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 $[1,1]$.