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Let $f$ and $g$ be two polynomials (polynomial functions in $n$ variables); if in some localization of the ring $k[X_1,\ldots, X_n]$ exists the class $\frac{f}{g}$, it defines in a unique way the function $p\mapsto f(p){(g(p))}^{-1}$. Infact $\frac{f}{g}=\frac{f'}{g'}$ as equivalence classes if and only if they define the same function. The problem is the following: suppose that $\frac{f}{g}=\frac{f'}{g'}$ but there is a point $q$ such that $g(q)\neq0$ and $g'(q)=0$, then the function $p\mapsto \frac{f'(p)}{q'(p)}$ isn't well defined. For example $\frac{X^2}{X}=X$ in $k[X]_X$ (localizaztion of the ring $K[X]$) so the two associated functions should be the same, but clearly in $0$ there is some problem. In some textbooks it seems that the two functions must be considered however the same.

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Well, geometrically, if you localize at X , you're left with all primes NOT containing X . So here, at "0" doesn't happen, since (X) is not a prime ideal of $$k[X]_X$$. –  Dedalus May 26 '12 at 7:58
    
So are the two functions $\frac{X^2}{X}$ and $X$ the same? –  Dubious May 26 '12 at 9:05
    
I try to make more clear the question: Suppose that $\frac{f}{g}=\frac{f'}{g'}$ in $S^{-1}K[X_1,\ldots X_n]$ where $g,g'\in S$. With the canonical equivalence relation in $S^{-1}\times K[X_1,\ldots X_n]$ we have $(f,g)\sim (f',g')$; introducing an ''evaluation at some point'' as $p\mapsto f(p)g(p)^{-1}$, we have a function. It is easy to show that $(f,g)\sim (f',g')$ iff the associated functions are the same (so coincide in all points). But what about the points where $g(p)=0$ or $g'(p)=0$? –  Dubious May 26 '12 at 9:14

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up vote 3 down vote accepted

Suppose $V\subset \mathbb A^N_k $ is an affine irreducible algebraic variety over an algebraically closed field $k$.
Associated to $V$ are the ring of regular functions $\mathcal O(V)$ and its fraction field $Rat(V)= Frac(\mathcal O(V)$.
The ring $\mathcal O(V)$ is exactly the ring of regular functions $V\to k$, the functions algebraic geometry is concerned with.
But what you have to keep in mind is that $Rat(V)$ does not consist of functions : it is a field formally constructed from $\mathcal O(V)$ by algebraic wizardry, period.
However to each $\phi \in Rat(V)$, you can associate a non-empty open subset $\emptyset \neq U\subset V$ and a regular function $\tilde {\phi}:U\to k$. The relation between $\phi$ and $\tilde {\phi}$ is the following:

For every $p\in U$ there exist $f,g\in \mathcal O(V)$ with $g(p)\neq 0$ and $\tilde {\phi}(p)=\frac {f(p)}{g(p)}\in k$

Note carefully that $f$ and $g$ vary according to the choice of $p\in U$ see this recent question
The set $U$ is the set of all $p\in V$ such that $\phi\in \mathcal O_{V,p}$ .

In your example the function $\tilde {\phi}$ associated to $ \phi=\frac {X^2}{X}$ is exactly the function $X\in \mathcal O (V)$ since $\frac {X^2}{X}=X\in Rat(V)$ (and of course $U$ is the whole of $V$) .

And please, please, please, never even think of pronouncing the words "limit" or "real value" or "L'Hospital rule": leave them to our Analysis friends ...

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Very, very well! the key point is that $\frac{f}{g}$ formally it is not a function but simply a couple of polynomial functions. –  Dubious May 26 '12 at 9:42
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@Galoisfan: Almost right. "Polynomial function" is also the wrong idea -- you just want to work with polynomials. This is incredibly important in some situations, e.g. when working over finite fields. Over $\mathbb{F}_2$, x^2 and x are obviously different polynomials, but equal polynomial functions. –  Hurkyl May 26 '12 at 11:42

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