SOOCHOW JOURNAL OF MATHEMATICS Volume 32, No. 2, pp. 241-243, April 2006 DIVISIBLE IS INJECTIVE BY SHALOM FEIGELSTOCK Abstract. It is well known that injective modules are divisible. A result attributed to Cartan and Eilenberg states that if R is a commutative integral domain then a torsion free R-module is injective if and only if it is divisible. In this note the same result is obtained for a larger class of rings, namely rings R which embed in a ring S in which every element of R is invertible. R is a ring with unity, and is a subring of a ring S: All modules are unital left R-modules. The injective hull of a module M will be denoted by I(M). Definition. An R-module M is R-torsion free if rm = 0 for r 2 R and m 2 M, then either r = 0 or m = 0: Definition. An R-module M is R-divisible if rM = M for all 0 =6 r 2 R. Lemma 1. If every non-zero element of R is invertible in S then S is an injective R-module. Proof. Let 0 =6 x 2 I(S): There exists 0 =6 r 2 R such that rx 2 S; so x = r−1(rx) 2 S: Corollary 2. If S has no zero-divisors and if S is an R-divisible R-module then S is an injective R-module. Proof. Let 0 =6 r 2 R. There exists s 2 S such that rs = 1: Therefore srs = s and so (sr − 1)s = 0: Since S has no zero-divisors sr = 1; and S is injective by Lemma 1. Received October 19, 2004; revised December 14, 2004. AMS Subject Classification. 16D50. Key words. injective module, divisible module. 241 242 SHALOM FEIGELSTOCK Theorem 3. If every non-zero element of R is invertible in S then I(R) is a subring of S; and is a ring of quotients for R: Proof. Let x; y 2 I(R). There exists 0 =6 r 2 R such that rxy 2 I(R): Since I(R) is R-divisible, rxy = rz for some z 2 I(R): Elements of R are not zero-divisors in S so xy = z; i.e., I(R) is a subring of S. Let 0 =6 a 2 R. The map f : Ra ! I(R) defined by f(xa) = x for all x 2 R is an R-homomorphism, so it extends to an R-homomorphism g : R ! I(R) by [2] Proposition 3.17. Therefore 1 = f(a) = g(a1) = ag(1); and so I(R) is a ring of quotients for R: Lemma 4. If every non-zero element of R is invertible in S then every R-divisible R-torsion free module M is an I(R)-module. Proof. For x 2 I(R) there exists 0 =6 r 2 R such that rx 2 R: For m 2 M define xm to be the unique m0 in M satisfying rxm = rm0: Theorem 5. If every non-zero element of R is invertible in S then an R-torsion free module M is injective if and only if M is R-divisible. Proof. If M is injective then it is well-known that M is R-divisible, ([1], Proposition 1.2 or [2], (Corollary 3.17)'). Let M be R-divisible, and let 0 =6 x 2 I(M): There exists 0 =6 r 2 R such that rx 2 M. It follows from Theorem 3 and Lemma 4 that x = r−1(rx) 2 M: Corollary 6. Let R be a ring with unity. If R is either a commutative integral domain, a torsion free ring such that QR is a division ring, a Noetherian ring without zero-divisors, or if R is a uniform R-module, then a torsion free module M is injective if and only if M is R-divisible. Proof. If R is a Noetherian ring without zero-divisors, or a uniform R- module then Ra \ Rb =6 0 for all non-zero a; b 2 R; i.e., R satisfies the left Ore condition, and so R can be embedded in a division ring. The other cases are obvious. Since a direct sum of R-divisible modules is R-divisible, Theorem 5 yields: Corollary 7. If every non-zero element of R is invertible in S then a direct sum of R-torsion free injective modules is injective. DIVISIBLE IS INJECTIVE 243 References [1] H. Cartan and S. Eilenberg, Homological Algebra, Princeton, 1956. [2] T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Math., Springer-Verlag, Berlin-Heidelberg, New York, 1998. Department of Mathematics, Bar-Ilan University, Ramat-Gan, Israel. E-mail: [email protected].