Dirceu BAGIO and Antonio PAQUES Abstract. We deal with the following variant of the primitive element theorem: any commutative strongly separable extension of a commu- tative ring can be embedded in another one having primitive element. This statement holds for connected strongly separable extension of com- mutative rings which are either local or connected semilocal. We show that it holds for a more general family of rings, that is, for connected commutative rings whose quotient ring by the corresponding Jacobson radical is von Neumann regular and locally uniform. Some properties of the (connected) separable closure of such rings are also given as an application of this result.
I NTRODUCTION
Throughout this paper by ring we mean a commutative ring with identity element.  By a connected ring we mean a ring whose unique idempotents are 0 and 1. Let SニR be a ring extension. We say that S has a primitive element over R if there exists α ∈ S such that S = R [ α ]. The existence of primitive elements for strongly separable extensions has been extensively studied by several authors (see, for instance, [1, 9, 10, 11, 12, 20, 22]).  It holds for fields and, more generally, for rings with many units [15] under certain restrictive conditions on the cardinality of their residue fields (see [20]). For instance, any strongly separable extension S of a semilocal ring R , with constant rank over R , has a primitive element over R if and only if | R/m | ≥ rank RS , for every maximal ideal m of R. Our aim in this paper is concerned with a variant of the primitive element theorem. Indeed, we are interested in the following question: the assertion every strongly separable extension S of a ring R can be embedded (*) into another one having primitive element holds without any restriction on the cardinality of the residue fields of R ? This question has been affirmatively answered for connected strongly sep- arable extension of R in the case that R is either local [18] or connected semilocal [2]. We prove in Section 2 that the assertion (*) is also true in a more general situation, that is, for connected strongly separable extensions of a connected
D. BAGIO AND A. PAQUES
ring R whose quotient ring by its Jacobson radical is von Neumann regular and locally uniform. As an application of this main result we present in Section 3 some interesting properties of the (connected) separable closure of such a ring. The notion of locally uniform ring, which we introduce in Section 1, is a slight extension of the notion of uniform ring as considered in [4].
P RELIMINAIRES
1.
In all of this paper we will be employing freely the ideas and results of [23] on boolean spectrum and boolean localization of a ring (see also [14, 21]). We begin by introducing the terminology we will need. For any ring R let B(R) denotes the boolean ring of all idempotents of R and Spec(B(R)) denotes the boolean spectrum of R consisting of all prime (equivalently maximal) ideals of B ( R ).  A base for a topology on Spec(B(R)) is given by the family of basic open sets {U e | e ∈ B(R)}, where U e = {x ∈ Spec(B(R)) | 1 − e ∈ x} . This base defines a compact, totally disconnected, Hausdorff topology on Spec(B(R)). By localization of R at x , for each x ∈ Spec(B(R)), we mean the quotient ring R x = R/I(x) where I(x) denotes the ideal of R generated by the elements of x .  By [23, 2.13] R x is a connected ring.  For any R-module M , M x = M S R R x = M/I(x)M . For any element a ∈ M , a x denotes the image of a in M x . For every R -module homomorphism f : M → N , the corresponding induced R x -homomorphism f x : M x → N x is given by fx = f ∅ Rx . We say that a ring R is locally uniform if for each x ∈ Spec(B(R)) and each finite subset F of R there exist an idempotent e = e(x,F) ∈ R and a collection of ring isomorphisms φ y : R y → R x such that x ∈ U e and φιν (αι) = α x , for every α ∈ F and γ ∈ U e .  Üniform rings as introduced in [4] are locally uniform but the converse is not true as it will be shown in the following first example.  We denote by J(R) the Jacobson radical of the ring R. Example 1.1   Let R be a semilocal ring with at least two maximal ideals such that the corresponding residue fields are not isomorphic. Put R ′ = R/J(R). Note that Spec(B(R')) is finite. Thus, in order to verify that R' is locally uniform, given any x ∈ Spec(B(R ′ )) and any finite subset F of R' it is enough to take e = e(x, F) ∈ B(R') such that U e = {x} and the identity isomorphism id x : R ′ x → R ′ x . On the other hand R ′ x is a field. So, I(x) = m/J(R) for some maximal ideal m of R and Rf = R/m. Therefore, it follows from the assumption on the maximal ideals of R that R ′ is not uniform. We observe that the semilocal ring given in Example 1.1 is a particular example of rings R such that R/J(R) is von Neumann regular and locally uniform. In the following example we give a way to construct (connected) rings with this same property and with infinitely many maximal ideals. Example 1.2  Let p ∈ Z be a prime integer and R 0 = Z ( p ) be the localization of Z at pZ . Following Hasse [8] there exists a quadratic extension K 1 of the rational number field K 0 = Q such that pR1 = q1q2, where R1 denotes the integral closure of R 0 in K 1 and q 1 and q 2 are the unique maximal ideals of R1 over q0 = p2(p). Again by the same result due to Hasse there exists a quadratic extension K 2 of K 1 such that q i R 2 = q i 1q i 2 where R 2 denotes the integral closure of R 1 in K 2 and q i1 and q i2 are the unique maximal ideals of R 2 over q i , i = 1,2.  Applying this same argument successively we will get a tower of rings R 0 ⊆ R 1 ⊆ R 2 ⊆ ⋅⋅ ⊆ R , where R = ∪ j ≥ 0 R j is the integral closure of R 0 in K = U i>0 K j . From the construction of R it is easy to see that: (i) R has infinitely many maximal ideals and all of them are over qo; (ii) R/pR is a ring of Krull dimension zero and (iii) R/p ~ Z/pZ for every prime ideal p of R. It is a consequence of the assertions (i) and (ii) that R' = R/J(R) ~ (R/pR)/(J(R)/pR) = (R/pR)/J(R/pR) is von Neumann regular by [7, Lemma 1].  And it follows from assertion (iii) that R ′ z ≥ Z /pZ for all z ∈ Spec(B(R')). Put R'z = {0 z , 1 z , . . . , (p − 1) z } . Now take x ∈ Spec(B(R')) and F = {α 1 , . . . ,α n } a finite set of elements of R ′ . Assume that ( aj ) x = ( ij ) x , with 0 ≤ ij ≤ p − 1 and 1 ≤ j ≤ n . Thus, for each j there is an idempotent e j ∈ R ′ such that x ∈ U e j and ( aj )y = ( ij )y for every y ∈ U e j [23, 2.9]. Let U e = ∩ 1≤j≤n U e j , with e = ∩ 1≤j≤n e j , and φy : R ′ y → R ′ x be such that φy ( iy ) = ix , for every y ∈ U e . Clearly φy is a ring isomorphism and φy((αj)y) = φy((ij)y) = (ij)x = (αj)x, for all y ∈ Ue and 1 ≤ j ≤ n. Therefore R ′ is locally uniform.
2. THE MAIN THEOREM
Let R ⊆ S be a ring extension. We say that S is a strongly separable extension of R if S is separable as R-algebra and finitely generated and projective as R -module. If for any finite subset N ⊂ S there exists a subal- gebra L of S which contains N and is a strongly separable extension of R, we say that S is a locally strongly separable extension of R. We say that a connected ring is separably closed if its unique connected strongly separable extension is itself. We will denote by Ω( R ), up to isomorphism, the (con- nected) separable closure of a connected ring R , that is, Ω( R ) is a locally strongly separable extension of R which is connected and separably closed.
D. BAGIO AND A. PAQUES
For more about the (connected) separable closure of a connected ring we refer to [9, 14]. Theorem 2.1 below provides a generalization of the primitive element theorem and it was already stated for local rings [18] and for connected semilocal rings [2]. In this paper we extend it to the setting of the connected rings R such that R/J(R) is von Neumann regular and locally uniform. A polynomial f(X) ∈ R[X] is said to be separable over R if it is monic and R[X]/(f(X)) is a separable R-algebra.  A monic polynomial f(X) ∈ R [ X ] is defined to be indecomposable in R [ X ] if whenever there exist monic polynomials g(X), h(X) ∈ R[X] such that f(X) = g(X)h(X) it follows that g(X) = 1 or h(X) = 1. Theorem 2.1  Let R be a connected ring and S ⊆ Ω(R) be a strongly separable extension of R.  Assume that R/J(R) is von Neumann regular and locally uniform. Then there exist a polynomial f(X) ∈ R[X] and α ∈ Ω(R) such that: (i) f(X) is separable and indecomposable, (ii) f(α) = 0 and R[α] ~ R[X]/(f(X)). (iii) Ｓ ⊆ R [α] . Proof. Let R' = R/J(R), S' = S/J(S). Note that R'x is a field for every x ∈ Spec(B(R')) . Let Y = { x ∈ Spec(B(R'))||R c | < ∞}. The proof will be divided in two parts. Firstly assume that Y = 0.  Then S r has primitive element over R r for all x ∈ Spec(B(R')) [9, Lemma 3.1].  Let α'(x) ∈ S' be such that S ′ x = R ′ x [ α ′ ( x ) x ] = R ′ [ α ′ ( x )] x . So, for each x ∈ Spec ( B ′ )) there exists an idempotent e(x) ∈ R' such that x ∈ Ue(x) and S'e(x) = R'[c'(x)]e(x) [23, 2.8 and 2.11].  Applying compactness arguments we obtain elements α ′ 1 , . . . , α ′ , ∈ S ′ and orthogonal idempotents e 1 , . . . , e r ∈ R ′ such that ∑ 1 ≤ i ≤ r e i = 1 and S'e i = R'[α']e i . Then for α' = ∑ 1<i<r α'ie i we have S' =̃R'[α'] and by Nakayama's lemma S = R [ α ] with α ∈ S such that α ′ = α + J ( S ). Finally by [16, Theorem 3.3] there exists a separable and indecomposable polynomial f(X) ∈ R[X] such that f(α) = 0 and S ～ R[X]/(f(X)) . Now consider Y̸= ∅. Note that S is free as R-module [6, Theorem 2.10] of constant rank n say. Let p ∈ Z be a prime integer not divisor of n . From now on we will proceed by steps. Claim 1. For each x ∈ Spec(B(R')) there exist an idempotent e(x) in R' and a monic polynomial g(X) ∈ R ′ [X] of degree p such that: (i) g(X)z is separable over R z for all z ∈ Ue(x), (ii) g(X) z is separable and indecomposable in R ′ z [ X ] for all z ∈ Ue (x), whenever x ∈ Y , (iii) Ue(x) ∩ Ue(y) = ∅ whenever x̸∈ Y and y ∈ Y . Clearly there exists a separable polynomial of degree p in R x [ X ] for every x ∈ Spec(B(R')).  And we may assume that it is also indecomposable if x ∈ Y because, in this case, R r is a finite field. Take g(X) = a 0 + a 1 X + · · · + a p − 1 X p − 1 + X p ∈ R ′ [X] such that g(X) x ∈ R x [ X ] is such a polynomial. Consequently the discriminant d( g ( X ) x ) = d(g(X)) x of g(X) x is a unit in R x . So, (d(g(X)) λ ) x = 1 x for some λ ∈ R ′ . By [23, 2.9] there exists an idempotent e 1 ∈ R ′ such that x ∈ U e 1 and (d(g(X)))e1 = e1. On the other hand, there exist by assumption an idempotent e 2 ∈ R ′ and rings isomorphisms φ z : R ′ z → R z such that x ∈ U e z and φ z (( α j ) z ) = ( α j ) x for all z ∈ Ue 2 and 0 ≤ j ≤ p − 1. It is enough to take Ue(x) = Ue1 ∩ Ue2  with e(x) = e1e2. Claim 2.  There exists a monic polynomial t(X) ∈ R[X] of degree p, which is separable over R and indecomposable in S [ X ]. By Claim 1 and the usual compactness argument we can insure that there are pairwise orthogonal idempotents e1 = e(x1),...,er = e(xr) in R' and monic polynomials of degree p g1(X),...,gr(X) in R'[X] such that E 1<i<r ei = 1 and e igi (X) is separable over eiR' for all 1 ≤ i ≤ r . And, in addition, e i g i ( X ) is indecomposable over e i R ′ for those i such that x i ∈ Y . Let g(X) = ∑ 1< i<r e i g i (X) and t(X) ∈ R [X] be monic and such that g(X) = t(X) modulo J(R)[X].  By construction t(X) is of degree p and separable over R. In order to verify that t ( X ) is indecomposable in S [ X ] take y ∈ Y . Note that g(X)y = gi(X)y is indecomposable in Riy[X] for some i such that y ∈ Ue,.  Furthermore, R', = R'/I(y) is a field then I(y) = m/J(R) for some maximal ideal m of Ř and R ly ~ R/m. So t ( X ) is indecomposable modulo m[X].  Let M be a maximal ideal of S over m.  Since rank R/mS/M = rankRS = n and by assumption p does not divide n then the claim follows. From now on let q = min{|R'z||x ∈ Y} and assume that the prime integer p above considered also satisfies ☞= q ≥ n. By [9, Theorem 1.1] we can also assume that S is a Galois extension of R in the sense of [3]. Set T = S[X]/(t(X)). It easily follows from the properties of t(X) that T is a connected strongly separable extension of R. Claim 3. T has a primitive element over R.
D. BAGIO AND A. PAQUES
By Nakayama's lemma and boolean localization it is enough to show that T x has a primitive element over R x for every x ∈ Spec(B(R)), where T' = T/J(R)T = T/J(T). It is clear that T x has a primitive element over R x if x F Y . So let x ∈ Y and M 1 , . . . , M s be the maximal ideals of S over m, where m is such that R'x ~ R/m. Recall that t(X) = g(X) modulo J(R)[X] as constructed above. By the same arguments used in Claim 2 t(X) is indecomposable in S/M j [X], for all 1 ≤ j ≤ s. By [16, Theorem 3.5] there is a unique maximal ideal M ′ , of T over Mj and | T/Mj : S/Mj | = p, for each 1 ≤ j ≤ s. On the other hand, we have S/Mj ~ S/M1 since as assumed above S is a Galois extension of R. Therefore, [T/M', : R/m] = p[S/Mj : R/m] = p [S/M1 : R/m]. Let F g denote the finite Galois field with q elements and N g ( m ) the num- ber of all monic and indecomposable polynomials of degree m in F q [ X ]. By [13, Theorem 3.25] we have Nq(p) = q =q.  Hence, if | R r | = q x then Nqz (p) = qz-qz > qz-q = Nq(p). If [ S/M 1 : R/m] = 1 then Nqz (p [ S/M 1 : R/m]) = Nqz (p) ≥ Nq (p) ≥ n. If [S/M1 : R/m] ≥ 2 then by [10, Lemma 1.2] we have Nqz (p [S/M1 : R/m]) ≥ Nqz (p) Nqz ([S/M1 : R/m]) ≥ Nqz (p) ≥ Nq (p) ≥ n. Thus N qz (p [S/M1 : R/m]) ≥ n ≥ s and there exist distinct separable and indecomposable polynomials h 1 (X),...,h s (X) of degree p [S/M 1 : R/m] in R/m X . By [16, Theorem 2.1] mT = ∩ 1<j<s M ′j.  Also T/M ′j = R/m[X]/(h j (X)) for all 1 ≤ j ≤ s.  So by chinese remainder theorem we have T/mT = ⊕ 1<j<s T/M ′ j ~ R/m[X]/(h(X)), with h(X) = h 1 (X) ··· h s(X). Finally, by observing that T x = T ′ / I ( x ) T ′ = (T ′ J ( T ))/(mT ′ J ( T )) ~ T/mT the claim follows. The conclusion of the proof of Theorem 2.1 follows now from Claim 3 and [16, Theorem 3.3]. Remark 2.2  In Theorem 2.1 the ring extension T = R[α]ニS also satisfies rankRT = p f rankRS , with p a prime integer and either c = 0 or c = 1, in the following two cases: R is semilocal [2, Theorem 2.1.1] or S is a Galois extension of R. Remark  2.3 Corollaries 1.2 and 1.4 of [18] have natural corresponding extensions, with similar proofs, to the setting of connected rings R such that R/J(R) is von Neumann regular and locally uniform. 3. More about separable closures For any ring R we will denote by Max(R) the set of all maximal ideals of R. Our purpose in this section is twofold. Firstly we will give necessary and sufficient conditions in order to Max(R) and Max(Ω(R)) have the same cardinality (Theorem 3.1). Secondly we will present an interesting charac- terization of Aut R (Ω( R )) (Corollary 3.6).  We will do that in the setting of connected rings R such that R/J(R) is von Neumann regular and lo- cally uniform. In particular, our first result is an improved and generalized version of Theorem 1.5 of [17]. Given a ring extension SニR and a maximal ideal M of S we de- note by D(M) the decomposition group of M, that is, D(M) = {σ ∈ Aut R ( S ) | σ ( M ) = M } . For S integral over R let ψ denotes the contrac- tion map from Max(S) onto Max(R), that is, ψ(M) = M ∩ R for all M ∈ Max(S). Theorem 3.1   Let R be a connected ring such that R/J(R) is von Neumann regular and locally uniform. Then the following statements are equivalent: (i) The map ψ : Max(Ω(R)) → Max(R) is bijective. (ii) D(M) = Aut R (Ω(R)), for every M ∈ Max (Ω(R)). (iii) If f (X) is separable and indecomposable in R [ X ] then f (X) = f (X)+ mR[X] is separable and indecomposable in R/m[X] for every m ∈ Max(R). Proof. (i)⇒(ii)  Take M ∈ Max(Ω(R)) and σ ∈ Aut R (Ω(R)). Then σ(M) ∩ R = σ(M ∩ R) = M ∩ R and consequently σ(M) = M. (ii)⇒(i) If M1,M2 ∈ Max(Ω(R)) satisfy M1 ∩ R = M2 ∩ R then there exists σ ∈ Aut R (Ω(R)) such that σ (M1) = M2 [16, Lemma 2.2]. The result follows by the assumption. (i)⇒(iii)  Let f(X) ∈ R[X] be separable and indecomposable. The separa- bility of f ( X ) over R/m is clear, for all m ∈ Max ( R ). Put T = R [ X ]/( f ( X )). Clearly T is a connected strongly separable extension of R and by [5, The- orem III.3.3] we may assume that T is contained in Ω( R ).  Thus it fol- lows by the assumption that for each m ∈ Max(R) there exists a unique M ∈ Max ( T ) such that M ∩ R = m. Consequently f ( X ) is indecomposable in R/m[X] for all m ∈ Max(R) by [16, Theorem 3.5]. (iii)⇒(i) Let M1,M2 ∈ Max(Ω(R)) such that M1 ∩ R = M2 ∩ R = m. By [16, Theorem 2.1] we have mΩ(R) ⊆ M 1 ∩ M 2 . Thus, if mΩ(R) = M 1 the result follows. Assume that mΩ(R) ∑ M 1 . Then there exist z ∈ M 1 \mΩ(R) and a strongly separable extension S of R such that z ∈ S ⊆ Ω(R). Note that
D. BAGIO AND A. PAQUES
z ∈ (S ∩ M 1 ) \ mS , so mS ⊆ S ∩ M 1 . Since S ∩ M 1 is a maximal ideal of S over m it follows from [16, Theorem 2.1] that S contains at least two maximal ideals over m.  On the other hand R/J(R) is von Neumann regular and locally uniform, so there exist α ∈ Ω( R ) and a separable and indecomposable polynomial f(X) ∈ R[X] such that f(α) = 0 and S ⊆ R[α] ～ R[X]/(f(X)) by Theorem 2.1. Consequently it follows from [16, Theorem 3.5] that R[α] contains a unique maximal ideal over m, which is a contradiction. For our second result mentioned above we need some preparation and we start with the following lemmas. Let R be a von Neumann regular ring. Then every locally Lemma 3.2 strongly separable extension of R is von Neumann regular. Proof.  Let T be a locally strongly separable extension of R. Take a ∈ T. Thus T contains a strongly separable extension S of R such that a ∈ S . Note that if S is von Neumann regular then a = a²b for some b ∈ S and consequently T is also von Neumann regular. Hence, it is enough to prove that S is von Neumann regular. For every maximal ideal m of R, S m is a separable extension of R m and R m is a field. So S m is a finite direct sum of fields and consequently a von Neumann regular ring. The required follows from [7, Lemma 1]. ■ Lemma 3.3 Let R be a connected ring, I ⊆ R an ideal and T a locally strongly separable extension of R. Then IT ( ) R = I. Proof. Clearly I ⊆ IT ∩ R.  Now take c ∈ IT ∩ R.  Then c ∈ R and c = ∑ 1≤i≤n α i b i ∈ R with α i ∈ I and b i ∈ T . Consider S ⊆ T a strongly separable extension of R containing bi , 1 ≤ i ≤ n. So c ∈ IS ∩ R. Since R is a direct summand of S [5, Corollary III.2.3] we have IS = I ⊕ I N for some R-module N and c = a + b with a ∈ I and b ∈ IN .  Consequently ■ b = c − a ∈ R ∩ N = 0 and c ∈ I . Lemma 3.4 Let R be a connected ring, I ⊆ R an ideal such that R/I is von Neumann regular. Then Ω(R)/IΩ(R) also is von Neumann regular. Proof. By Lemma 3.2 it is enough to prove that Ω(R)/IΩ(R) is a lo- cally strongly separable extension of R/I . It follows from Lemma 3.3 that Ω(R)/IΩ(R) is an extension of R/I.  Let a 1 + IΩ(R),...,α n + IΩ(R) ∈ Ω(R)/IΩ(R). Then there exists a strongly separable extension S of R such that α 1 , . . . , α n ∈ S ⊆ Ω( R ). Clearly S/IS is a strongly separable extension of R/I.  On the other hand, Ω(R) is a locally strongly separable exten- sion of S [19, Proposition 2] and so IΩ(R) ∩ S = (IS)Ω(R) ∩ S = IS by Lemma 3.3. Therefore Ω( R )/IΩ( R ) is an extension of S/IS and the result follows. Theorem 3.5   Let R be a connected ring such that R/J(R) is von Neu- mann regular and m ∈ Max(R).  Then Ω(R/m) = Ω(R)/M for all M ∈ Max(Ω(R)) such that M ( R = m. Proof.  Let M ∈ Max(Ω(R)) such that M ∩ R = m. Claim 1. Ω(R)/M is a locally strongly separable extension of R/m. Take a 1 + M,...,an + M ∈ Ω(R)/M.  Then a 1 ,...,an ∈ S for some strongly separable extension S of R contained in Ω(R). Note that Ω(R) is integral over S , so M ∩ S is a maximal ideal of S over m. Moreover, S m is a strongly separable extension of R m and consequently a semilocal ring whose maximal ideals are in bijective correspondence with the maximal ideals of S over m. Hence there is only finitely many maximal ideals of S over m and S/MNS is a direct summand of S/mS by [16, Theorem 2.1] and chinese remainder theorem. Therefore S/M ∩ S is a strongly separable extension of R/m and the claim follows.
Claim 2.   Ω(R)/M is separably closed.
Put R' = R/J(R), Ω(R)' = Ω(R)/J(R) and M' = M/J(R)Ω(R). By Lemma 3.4 Ω( R ) ′ is von Neumann regular, so M ′ = I( x ) for some x ∈ Spec(B(Ω(R)′)). Therefore Ω(R)′ x = Ω(R)′ / M′ = Ω(R) / M. Let T be a connected and strongly separable extension of Ω( R )/ M. Then T ~ Ω(R)'/M'[X]/(f(X)) for some separable and indecomposable polyno- mial f(X) ∈ Ω(R)′/Ω(′[X]. Let g1(X) ∈ Ω(R) ′ [X] be a monic polynomial such that g1(X) x = f(X) . It follows from the separability of g1(X)x and from [23, 2.9] that there exists an idempotent e 1 ∈ Ω(R) ′ such that x ∈ U e 1 and e 1 g 1 (X) is separable over e 1 Ω( R) ′ . Put Y = Spec(B(Ω(R)′))\U e 1 and take g(X) ∈ Ω(R)′[X] a monic polyno- mial such that deg(g(X)) = deg(f(X)) and g(X)y ∈ Ω(R')y[X] is separable, for each y ∈ Y . Since Y is an open set, it follows again from the separability of g(X)y and from [23, 2.9] that there exists for each y ∈ Y an idempo- tent e(y) ∈ Ω(R) ′ such that y ∈ Ue(y) ⊆ Y and e(y)g(X) is separable over e(y)Ω(R)′. Now by compactness arguments we get pairwise orthogonal idempotents e2, . . . , en ∈ Ω(R) ′ and polynomials g2(X), . . . , gn(X) ∈ Ω(R) ′ [X] such that e1 + e2 + · · · + en = 1, deg(gi(X)) = deg(f(X)) and eigi(X) is separable over eiΩ(R)′, for all 2 ≤ i ≤ n. Consequently g(X) = e1g1(X) + e2g2(X) + · · · +
D. BAGIO AND A. PAQUES
engn(X) is separable over Ω(R) ′ and deg(g(X)) = deg(f(X)). Furthermore, e 1g1 (X) (and, consequently, also g(X)) is indecomposable in Ω(R) ′ [X]. Let h(X) ∈ Ω(R)[X] a monic polynomial such that g(X) = h(X) mod- ulo J(R)Ω(R). By construction h(X) is separable and indecomposable in Ω(R)[X]. So Ω(R)[X]/( h (X)) is a connected strongly separable extension of Ω(R).  Hence deg(f(X)) = deg(h(X)) = 1 and T = Ω(R)/M.  The proof is complete.口Corollary 3.6  Let R be a connected ring such that R/J(R) is von Neu- mann regular, m ∈ Max(R) and M ∈ Max(Ω(R)) satisfying M ∩ R = m. Then D(M) ~ Aut R/m(Ω(R/m)). If in addition R/J(R) is locally uniform then Aut R (Ω(R)) ~ Aut R /m(Ω(R/m)), for all m ∈ M αx (R). Proof.  It follows from [16, Theorem 2.7] and Theorems 3.1 and 3.5. ■
A CKNOWLEDGEMENT
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