ag.algebraic geometry - Rational map defined over K leads to algebra question

Hello,

Concrete algebraic question : Let $K$ be a perfect field, $bar{K}$ a fixed algebraic closure and let $f in bar{K}[x_1,ldots,x_n]$. I was wondering when there exists another polynomial (non-zero) $g in bar{K}[x_1,ldots,x_n]$ such that $fg in K[x_1,ldots,x_n]$ ?

Another formulation would be, when $f cdot bar{K}[x_1,ldots,x_n] cap K[x_1,ldots,x_n]$ is stricly larger than {0} ?

My motivation : Let $V subseteq mathbb{P}^n(bar{K}) $ be a variety defined over $K$, i.e., its ideal $I(V)$ can be generated (over $bar{K}[x_1,ldots,x_n]$) by polynomials in $K[x_0,ldots,x_n]$.
Let $phi = [f_0, ldots ,f_n] : V_1 to V_2$ be a rational map between projective varieties defined over $K$. The arithmetic of elliptic curves, by Silverman, say $phi$ is defined over $K$ when there exists some $lambda in bar{K}^{ast}$ such that the $lambda f_0, ldots, lambda f_n in K(V_1) = frac left( K[x_0,ldots,x_n] / I(V) right)$, where $I(V)$ is generated by polynomials in $K[x_0,ldots x_n]$.

Hence if $phi = [i, i] : mathbb{P}^1(mathbb{C}) to mathbb{P}^1(mathbb{C})$ and $psi : [X-i,X-i] : mathbb{P}^1(mathbb{C}) to mathbb{P}^1(mathbb{C})$, I hope I'm not wrong if I say they are not the same rational maps (the first being defined over $mathbb{R}$ but not the second one). But they both are the morphism $[1,1]$.