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The problem with viewing this as only a computation in the ring `R(x)` (I think that's what you mean, not "the ring of rational fractions on R[x]"—maybe you mea

by JadeNB 8y ago

The problem with viewing this as only a computation in the ring `R(x)` (I think that's what you mean, not "the ring of rational fractions on R[x]"—maybe you meant the fraction field of `R[x]`?) is that, while all the equalities of rational functions make perfectly good sense in `R(x)`, there is no "plug in `x = 1`" function on `R(x)`. One must 'promote' the equality in `R(x)` to one in `R[x]` by clearing denominators, and only then plug in `x = 1`. (Of course this is what you do, and I agree that it's the right, or at least a good, way to explain it; but I think that it's important to emphasise explicitly how the algebraic point of view avoids any concern with taking limits.)

gizmo686 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. 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.
edflsafoiewq 8y ago
Ahhh, apparently the ring of rational fractions on R means the field of fractions on R[x]. I did mean R(x). Why don't I have a "plug in x=1" (partial) function R(x) -> R? And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right?
- JadeNB 8y ago
  > Why don't I have a "plug in x=1" (partial) function R(x) -> R? Precisely for the reason that you say: mathematicians in the main tend to avoid the language of partial functions, so I would say not "there is a partial function R(x) \to R" but "there is a function R[x]_{(x - 1)} \to R" (where R[x]_{(x - 1)} is a localisation). > And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right? My language was somewhat fanciful. I don't think that one can say that an equation in R(x) is an equation in R[x], for essentially syntactic reasons; but, just as you said, if the rational function a/b equals the rational function c/d (a/b = c/d in R(x)), then a * d = b * c in R[x]. That's what I meant by 'promoting' the equality (although it is not great, and certainly not technical, language).

by JadeNB 8y ago

gizmo686 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. 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.

edflsafoiewq 8y ago

Ahhh, apparently the ring of rational fractions on R means the field of fractions on R[x]. I did mean R(x). Why don't I have a "plug in x=1" (partial) function R(x) -> R? And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right?

JadeNB 8y ago
> Why don't I have a "plug in x=1" (partial) function R(x) -> R? Precisely for the reason that you say: mathematicians in the main tend to avoid the language of partial functions, so I would say not "there is a partial function R(x) \to R" but "there is a function R[x]_{(x - 1)} \to R" (where R[x]_{(x - 1)} is a localisation). > And what do you mean by "promote"? An equation in R(x) "is" a (set of) equations in R[x] right? My language was somewhat fanciful. I don't think that one can say that an equation in R(x) is an equation in R[x], for essentially syntactic reasons; but, just as you said, if the rational function a/b equals the rational function c/d (a/b = c/d in R(x)), then a * d = b * c in R[x]. That's what I meant by 'promoting' the equality (although it is not great, and certainly not technical, language).