Classification of finitely generated bimodules over “skew PID”s?

Let R be a noncommutative ring with a 1 and no zero divisors, such that all (two-sided) ideals of R are principally generated. Is there a classification theorem for finitely generated bimodules over R? What if we strengthen the condition to all (two-sided) ideals of R are principally generated by an element of the center of R?

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I assume "principally generated" means that the ideal $I$ is $RaR$ for some $a/in R$? This theorem would presumably generalize a theorem about (commutative) PIDs, which would in turn generalize a theorem about (commutative fields. Do you know a classification theorem for f.g. bimodules over a field? – Tom Goodwillie Sep 30 '10 at 1:00
The definition is $I$ is the intersection of all ideals that $a$ is a member of, for some $a\in R$, which I think comes out to $RaR+RaR+...+RaR$. Also, I don't, although the first thing that comes to mind is that if the field $F$ is a prime field, then they are all isomorphic to exactly one of $\{F^0,F^1,F^2,F^3,...\}$. – Ricky Demer Sep 30 '10 at 1:17
Yes, that's what I meant by RaR. – Tom Goodwillie Sep 30 '10 at 3:03
An $R$-bimodule is the same as a module for $R\otimes R$. If $F$ is a field transcendental over the prime field then $F\otimes F$ is a pretty complicated ring. – Tom Goodwillie Sep 30 '10 at 3:04
Sorry, this is definitely not the point of the question but: is it really a good idea to describe a noncommutative ring all of whose two-sided ideals are principal as a "skew PID"? This for instance makes every simple algebra into a skew PID. I would have rather thought that the correct analogues were left-PIDs and right-PIDs. Or am I mistaken? (I am shaky on noncommutative rings, although I have some motivations to learn.) – Pete L. Clark Sep 30 '10 at 5:14