Let $V/K$ be a variety defined over a field $K$ Let $K[X]$ be the polynomial ring with $n$ variables $X_1,...,X_n$. We define the affine coordinate ring of $V/K$ to be $$K[V]=K[X]/I(V/K)$$ where $I(V/K)=\{f\in K[X]:f(P)=0 \text{ for all }P\in V\}$. In Silverman's book on elliptic curves, he states that "since an element $f\in K[V]$ is well-defined up to adding a polynomial vanishing on $v$, it induces a well-defined function $f:V\rightarrow K$".
My question is what is the well-defined function? How does it come about?
Let $f$ be an element of $K[V]=K[X]\ /\ I(V/K)$, and let us lift it to a "true" polynomial $F\in K[X]$. Using this one choice, we can define a map $V\to K$, by mapping a point $P\in V$ to $F(P)\in K$. We then define $$ f(P):=F(P)\ , $$ and the question is, why this definition does not depend on the choice of the lift $F$ of $f$. Well, let us take an other lift $G\in K[X]$ of $f$. Then by definition of $K[V]=K[X]\ /\ I(V/K)$, the "true" polynomials $F,G$ differ by a polynomial $(F-G)$ in the ideal $I(V/K)$ defined by the variety $V$. in particular, for the one chosen point $P$ above we have $0=(F-G)(P)=F(P)-G(P)$. This means that $F(P)=G(P)$, so the definition $f(P):=F(P)$ does not depend on the special lift / representative.