Almost clean rings and arithmetical rings

It is shown that a commutative B\'ezout ring $R$ with compact minimal prime spectrum is an elementary divisor ring if and only if so is $R/L$ for each minimal prime ideal $L$. This result is obtained by using the quotient space $\mathrm{pSpec} R$ of the prime spectrum of the ring $R$ modulo the equivalence generated by the inclusion. When every prime ideal contains only one minimal prime, for instance if $R$ is arithmetical, $\mathrm{pSpec} R$ is Hausdorff and there is a bijection between this quotient space and the minimal prime spectrum $\mathrm{Min} R$, which is a homeomorphism if and only if $\mathrm{Min} R$ is compact. If $x$ is a closed point of $\mathrm{pSpec} R$, there is a pure ideal $A(x)$ such that $x=V(A(x))$. If $R$ is almost clean, i.e. each element is the sum of a regular element with an idempotent, it is shown that $\mathrm{pSpec} R$ is totally disconnected and, $\forall x\in\mathrm{pSpec} R$, $R/A(x)$ is almost clean; the converse holds if every principal ideal is finitely presented. Some questions posed by Facchini and Faith at the second International Fez Conference on Commutative Ring Theory in 1995, are also investigated. If $R$ is a commutative ring for which the ring $Q(R/A)$ of quotients of $R/A$ is an IF-ring for each proper ideal $A$, it is proved that $R_P$ is a strongly discrete valuation ring for each maximal ideal $P$ and $R/A$ is semicoherent for each proper ideal $A$.


Introduction
In this paper we consider the following two questions: The first question was posed by M. Henriksen in [17] in 1955, and the second by M.D. Larsen, W.J Lewis and T.S. Shores in [20] in 1974.
In Section 3 we prove that these two questions are equivalent but they are still unsolved.
To show this equivalence, we use the quotient space pSpec R of Spec R modulo the equivalence generated by the inclusion, where R is a commutative ring. When R is a Gelfand ring, i.e. each prime ideal is contained in only one maximal, pSpec R is Hausdorff and homeomorphic to Max R (Proposition 2.1). On the other hand, if each prime ideal contains a unique minimal prime, then pSpec R is Hausdorff and there is a continuous bijection from Min R into pSpec R which is a homeomorphism if and only if Min R is compact. There is also a continuous surjection τ R : pSpec R → Spec B(R), where B(R) is the Boolean ring associated to R, and τ R is a homeomorphism if and only if pSpec R is totally disconnected. In this case, it is possible to get some interesting algebraic results by using Lemma 2.9.
A ring R is said to be clean (respectively almost clean (see [24])) if each element of R is the sum of an idempotent with a unit (respectively a regular element). In Section 4 we show that the total disconnectedness of pSpec R is necessary if the ring R is almost clean. Recall that a ring R is clean if and only if R is Gelfand and Max R totally disconnected: see [25,Theorem 1.7], [22,Corollary 2.7] or [9,Theorem I.1]. Almost clean rings were introduced by McGovern in [24] and studied by several authors: Ahn and Anderson [1], Burgess and Raphael [4] and [3], Varadarajan [30]. If Q is the quotient ring of R and if each prime ideal of R contains a unique minimal prime, we show that pSpec R and pSpec Q are homeomorphic and B(R) = B(Q); moreover, if R is arithmetical, then R is almost clean if Q is clean, and the converse holds if Q is coherent.
In Section 5 we give partial answers to some questions posed by Facchini and Faith at the second International Fez Conference on Commutative Ring Theory in 1995 [12]. If R is fractionally IF, it is shown that R/A is semicoherent for each ideal A and R P is a strongly discrete valuation ring for each maximal ideal P . We give an example of a finitely fractionally self FP-injective ring which is not arithmetical; recall that Facchini and Faith proved that each fractionally self FP-injective ring is arithmetical. It is also proven that any ring which is either clean, coherent and arithmetical or semihereditary is finitely fractionally IF. However, there exist examples of clean coherent arithmetical rings with a non-compact minimal prime spectrum; recall that the author proved that Min R/A is compact for any ideal A of a fractionally self FP-injective ring R.
All rings in this paper are associative and commutative with unity, and all modules are unital.We denote respectively Spec R, Max R and Min R, the space of prime ideals, maximal ideals and minimal prime ideals of R, with the Zariski topology. If A a subset of R, then we denote V (A) = {P ∈ Spec R | A ⊆ P } and D(A) = Spec R \ V (A).

A quotient space of the prime spectrum of a ring
If R is a ring, we consider on Spec R the equivalence relation R defined by LRL ′ if there exists a finite sequence of prime ideals (L k ) 1≤k≤n such that L = L 1 , L ′ = L n and ∀k, 1 ≤ k ≤ (n − 1), either L k ⊆ L k+1 or L k ⊇ L k+1 . We denote by pSpec R the quotient space of Spec R modulo R and by λ R : Spec R → pSpec R the natural map. The quasi-compactness of Spec R implies the one of pSpec R, but generally pSpec R is not T 1 : see [21, Propositions 6.2 and 6.3]. However: Proposition 2.1. The following conditions are equivalent for a ring R: (1) the restriction of λ R to Max R is a homeomorphism; (2) the restriction of λ R to Max R is injective; (3) R is Gelfand. In this case pSpec R is Hausdorff.
(ii) ⇒ (iii). If a prime ideal is contained in two maximals ideals P 1 and P 2 we get that λ R (P 1 ) = λ R (P 2 ).
(iii) ⇒ (i). If L is a prime ideal we denote by µ(L) the unique maximal ideal containing L. It is easy to verify that µ(L) = µ(L ′ ) if LRL ′ . So, µ induces a map µ : pSpec R → Max R. We easily show thatμ −1 = λ R | Max R . By [ Proof. If L and L ′ are prime ideals of T such that L ⊆ L ′ then a ϕ(L) ⊆ a ϕ(L ′ ). Hence, if x ∈ pSpec T , we can put b ϕ(x) = λ R ( a ϕ(L)) where L ∈ x. Since λ R and a ϕ are continuous, so is b ϕ.
An exact sequence 0 → F → E → G → 0 is pure if it remains exact when tensoring it with any R-module. Then, we say that F is a pure submodule of E. By [13,Proposition 8.6] F is a pure submodule of E if every finite system of equations with coefficients r j,i ∈ R and unknowns x 1 , . . . , x n , has a solution in F whenever it is solvable in E. The following proposition is well known.
Proposition 2.4. Let A be an ideal of a ring R. The following conditions are equivalent: (1) A is a pure ideal of R; (2) for each finite family (a i ) 1≤i≤n of elements of A there exists t ∈ A such that a i = a i t, ∀i, 1 ≤ i ≤ n; (3) for all a ∈ A there exists b ∈ A such that a = ab; Moreover, if A is finitely generated, then A is pure if and only if it is generated by an idempotent.
. If a ∈ A, 1 is solution of the equation ax = a. So, this equation has a solution in A.
(iii) ⇒ (ii). Let a 1 , . . . , a n be elements of A. We proceed by induction on n. There exists t ∈ A such that a n = ta n . By induction hypothesis there exists s ∈ A such that a i − ta i = s(a i − ta i ), ∀i, 1 ≤ i ≤ (n − 1). Now, it is easy to check that (ii) ⇒ (i). We consider the following system of equations: Assume that (c 1 , . . . , c n ) is a solution of this system in R. There exists s ∈ A such that a j = sa j , ∀j, 1 ≤ j ≤ p. So, (sc 1 , . . . , sc n ) is a solution of this system in A.
We set 0 P the kernel of the natural map R → R P where P ∈ Spec R.
Lemma 2.5. Let R be a ring and let C a closed subset of Spec R. Then C is the inverse image of a closed subset of pSpec R by λ R if and only if C = V (A) where A is a pure ideal. Moreover, in this case, A = ∩ P ∈C 0 P .
Proof. Let A be a pure ideal, and let P and L be prime ideals such that A ⊆ P and L ⊆ P . Since A is pure, for each a ∈ A there exists b ∈ A such that a = ab.
Suppose that C is the inverse image of a closed subset of pSpec R by λ R . We put A = ∩ P ∈C 0 P . Let b ∈ B and P ∈ C. Then C contains each minimal prime ideal contained in P . So, the image of b, by the natural map R → R P , belongs to the nilradical of R P . It follows that there exist 0 = n P ∈ N and s P ∈ R \ P such that s P b nP = 0. Hence, ∀L ∈ D(s P ) ∩ C, b nP ∈ 0 L . A finite family (D(s Pj )) 1≤j≤m covers C. Let n = max{n P1 , . . . , n Pm }. Then b n ∈ 0 L , ∀L ∈ C, whence b n ∈ A. We deduce that C = V (A). Now, we have If {x} is closed in pSpec R we denote by A(x) the pure ideal of R for which x = V(A(x)). A topological space is called totally disconnected if each of its connected components contains only one point. Every Hausdorff topological space X with a base of clopen neighbourhoods is totally disconnected and the converse holds if X is compact (see [16,Theorem 16.17]). Proposition 2.7. Let R be a ring. Then the following conditions are equivalent: (1) pSpec R is totally disconnected; (2) for each x ∈ pSpec R, {x} is closed and A(x) is generated by idempotents.
(i) ⇒ (ii). Let x ∈ pSpec R and a ∈ A(x). There exists b ∈ A(x) such that a = ab.
) is open and contains x. The condition pSpec R is totally disconnected implies that there exists an idempotent e such that x ⊆ D(e) ⊆ λ ← R (U ) ⊆ D(1 − b). If follows that e ∈ R(1 − b). So ea = 0 and consequently a = a(1 − e). From x ⊆ D(e) and e(1 − e) = 0 we deduce that (1 − e) ∈ A(x) by Lemma 2.5.
For any ring R, B(R) is the set of idempotents of R. For any e, e ′ ∈ B(R) we put e ⊕ e ′ = e + e ′ − ee ′ and e ⊙ e ′ = ee ′ . With these operations B(R) is a Boolean ring. The space Spec B(R) is denoted by X(R). Then X(R) is Hausdorff, compact and totally disconnected. If x ∈ X(R) the stalk of R at x is the quotient of R by the ideal generated by the idempotents contained in x.
Proposition 2.8. Let R be a ring. The following assertions hold: (1) there exists a surjective continuous map τ R : pSpec R → X(R); (2) pSpec R is totally disconnected if and only if τ R is a homeomorphism. In this case, for each x ∈ pSpec R, R/A(x) is the stalk of R at τ R (x).
Proof. (i). If L and L ′ are prime ideals of R, L ⊆ L ′ , then L ∩ B(R) = L ′ ∩ B(R) since each prime ideal of B(R) is maximal. So, τ R is well defined. It is easy to check that for any idempotent e ∈ R, τ ← R (D(e)) = λ → R (D(e)). Hence τ R is continuous. For each x ∈ X(R), if L is a maximal ideal containing all elements of x, then The last assertion is a consequence of Proposition 2.7 and Lemma 2.5.
The following lemma will be useful to show some important results of this paper. Lemma 2.9. Let R be a ring such that pSpec R is totally disconnected. Then any R-algebra S (which is not necessarily commutative) satisfies the following condition: let f 1 , . . . , f k be polynomials over S in noncommuting variables x 1 , . . . , x m , y 1 , . . . , y n . Let a 1 , . . . , a m ∈ S. Assume that, ∀x ∈ pSpec R there exist b 1 , . . . , b n ∈ S such that: Then there exists a finitely generated ideal A ⊆ A(x) such that There exists an idempotent e x such that Hence A finite family (λ → R (D(e xj ))) 1≤j≤p covers pSpec R. We may assume that (e xj ) 1≤j≤p is a family of orthogonal idempotents. We put d ℓ = e x1 b x1,ℓ +· · ·+e xp b xp,ℓ , ∀ℓ, 1 ≤ ℓ ≤ n. Then f i (a 1 , . . . , a m , d 1 , . . . , d n ) = 0, ∀i, 1 ≤ i ≤ k.
We denote by gen M the minimal number of generators of a finitely generated R-module M . The following proposition is an example of an algebraic result that can be proven by using Lemma 2.9. Recall that the trivial extension It is convenient to identify R ⋉ M with the R-module R ⊕ M endowed with the following multiplication: (r, x)(s, y) = (rs, ry + sx), where r, s ∈ R and x, y ∈ M . Proposition 2.10. Let R be a ring such that pSpec R is totally disconnected. Let M be a finitely generated R-module and F a finitely presented R-module. Then: . . , f n } be a spanning set of F with the following relations: Thus E 1 has a solution modulo A(x)S for each x ∈ pSpec R. By Lemma 2.9 E 1 has a solution x j,i , y i , z i,j , 1 ≤ j ≤ p, 1 ≤ i ≤ n in S. It is easy to check that is an easy consequence of (i). (iii). Let the notations be as in (i). We assume that m 1 , . . . , m p verify the following relations: It follows that there exist w k,l ∈ R such that: We deduce that Conversely, if there exists an epimorphism φ : . . , m p , from the relation (rel) we get a relation (rel1) which is a linear combination of the relations j=p j=i d k,j m j = 0. By using the equalities 1, we get that (rel) is a linear combination of the relations We put E = E 1 ∪ E 2 . As in (i) and by using the above observation we show that E has a solution. We define φ : F → M as in (i), and by using the fact that E 2 has a solution, we prove that φ is injective by using the above observation.

Hermite rings and elementary divisor rings
We say that R is an elementary divisor ring if for every matrix A, with entries in R, there exist a diagonal matrix D and invertible matrices P and Q, with entries in R, such that P AQ = D. Then we have the following implications: elementary divisor ring ⇒ Hermite ring ⇒ Bézout ring ⇒ arithmetical ring; but these implications are not reversible: see [14] or [5]. Recall that a ring R is (semi)hereditary if each (finitely generated) ideal is projective. If F is a submodule of a module E and x an element of E, then the ideal {r ∈ R | rx ∈ F } is denoted by (F : x).
The following was already proved, see [ (ii) is an immediate consequence of (i) because Min R is compact if R is semihereditary by [26,Proposition 10].
The following example shows that pSpec R is not generally totally disconnected, even if R is arithmetical. Example 3.3. Consider [31, Example 6.2 (due to Jensen)] defined in the following way: let I be a family of pairwise disjoint intervals of the real line with rational endpoints, such that between any two intervals of I there is at least another interval of I; let R be the ring of continuous maps R → R which are rational constant by interval except on finitely many intervals of I on which it is given by a rational polynomial. It is easy to check that R is a reduced indecomposable ring. It is also Bézout (left as an exercise!).
Then (0 : f ) is not finitely generated, whence R is not semihereditary. So, pSpec R is an infinite set and a compact connected topological space.
Theorem 3.4. Let R be a ring such that pSpec R is totally disconnected. Assume that R/A(x) is Bézout for each x ∈ pSpec R. Then R is an elementary divisor ring if and only if so is R/L, for each minimal prime ideal L.
Proof. Only "if" requires a proof. By Theorem 3.1 R is Hermite. Let x ∈ pSpec R and R ′ = R/A(x). Then R ′ has a unique minimal prime ideal. Let L be the minimal prime ideal of R such that L/A(x) is the minimal prime of R ′ . Thus L/A(x) is contained in the Jacobson radical J (R ′ ) of R ′ . So, R ′ /J (R ′ ) is an elementary divisor ring since it is a homomorphic image of R/L. By [17, Theorem 3] a Hermite ring S is an elementary divisor ring if and only if so is S/J (S). Hence R/A(x) is an elementary divisor ring. Let a, b, c ∈ R such that Ra + Rb + Rc = R. We consider the polynomial equation in variables X, Y, S, T : aSX + bT X + cT Y = 1. By [15,Theorem 6], this equation has a solution modulo A(x), ∀x ∈ pSpec R. So, by Lemma 2.9 there is a solution in R. We conclude by [15,Theorem 6].
With a similar proof as in Corollary 3.2, we get Corollary 3.5. The second condition shows that the two questions 1.1 and 1.2 have the same answer. The third assertion can be also deduced from [28,Corollary].

Almost clean rings
In [24,Proposition 15], McGovern proved that each element of a ring R is the product of an idempotent with a regular element if and only if R is a PP-ring, i.e. each principal ideal is projective, and he showed that each PP-ring is almost clean ( [24,Proposition 16]). The aim of this section is to study almost clean rings.
In the sequel, if R is a ring, R(R) is its set of regular elements of R and By [2, Corollaire 2 p.92] each zero-divisor is contained in an element of Φ R . So, the following proposition is obvious.
Proposition 4.1. Let R be a ring. The following conditions are equivalent: (1) For each a ∈ R, either a or (a − 1) is regular.
(2) R is almost clean and indecomposable.
Corollary 4.2. Let R be an arithmetical ring, Q its ring of fractions and N its nilradical. Then: (1) R is almost clean and indecomposable if and only if Q is a valuation ring; (2) R/A is almost clean and indecomposable for each ideal A ⊆ N if and only if N is prime and uniserial.
Proof. (i). Assume that R is almost clean and indecomposable. Then if L, L ′ ∈ Φ R then there exists a maximal ideal P such that L + L ′ ⊆ P . Since R P is a valuation ring, either L ⊆ L ′ or L ′ ⊆ L. By [2, Corollaire p.129] Φ R is homeomorphic to Spec Q. It follows that Q is local. Conversely, Φ R contains a unique maximal element.
(ii). First, assume that R/A is almost clean and indecomposable for each ideal A ⊆ N . Since Q is a valuation ring then N is prime. By way of contradiction suppose ∃a, b ∈ N such that neither divides the other. We may assume that Ra ∩ Rb = 0. Let A and B be maximal submodules of Ra and Rb respectively. We may replace R by R/(A + B) and assume that Ra and Rb are simple modules. Let L and P be their respective annihilators. Since R L is a valuation ring and R L a = 0, we have R L b = 0. So, L = P . It follows that ∃c ∈ L such that (1 − c) ∈ P . Neither c nor (1 − c) is regular. This contradicts that R is almost clean.
Conversely, suppose that N is prime and uniserial. Then, if A is an ideal contained in N , N/A is also uniseriel. So, the ring of fractions of R/A is a valuation ring: see [29, p.218, between the definition of a torch ring and Theorem B].
Following Vámos [29], we say that R is a torch ring if the following conditions are satisfied : (1) R is an arithmetical ring with at least two maximal ideals; (2) R has a unique minimal prime ideal N which is a nonzero uniserial module. We follow T.S. Shores and R. Wiegand [27], by defining a canonical form for an R-module E to be a decomposition E ∼ = R/I 1 ⊕ R/I 2 ⊕ · · · ⊕ R/I n , where I 1 ⊆ I 2 ⊆ · · · ⊆ I n = R, and by calling a ring R a CF-ring if every direct sum of finitely many cyclic modules has a canonical form. Proof. By [27, Theorem 3.12] every CF-ring is arithmetical and a finite product of indecomposable CF-rings. If R is indecomposable then R is either a domain, or a local ring, or a torch ring. By Corollary 4.2 R is almost clean.
By Proposition 2.8 there is some similarity between [3, Theorem 2.4] and the following theorem. (1) R is almost clean; (2) pSpec R is totally disconnected and ∀r ∈ R, ∀x ∈ pSpec R, ∃s x ∈ R(R) such that either r ≡ s x modulo A(x) or (r − 1) ≡ s x modulo A(x); (3) pSpec R is totally disconnected and for each x ∈ pSpec R, R/A(x) is almost clean. Then (i) ⇔ (ii) ⇒ (iii) and the three conditions are equivalent if every principal ideal of R is finitely presented.
Proof. (i) ⇒ (ii). Let x and y be two distinct points of pSpec R. Let P and P ′ be two minimal prime ideals of R such that P ∈ x and P ′ ∈ y. There is no maximal ideal containing P and P ′ . So, P + P ′ = R and there exist a ∈ P and a ′ ∈ P ′ such that a + a ′ = 1. We have a = s + e where s is regular and e idempotent. Since s / ∈ P we get that e / ∈ P . It follows that for each L ∈ x, (1 − e) ∈ L and e / ∈ L. So, x ⊆ D(e). We have a ′ = −s + (1 − e). In the same way we get y ⊆ D(1 − e). Therefore x and y belong to disjoint clopen neighbourhoods. Hence pSpec R is totally disconnected. Now, let r ∈ R and z ∈ pSpec R. We have r = s + e where s is regular and e idempotent. If z ⊆ V (e) then e ∈ A(z) and r ≡ s modulo A(z); and, if z ⊆ V (1 − e) then (1 − e) ∈ A(z) and (r − 1) ≡ s modulo A(z).
(ii) ⇒ (i). Let a ∈ R. Let Q be the ring of fractions of R and let S = R ⋉ Q be the trivial extension of R by Q. We consider the following polynomial equations in S: E 2 = E, E + X = (a, 0) and XY = (0, 1). Let x ∈ pSpec R. If a ≡ s x modulo A(x) where s x is a regular element of R, then E = (0, 0), X = (s x , 0), Y = (0, 1/s x ) is a solution of these polynomial equations modulo A(x)S; if (a − 1) ≡ s x modulo A(x), we take E = (1, 0). So, by Lemma 2.9, these equations have a solution in S: E = (e, q), X = (s, u), Y = (t, v). From E 2 = E we deduce that e 2 = e and (2e − 1)q = 0. So, q = 0 since (2e − 1) is a unit. From E + X = (a, 0) we deduce that u = 0, and from XY = (0, 1) we deduce that sv = 1. Hence s is a regular element of R and a = e + s. We conclude that R is almost clean.
(iii) ⇒ (ii). We assume that each principal ideal is finitely presented. Let x ∈ pSpec R. It remains to show that any regular element a modulo A(x) is congruent to a regular element of R modulo A(x). Since (0 : a) is finitely generated and A(x) is generated by idempotents, there exists an idempotent e ∈ A(x) such that (0 : a) ⊆ Re. Now, it is easy to check that a(1 − e) + e is a regular element.
The following examples show that the conditions (i) and (iii) are not generally equivalent.
Example 4.5. [3, Examples 2.2 and 2.9.(i)] are non-almost clean arithmetical rings (the second is reduced) with almost clean stalks. These are defined in the following way: let D be a principal ideal domain and ∀n ∈ N, let R n be a quotient of D by a non-zero proper ideal I n ; let R be the set of elements r = (r n ) n∈N of Π n∈N R n which satisfy ∃m r ∈ N and ∃d r ∈ D such that ∀n ≥ m r , r n = d r + I n . We put e m = (δ m,n ) n∈N , ∀m ∈ N. It is easy to check that the points of pSpec R are: The following example shows that the condition "each principal ideal is finitely presented" is not necessary if R is almost clean.  (1) pSpec Q and pSpec R are homeomorphic; (2) each idempotent of Q belongs to R; (3) if Q is clean then R is almost clean.
Proof. (i). If ϕ : R → Q is the natural map then Min R ⊆ Φ R = Im a ϕ. It follows that each prime ideal of Q contains only one minimal prime and b ϕ is bijective. Moreover, since pSpec Q and pSpec R are compact, b ϕ is a homeomorphism.
(ii). Let e an idempotent of Q. Then λ → Q (D(e)) is a clopen subset of pSpec Q, whence its image by b ϕ is a clopen subset of pSpec R and consequently it is of the form λ → R (D(e ′ )) where e ′ is an idempotent of R. But the inverse image of D(e ′ ) ⊆ Spec R by a ϕ is D(e ′ ) ⊆ Spec Q. So, e = e ′ ∈ R.
(iii). Assume that Q is clean. Let r ∈ R. Then r = q + e where q is a unit of Q and e an idempotent. Since e ∈ R, q is a regular element of R.
Corollary 4.8. Let R be an almost clean arithmetical ring and Q its ring of fractions. If Q is coherent then Q is clean. In this case Q is an elementary divisor ring.
Proof. Recall that an arithmetical ring is coherent if and only if each principal ideal is finitely presented because the intersection of any two finitely generated ideals is finitely generated by [27,Corollary 1.11]. By Proposition 4.7 we may assume that pSpec Q = pSpec R. This space is totally disconnected by Theorem 4.4. Let x ∈ pSpec Q and let A(x) be the pure ideal of R such that x = V (A(x)). Then x = V (QA(x)). If s is a regular element of R then s + A(x) is a regular element of R/A(x). So, R/A(x) and Q/QA(x) have the same ring of fractions which is a valuation ring by Corollary 4.2. Hence Q/QA(x) is almost clean, ∀x ∈ pSpec Q. By Theorem 4.4 Q is almost clean too. We conclude that Q is clean since each regular element is a unit.
The last assertion is a consequence of [9, Theorem I.1 and Corollary II.2].
We don't know if the assumption "Q is coherent" can be omitted. The following example shows that the conclusion of the previous corollary doesn't hold if R is not arithmetical, even if R has a unique minimal prime ideal. Example 4.9. Let K be a field, D = K[x, y] (x,y) where x, y are indeterminates, E = D/Dx ⊕ D/Dy and R the trivial extension of D by E. Since R contains a unique minimal prime ideal, Q is indecomposable. We put r = x 0 0 x and s = y 0 0 y . Clearly r and s are zerodivisors but r + s is regular. It follows that r r+s and s r+s are two zerodivisors of Q whose sum is 1. So, Q is not almost clean.

Fractionally IF-rings
Let P be a ring property. We say that a ring R is (finitely) fractionally P if the classical ring of quotients Q(R/A) of R/A satisfies P for each (finitely generated) ideal A. In Some preliminary results are needed. As in [23] a ring R is said to be semicoherent if Hom R (E, F ) is a submodule of a flat R-module for any pair of injective R-modules E, F . An R-module E is FP-injective if Ext 1 R (F, E) = 0 for any finitely presented R-module F, and R is self FP-injective if R is FP-injective as R-module. We recall that a module E is FP-injective if and only if it is a pure submodule of every overmodule. If each injective R-module is flat we say that R is an IF-ring. By [6, Theorem 2], R is an IF-ring if and only if it is coherent and self FP-injective. Proof. If R is coherent then Hom R (E, F ) is flat for any pair of injective modules E, F by [13, Theorem XIII.6.4(b)]; so, R is semicoherent. Conversely, let E be the injective hull of R. Since R is a pure submodule of E, then, for each injective R-module F , the following sequence is exact: By using the natural isomorphisms Hom R (F ⊗ R B, F ) ∼ = Hom R (F, Hom R (B, F )) and F ∼ = Hom R (R, F ) we get the following exact sequence: So, the identity map on F is the image of an element of Hom R (F, Hom R (E, F )). Consequently the following sequence splits: It follows that F is a direct summand of a flat module. So, R is an IF-ring.
Corollary 5.2. Let R be a ring. Assume that its ring of quotients Q is self FPinjective. Then R is semicoherent if and only if Q is coherent.
Proof. If R is semicoherent, then so is Q by [23,Proposition 1.2]. From Proposition 5.1 we deduce that Q is coherent. Conversely, let E and F be injective R-modules. It is easy to check that the multiplication by a regular element of R in Hom R (E, F ) is injective. So, Hom R (E, F ) is a submodule of the injective hull of Q ⊗ R Hom R (E, F ) which is flat over Q and R because Q is an IF-ring.
Proof. Assume that A is not prime and let A ′ = AR A ♯ . Then R A ♯ /A ′ is the ring of quotients of R/A. So, by [7, Théorème 2.8], R A ♯ /A ′ is self FP-injective because each non-unit is a zero-divisor. By [8,Corollary II.14] it is coherent if and only if A ′ is principal. So, we conclude by Corollary 5.2.
A valuation ring R is called strongly discrete if there is no non-zero idempotent prime ideal.
Corollary 5.4. Let R be a valuation ring. Then R/A is semicoherent for each ideal A if and only if R is strongly discrete.
Proof. Assume that R is strongly discrete. Each ideal A is of the form A = aL, where L is a prime ideal and a ∈ R. Clearly L = A ♯ . Then, L 2 = L implies that AR L is principal over R L . Since A is the inverse image of AR L by the natural map R → R L , R/A is semicoherent by Corollary 5.3.
Conversely, let L be non-zero prime ideal, let A = aLR L , where 0 = a ∈ R L and let A ′ be the inverse image of A by the natural map R → R L . Clearly L = (A ′ ) ♯ . Since R/A ′ is semicoherent, A is principal over R L by Corollary 5.3. It follows that L is principal over R L . So, L = L 2 . Now, we can prove one of the main results of this section.
Theorem 5.5. Let R be a fractionally IF-ring. Then, R/A is semicoherent for each ideal A and R P is a strongly discrete valuation ring for each maximal ideal P .
Proof. By [12, Theorem 1] R is arithmetical because it is fractionally self FPinjective. Let P be a maximal ideal and let A be an ideal of R P . If B is the kernel of the following composition of natural maps R → R P → R P /A, then Q(R P /A) = Q(R/B) is an IF-ring. We conclude by Corollaries 5.3 and 5.4.
It is obvious that each von Neumann regular ring is a fractionally IF-ring. Moreover: Proposition 5.6. Let R be an arithmetical ring which is locally strongly discrete. Then R is a fractionally IF-ring in the following cases: (1) R is fractionally semilocal; (2) R is semilocal; (3) R is a Prüfer domain of finite character, i.e. each non-zero element is contained in but a finite number of maximal ideals.
Proof. (i). We may assume that R = Q(R). By [12,Lemma 7], for each maximal ideal P , R P = Q(R P ). It follows that R P is self FP-injective. Moreover it is coherent by Corollary 5.4 and Proposition 5.1. Since Π P ∈Max R R P is a faithfully flat R-module and an IF-ring, we deduce that R is IF too.
(iii) follows from (ii) since R/A is semilocal for each non-zero ideal A.
Question 5.7. What are the locally strongly discrete Prüfer domains which are fractionally IF?
The following example shows that an arithmetical ring which is locally Artinian is not necessarily fractionally IF.
Example 5.8. Let K be a field, V = K[X]/(X 2 ) where X is an indeterminate and let x be the image of X in V . For each p ∈ N we put R 2p = V and R 2p+1 = V /xV ∼ = K. Let S = n∈N R n , J = n∈N R n and let R be the unitary V -subalgebra of S generated by J. For each n ∈ N we set e n = (δ n,p ) p∈N and we denote by 1 the identity element of R. Let P be a maximal ideal of R: • either J ⊆ P ; in this case P = P ∞ = J + xR and R P∞ = R/J ∼ = V ; • or ∃n ∈ N such that e n / ∈ P ; in this case P = P n = R(1 − e n ) + Rxe n , R Pn = R/R(1 − e n ) ∼ = V if n is even and R Pn ∼ = K if n is odd.
Then, for each maximal ideal P , R P is an artinian valuation ring. Now it is easy to check that (0 : x1) = Rx1 + n∈N Re 2n+1 . So, R is not coherent.
The following proposition is a short answer to another question posed by Facchini and Faith in [12, question Q3 p.301].
Proposition 5.9. There exists a non-arithmetical zero-Krull-dimensional ring R which is finitely fractionally self FP-injective.
Proof. Let V be the artinian valuation ring of Example 5.8, S = V N and J = V (N) . Let y = (y n ) n∈N , z = (z n ) n∈N ∈ S such that, ∀p ∈ N, y 2p = z 2p+1 = x and y 2p+1 = z 2p = 0, and let R be the unitary V -subalgebra of S generated by y, z and J. The idempotents (e n ) n∈N are defined as in Example 5.8. Let P ∈ Max R: • either J ⊆ P ; in this case P = P ∞ = J + yR + zR and R P∞ = R/J ∼ = K[Y, Z]/(Y, Z) 2 ; • or ∃n ∈ N such that e n / ∈ P ; in this case P = P n = R(1 − e n ) + Rxe n , Clearly, R P∞ is not a valuation ring. So, R is not arithmetical. First, we show that R is a pure submodule of S. It is sufficient to prove that R P is a pure submodule of S P for each maximal ideal P . It is obvious that R Pn ∼ = S Pn ∼ = e n S ∼ = V . It remains to be shown that R P∞ is a pure submodule of S P∞ ∼ = S/J. We consider the following equations: where r i,j , s i ∈ R, ∀i, 1 ≤ i ≤ p, ∀j, 1 ≤ j ≤ m. When these equations have a solution in S, we must prove they have a solution in R too. This can be done by using the basis {1, y, z, e n , xe n | n ∈ N} of R over K. Consequently R is pure in S. Now, let A be a finitely generated ideal of R. Then R/A is a pure submodule of S/SA. We have S/SA ∼ = n∈N (R/A) Pn . For each n ∈ N, (R/A) Pn is self injective, whence it is an injective (R/A)-module. We deduce that S/SA is injective over R/A. Hence R/A is self FP-injective.
Finally, for the second question posed by Facchini and Faith in [12, question Q2 p.301], we shall prove Theorem 5.11. The following lemma is needed. Lemma 5.10. Let R be a clean ring such that R = Q(R) and (0 : a) is finitely generated for each a ∈ R. Then, for each maximal ideal P , R P = Q(R P ).
Proof. Let P be a maximal ideal of R. By way of contradiction, suppose that R P contains a regular element which is not a unit. So, ∃a ∈ P such that (0 P : a) = 0 P (since R is clean, R P = R/0 P by [9, Proposition III.1]). It follows that (0 : a) ⊆ 0 P . Since (0 : a) is finitely generated and 0 P is generated by idempotents, there exists an idempotent e ∈ 0 P such that (0 : a) ⊆ Re. Now, it is easy to check that a(1 − e) + e is a regular element contained in P . This contradicts that R = Q(R).
Theorem 5.11. The following assertions hold: (1) let R be an almost clean coherent arithmetical ring. Assume that R/A(x) is either torch or local or a domain ∀x ∈ pSpec R. Then R is finitely fractionally IF; (2) each clean coherent arithmetical ring is finitely fractionally IF; (3) each semihereditary ring is finitely fractionally IF; (4) let R be a zero-Krull-dimensional ring or a one-Krull-dimensional domain.
Then R is finitely fractionally IF if and only if R is coherent and arithmetical.
Proof. (i). Let A be a finitely generated ideal of an almost clean coherent arithmetical ring R and let N be the nilradical of R. Since each prime ideal contains only one minimal prime, by [ On the other hand, Re is an almost coherent arithmetical ring and Ae is a finitely generated ideal contained in N e. Let T = (Re/Ae) and let φ : R → T be the natural epimorphism. Then pSpec T is totally disconnected because it is homeomorphic to pSpec (Re). Let x ∈ pSpec T . Then T /A(x) is the quotient of R/A( b φ(x)) modulo an ideal contained in the minimal prime of R/A( b φ(x)). By Corollary 4.2 T /A(x) is almost clean. We deduce that T is coherent and almost clean by Theorem 4.4. By Corollary 4.8 T ′ = Q(T ) is clean. By Lemma 5 T ′ P = Q(T ′ P ) for each maximal ideal P of T ′ . We deduce that T ′ P is IF because it is a valuation ring. So, since T ′ is locally IF, it is IF too. Hence Q(R/A) is IF.
(ii) and (iii). If R is either clean, coherent and arithmetical or semihereditary, then R satisfies the conditions of (i). Hence R is finitely fractionally IF.
(iv). Observe that Q(R/A) = R/A for each non-zero proper ideal A. First assume that R is arithmetical and coherent. We deduce that R is finitely fractionally IF from (ii) and (iii).
Conversely, let P be a maximal ideal of R and A a finitely generated ideal of R P . There exists a finitely generated ideal B of R such that A = B P . So, R P /A ∼ = (R/B) P . Since R/B is IF, so is R P /A by [7, Proposition 1.2]. Hence we may assume that R is local and we must prove that R is a valuation ring. If not, there exist a, b ∈ R such that a / ∈ Rb and b / ∈ Ra. The coherence of R/(ab) implies that Ra ∩ Rb is finitely generated. It follows that R/(Ra ∩ Rb) is IF. We may assume that Ra ∩ Rb = 0. By [19, Corollary 2.5] A = (0 : (0 : A)) for each finitely generated ideal A. We deduce that 0 = Ra ∩ Rb = (0 : (0 : a) + (0 : b)). Then (0 : a) + (0 : b) is a faithful finitely generated proper ideal. By [19,Corollary 2.5] this is not possible. Hence R is a valuation ring.
If R is fractionally self FP-injective, then, by [8, Theorem III.1] Min R/A is compact for each proper ideal A. The following example shows that this is not true if R is finitely fractionally IF, even if R is a coherent clean arithmetical ring.
Example 5.12. Let D be a valuation domain. Assume that its maximal ideal P ′ is the only non-zero prime and it is not finitely generated. Let 0 = d ∈ P ′ . We put V = D/dD, P = P ′ /dD and R = V N . It is easy to check that R is clean, Bézout and coherent. So, by (ii) of Theorem 5.11 R is finitely fractionally IF. Since P is not finitely generated, ∀n ∈ N, ∃b n ∈ P such that b n n = 0. We set b = (b n ) n∈N . Let N be the nilradical of R. If there exists c = (c n ) n∈N ∈ R such that (b − bcb) ∈ N , then ∃m ∈ N such that b m (1 − cb) m = 0. If n ∈ N, n ≥ m we get that b n n (1 − c n b n ) n = 0. Clearly there is a contradiction. So, R/N is not Von Neumann regular. Now, let c ∈ R such that (N : c) = N . We shall prove that c is a unit. By way of contradiction, suppose ∃k ∈ N such that c k ∈ P . We put e k = (δ k,n ) n∈N . Then ce k = c k e k ∈ N . It follows that e k ∈ N , which is absurd. So, ∀k ∈ N, c k / ∈ P . Therefore c is a unit and R/N is equal to its quotient ring. By [11,Theorem 5] a reduced arithmetic ring S is semihereditary if and only if Q(S) is von Neumann regular. By [26, Proposition 10] a reduced arithmetic ring S is semihereditary if and only if Min S is compact. Consequently Min R/N is not compact . Hence Min R is not compact too. (If P ′ is finitely generated by p and R = n∈N D/p n+1 D, then R is clean, arithmetical, coherent and finitely fractionally IF, but Min R is not compact. We do the same proof by taking b n = p + p n+1 D, ∀n ∈ N.)