PIF-Property in Pullbacks

Transcription

1 International Mathematical Forum, 5, 2010, no. 33, PIF-Property in Pullbacks Chahrazade Bakkari Department of Mathematics, Faculty of Science and Technology of Fez Box 2202, University S. M. Ben Abdellah Fez, Morocco Abstract In this paper we introduce and investigate a class of those rings in which every projective ideal is free. We establish the transfer of this notion to pulbacks and localizations and then generate new and original families of rings satisfying this property. Mathematics Subject Classification: 13D05, 13D07, 16E05, 16E10, 16E30 Keywords: PIF-ring, projective ideal, free ideal, pullbacks, localization 1 Introduction Throughout this paper, all rings are commutative with identity elements, and all modules are unitary. A ring R is called a PIF-ring if every projective ideal is free. A local rings and Bézout domains are examples of PIF-rings. Also, every von Neumann regular ring which is not a field is example of non PIF-rings. This article determined when the classical D + M construction (in which the ambient domain K + M is a valuation domain) produces a PIF-rings. Our main result, Theorem 2.1, studies the transfer of PIF-property between a general D + M construction and the associated ring D. See for instance [1, 5, 6, 7]. Our aim in this paper is to investigate the possible transfer of the PIFproperty to pullbacks ((D + M)-constructions) and to localizations. It is convenient to use local to refer to (not necessarily Noetherian) ring with a unique maximal ideal. Also, unadorned tensor products are generally taken over the implicit base ring.

2 1626 Chahrazade Bakkari 2 Main Results This section is devoted to the transfer of PIF-property to pullbacks and localizations. In the following main result (Theorem 2.1), we adopt the following riding assumptions and notations: T is a domain of the form T = K + M, where K is a field and M is a nonzero maximal ideal of T ; D is a subring of K such that qf(d) =K; and R = D + M (For more details, see [1, 5, 6, 7]. Theorem 2.1 Let T and R be as above and assume that T is a PIF-ring. Then R is a PIF-ring if and only if so is D. Proof. Assume that R is a PIF-ring and let I be a proper projective ideal of D. Hence, J := I D R(= I + M) is a projective ideal of R and so J = R(a + m) for some a D and m M since R is a PIF-ring. We claim that a 0. Deny. Then I + M(= J) =Rm M and so I =0,a contradiction. Therefore, J = I + M = R(a + m) =Da + M (since am = M for each a K) and so I = Da, as desired. This means that D is a PIF-ring. Conversely, assume that D and T are PIF-rings and let J be a proper projective ideal of R. Then J := n i=1 R(a i + m i ) for some non negative integer n, a i D and m i M since J is a projective ideal of the domain R. Two cases are possible: Case 1: a i 0 for some i =1,...,n. In this case, J = n i=1 Da i + M (since am = M for each a K). I := n i=1 Da i. We claim that I is a projective ideal of D. Indeed, for any D-module N, we have by [2, p.118], Set Ext D (I,N D R) = Ext R (I D R, N D R)=0 On the other hand, N is a direct summand of N D R since D is a direct summand of R. Therefore, Ext D (I,N) = 0 for every D-module N. This means that I is a projective ideal of D. Therefore, I = Da for some a( 0) I since D is a P -ring and so J = Ra, as desired. Case 2: a i = 0 for each i =1,...,n. In this case, J := n i=1 Rm i ( M). Hence, JT = n i=1 Tm i (= J R T )isa projective ideal of T (since T is R-flat) and so JT = Tm for some m M (since T is a P -ring). But Mm = JTM = JM J JT = Tm = Km+Mm.

3 PIF-property in pullbacks 1627 Hence, J := I + Mm for some D-submodule I of Km. We claim that I 0. Deny. Then J = Mm and so Tm = JT = TMm = Mm which means that T = M (since T is a domain), a contradiction. Hence, I 0. Hence, for each i =1,...,n, we can write m i =(d i /d)m +(n i /d)m, where d i D, d D {0}, n i M, (d i /d)m I( Km) and (n i /d)m Mm. Therefore, J = n i=1 Rm i = (m/d)[ n i=1 R(d i + n i )] = (m/d)[ n i=1 Rd i ] = (m/d)[ n i=1 Dd i + M] since d i 0 for some i = 1,...,n (since 0 I = ni=1 D(d i /d)m). But J(= (m/d)[ n i=1 Dd i D]) is a projective ideal of R; hence ni=1 Dd i + M is a projective ideal of R and so n i=1 Dd i is a projective ideal of D as in the case 1. Therefore, n i=1 Dd i = Da for some a D {0} since D is a P -ring and so J =(m/d)(da + M) =(m/d)ra = R(am/d) is a principal ideal and this completes the proof of Theorem 2.1. Recall that a ring R is called coherent if every finitely generated ideal of R is finitely presented. Examples of coherent rings are Noetherian rings, Boolean algebras, von Neumann regular rings, valuation rings, and Bézout domains. See for instance [6]. Theorem 2.1 enriches the literature with new examples of non-local PIFdomains and quite far from being Bézout domains, as shown below. Example 2.2 Let Z be the ring of integers, Q = qf(z) and let R be the real numbers. Let T := Q + XR[[X]] be a local domain, where XR[[X]] is its maximal ideal, and let S := Z + XR[[X]].Then: 1) S is a PIF-ring by Theorem ) S is not local by [1, Theorem 25.1(3)] since Z is not local. 3) S is not coherent by [6, Theorem 2.8(1)]. In particular R is not Noetherian. Now, we show that the localization of a PIF-ring is not in general a PIFring. For this, we need the notion of an amalgamated duplication of a ring R along an ideal. The amalgamated duplication of a ring A along an ideal I, introduced by D Anna and Fontana and denoted by A I(see for instance [3, 4]), is the following subring of A A (endowed with the usual componentwise operation): A I= {(a, a + i)/a A, i I}. Now, we are able to show that the localization of a PIF-ring is not in general a PIF-ring.

4 1628 Chahrazade Bakkari Proposition 2.3 Let A be a local ring, I be a regular proper ideal of A and let R := A I be the amalgamated duplication of a ring A along I. Then: 1) R(= A I is a local ring. In particular, R is a PIF-ring. 2) Let S be the set of regular elements of R (which is a multiplicative set of R). Then S 1 R(:= Q(R), the total ring of quotient of R) is not a PIF-ring. Proof. 1) R is a local ring by [4, Theorem 3.5(a)(v)] since A is a local ring. In particular, R is a PIF-ring. 2) By [4, Corollary 3.3(d)], we have that Q(R)(:= S 1 (A I)) = Q(A) Q(A) and it remains to show that Q(A) Q(A) is not a PIF-ring. The ideal J := Q(A) 0 is projective since: (Q(A) 0) (0 Q(A)) = Q(A) Q(A) =Q(R). But J is not free since (0, 1 Q(A) )J =(0, 1 Q(A) )(Q(A) 0) = (0, 0), as desired. Corollary 2.4 Let A be a local domain, K := qf(a), I be a proper ideal of A and let R := A I be the amalgamated duplication of a ring A along I. Then: 1) R is a local ring. In particular, R is a PIF-ring. 2) Q(R) =K K and is not a PIF-ring. Also, the converse is not always true as the following example shows. Example 2.5 Let R be a von Neumann regular ring which is not a field. Then: 1) R is not a PIF-ring. 2) R M is a PIF-ring for each maximal ideal M of R (since R M is a field). References [1] E. Bastida and R. Gilmer; Overrings and divisorial ideals of rings of the form D + M, Michigan Math. J. 20, 79-95, (1973). [2] H. Cartan and S, Eilenberg; Homological algebra, Princeton University press, (1956).

5 PIF-property in pullbacks 1629 [3] M. D Anna and M. Fontana, The amalgamated duplication of a ring along a multiplicative-canonical ideal, Ark. Mat. 45(2007), no. 2, [4] M. D Anna and M. Fontana, An amalgamated duplication of a ring along an ideal: the basic properties, J. Algebra Appl. 6 (2007), no. 3, [5] D.E. Dobbs and I. Papick; When is D + M coherent?, Proc. Amer. Math. Soc. 56, 51-54, (1976). [6] S. Glaz; Commutative Coherent Rings, Springer-Verlag, Lecture Notes in Mathematics, 1371, (1989). [7] L. Gruson and M Raynaud, Criteres de platitude et de projectivite. Techniques de platification d un module, Inv. Math. 13, 1-89, (1971). Received: January, 2010