Value of polynomials in quotient ring Let $K$ be a field. If we have a polynomial ring, $K[X_1,...,X_n]$, and an ideal $I$, we can form the quotient ring, $$K[X_1,...,X_n]/I.$$ For a given ideal, $I$, if we take an element of this quotient ring, when is it well defined as a function? 
For instance, consider $$\mathbb{C}[X]/(X^2 -1).$$ Then, for example, in this ring, we have that (the images of) $f(X) = 4(X^2 -1) + 2X$, and $g(X) = 2X$ are equal. However, $\overline{F}(p) = \overline{G}(p)$ if and only if $p = \pm 1$. Of course, the "canonical" representation of this element in $\mathbb{C}[X]/(X^2 -1)$ is $\overline{g}(X)$, but $\overline{f}(X)$ is an equally valid representation.
However, if we let $V \subset \mathbb{A}^n(K)$, and define $$I(V) = \{F \in K[X_1,...,X_n] : F(p) = 0, \forall p \in V\},$$ then elements of $$\Gamma(V) = K[X_1,...,X_n]/I(V)$$ are well defined on $V$, since if $f,g \in K[X_1,...,X_n]$ such that $\overline{f}=\overline{g}$, then $f-g \in I(v)$, so $f(g) = g(p)$. I am also aware that there is an isomorphism between $K[X_1,...,X_n]/I(v)$ and polynomial functions from $V$ to $K$.
I've come across the above problem whilst studying algebraic geometry (local rings, poles, etc.). I feel my problem is a matter of perception. Should elements of $K[X_1,...,X_n]/I$ just be viewed formally? Does it only work out nicely in the latter case because of the stated isomorphism?
 A: The answer has already been given by sebigu (and yourself), but let me elaborate a little bit.
Take an ideal $I \subset K[X_1,\ldots,X_n]$, consider the quotient ring $R = K[X_1,\ldots,X_n]/I$. Take an element $f\in R$.
Well, $f$ is just an element of $R$, nothing more. But $f$ defines a function $\Phi: V(I) \rightarrow K$, where $V(I)\subset\mathbb A^n$ is the affine algebraic subset defined by $I$. How? Choose a representative $F$ of $f$ in $K[X_1,\ldots,X_n]$ ($F$ is just a polynomial with coefficients in $K$); for every $x = (x_1,\ldots,x_n)\in V(I)$, put the coordinates of $x$ in the polynomial $F$, and get a "value" $F(x)\in K$.
A priori, $F(x)$ might depend on your choice of $F$. But if you take another representative $F'$ of $f$, and call $G = F-F'$ their difference. By definition of quotient ring, $G$ belongs to the ideal $I$. By definition of $V(I)$, $G(x) = 0$, since $x \in V(I)$. Therefore, $F(x) = F'(x)$ and the "value" does not depend on the choice of the representative $F$. The function $\Phi:V(I) \rightarrow K$ is well-defined.
I think your confusion arises from identifying $f$ and $\Phi$ (often they will be denoted by the same symbol!): keep in mind that $f$ is an element of the quotient ring, $\Phi$ is a function. They are quite different things!
In your example, $f = 4(X^2-1) + 2X$ and $g = 2X$ are the same element of $\mathbb C[X]/(X^2-1)$: $f = g$ (because their difference lies in the ideal $I=(X^2-1)$). This element $f$ defines a function $\Phi:V(I) \rightarrow \mathbb C$.
What is $V(I)$ here? Well, $V(I)\subset \mathbb A^1$ is just the union $\{-1\}\cup\{1\}$. The function $\Phi$ is well-defined in these two points and nowhere else.
Hope this helped.
