 Advances in Pure Mathematics, 2012, 2, 39-40 http://dx.doi.org/10.4236/apm.2012.21009 Published Online January 2012 (http://www.SciRP.org/journal/apm) A Note on the (Faith-Menal) Counter Example R. H. Sallam Mathematics Department, Faculty of Science, Helwan University, Cairo, Egypt Email: rsallams@hotmail.com Received September 1, 2011; revised October 15, 2011; accepted October 22, 2011 ABSTRACT Faith-Menal counter example is an example (unique) of a right John’s ring which is not right Artinian. In this paper we show that the ring T which considered as an example of a right Johns ring in the (Faith-Menal) counter example is also Artinian. The conclusion is that the unique counter example that says a right John’s ring can not be right Artinian is false and the right Noetherian ring with the annihilator property rl(A) = A may be Artinian. Keywords: John’s Ring; Artinian and Noetherian Rings 1. Introduction A ring R is called right John’s ring if it is right Noethe-rian and every right Ideal A of R is a right annihilator i.e. rl(A) = A for all right ideals A of R. John’s (, Theorem 1) by using a result of Kurshan (, Theorem 3.3), showed that a right Noetherian ring is right Artinian provided that every right ideal is a right annihilator. Ginn  showed that Kurshan result was false. Ginn’s example does not provide a counter example to John’s theorem. Therefore the validity of John’s theorem was doubtful. Faith-Menal counter example proved that there is an example (unique) of a right John’s ring which is not right Artinian. Here in this paper we prove the false of the Faith- Menal counter example by proving that the considered non-Artinian right john’s ring is in fact right Artinian. So the John’s theorem may be true see . All rings considered in this paper are associative rings with identity. We recall the Faith-Menal counter example in Section 1 and we prove that it is false in Section 2. 2. Section 1: The Counter Example Example 8.16 (Faith-Menal) . Let D be any countable, extentially closed division ring over a field F, and let R = DFFTRD(x). Then T(R, D) is a non-Artinian right John’s ring. Proof Cohn shows that R is simple, principle right ideal do-main that is right V ring (Theorems 8.4.5 & 5.5.5 ) and D is an R-R bimodule such that DR is the unique simple right R module. Hence T(R, D) is a right John’s ring by Theorem 8.15 (in this book). But T(R, D) is not Artinian because if it were then R would also right Artinian and hence a field which is a contradiction. Here and T0,d R For more information about this example see . 3. Section 2: A Note on the Counter Example Theorem The right John’s ring T(R, D) defined in the counter ex-ample is Artinian. Proof Recall the following (Exercises 10.7 ): Let φ: D→R be a ring homomorphism and let M be a right R-module (or left R-module) then 1) Via φ M is right D-module; 2) If MD is Artinian or Noetherian then so is MR; 3) If R is finite dimensional algebra (via φ) over a field D then the following is equivalent: a) MR is Artinian and Noetherian b) MR is finitely generated c) MD is finite dimension 1) Consider the ring homomorphism φ: D → R defined by φ (d) = d1f. Every R-module homomorphism is a D-homomorphism via φ. Since D is a division ring so it will be semisimple ring and hence every right D-module is semisimple (Corollary 8.2.2 ). 2) If the ring TRD is John’s ring (as in the coun- ter example above) then it is Noetherian and hence R and D are Noetherian. As D is right Noetherian ring then every finitely generated right D module is Noetherian (6.1.3 ). 3) Every finitely generated right D-module M is semi- simple so it is right Artinian. Copyright © 2012 SciRes. APM R. H. SALLAM 40 4) Since every finitely generated right D-module M is Artinian and Noetherian then M is Artinian and Noethe-rian as an R-module. 5) Now since R is simple principle ideal domain then R is a finite dimensional k-algebra where k is a subring of the center of R with identity 1R and hence R is a finite dimensional algebra over the field Z (D) the center of D. Applying the above equivalence we get that every finitely generated right R-module is Artinian and hence R is right Artinian and this imply that T is also Artinian. 4. Conclusion The conclusion is that the unique example that says a right John’s ring can not be right Artinian is false. REFERENCES  B. Johns, “Annihilator Conditions in Noetherian Rings,” Journal of Algebra, Vol. 49, No. 1, 1977, pp. 222-224. doi:10.1016/0021-8693(77)90282-4  R. P. Kurshan, “Rings Whose Cyclic Modules Have Finitely Generated Socle,” Journal of Algebra, Vol. 15, No. 3, 1970, pp. 376-386. doi:10.1016/0021-8693(70)90066-9  S. M. Ginn, “A Counter Example to a Theorem of Kur-shAn,” Journal of Algebra, Vol. 40, No. 1, 1976, pp. 105-106. doi:10.1016/0021-8693(76)90090-9  W. K. Nicholson and M. F. Yousif, “Quasi-Frobenius Rings,” Series Cambridge Tracts in Mathematics, No. 158, 2003.  F. W. Anderson and K. R. Fuller, “Rings and Categories of Module,” Springer Verlag, New York, 1991.  F. Kasch, “Modules and Rings, London Mathematical Society Monographs,” Vol. 17, Academic Press, New York, 1982. Copyright © 2012 SciRes. APM