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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 $[1,1]$.

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It's OK, even w/o perfectness: if char. is $p > 0$ then multiplying $f$ against $f^{p^n - 1}$ for big $n$ yields $f^{p^n}$ with coeffs in $K_ s$ inside $\overline{K}$. In general (any char.), now all coeffs in finite Galois ext'n $L/K$. Multiplying against conj's under non-triv. el'ts of ${\rm{Gal}}(L/K)$ yields $fg$ w/ coeffs lie in $K$. But motivation is misplaced: the def'n you cite is wrong. The right def'n is all $f_i/f_j \in K(V)$ for one $j$ s.t. $f_j \ne 0$. So in final para. $\phi = \psi = [1,1]$. The alg. geom. in AEC is clunky; it will all become cleaner when you learn schemes. –  BCnrd Jun 29 '10 at 3:05
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