3 ms·

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

• 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).