A Note on the (Faith-Menal) Counter Example

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.


Introduction
A ring R is called right John's ring if it is right Noetherian 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 ( [1], Theorem 1) by using a result of Kurshan ( [2], Theorem 3.3), showed that a right Noetherian ring is right Artinian provided that every right ideal is a right annihilator.
Ginn [3] 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 [1].
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. For more information about this example see [4].

Section 2: A Note on the Counter Example Theorem
The right John's ring T(R, D) defined in the counter example is Artinian.

Proof Recall the following (Exercises 10.7 [5]):
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 M D is Artinian or Noetherian then so is M R ; 3) If R is finite dimensional algebra (via φ) over a field D then the following is equivalent: a) M R is Artinian and Noetherian b) M R is finitely generated c) M D is finite dimension 1) Consider the ring homomorphism φ: D → R defined by φ (d) = d  1 f .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 [3]).
2) If the ring T R D   is John's ring (as in the counter 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[6]).
3) Every finitely generated right D-module M is semisimple so it is right Artinian.4) Since every finitely generated right D-module M is Artinian and Noetherian then M is Artinian and Noetherian 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 1 R 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.
Example 8.16 (Faith-Menal)[4].Let D be any countable, extentially closed division ring over a field F, and let R = D F  F T R D (x).Then T(R, D) is a non-Artinian right John's ring.Proof Cohn shows that R is simple, principle right ideal domain that is right V ring (Theorems 8.4.5 & 5.5.5[4]) and D is an R-R bimodule such that D R 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 T 0,d R   