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>> No.6940521 [View]
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6940521

>>6938452
Incorrect, OP is not dividing by zero. You can divide by h in the ring of polynomials in two variables because the numerator is a multiple by h. And then you can apply the evaluate-h-at-0 ring homomorphism from [the ring of polynomials in x and h] to [the ring of polynomials in x]. This is perfectly valid; at no point are we dividing by zero.

>>6938446
Using continuity is cheating. The point of this is to do everything algebraically. Indeed the definition OP gave is one I really like (for polynomials) since it doesn't depend on the scalar field or any notion of continuity, and abstracts the notion of "cancel and set h equal to 0" into a logically valid form. Moreover, this cannot work as a definition of the derivative, since you defined F(x,0) in terms of the derivative.

So here's OP's question: given any field K, for which rational functions f in K(x) does the quotient [f(x+h)-f(x)]/h (a priori as an element of K(x,h)) exist in K(x)[h]? Given such an f, computing the quotient is just a matter of expanding out and then factoring out the h to cancel.

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