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Help needed!

Def. Let $F(x _ 1,\dots,x _ n),G(x _ 1,\dots,x _ n)$ be two polynomials in $K[x _ 1,\dots,x _ n].$ $F,G$ are indecomposable iff there are no $u(x)\in K[x]$ with $\operatorname{deg}(u)\ge 2$ satisfying $F=u(H)$ for any $H\in K[x _ 1,\dots,x _ n]$.

Prop. If $F,G$ are two indecomposable polynomial and exist one variable polynomials $U _ 1(x),U _ 2(x)$ s.t. $U _ 1(F)=U _ 2(G)$, then $G=aF+b$ for $a,b\in K$.

Is there an elementary proof to this proposition? Note that $F,G$ has at least two variables or this proposition is trivially satisfied. Thanks alot!

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  • $\begingroup$ Is there a non-elementary proof you are aware of? $\endgroup$
    – Igor Rivin
    Oct 5, 2011 at 13:35
  • $\begingroup$ Also, I assume that by "trivially satisfied" you mean "false", since you can have $U_1 = G\quad U_2 = F.$ $\endgroup$
    – Igor Rivin
    Oct 5, 2011 at 14:23
  • $\begingroup$ Igor Rivin: The only proof I know of is by showing that $K(F,G)$ has a single generator $w$, and then by arguing that $w$ could be taken to be a polynomial. (Which is basically the proof given below in the answer). However, I was told that a "simple and direct proof of just a few lines" existed. So I was curious to what other approaches could be made? $\endgroup$ Oct 5, 2011 at 16:09
  • $\begingroup$ Ancheng: since you were aware of the proof in that reference (Lemma 5.2 of math.univ-lille1.fr/~pde/bodin-debes-najib.pdf) I now withdraw the answer. $\endgroup$ Oct 5, 2011 at 17:05

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