It is proven that every commutative semihereditary Bezout ring in which any regular element is Gelfand (adequate), is an elementary divisor ring.Keywords: Bezout ring, adequate ring, elementary divisor ring, semihereditary ring, provided by Gillman and Henriksen [2]. In [13], it was proved that a commutative Bezout ring is Hermite iff it is a ring of stable range 2.A von Neumann regular ring is a ring R such that for every a ∈ R there exists an x ∈ R such that a = axa. A ring R is said to be semiregular if R/J(R) is regular and idempotents lift modulo J(R), where J(R) denotes the Jacobson radical of R [9].An element a ∈ R is called clean if a can be written as the sum of a unit and an idempotent. If each element of R is clean, then we say R is a clean ring [9].A clean ring is a Gelfand ring. Recall that a ring is called a Gelfand ring if wheneverA ring is called a P M-ring if every prime ideal is contained in a unique maximal ideal [1].It had been asserted that a commutative ring is a Gelfand ring iff it is a P M-ringA ring is called a P M * -ring if every nonzero prime ideal is contained in a unique maximal ideal [14].An element a of the ring R is called an atom if from decomposition a = b · c follows that b or c is a unit of R.The group of units of a ring R will be denoted by U(R), the Jacobson radical of a ring R will be denoted by J(R). For a ring R, spec(R) (mspec(R)) denotes the collection of prime (maximal) ideals of R, and spec(a) = {P ∈ spec(R)|a ∈ P }, mspec(a) = {M ∈ mspec(R)|a ∈ M}.We start with trivial statements that are of a technical nature. Proposition 2. Let R be a Bezout ring. Then element a ∈ R is an atom iff the factor ring R/aR is a field.Proof. Let a ∈ R be an atom. Denoteb = b + aR for some b ∈ R andR = R/aR. Letb =0, then b / ∈ aR. Since R is a Bezout ring, we obtain aR + bR = dR and a = a 0 d, b = b 0 d, au + bv = d for some d, a 0 , b 0 , u, v ∈ R. Since a is an atom, we have d ∈ U(R) or a 0 ∈ U(R). If d ∈ U(R), we haved ∈ U(R) and thenb ∈ U(R), wherē d = d + aR. If a 0 ∈ U(R), we have b = aa −1 0 b 0 ∈ aR and this is impossible, sinceb =0. So we proved thatR is a field.IfR is a field, then aR ∈ mspec(R) and, obviosly, a is an atom in R. Proposition is proved. Now let R be a ring and let a ∈ R \ 0. Let us denoteR = R/aR andb = b + aR. Proposition 3. Let R be a ring. Thenb ∈ U(R) iff aR + bR = R. Proof. Letb ∈ U(R), thenbv =1, wherev = v + aR, and bv − 1 ∈ aR i.e. aR + bR = R.If aR + bR = R, then au + bv = 1 for some u, v ∈ R andbv =1. Proposition is proved.Regarding the concept of a stable range of the ring, one can distinguish a stable range of 1 and 2.A ring R is said to have stable range 1 if for all a, b ∈ R such that aR + bR = R, there exists y ∈ R such that (a + by)R = R.A ring R is said to have stable range 2 if for all a, b, c ∈ R such that aR+bR+cR = R, there exists x, y ∈ R such that (a + cx)R + (b + cy)R = R.Proposition 4. Let R be a ring of stable range 2 and let aR + bR = dR. Then there exist elements a 0 , b 0 ∈ R such that a = a 0 d, b = b 0 d and a 0 R + b 0 R = R.
