Let $F$ be any infinite field, $U\subset F^n$ be an open, dense (in Zariski topology) subset,
$x\_1,x\_2,…,x\_n$ be an algebraic independent system of variables over $F$ , $f,f\_1,f\_2,…,f\_n \in F(x\_1,x\_2,…,x\_n)$ be rational functions and $g:F^n\rightarrow F$ be a function(note that the rationality of the $g$ is not given). If $f(u)=g(f\_1(u), f\_2(u),…, f\_n(u))$ for any $u\in U$ can one say that $f(x)\in F(f\_1(x), f\_2(x),…,f\_n(x))$? In other words does it imply the "rationality" of $g$?



Is g( ) rational if it looks that way on a large rational subset?
No. Take $F$ to be the algebraic closure of a finite field of characteristic $p>0$. Now, let $n=1, U = F, f\_1(x)=x^p, f(x)=x$. Then we can take $g(x) = x^{1/p}$, which is a well-defined function in the set-theoretic sense, but is not rational. Or, if you prefer, $x \notin F(x^p)$.


