# On indecomposable Polynomials.

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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Is there a non-elementary proof you are aware of? – Igor Rivin Oct 5 '11 at 13:35
Also, I assume that by "trivially satisfied" you mean "false", since you can have $U_1 = G\quad U_2 = F.$ – Igor Rivin Oct 5 '11 at 14:23
Ancheng Lau: I took the liberty of editing your question and improve the latex. Should you like to learn how to do it, please edit it again and check the latex command I have inserted. Note that the effort of writing one's question in nice and readable form is always welcomed and appreciated. – Pietro Majer Oct 5 '11 at 15:08
Pietro Major: Thankyou for your answers and suggestions. I would be more careful of the latex form for future questions. – Ancheng Lau Oct 5 '11 at 16:01
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? – Ancheng Lau Oct 5 '11 at 16:09