# Gaussian properties of total rings of quotients, S Bazzoni, S Glaz

Tags: Gaussian ring, maximal ideal, Theorem, Journal of Algebra, arithmetical ring, conditions, ring, weak global dimension, condition, fer, zero divisors, local ring, commutative ring, finitely generated, von Neumann, quotients, total quotient ring, semihereditary ring, commutative rings, Noetherian Gaussian, Lemma, S. Glaz, global dimension, Bazzoni
Content: Journal of Algebra 310 (2007) 180­193
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(1) R is semihereditary. (2) The weak global dimension of R is at most one. (3) R is an arithmetical ring. (4) R is a Gaussian ring. (5) R is a Prьfer ring. In  and  it is proved that each one of the above conditions implies the next one, and examples are given to show that in general the implications cannot be reversed. Moreover, an investigation is carried out to see which conditions may be added to some of the preceding properties in order to reverse the implications. In this article we push further the analysis of the five Prьfer-like conditions listed above through relating the property of a ring with the property of its total ring of quotients. In particular, in Section 3 we show that a Prьfer ring R satisfies one of the five conditions if and only if the total ring of quotients Q(R) of R satisfies that same condition. This implies, that for a Prьfer ring with von Neumann total ring of quotients the five conditions are all equivalent. This is as far as one can go in requiring the equivalence of all the five conditions, when Q(R) is not a field. In Section 4 we generalize some results obtained by Tsang , by giving other characterizations of a local Gaussian ring R, and illustrating some properties of the annihilators of the elements of R. In  the interest is primarily on the homological properties of the class of Gaussian rings. In particular, it is shown that if R is a coherent Gaussian ring, then the small finitistic dimension of R is at most one. In Section 5 we generalize this result by proving that it holds for any Gaussian ring. In Section 6 we consider THE PROBLEM of determining the possible values for the weak global dimension of a Gaussian ring. In  it is shown that for a coherent Gaussian ring the possible values for the weak global dimension of R are 0, 1, or . We conjecture that the same is true for every Gaussian ring, and prove the conjecture in certain cases. WE Note that it follows from Osofsky  that arithmetical rings have weak global dimension at most one or . We prove that the same holds for every Gaussian ring R which admits a maximal ideal m such that the localization Rm has nilpotent radical. Throughout the paper R will always denote a commutative ring with identity and Q(R) will denote the total ring of quotients of R. 2. Preliminaries We recall the definitions of the five classes of rings mentioned in the introduction. Definition 1. A ring R is called semihereditary if every finitely generated ideal of R is projective. Since, a finitely generated ideal over a domain is invertible if and only if it is projective, the class of semihereditary rings provides an extension of the class of Prьfer domains to rings with zero divisors. Definition 2. Denote by w.gl.dim R the weak (or flat) global dimension of a ring R. Then w.gl.dim R 1 if and only if every ideal of R is flat, or equivalently, if and only if every finitely generated ideal of R is flat.
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Semihereditary rings have weak global dimension at most one. In fact, they are exactly the rings R with w.gl.dim R 1, which are coherent (see ). The class of rings of weak global dimension at most one can also be considered to be an extension of the class of Prьfer domains to rings with zero divisors. To see this, recall that for a ring R, w.gl.dim R 1 if and only if every localization of R at a maximal ideal is a valuation domain (see ). At this point it is worth mentioning the characterization of semihereditary rings given by Endo in . Theorem 2.1. (See .) A ring R is semihereditary if an only if w.gl.dim R 1 and Q(R) is von Neumann regular. L. Fuchs  introduced the class of arithmetical rings. Arithmetical rings were also studied in [13,14]. Definition 3. A ring R is arithmetical if the lattice of the ideals of R is distributive. Arithmetical rings can be characterized by the property that in every localization at a prime (maximal) ideal, the lattice of the ideals is linearly ordered. Therefore the class of arithmetical rings provides another extension of the class of Prьfer domains. Moreover, by the previous remarks, if w.gl.dim R 1, then R is an arithmetical ring. The focus of this article is on Gaussian rings introduced by Tsang in , which provide another class of rings extending the class of Prьfer domains to rings with zero divisors. Definition 4. If R is a ring, and x is an indeterminate over R, the content c(f ) of a polynomial f R[x] is the ideal of R generated by the coefficients of f . A polynomial f R[x] is called a Gaussian polynomial if c(f g) = c(f )c(g) for every polynomial g R[x], and a ring R is called a Gaussian ring if every polynomial f R[x] is Gaussian. Tsang proved, among many other results, that if the content ideal of a polynomial f with coefficients in R is invertible, or more generally locally principal, then f is a Gaussian polynomial. Thus any arithmetical ring is a Gaussian ring. Tsang , and independently Gilmer  proved that a domain R is Gaussian if and only if it is Prьfer. The Gaussian property is a local property, namely a ring R is Gaussian if and only if every localization of R at a prime (maximal) ideal is Gaussian. We will make frequent use of several equivalent characterizations of a local Gaussian ring, which we summarize in Theorem 2.2. The basic ideas behind the proofs go back to Tsang's Unpublished PhD thesis . We sketch some of the proofs here for the reader convenience. Theorem 2.2. (See .) Let (R, m) be a local ring with maximal ideal m. The following conditions are equivalent. (a) R is a Gaussian ring. (b) If I is a finitely generated ideal of R and (0 : I ) is the annihilator of I , then I /I (0 : I ) is a cyclic R-module. (b ) Condition (b) for two generated ideals.
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(c) For any two elements a, b in R, there exists d in the annihilator of (a, b) such that (a, b) = (a, d) or (a, b) = (b, d). Moreover, d can be chosen so that b d + aR, or a d + bR, respectively. (d) For any two elements a, b in R, the following two properties hold: (i) (a, b)2 = (a2) or (b2), (ii) if (a, b)2 = (a2) and ab = 0, then b2 = 0 Proof. Tsang proves the equivalence (a) (b) and (a) (b ). Condition (c) is easily seen to be just a reformulation of condition (b ). The implication (a) (d) appears in Tsang's thesis and the equivalence (a) (d) has been noted by Lucas in . 2
As a consequence of this theorem we obtain that if (R, m) is a Noetherian ring, then R is Gaussian if and only if R/(0 : m) is an arithmetical ring. Tsang also showed that the prime ideals of a local Gaussian ring (R, m) are totally ordered by inclusion, thus the nilradical is the unique minimal prime ideal of R. It follows that a local Gaussian ring modulo its nilradical is a valuation domain. In particular a reduced local Gaussian ring is a valuation domain. Definition 5. R is a Prьfer ring if and only if every finitely generated regular ideal of R is invertible.
From the remarks following Definition 4 we conclude that Gaussian rings are Prьfer rings. Gaussian rings were also considered in  and . In summary (see also [8,9]), we have the following implications among the five Prьfer-like conditions considered in the introduction: (1) (2) (3) (4) (5).
3. The total ring of quotients
In this section we prove that if the total ring of quotients of a Prьfer ring R is Gaussian or arithmetical, then the same holds for R. As a corollary we obtain necessary and sufficient conditions on Q(R) for reversing all the implications of the five Prьfer-like conditions considered in the introduction. We proceed by developing several notions that will be used in our proofs, and recalling some definitions found in , ,  or .
Definition 6. (See [10,11].) Let P be a prime ideal of R. The large quotient ring of R with respect to P , denoted by R[P ], consists of the elements x Q(R) such that xs R for some element s R \ P . For every ideal I R, I denotes the set of elements x Q(R) such that xs I for some element s R \ P . Clearly R R[P ] Q(R) and Q(R[P ]) = Q(R). We also have, I I R[P ] and if P is a prime ideal of R, then P is a prime ideal of R[P ]. Moreover, in [11, p. 415] it is proved that the operation is a one-to-one inclusion preserving correspondence between prime ideals of R contained in P and prime ideals of R[P ] contained in P .
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Definition 7. (See [10,15].) Let P be a prime ideal of R. The core C(P ) of P is the set of all elements b R with the property that for each regular element r R there exists an element s R \ P such that bs/r R. Thus C(P ) = b R b/r R[P ] for every regular element r R . Let P be a prime ideal of R, using ideas of  and  we show: Lemma 3.1. If P consists of zero divisors, then C(P ) = R. If P is a regular prime ideal, then C(P ) is an ideal of R contained in P , and it consists of zero divisors. Proof. If P consists of zero divisors, then every regular element r of R is not in P ; hence for every b R, b = br/r R implies b C(P ). If P is a regular ideal and b C(P ), pick a regular element r P . Then bs/r R for some s R \ P implies bs rR P , hence b P . To prove that C(P ) consists of zero divisors, assume that a regular element r C(P ). Then, r2 is regular and rs/r2 R for some s R \ P implies s rR P , a contradiction. 2 Another important notion when dealing with Prьfer rings is the notion of a Manis valuation recalled below. Definition 8. (See [10,12,15,17].) Let K be a commutative ring. A (Manis) valuation on K is a pair (v, ) where is a totally ordered abelian group and v is a map from K onto satisfying the following properties: (1) v(xy) = v(x) + v(y). (2) v(x + y) min{v(x), v(y)}. (3) v(1) = 0 and v(0) = . If (v, ) is a valuation on K, then Rv = {x K | v(x) 0} is a subring of K, and Pv = {x K | v(x) > 0} is a prime ideal of Rv. Moreover, A = {x K | v(x) = } is a prime ideal both of R and K. Definition 9. Let R be a ring with total ring of quotients Q(R). If P is a prime ideal of R such that R = Rv and P = Pv for some valuation (v, ) on Q(R), then the pair (R, P ) is called a Manis valuation ring. It is known that (R, P ) is a Manis valuation ring if and only if for every x Q(R) \ R there exists y P such that xy R \ P . It follows that if (R, P ) is a Manis valuation ring and r is a regular element of R, then v(r) = . In fact, v(r-1) = -v(r). Lemma 3.2. Let P be a proper prime ideal of R and assume (R[P ], P ) is a Manis valuation ring with valuation (v, ). The following hold: (1) If the prime ideal P consists of zero divisors, then R[P ] coincides with Q(R) and (v, ) is a trivial valuation, namely v(x) = 0 for every x R[P ] \ P and v(y) = for every y P . (2) If P is a regular prime ideal, then C(P ) = {b R | v(x) = }; hence C(P ) = v-1() R is a prime ideal of R.
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Proof. (1) Let x Q(R), x = a/r for some elements a, r R with r regular. Then r / P , hence xr R implies x R[P ]. It remains to show that v(y) = for every y P . Assume v(y) = . Since v is a surjective map, there exists x Q(R) = R[P ] such that v(x) = - . But v(x) 0 for every x R[P ], implies = 0 contradicting y P . (2) If r is a regular element of P , then r is a regular element of P . Thus 0 < v(r) = , namely is not the trivial group. Let b C(P ); then for every regular element r R, b/r R[P ]. So v(b) v(r) for every regular element r R. We show now that for every element 0 < there is a regular element r R such that v(r) , and conclude that v(b) = . Let 0 < ; since v is a surjective map, there exists an element x Q(R) such that v(x) = - . Since x is of the form a/r for some element a R and some regular element r R, we have v(r ) = v(a) + . Conversely, assume b R is such that v(b) = . Then, for every regular element r R, v(b/r) = and v(r) = . So b/r R[P ], namely b C(P ). 2 In [10, Theorem 13] Griffin characterizes Prьfer rings by means of fifteen equivalent conditions which are the generalizations of analogous conditions on Prьfer domains. We will use Griffin's condition stating that a ring R is Prьfer if and only if (R[P ], P ) is a Manis valuation ring. We are now in a position to prove our main result concerning the Gaussian property of a total ring of quotients. Theorem 3.3. Let R be a Prьfer ring. Then R is Gaussian if and only if Q(R) is Gaussian. Proof. If R is a Gaussian ring, so is Q(R) since it is a localization of R. To prove sufficiency it suffices to show that every localization of R at a maximal ideal P of R is Gaussian. If P is not regular, then P Q(R) is a proper prime ideal of Q(R), and it is immediate to check that RP is the localization of Q(R) at P Q(R). It follows that RP is Gaussian, since by hypothesis Q(R) is Gaussian. Let P be a regular maximal ideal of R. By Tsang's characterization of a local Gaussian ring, Theorem 2.2, we have to prove that given two elements a, b R, the ideal (a, b)RP is of the form (a, d)RP or (b, d)RP , for some element d which annihilates (a, b)RP . Now R is a Prьfer ring, so by [10, Theorem 13] (R[P ], P ) is a Manis valuation ring. If a, b are not both in the core of P , we can apply [10, Lemma 5] to conclude that (a, b)RP is a principal ideal of RP , so it is generated by a or b and we are done. Thus, assume a, b are both in the core C of P . By Lemmas 3.2 and 3.1, C is a prime ideal of R and it consists of zero divisors. Thus RC is Gaussian, since it is the localization of Q(R) at the prime ideal CQ(R). Therefore, by Tsang's Theorem 2.2(c) we can assume that (a, b)RC = (a, d)RC , where d is an element of R which annihilates (a, b)RC ; moreover, we can choose d such that b d + aRC . Claim (a). Every element of R in the annihilator of (a, b)RC , belongs to the annihilator of (a, b)RP in RP . In fact, let d R be in the annihilator of (a, b)RC . Then, there exists t R \ C such t da = 0 = t db. Since v(t ) = , we can choose z Q(R) such that v(z) = -v(t ). Then v(zt ) = 0, hence there exists s R \ P such that szt R \ P . So, szt da = 0 = szt db implies d(a, b)RP = 0 in RP . We can write b = (r/t)a + d, for some r R and t R \ C.
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We consider two cases: First case: v(r) v(t). Arguing as above there are z Q(R) and s R \ P such that szt R \ P . Moreover, v(zr) 0; so there is s R \ P such that s zr R. Then s0 = s szt R \ P and we have s0b = (s zr)sa + s sztd. This shows that b (a, d)RP and also d (a, b)RP . So we conclude that (a, b)RP = (a, d)RP with d(a, b)RP = 0. Second case: v(r) < v(t). Let y Q(R) be such that v(y) = -v(r); then v(yr) = 0 and v(yt) > 0. So there exist s, s R \ P such that syr R \ P and s yt R. Then, s0 = s syr R \ P and s0a = (s yt)sb - (s yt)sd. This shows that a (b, cd)RP , where c = s yts R and cd (a, b)RP . Hence, (a, b)RP = (b, cd)RP . Clearly cd belongs to the annihilator of (a, b)RC in RC , hence, by claim (a) cd annihilates (a, b)RP . So we have shown that given two elements a, b R, the ideal (a, b)RP is of the form (a, d)RP or (b, d)RP , for some element d which annihilates (a, b)RP . 2 We consider now the case of arithmetical rings. The following easy lemma will be useful. Lemma 3.4. Let P be a prime ideal of a ring R. Then, the total ring of quotients of the localization of RP at the prime ideal P , is a localization of Q(R) with respect to a multiplicative subset of R. Proof. Let T be the multiplicative set of the regular elements of R and let S be the multiplicative set R \ P . Let U = {a R | a/1 is a regular element of RP }. U is a multiplicative subset of R and it is immediate to check that U T . Consider the subset U RP = {a/s | a U, s S} of RP . U RP is the set of regular elements of RP . In fact, r/s is a regular element of RP if and only if r/1 = (r/s)s is a regular element of RP , since s is invertible in RP . The ring of quotients Q(RP ) of RP is the localization of RP at the multiplicative set U RP . Thus Q(RP ) is the localization of R first at S and then at U RP . So Q(RP ) is also the localization of R at the multiplicative set U S. As noted above U contains the set of regular elements of R; thus RU is a localization of RT = Q(R). We conclude that Q(RP ) is a localization of Q(R) More precisely, RU = (RT )(URT ) and Q(RP ) = R(US) = (RU )(SRU ) = (RT )(USRT ). 2 Proposition 3.5. Let R be a Gaussian ring. Then R is arithmetical if and only if the total ring of quotients Q(R) of R is arithmetical. Proof. Necessity is obvious, since Q(R) is a localization of R. To prove sufficiency, it is enough to show that for every maximal ideal P of R, RP is an arithmetical ring. RP is a Gaussian ring and by Lemma 3.4, Q(RP ) is a localization of the arithmetical ring Q(R), hence it is arithmetical, too. Thus, without loss of generality, we can assume that R is a local ring. As noted by Tsang , the lattice of the prime ideals of a local Gaussian ring is linearly ordered; therefore the set of zero divisors Z(R) of R is a prime ideal L and the total ring of quotients of R is RL. Given two elements a, b R we will prove that the ideal (a, b) is principal, and therefore conclude that R is arithmetical, as desired. If one of the two elements is regular, then the ideal (a, b) is principal, since R is a local Gaussian ring, hence a local Prьfer ring. Assume a, b are both zero divisors and consider the ideal (a, b)RL of RL. Since, by hypothesis, RL = Q(R) is
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arithmetical, we have, say, b = a(r /r) where r is a regular element of R. Consider the ideals (a, r) and (b, r) in R. They are regular ideals, hence, by the previous remark they are principal. We conclude that (a, r) = (b, r) = (r). Thus, a = rc, b = rd for some elements c, d in R. We have rb = r2d, ar = r rc and, using rb = ar , we conclude that r2d = r rc. The regularity of r implies that b = rd = r c. Hence (a, b) = (rc, r c) = c(r, r ). Again by the regularity of r and by the fact that R is a local Prьfer ring, the ideal (r, r ) is principal. We conclude that (a, b) is also principal. 2 Theorem 3.6. Let R be a Prьfer ring. Then R is arithmetical if and only if the total ring of quotients Q(R) of R is arithmetical. Proof. The necessary condition follows by the fact that Q(R) is a localization of R. For the converse, note that if Q(R) is arithmetical, then it is also Gaussian. By Theorem 3.3 R is Gaussian. Thus, to conclude it is enough to appeal to Proposition 3.5. 2 We remark that Theorem 3.6 can also be deduced from a result of Griffin [10, Theorem 19]. We now make use of the results found so far to clarify the exact relation between each of the Prьfer-like conditions on the ring R, and the corresponding condition on its total ring of quotients Q(R). The implication in one direction is summarized in the result below: Theorem 3.7. If R is a ring satisfying any of the five Prьfer-like conditions mentioned in the introduction, then Q(R), the total ring of quotients of R, satisfies the same Prьfer-like condition. Proof. Conditions (1)­(4) are inherited by localizations, hence if R satisfies one of them, the same holds for Q(R), since it is a localization of R. Moreover, any total ring of quotients is a Prьfer ring. 2 Concerning the converse of Theorem 3.7, we note that none of the five Prьfer-like conditions on the total ring of quotients of a ring R implies the same condition on the ring R. In fact, for condition (5), note that any total ring of quotients is a Prьfer ring while there are non-Prьfer rings, even Noetherian ones. In  it is shown that if R is a local Noetherian reduced ring which is not a domain, then Q(R) is von Neumann regular. By Endo's Theorem 2.1, Q(R) is semihereditary, hence it satisfies all the five conditions above, while R is not a Prьfer ring, so its does not satisfy any of the five conditions. In the examples below we show that the implications (1) (2) (3) (4) (5) cannot be reversed, without additional conditions, even if all rings involved are total rings of quotients. Example 3.8. A non-Gaussian total ring of quotients. Let R = k[X, Y ]/(X, Y )3 where k is a field, X, Y are indeterminates over k. R coincides with its total ring of quotients, so it is a Prьfer ring, but it is not Gaussian. In fact, the maximal ideal of R is finitely generated, but its square is not principal. Example 3.9. A Gaussian total ring of quotients which is not arithmetical. Let R = k[X, Y ]/(X, Y )2 where k is a field, X, Y are indeterminates over k. R coincides with its total ring of quotients; R is Gaussian, since it is local and the maximal ideal has square zero, but is clearly not arithmetical.
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Example 3.10. An arithmetical total ring of quotients with w.gl.dim > 1. It is immediate to see that R = k[X]/(X)2 where k is a field and X is an indeterminate over k satisfies the desired conditions. Note that the rings considered in the preceding examples are all Noetherian. To find an example of a non-semihereditary total quotient ring with w.gl.dim 1 we need to leave the class of Noetherian rings, since any coherent ring with w.gl.dim 1 is semihereditary. Example 3.11. A non-semihereditary total ring of quotients with w.gl.dim 1. Consider an example of a non-semihereditary ring R with w.gl.dim R 1, as for instance the example produced in . The total ring of quotients of R, Q(R), has w.gl.dim Q(R) 1 and cannot be semihereditary, otherwise by Endo's characterization of semihereditary rings (Theorem 2.1), Q(R) would be von Neumann regular and thus the ring R would be semihereditary. At this point, using the results proved so far in this section, and other known results, we have a complete understanding of the effect that the assumption of one of the Prьfer-like conditions on the total ring of quotients of a ring R has on the ring itself. Theorem 3.12. Let R be a ring with total ring of quotients Q(R). The following conditions hold: (i) R is a semihereditary ring if and only if R is a Prьfer ring and Q(R) is a semihereditary ring. (ii) R has weak global dimension at most one if and only if R is a Prьfer ring and Q(R) has weak global dimension at most one. (iii) R is an arithmetical ring if and only if R is a Prьfer ring and Q(R) is an arithmetical ring. (iv) R is a Gaussian ring if and only if R is a Prьfer ring and Q(R) is a Gaussian ring. (v) Let (n) = (1), (2), (3) or (4). Then R satisfies condition (n) if and only if R satisfies condi- tion (n + 1) and Q(R) satisfies condition (n). Moreover, if the total ring of quotients of R is von Neumann regular, then all the five conditions above are equivalent on R. Proof. (i) The necessary condition has been proved in Theorem 3.7. For the converse, note that, if a total ring of quotients is semihereditary, then it is von Neumann regular by Endo's Theorem 2.1. Moreover, by [10, Theorem 20], a ring R is semihereditary if and only if R is Prьfer and Q(R) is von Neumann regular. (ii) By Theorem 3.7 only the sufficiency has to be proved. Recall that in [9, Theorem 2.2] it is proved that a ring has weak dimension less or equal 1 if and only if it is a reduced Gaussian ring. Assume that w.gl.dim Q(R) 1; then Q(R) is Gaussian and reduced, so R is reduced, too. By Theorem 3.3 R is Gaussian, thus it has weak dimension less or equal 1. So (ii) follows. (iii) Follows from Theorem 3.6 and (iv) is Theorem 3.3. (v) Let (n) = (1), (2), (3) or (4). If R satisfies condition (n + 1), then R is a Prьfer ring, so (v) follows by the previous conditions (i)­(iv). By [10, Theorem 20] a total ring of quotients is von Neumann regular if and only if it is semihereditary. So the last statement follows by the implications (1) (2) (3) (4) and part (v). 2
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a(d - ay) = 0 and (d - ay)2 = -2ayd = 0, since D2 = 0. Letting c = d - ay we clearly have (a, d) = (a, c) and c annihilates (a, d). So Tsang characterization in Theorem 2.2(c) is verified and R is Gaussian. If R satisfies (1) and (3), then R satisfies also (1) and (2) so R is Gaussian. 2 Example 4.2. This example shows that condition (1) is not enough to conclude that R is Gaussian. Let k be a field X, Y two indeterminates over k. Let R = k[X, Y ]/ X3, Y 2, X2Y and denote by x, y the images of X, Y in R. It is immediate to check that D = (x2, y), so D2 = 0 and R/D = k[X]/(X2) is arithmetical. But R is not Gaussian, since the annihilator of the maximal ideal (x, y) of R is (x, y)2 and R/(x, y)2 is not arithmetical. So by Tsang's characterization of local Noetherian Gaussian rings we conclude that R is not Gaussian. In all what follows we let D = {x R | x2 = 0}. Lemma 4.3. Assume that (R, m) is a local Gaussian ring. The following hold: (1) If a m \ D, then (0 : a) D. (2) If m is a nil ideal, then, for every element a m \ D, we have (D : a) D. (3) If m is a nil ideal and Dm = 0, then m4 = 0. Proof. (1) Let ab = 0. Since a2 = 0, we conclude by Theorem 2.2(d) that b2 = 0, hence b D. (2) Let a m \ D. Since m is a nil ideal there exists a minimum integer n > 1 such that 0 = an D. Then an-1 (D : a) and an-1 / D. (3) We first note that the condition Dm = 0 implies that for every a m \ D, (0 : a) = D. In fact, by hypothesis Da = 0 and by (1), (0 : a) D. Let n be the minimum integer such that 0 = an D, then n > 1 and (0 : an) = m, because anm Dm = 0. Since an-1 / D we have (0 : an-1) = D. Hence m = (0 : an) = ((0 : an-1) : a) = (D : a). Thus for every a m, am D, so m2 D and we conclude that m4 = 0. 2 Lemma 4.4. Assume that (R, m) is a local Gaussian ring with nil radical m. If m is not nilpotent, then m = m2 + D and m2 = m3. Proof. Consider the ring R/D. By Theorem 4.1, R/D is arithmetical and its maximal ideal m/D is not nilpotent. In fact, if (m/D)n = 0, for some n, then mn D and so m2n = 0 contrary to our hypothesis. Thus, by [5, X, 6 p. 357], m/D is idempotent, namely m2 + D = m. Multiplying this equality by m and by D we obtain m2 = m3 + mD and mD = m2D. So mD m3 and m3 + mD = m3 implies m2 = m3. 2 Lemma 4.5. Assume that (R, m) is a local Gaussian ring with nil radical m. If m is not nilpotent, then there exists an element d D such that D (0 : d) m. Proof. By Lemma 4.3(3), there exists d1 D such that (0 : d1) m. Let a m, a / (0 : d1); then 0 = ad1 D and a / D. Now, (0 : ad1) = ((0 : d1) : a) (D : a). Then by Lemma 4.3(2),
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(0 : ad1) D. Assume, by way of contradiction, that (0 : ad1) = m, for every a m, a / (0 : d1). Consider the local ring R = R/(0 : d1) with maximal ideal m = m/(0 : d1). R is such that the annihilator of every 0 = a m is m; hence m2 = 0Ї, that is m2 + (0 : d1) (0 : d1). Thus m2 + D (0 : d1) and by Lemma 4.4 we conclude that m (0 : d1), a contradiction. 2 5. The finitistic projective dimension of a Gaussian ring Denote by mod R the class of R-modules with a projective resolution consisting of finitely generated projective modules and by p.dR M THE PROJECTive dimension of the R-module M. Recall that the small finitistic dimension of R is defined by fP.dim R = sup{p.dR M | p.dR M < , M mod R}. In [9, Theorem 3.2] it is proved that if R is a coherent Gaussian ring, then fP.dim R 1. In the proof of that theorem the coherence of R is used only in quoting [7, Corollary 3.1.4] which is formulated for a coherent ring. We show how to adapt the proof of [9, Theorem 3.2] without assuming the coherence of R. The following lemma generalizes [7, Theorem 3.1.2]. Lemma 5.1. Let R be a ring and let I be an ideal contained in the Jacobson radical of R. If M mod R and TorRp (R/I, M) = 0 for every p 1, then: p.dR M = p.dR/I (M/I M). Proof. The proof of [7, Theorem 3.1.2] is by induction and relies on the fact that the syzygies of a finitely presented module over a coherent ring are again finitely presented. If a module M mod R, then its syzygies modules are again in mod R; thus the same proof of [7, Theorem 3.1.2] carries out in our hypotheses. 2 Lemma 5.2. Let R be a ring and let I be an ideal contained in the Jacobson radical of R. Then fP.dim R fP.dim R/I + w.dimR R/I. Proof. The proof is exactly the same as the proof of [7, Theorem 3.1.3]. In fact, assuming M mod R instead of M finitely presented the proof goes on without the hypothesis of coherence of the ring. 2 Proposition 5.3. Let R be a Gaussian ring, then fP.dim R 1. Proof. First we assume that R is local with maximal ideal m. In case m consists of zero divisors, then the proof of [9, Theorem 3.2, Case 1] shows that fP.dim R = 0. If m contains a regular element a, the proof of Case 2 in [9, Theorem 3.2] shows that fP.dim(R/aR) = 0. So by Lemma 5.2, fP.dim R w.dimR R/aR and, since a is a regular element, we have w.dimR R/aR p.dR R/aR 1. Assume R is not local. For every R-module M and every maximal ideal m of R we have p.dRm Mm p.dR M; moreover, if M mod R, then clearly Mm mod Rm. Thus if M mod R has finite projective dimension it follows that p.dRm Mm fP.dim Rm 1. Since
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for every module M mod R the weak and the projective dimension coincide and w.dimR M = sup{w.dimRm Mm | m Max R}, we conclude that fP.dim R 1 as desired. 2 6. The weak global dimension of a Gaussian ring
In  it is proved that the weak global dimension of a coherent Gaussian ring is either infinite or at most one. We first note that the same conclusion holds in the more general case of a Prьfer coherent ring.
Proposition 6.1. Let R be a coherent Prьfer ring. Then w.gl.dim R = 0, 1, or .
Proof. Assume that w.gl.dim R = n < . Then, every finitely generated ideal of R has finite projective dimension, that is R is a regular ring. By [7, Corollary 6.2.4], Q(R) is von Neumann regular, thus by Theorem 3.12 the five Prьfer-like conditions are equivalent on R. We conclude that w.gl.dim R 1. 2
We would like to extend the above result to an arbitrary Gaussian ring. Since w.gl.dim R = sup{w.gl.dim Rm | m Max R}, it is enough to prove that a local Gaussian ring has w.gl.dim = 0, 1, or . Moreover, by [9, Theorem 2.2], every reduced Gaussian ring has weak global dimension at most one. Thus, we can consider only non-reduced local Gaussian rings. Furthermore, recalling that the prime ideals in a local Gaussian ring R are linearly ordered, the nilradical n of R is a prime ideal and w.gl.dim R w.gl.dim Rn. So we can restrict our investigation to the case of a local Gaussian ring (R, m) such that the non-zero maximal ideal m coincides with the nilradical of R. We consider first the case in which the maximal ideal is nilpotent.
Lemma 6.2. Let (R, m) be a local ring which is not a field. Then w.dimR(R/m) = w.dimR m+ 1. Proof. Consider the exact sequence 0 m R R/m 0. Then w.dimR(R/m) = w.dimR m + 1 or R/m is flat. Assume by way of contradiction that R/m is flat; then m is pure in R. Hence am = aR m = aR, for every a m. By Nakayama's Lemma, am = aR implies a = 0, a contradiction. 2
Proposition 6.3. Let (R, m) be a local ring with non-zero nilpotent maximal ideal. Then w.dimR m = .
Proof. Let n be the nilpotency index of m. We prove that for every 1 k < n, w.dimR mn-k = w.dimR m + 1. So for k = n - 1 we get w.dimR m = w.dimR m + 1 which yields w.dimR m = . Let k = 1. Then mn-1m = 0 so 0 = mn-1 is isomorphic to a direct sum of copies of R/m. By Lemma 6.2, we conclude that w.dimR mn-1 = w.dimR m + 1. Let 1 h < n be the maximum integer such that w.dimR mn-k = w.dimR m + 1 for every k h and assume, by way of contra- diction, that h < n - 1. Consider the exact sequence
0
mn-h
mn-(h+1)

mn-(h+1) mn-h

0.
()
The term mn-(h+1)/mn-h is a non-zero semisimple, thus its weak dimension is equal to w.dimR m + 1. By assumption w.dimR mn-h = w.dimR m + 1. Thus, from the long ex-
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act sequence associated to () by tensoring with an arbitrary module X, we infer that w.dimR mn-(h+1) = w.dimR m + 1; contradicting the maximality of h. 2 Theorem 6.4. Let R be a Gaussian ring admitting a maximal ideal m such that the nilradical of the localization Rm is a non-zero nilpotent ideal. Then w.gl.dim R = . Proof. Let m be a maximal ideal of R such that Rm has a non-zero nilpotent nilradical. Since Rm is a Gaussian ring, the nilradical of Rm is a prime ideal, hence of the form nRm for some prime ideal n of R. Thus, the maximal ideal of the localization of R at n is non-zero and nilpotent. By Proposition 6.3, w.gl.dim Rn = . Since w.gl.dim R w.gl.dim RS for every localization RS of R we get the desired conclusion. 2 We were not able to prove that in general the weak global dimension of any Gaussian ring is either 0, 1, or . This is true for every arithmetical ring; in fact, Osofsky in  proved that an arithmetical local ring with zero divisors has infinite weak global dimension. Thus if R is an arithmetical ring such that every localization of R at a maximal ideal is a domain, then by  w.gl.dim R 1; otherwise there is a localization of R with infinite weak global dimension and the same holds true for R. With this evidence we formulate the following conjecture. Conjecture. The weak global dimension of a Gaussian ring R is 0, 1, or . References  D.D. Anderson, Another generalization of principal ideal rings, J. Algebra 48 (1997) 409­416.  D.D. Anderson, V. Camillo, Armendariz rings and Gaussian rings, Comm. Algebra 26 (1998) 2265­2272.  S. Endo, On semi-hereditary rings, J. Math. Soc. Japan 13 (1961) 109­119.  L. Fuchs, Ьber die Ideale arithmetischer Ringe, Comm. Math. Helv. 23 (1949) 334­341.  L. Fuchs, L. Salce, Modules over Non-Noetherian Domains, Math. Surveys Monogr., American Mathematical So- ciety, Providence, 2001.  R. Gilmer, Multiplicative Ideal Theory, Marcel Dekker, 1972.  S. Glaz, Commutative Coherent Rings, Lecture Notes in Math., vol. 1371, Springer-Verlag, 1989.  S. Glaz, Prьfer conditions in rings with zero divisors, in: Lect. Notes Pure Appl. Math., vol. 241, CRC Press, 2005, pp. 272­282.  S. Glaz, Weak dimension of Gaussian rings, Proc. Amer. Math. Soc. 133 (2005) 2507­2513.  M. Griffin, Prьfer rings with zero divisors, J. Reine Angew. Math. 239/249 (1970) 55­67.  M. Griffin, Valuation and Prьfer rings, Canad. J. Math. 26 (1974) 412­429.  J.A. Huckaba, Commutative Rings with Zero Divisors, Monographs and Textbooks Ser. in Pure and Appl. Math., vol. 117, Marcel Dekker, 1988.  C.U. Jensen, A remark on arithmetical rings, Proc. Amer. Math. Soc. 15 (1964) 951­954.  C.U. Jensen, Arithmetical rings, Acta Math. Hungar. 17 (1966) 115­123.  M. Larsen, P. McCarthy, Multiplicative Theory of Ideals, Academic Press, New York, 1971.  G. Lucas, The Gaussian property for rings and polynomials, preprint.  M.E. Manis, Extension of valuation theory, Bull. Amer. Math. Soc. 73 (1967) 735­736.  B. Osofsky, Global dimension of commutative rings with linearly ordered ideals, J. London Math. Soc. 44 (1969) 183­185.  H. Tsang, Gauss's lemma, PhD thesis, University of Chicago, 1965.

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