Definition. Let $V_1$ and $V_2 \subseteq \mathbb{P}^n$ be projective varieties. A rational map from $V_1$ to $V_2$ is a map of the form $$\varphi : V_1 \rightarrow V_2, \qquad \varphi = [f_0,\ldots,f_n]$$
where the functions $f_0,...,f_n ∈ K(V_1)$ have the property that for every point $P ∈ V_1$ at which $f_0,...,f_n$ are all defined, $$\varphi (P) = f_0(P),...,f_n(P).$$
(p.11, The Arithmetic of Elliptic Curves, Joseph H. Silverman)
On the next page, the remark explains that you can often multiply by polynomials to knock out the problematic points wher the $f_i$'s are undefined.
It seems to me that we can always do this. For instance, the map $$\varphi : \mathbb{P}^1 \rightarrow \mathbb{P}^1, \qquad [x,y] \mapsto [A(x,y)/B(x,y),C(x,y)/D(x,y)]$$ seems to be the same as the map
$$\psi: \mathbb{P}^1 \rightarrow \mathbb{P}^1, \qquad [x,y] \mapsto [A(x,y)D(x,y),C(x,y)B(x,t)].$$
If so, it's not clear to my why we bother allowing the $f_i$ to be rational functions, when they could easily be assumed to be polynomials via the above trick.
What am I not understanding?
To "knock out" a problematic point you have to multiply by some other function. Unfortunately the same function may not be able to knock out all the problematic points.
Actually a rational map is an equivalence class of what you have described: Let $U$ be a (Zariski) open set got by avoiding those problematic points, for a given $\phi$. Let $\psi$ be another collection of rational functions $g_i$'s with $W$ as corresponding open set where everything is well-defined. Say that $\phi$ and $\psi$ are equivalent if $U\cap W$ is nonempty and $\phi$ and $\psi$ agree on that intersection.