Riemann-Roch for ARBITRARY Function Fields

I know that on an algebraic function field in one variable over any base field, there is a good divisor theory for it and a Riemann-Roch Theorem; in particular, there is a 'good' notion of 'genus'. (Which, I guess, is in general different from the corresponding notions in the geometric sense.) References are e.g. Chevalley's book and Deuring's book on algebraic function field in one variable, and Fried&Jarden's "Field Arithmetic".

Is there a generalization of these things to an arbitrary function field, which is of arbitrary transcendental degree over an arbitrary base field? Thanks!

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Function fields is not the right language to do these things in higher dimensions. There is a Riemann-Roch theorem in higher dimensions in the language of varieties, due to Hirzenbruch and generalized further by Grothendieck. – Felipe Voloch Mar 12 '11 at 23:50
Thanks, Felipe. Yes, I am aware of those generalizations. But is there any 'justification' why it's not the right language to use function fields? – Jizhan Hong Mar 13 '11 at 6:44
Dear Jizhan, one justification for abandoning function fields is that in higher dimensions the function field no longer determines the variety, even it that variety is assumed smooth. For example, given a field $k$, $\mathbb P^2_k$ and $\mathbb P^1_k \times \mathbb P^1_k$ are not isomorphic varieties but they have the same function field, namely $k(x,y)$. – Georges Elencwajg Mar 13 '11 at 9:55
Georges, thanks! I see your point. My question was actually more like: REGARDLESS whether it makes sense in the context of varieties or not, is there a 'good' generalization of them from the 1-variable case to the multi-variable case (and maybe a good generalization would turn out to be useful in geometry eventually as well, but that's something else). You know, just like the one-variable case, in general, the notion of genus could be different for a variety and its function field. I guess that's also why I didn't mention 'varieties' in the question. – Jizhan Hong Mar 13 '11 at 12:50
Here's a small observation. In characteristic $0$, the answer is yes in principle, but it may be hard to extract a concrete statement. To see this, note that every function field is the function field of a smooth projective variety (by Hironaka), and that the arithmetic is birationationally invariant. – Donu Arapura Mar 13 '11 at 14:53