Are Semiprimitive (Jacobson Semisimple, J-Semisimple) Rings Dedekind Finite?

Question

Unfortunately there exists some ambiguity in the term semisimple. We have the common notion of semisimple, that of Artinian semisimple, and the less common one, that of Jacobson semisimple (not less common as in this object is less common, but less common as in it's referred to as simply "semisimple" less often). I encountered a problem which asked to show that a semisimple ring is Dedekind finite. This is fine if we have the common definition as then the ring is by definition Artinian and the whole thing follows from there:

If $ab=1$ then we can look at $bR\supset b^2R \supset \ldots$ and by DCC we have that $b^nR=b^{n+1}R$ so $b^n=b^{n+1}r$ and multiplying by $a^n$ on the left we get $1=br$ and of course $a=abr=r$ so $ba=1$.

However, if by chance the problem had intended the less common usage, prove that a semiprimitive $R$ is Dedekind finite, I'm not sure how to do this, and consequently I'm not sure if it's true or not. A semiprimitive ring is one in which the Jacobson radical is trivial (hence the other names). This makes using theorems like $R$ is Dedekind finite iff $R/J(R)$ is Dedekind finite useless. Does anyone have a counterexample or proof of this? Or a reference? Thanks so much!

Answer 1

Here is a way to construct an example of a semiprimitive ring that is not Dedekind finite.

First, use the result of

Shepherdson, J. C. Inverses and zero divisors in matrix rings. Proc. London Math. Soc. (3) 1, (1951). 71-85.

to create a ring $R$ which has no zero divisors, but has the property that $M_2(R)$ is not Dedekind finite. [Roughly, the idea of his paper is to create an algebra over a field that has a generic pair of $2\times 2$ matrices $A$ and $B$ such that $AB=I$. It turns out that the relation $BA=I$ is not forced. You prove that there are no zero divisors by introducing a normal form for elements.]

Now use the result of

Cohn, P. M. Simple rings without zero-divisors, and Lie division rings. Mathematika 6 1959 14-18.

to embed $R$ into a simple ring $S$. The ring $M_2(S)$ is semiprimitive, since it is a simple ring, but it is not Dedekind finite since it has a subring $M_2(R)$ that is not.

Answer 2

No, semiprimitive rings are not necessarily Dedekind finite. You can find examples like this at the Database of Ring Theory.

The canonical example of a ring in which there exists $a,b$ such that $ab=1$ but $ba\neq 1$ is the ring of linear transformations of an infinite dimensional vector space, where the transformations are written on the left side of vectors. It has Jacobson radical zero. Actually more is true: it is right primitive and von Neumann regular, both of which imply it has Jacobson radical zero.