to aritrary -ideal semigroups. As already shown in Section 2 there exists a close connection etween π -staility and the structure of idempotents of the -ideal semigroup. As pointed out in [9] and [17] the idempotents of the ideal semigroup of a noetherian integral domain contain the identity. Unfortunately, there is no analogous result for general -ideal semigroups. In Section 5 it will e shown that the idempotents of the v-ideal semigroup of a Mori domain need not contain the identity. It is natural to ask for conditions on R that enforce the -idempotents to e trivial (i.e. they contain the identity). The first main result and its corollary deal with this prolem. Theorem 4.1. Let R e an integral domain, K a field of quotients of R, a star operation on R and I, J F (R) such that (IJ) = I. 1. If F R/R {0} and R is a Dedekind domain, then J R = R. 2. Suppose that one of the following conditions is satisfied: a. F R/R {0} and R is a semilocal principal ideal domain.. R is a -noetherian G-domain. c. R is -noetherian and dim(r) = 1. Then JR(I) = R(I). 3. If R is -noetherian and -max(r) = spec 1 (R), then (JR(I)) = R(I). Proof. 1. Let F R/R {0}, and R a Dedekind domain. Of course, I R, IJ R F ( R). Since I R and IJ R are invertile this implies that (I R) v R = I R and (IJ R) v R = IJ R. It follows that IJ R = ( R : ( R : IJ R)) = ( R :( R :(IJ R) )) = ( R :( R :(I R) )) = ( R :( R :I R)) = I R. Since I R is invertile, it follows that J R = R. 2.a. Let F R/R {0} and let R e a semilocal principal ideal domain. Then R = R. By 1. it follows that JR(I) R = J R = R, hence JR(I) M for all M max( R). Since R is semilocal, this implies that JR(I) M max( R) M = R\ R, hence JR(I) R. Of course, R/R(I) is an integral extension, hence R R(I) = R(I). This implies that JR(I) R = JR(I) R(I) R = JR(I) R(I), hence JR(I) = R(I). 2.. Let R e a -noetherian G-domain. Without restriction we may assume that R is not a field. It follows that (R(I) : JR(I)) = (I : IJ) = (I : (IJ) ) = R(I), hence JR(I) (JR(I)) vr(i) = R(I). Since R(I) F (R) is an overring of R, it follows y Lemma that R(I) is [R(I)]-noetherian, hence

10 10 ANDREAS REINHART R(I) is a Mori domain. If R(I) = K, then K F (R), hence R = K, a contradiction. Consequently, Lemma implies R(I)\R(I) = P v-max(r(i)) P, and for all P v-max(r(i)), it follows that P {0}. Therefore P R {0} for all P v-max(r(i)), hence {0} = M spec(r),m {0} M R P v-max(r(i)) P. If 0 x P v-max(r(i)) P, then {P v-spec(r(i)) x P } v-max(r(i)), and thus v-max(r(i)) is finite y Lemma Assume that JR(I) R(I). Then there is some Q max(r(i)) such that JR(I) Q. Since Q R(I)\R(I) = P v-max(r(i)) P, it follows that there exists some P v-max(r(i)) such that Q P, hence Q = P and R(I) = (JR(I)) vr(i) Q vr(i) = Q, a contradiction. 2.c. Let R e -noetherian and dim(r) = 1. If P max(r), then R P is a P -noetherian G-domain y Lemma Since I P, J P F P (R P ), and I P = ((IJ) ) P = (I P J P ) P, we otain J P R(I P ) = R(I P ) y 2... Hence Lemma implies that JR(I) = P max(r) (JR(I)) P = P max(r) J P R(I P ) = P max(r) R(I P ) = P max(r) R(I) P = R(I). 3. Let R e -noetherian and -max(r) = spec 1 (R). If P spec 1 (R), then R P is P -noetherian and I P = ((IJ) ) P = (I P J P ) P y Lemma Since I P, J P F P (R P ) and dim(r P ) = 1, we otain J P R(I P ) = R(I P ) y 2.c.. Hence Lemma implies that ((JR(I)) ) P = (J P R(I P )) P = R(I P ) P = R(I P ) = R(I) P. Finally, we have (JR(I)) = P spec 1 (R) ((JR(I)) ) P = P spec 1 (R) R(I) P = R(I) y Lemma Corollary 4.2. Let R e an integral domain and a star operation on R. 1. If R is a Dedekind domain and F R/R E(F (R)), then R = R. 2. Suppose that one of the following conditions is satisfied: a. F R/R {0} and R is a semilocal principal ideal domain.. R is a -noetherian G-domain. c. R is -noetherian and -max(r) = spec 1 (R). Then E(F (R)) = {I F (R) I = R(I)}. Proof. 1. Let R e a Dedekind domain and F R/R E(F (R)). Then it follows y Theorem that R = F R/R R R R, hence R = R. 2. This is an immediate consequence of Theorem and Theorem Theorem 4.3. Let R e an integral domain, K a field of quotients of R and a star operation on R such that R is -noetherian. Suppose that one of the following conditions is fulfilled: a. R is a semilocal principal ideal domain.. -max(r) = spec 1 (R) and R P is a Krull domain for all P spec 1 (R). Then R is π -stale. Proof. a. Let R e a semilocal principal ideal domain and I F (R). Then I R F ( R). Since R is a principal ideal domain, there exists some a K such that I R = a R, hence a 1 I R = R. Hence it follows that a 1 I M for all M max( R). Since R is semilocal, Lemma 3.2 implies that there exists some u N satisfying a u I u M max( R) M = R\ R. Consequently there exists some R a u I u, hence R 1 a u I u R. Let J = 1 a u (I u ). Then J F (R). Since R is -noetherian, there exists some finite suset E 1 a u I u satisfying J = (E). Since E is finite and E R, there exists some finitely generated R-sumodule M of K such that R[E] M, hence R[E] F (R). Let R = (R[E]). Then R F (R) and J R. Since R J, it follows for all m N that (J m ) (J m+1 ) and (J m ) ((R ) m ) = (R[E] m ) = R. There exists some x R such that xr R, hence for all m N it follows that x(j m ) x(j m+1 ) and x(j m ) R. Since R is -noetherian this implies that there exists some r N such that x(j r ) = x(j k ) for all k N r. This implies that (J r ) = (J 2r ). Consequently,

11 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 11 R J (J r ) R (J r ), hence (J r ) = R (J r ). This implies that r a ru (I ru ) = R (I ru ). Let n = ru N. Then R (I n ) = r a ru (I ru ) (I ru (R (I ru ) : I ru )) R (I ru ), hence (I n (R (I n ) : I n )) = R (I n ).. Let -max(r) = spec 1 (R) and let R P e a Krull domain for all P spec 1 (R). Let I F (R) and P = {P spec 1 (R) I P R P }. There is some s R such that si R and some s I\{0}. Let t = ss R. Then tr = ss R sir = si I. It follows that R is a Mori domain, hence Lemma implies that {P spec 1 (R) st P } is finite. Let us show that P {Q spec 1 (R) st Q}. Let Q P and assume that st R\Q, then st R Q, hence s, t R Q. This implies that I Q = si Q = (si) Q R Q = tr Q = (tr) Q I Q, hence I Q = R Q, a contradiction. It follows that P is finite. If P spec 1 (R), then R P is P -noetherian y Lemma , hence R P is a Mori domain. Therefore R P is a semilocal principal ideal domain y Lemma Since I P F P (R P ), it follows y a. and Lemma that there exists a least n P N such that (IP k (R P (IP k ):Ik P )) P = R P (IP k ) for all k N n P (note that n P = 1 if P P). If n = max({n Q Q spec 1 (R)} {1}), then it follows y Lemma that ((I n (R (I n ) : I n )) ) Q = (IQ n(r Q (IQ n) : In Q )) Q = R Q (IQ n) = R (I n ) Q for all Q spec 1 (R). Consequently, we have (I n (R (I n ) : I n )) = Q spec 1 (R) ((In (R n ) : I n )) ) Q = Q spec 1 (R) R (I n ) Q = R (I n ) y Lemma Corollary 4.4. Let R e an integral domain and a star operation on R. following conditions is fulfilled: 1. R is noetherian and -max(r) = spec 1 (R). 2. R is -noetherian, -max(r) = spec 1 (R) and F R/R {0}. Then R is π -stale. Assume that one of the Proof. 1. Let R e noetherian, -max(r) = spec 1 (R) and P spec 1 (R). Then R P is noetherian with dim(r P ) = 1. Therefore the theorem of Krull-Akizuki implies that R P is a Dedekind domain, hence R P is a Krull domain. Consequently, Theorem 4.3 implies that R is π -stale. 2. Let R e -noetherian, -max(r) = spec 1 (R), F R/R {0} and P spec 1 (R). Then R P is a Mori domain y Lemma Hence Lemma implies that F RP /R P = (F R/R ) P F R/R {0}, consequently R P is a Krull domain y Lemma Therefore R is π -stale y Theorem 4.3. Oviously, the conditions of Corollary 4.4. are satisfied for aritrary noetherian 1-dimensional integral domains. We were not e ale to decide whether -noetherian integral domains satisfying -max(r) = spec 1 (R) are π -stale. By Theorem and Corollary 4.2. it is sufficient to show that the -ideal semigroup of such integral domains is π-regular. 5. Examples The examples and counterexamples of this section are ased on surings of the ring of formal power series R[X ] over an integral domain R. For a power series f we denote y (f i ) i N0 the sequence of its coefficients, so that f = i N 0 f i X i. For d R we define R d = {f R[X ] d f 1 } and if, c R are such that d and c d, then we set I,c = {f R[X ] f 0, c f 1 }. Oserve that R is completely integrally closed if and only if R[X ] is completely integrally closed (for a proof see [8, Theorem 16]). Recall that if R[X ] is a Mori domain, that R is a Mori domain (since R[X ] K = R (where K is a field of quotients of R), this is a consequence of [1, Theorem 2.4]). First we study some ring theoretical properties of R d and investigate the elements of its v-ideal semigroup. Especially we investigate the completely integrally closed case, where the conductor turns out to e an idempotent of the v-ideal semigroup. In Lemma 5.3. and Lemma 5.4. we study most of the properties introduced in Section 2 with respect to the v-ideal semigroup of R d. Moreover we explicitly calculate

12 12 ANDREAS REINHART the set of v-idempotents in some special cases. The most important counterexamples are consolidated in Example Lemma 5.1. Let R e an integral domain, d R and K a field of quotients of R[X ]. 1. R d is an intermediate ring of R and R[X ], and K is a quotient field of R d. 2. For all, c R such that d and c d it follows that I,c = (, cx, X 2, X 3 ) Rd F v (R d ), (I,c ) 1 = I d c, d, R(I,c ) = {f R[X ] c f 1 } and (I k,c ) 1 = {f R[X ] d k f 1, d k 1 cf 0 } for all k N. Proof. 1. Of course, R R d R[X ]. Therefore it is sufficient to show that if f, g R d, then f + g, fg R d. Let f, g R d. Then f + g, fg R[X ], d (f 1 + g 1 ) = (f + g) 1 and d (f 0 g 1 + f 1 g 0 ) = (fg) 1, hence f + g, fg R d. Let K K e the quotient field of R d. Since X = X 3 X 2 K, it follows that R[X ] K, hence K = K. 2. Let, c R e such that d and c d. At first, let us show that I,c = (, cx, X 2, X 3 ) Rd. : Let f R[X ] e such that f 0 and c f 1. Then there are v, w R such that f 0 = v and f 1 = wc, hence f = v + wcx + ( i N 2,i 3 f ix i 2 )X 2 + f 3 X 3 (, cx, X 2, X 3 ) Rd. : Since, cx, X 2, X 3 I,c, it is sufficient to show that I,c is an R d -sumodule of R[X ]. Therefore let g, h I,c and let p, q R d. Then gp + hq R[X ], (g 0 p 0 + h 0 q 0 ) = (gp + hq) 0. It follows that c d p 1 and c d q 1. Consequently, c (g 1 p 0 +g 0 p 1 +h 1 q 0 +h 0 q 1 ) = (gp+hq) 1, hence gp+hq I,c. Since X 2 R d and X2 (, cx, X 2, X 3 ) Rd = (X 2, cx 3, X 4, X 5 ) Rd R d, it follows that (, cx, X 2, X 3 ) Rd F (R d ). Let k N. If g K is such that g(, cx, X 2, X 3 ) k R d R d, then g k R d R[X ] and gx 2k R d R[X ]. Hence g R[X ]. This implies that ((, cx, X 2, X 3 ) k R d ) 1 = {f R[X ] f(, cx, X 2, X 3 ) k R d R d }. If f R[X ], then X 2 f R d. Hence ((, cx, X 2, X 3 ) k R d ) 1 = {f R[X ] f k, f k 1 cx R d } = {f R[X ] d k f 1, d k 1 cf 0 }. Since d d, d c d and, c R are aritrary it follows that (, cx, X2, X 3 ) 1 R d = {f R[X ] d c f 0, d f 1} = I d c, and (I d,c ) v = (I d c, ) 1 = I d,c. Finally, it follows that R(I,c ) = {f R[X ] f(, cx, X 2, X 3 ) Rd I,c } = {f R[X ] f, fcx, fx 2, fx 3 I,c } = {f R[X ] c f 1 }. Lemma 5.2. Let R e a completely integrally closed domain, d R and K a field of quotients of R[X ]. 1. R d = R d = R[X ] = (1, X) Rd. 2. F Rd /R d = I d,d = (d, dx, X 2, X 3 ) Rd E(F v (R d )). 3. Let Q spec(r d ) e such that F Rd /R d Q. Then ht(q) X P spec 1 (R d ) (R d) P. 5. Let R d e a Mori domain and P spec 1 (R d ). Then (R d ) P is a discrete valuation domain. Proof. 1. Let us show that R[X ] = (1, X) Rd. : Let f R[X ]. Then f = f 1 X +( i N 0,i 1 f ix i )1 (1, X) Rd. : Clear. Since R[X ] is completely integrally closed, this implies that R[X ] R d R d R[X ] = R[X ], hence R d = R d = R[X ] = (1, X) Rd. 2. It follows y Lemma that I d,d = (d, dx, X 2, X 3 ) Rd F v (R d ). Let us show that F Rd /R d = I d,d. : Let f F Rd /R d. Then f R d, hence d f 1. Since fx R d, we have d (fx) 1 = f 0. : Since R d = R[X ], it follows y 1. that {d, dx, X 2, X 3 } F Rd /R d, hence (d, dx, X 2, X 3 ) Rd F Rd /R d. Therefore Lemma implies that (Id,d 2 ) 1 = {f R[X ] d d 2 f 0, d d 2 f 1 } = R[X ], hence ((F Rd /R d ) 2 ) v = (Id,d 2 ) v = (R[X ]) 1 = F Rd /R d. 3. Let h : R[X ] R e the canonical ringepimorphism and g = h Rd. Then g is a ringepimorphism. Let us show that Ker(g) = (dx, X 2, X 3 ) Rd. : Let f Ker(g). Then f 0 = 0. Of course, d f 1, hence f = f 1 X + X 2 ( i N 2,i 3 f ix i 2 ) + f 3 X 3 (dx, X 2, X 3 ) Rd. : Trivial. This implies that R d /(dx, X 2, X 3 ) Rd = R, consequently (dx, X 2, X 3 ) Rd spec(r d ). Since {0} (dx, X 2, X 3 ) Rd F Rd /R d Q, we have ht(q) 2.

13 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS Let P spec 1 (R d ). Then it follows y 2. and 3. that {d, dx, X 2, X 3 } P. Case1: d P : Since d, dx R d, we have X = d 1 dx (R d ) P. Case2: dx P : Since dx, dx 2 R d, it follows that X = (dx) 1 dx 2 (R d ) P. Case3: X 2 P : Since X 2, X 3 R d, we have X = X 2 X 3 (R d ) P. Case4: X 3 P : Since X 3, X 4 R d, it follows that X = X 3 X 4 (R d ) P. 5. Since X (R d ) P (y 4.), it follows y Lemma and 1. that (R d ) P = ( R d ) P = R[X ] P ((R d ) P ) P = (R d ) P, hence (R d ) P is completely integrally closed. Since (R d ) P is a Mori domain, this implies that (R d ) P is a Krull domain. Since (R d ) P is a Krull domain and dim((r d ) P ) = 1, it follows that (R d ) P is a Dedekind domain and since (R d ) P is local this implies that (R d ) P is a discrete valuation domain. Lemma 5.3. Let R e a completely integrally closed domain, d R and K a field of quotients of R[X ]. 1. {I,c, c R, d, c d} {I F v (R d ) F Rd /R d I and I 2 I}, and if R is a GCD-domain, then {I,c, c R, c d, GCD( d, ) = R } E(F v (R d )). 2. Let {I F v (R d ) F Rd /R d I and I 2 I} {I,c, c R, d, c d}. Then for all a R such that a 2 d it follows that a R. 3. If R/dR is finite, then {I I is an R d -sumodule of R d such that F Rd /R d I} is finite and F v (R d ) is almost complete. 4. If R/dR is finite, then for all I F v (R d ) such that F Rd /R d I R d it follows that I is πv-regular. 5. Let I F v (R d ) e such that F Rd /R d I R d. Then there exists some P R R such that I = {f R[X ] for all (, c) P it follows that d (f 1 + cf 0 )}. Proof. 1. Let, c R e such that d and c d and I = I,c. Then it follows y Lemma that I F v (R d ). Since d and c d, it follows y Lemma that F Rd /R d = {f R[X ] d f 0, d f 1 } {f R[X ] f 0, c f 1 } = I. Since I 2 = ( 2, cx, X 2, X 3, c 2 X 2, cx 3, cx 4, X 4, X 5, X 6 ) Rd and all generators of I 2 are products of elements of I and R d, it follows that I 2 I. Now let R e a GCD-domain, c and GCD( d, ) = R. It follows y Lemma that (I 2 ) 1 = {f R[X ] d f 1, d c f 0}. Since c, it follows that d c d, hence GCD( d c, ) = R. Since R is a GCD-domain, this implies that (I 2 ) 1 = {f R[X ] d f 1, d c f 0} = ( d c, d X, X2, X 3 ) Rd = I 1. Hence I = (I 2 ) v, and thus I E(F v (R d )). 2. Let a R e such that a 2 d and I = (a 2, a 2 X, ax + a, X 2, X 3 ) Rd. Of course, I F (R d ). By Lemma I 1 = (I a 2,a 2 + (ax + a)r d) 1 = (I a 2,a 2) 1 (ax + a) 1 R d = I d a 2, {f d K (ax + a)f a 2 R d } = {f R[X ] d a 2 f 0, d a 2 f 1, d a (f 0 + f 1 )}. Let us show that I 1 = ( d a, d a X, d a 2 (1 X), X 2, X 3 ) Rd. : Let f R[X ] e such that d a 2 f 0, d a 2 f 1 and d a (f 0 + f 1 ). Then there exist some r, s R such that f 0 + f 1 = d a r and f 0 = d a 2 s, hence f = f 0 + f 1 X + i N 2 f i X i = s d a 2 (1 X) + r d a X + i N 2 f i X i ( d a, d a X, d a 2 (1 X), X 2, X 3 ) Rd. : Trivial, since I 1 is an R d -module. Consequently, it follows y Lemma that I v = (I 1 ) 1 = (I d a, + d d a a (1 X)R 2 d ) 1 = (I d a, ) 1 d a ( d a (1 X)) 1 R 2 d = I a,a {f K d a (1 X)f R 2 d } = {f R[X ] a f 0, a f 1, a 2 (f 1 f 0 )}. Let us show that I v = I. : Let f R[X ] e such that a f 0, a f 1 and a 2 (f 1 f 0 ). Then there exist some r, s R such that f 1 = f 0 + a 2 r and f 0 = as. This implies that f = f 0 + f 1 X + i N 2 f i X i = s(ax + a) + ra 2 X + i N 2 f i X i I. : Trivial. Of course, I F v (R). Since a 2 d, it follows y Lemma and Lemma that F Rd /R d = I d,d I a2,a 2 = (a2, a 2 X, X 2, X 3 ) Rd I. Of course, I 2 = (I a2,a 2)2 + (a 3 X + a 3, a 4 X + a 4, a 2 X 2 + 2a 2 X + a 2, ax 3 + ax 2, ax 4 + ax 3 ) Rd (a 2, a 2 X, X 2, X 3 ) Rd I. Therefore there are some 0, c 0 R such

14 14 ANDREAS REINHART that 0 d, c 0 d and I = I 0,c 0. It follows y Lemma that ax + a I 0,c 0, hence c 0 a. Since c 0 X I = {f R[X ] a f 0, a f 1, a 2 (f 1 f 0 )}, it follows that a 2 c 0. This implies that a 2 a, hence a R. 3. Let R/dR e finite and M = {I I is an R d -sumodule of R d such that F Rd /R d I}. Let f : R/dR R/dR R d /F Rd /R d e defined y f((a + dr, + dr)) = (a + X) + F Rd /R d for all a, R. Let a, a 1,, 1 R e such that a + dr = a 1 + dr and + dr = 1 + dr. Then there exist some c 1, d 1 R such that a = a 1 + c 1 d and = 1 + d 1 d. Therefore Lemma implies that (a + X) + F Rd /R d = (a X) + (c 1 d + d 1 dx) + F Rd /R d = (a X) + F Rd /R d, hence f is well-defined. Now let g R d = R[X ]. Then y Lemma i N 2 g i X i F Rd /R d, hence f((g 0 + dr, g 1 + dr)) = (g 0 + g 1 X) + F Rd /R d = g + F Rd /R d, and thus f is surjective. Since R/dR R/dR is finite, this implies that R d /F Rd /R d is finite. Since there exists a ijection etween the set of all R d -sumodules of R d /F Rd /R d and M, it follows that M is finite. It follows y Lemma that E(F v (R d )) M, hence E(F v (R d )) is finite. This implies that F v (R d ) is almost complete. 4. Let R/dR e finite and I F v (R d ) such that F Rd /R d I R d. It follows y Lemma that F Rd /R d = ((F Rd /R d ) k ) v (I k ) v (( R d ) k ) v = R d for all k N. Since {I F v (R d ) F Rd /R d I R d } is finite, there are some r, s N such that r s and (I r ) v = (I s ) v. Hence I is πv-regular y Lemma Let P = {(, c) R R + cx I 1 }. Let us show that I = {f R[X ] for all (, c) P we have d (f 1 +cf 0 )}. : Let f I and (, c) P. Since +cx I 1, it follows that (+cx)f R d and hence d (( + cx)f) 1 = f 1 + cf 0. : Let f R[X ] e such that d (f 1 + cf 0 ) for all (, c) P. It is sufficient to show that fg R d for all g I 1. Let g I 1. Then g I 1 (R d : K F Rd /R d ) = R d = R[X ]. It follows y Lemma that i N 2 g i X i (X 2, X 3 ) Rd F Rd /R d I 1, hence g 0 + g 1 X I 1. Therefore (g 0, g 1 ) P, and thus d (g 0 f 1 + g 1 f 0 ) = (gf) 1. This implies that gf R d. Lemma 5.4. Let R e a completely integrally closed domain, d R and K a field of quotients of R[X ]. 1. The following assertions are equivalent: a. d R.. R d is completely integrally closed. c. F v (R d ) is a group. d. F v (R d ) is a Clifford semigroup. e. E(F v (R d )) = {F F v (R d ) F = R(F )}. f. F Rd /R d {F F v (R d ) F = R(F )}. 2. If d is a product of pairwise non-associated prime elements of R, then {I F v (R d ) F Rd /R d I and I 2 I} = {I,c, c R, d, c d}. 3. If d is a product of prime elements of R, then E(F v (R d )) = {I,c, c R, c d, GCD(, d ) = R } and F v (R d ) is almost complete. 4. If R[X ] is factorial and if d is a prime element of R or R/dR is finite, then F v (R d ) is complete. Proof. 1. a.. Let d R. Then R d = {f R[X ] d f 1 } = R[X ] = R d y Lemma , hence R d is completely integrally closed.. c. Clear. c. d. Trivial. d. a. Let F v (R d ) e a Clifford semigroup and I = F Rd /R d + XR d. It follows y Lemma that I = (d, X, X 2, X 3 ) Rd. We show that (I 2 ) v = (IF Rd /R d ) v = F Rd /R d. It follows y Lemma and that F Rd /R d = ((F Rd /R d ) 2 ) v (IF Rd /R d ) v (I 2 ) v = ({f R[X ] d d 2 f 1, d df 0 }) 1 = (R[X ]) 1 = F Rd /R d, hence (I 2 ) v = (IF Rd /R d ) v = F Rd /R d. Since I is v-regular, there exist some E E(I) and

15 STRUCTURE OF GENERAL IDEAL SEMIGROUPS OF MONOIDS AND DOMAINS 15 J F v (R d ) such that (IJ) v = E, hence (F Rd /R d J 2 ) v = (I 2 J 2 ) v = (E 2 ) v = E. This implies that (F Rd /R d E) v = ((F Rd /R d ) 2 J 2 ) v = (F Rd /R d J 2 ) v = E. It follows y Lemma that F Rd /R d E = (EF Rd /R d ) v ( R d F Rd /R d ) v = F Rd /R d, hence E = F Rd /R d. This implies that F Rd /R d E(I), hence F Rd /R d = (IF Rd /R d ) v = I. Finally, it follows y Lemma that X I = F Rd /R d = I d,d. Therefore d 1 and thus d R. c. e. Clear. e. f. This is an immediate consequence of Lemma f. a. Let F Rd /R d {F F v (R d ) F = R(F )}. Then y Lemma , 1 F Rd /R d = I d,d. It follows that d 1, hence d R. 2. Let n N and let (p i ) n i=1 e a finite sequence of pairwise non-associated prime elements of R such that d = n i=1 p i. Now we show the equality. : This follows y Lemma : Let I F v (R d ) e such that F Rd /R d I and I 2 I. Then I R(I) R d = R[X ]. By Lemma there exists some P R R such that I = {f R[X ] d ( 0 f 1 + c 0 f 0 ) for all ( 0, c 0 ) P }. Let M = {i [1, n] there is some ( 0, c 0 ) P such that p i c 0 }, N = {i [1, n] there is some ( 0, c 0 ) P such that p i 0 }, = i M p i and c = i N p i. Next we show that I = I,c. : Let f I. Then f 2 I 2 I. If ( 1, c 1 ) P, then d ( 1 (f 2 ) 1 + c 1 (f 2 ) 0 ). Therefore d ( 1 (2f 0 f 1 ) + c 1 f0 2 ), and thus d ( 1 f 0 f 1 + f 0 ( 1 f 1 + c 1 f 0 )). Since d ( 1 f 1 + c 1 f 0 ), it follows that d 1 f 0 f 1. Now we prove that f 0. It is sufficient to show that p i f 0 for all i M. Let i M. Then there exists some ( 0, c 0 ) P such that p i c 0. Case1: p i 0 : Since p i d 0 f 0 f 1, we have p i f 0 or p i f 1. If p i f 1, then since p i d ( 0 f 1 + c 0 f 0 ), we have p i f 0. Case2: p i 0 : Since p i d ( 0 f 1 + c 0 f 0 ), we have p i f 0. Now we prove that c f 1. It is sufficient to show that for all i N it follows that p i f 1. Let i N. Then there exists some ( 0, c 0 ) P such that p i 0. Since p i d 0 f 0 f 1, we have p i f 0 or p i f 1. If p i f 0, then p i f 1, since p i d ( 0 f 1 + c 0 f 0 ). : Since X 2, X 3 F Rd /R d I, we have to show that, cx I. Let us show that I. It is sufficient to show that for all i [1, n] and all ( 1, c 1 ) P it follows that p i c 1. Let i [1, n] and ( 1, c 1 ) P. Case1: For all ( 0, c 0 ) P it follows that p i c 0 : Of course, p i c 1 c 1. Case2: There exists some ( 0, c 0 ) P such that p i c 0 : Since i M, we have p i c 1. Let us show that cx I. It is sufficient to show that for all i [1, n] and all ( 1, c 1 ) P it follows that p i 1 c. Let i [1, n] and ( 1, c 1 ) P. Case1: For all ( 0, c 0 ) P it follows that p i 0 : Of course, p i 1 1 c. Case2: There is some ( 0, c 0 ) P such that p i 0 : Since i N, it follows that p i c 1 c. 3. Without restriction let n N, (p i ) n i=1 a finite sequence of pairwise non-associated primes of R and (k i ) n i=1 N[1,n] e such that d = n. Claim: (Iki+1 p ) i,1 v = I k p i for all i [1, n]. Proof of the claim: i=1 pki i Let i [1, n]. Then it follows y Lemma that (I ki+1 p i,1 ) 1 = {f R[X ] d p ki+1 i f 1, d p ki i f 0} = {f d d R[X ] f p k i 1, f i p k i 0 } = I d i p k d,, hence (I ki+1 i p k p ) i i,1 v = (I d p k d, ) 1 = I i p k k i p i i,pk i y Lemma i i i i i Next we show the equality. : Let I E(F v (R d )). Then we have F Rd /R d I R d y Lemma , hence there exists some P R R such that I = {f R[X ] d (f 1 + cf 0 ) for all (, c) P } y Lemma Let M = {i [1, n] p i f 0 for all f I}, l i = max{r [0, k i ] p r f 1 for all f I} for all i [1, n], = j M pkj j and c = n j=1 plj j. Next we show that c. Let i M. Then I I p i,1 y Lemma , and hence I = (I ki+1 ) v (I ki+1 p ) i,1 v = I k p i i,pk i y the claim. It follows that p ki i f 1 for all f I. Therefore i l i = k i, hence c. Of course, c d and since (p i ) n i=1 is a sequence of pairwise non-associated prime elements of R, we have GCD(, d ) = R. It remains to prove that I = I,c. : Let f I. If j M, then I I p k j j i,pk i i,pk j. Therefore p kj j f 0, hence j f 0. Of course, p lj j f 1 for all j [1, n], hence c f 1. : It follows y Lemma and Lemma