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You could view R[x] as a subring of R(x). At that point, once you have tranformed the problem into one contained entirely within R[x], evaluation at a point bec

You could view R[x] as a subring of R(x). At that point, once you have tranformed the problem into one contained entirely within R[x], evaluation at a point becomes a well defined function. I suppose you could view evaluation as a non total function on R(x)XR (or rather, a function on the annoying to write domain where it is defined), but I have never seen anyone do that explicitly.

JadeNB 8y ago
> You could view R[x] as a subring of R(x). At that point, once you have tranformed the problem into one contained entirely within R[x], evaluation at a point becomes a well defined function. Indeed, this is exactly the process that I meant to describe by: > One must 'promote' the equality in `R(x)` to one in `R[x]` by clearing denominators, and only then plug in `x = 1`. > I suppose you could view evaluation as a non total function on R(x)XR (or rather, a function on the annoying to write domain where it is defined), but I have never seen anyone do that explicitly. I agree that it would be strange, for many reasons. Mathematicians in the main tend to avoid the language of partial functions. An individual evaluation function, say at `x = c`, is naturally viewed as a function on the localisation R[x]_{(x - c)}; but, if one wants to view the evaluation as (informally) a "two-place" function, then I would say that it lives naturally on R[x] \times R.