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Is there a way to prove that the following polynomials $f,g,h$ do not divide each other in the polynomial ring $\mathbb{C}[x,y,z]$?

$$f(x,y,z)=x^5-yz, \ \ g(x,y,z)=y^2-xz, \ \ h(x,y,z)=z-xy$$

Anonymous
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2 Answers2

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Choose $(x,y,z)$ from $\{ (1,0,0), (0,1,0), (0,0,1) \}$ and note that exactly one of $f,g,h$ is 1 for each element and zero for the others. Hence then cannot divide one another.

To illustrate, suppose $f \mid g$, then if $f(x,y,z) = 0$ we must have $g(x,y,z) = 0$, but $g(0,1,0) = 1$ and $f(0,1,0) = 0$.

copper.hat
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  • +1 That's the slick way to do it - using evaluation to deduce polynomial indivisibility via integer indivisibility. Similarly one shows that $,x\nmid 1,$ in $,R[x],,$ so it's not a field. Given the comments on that answer, this way of proof seems to not be as well-known as it should be. One can go further and factor polynomials using factorization of integer evaluations (though not efficiently). – Bill Dubuque Jan 27 '15 at 22:47
  • @BillDubuque: I like the style & succinctness of your proof. It shows how elements of the underlying structure (in this case the presence of a non unit) percolate up to the 'containing' structure. – copper.hat Jan 27 '15 at 23:05
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$x^5-yz$ does not divide $y^2-xz$ because the former has degree $5$ in $x$ and the latter has degre $1$. Conversely $y^2-xz$ does not divide $x^5-yz$ because the former has degree $2$ in $y$ and the latter has degree $1$. Similarly $f$ and $g$ can't divide $h$.

$h$ can't divide $f$, nor $g$, because since $h(2,2,4)=0$, $f(2,2,4)$ and $g(2,2,4)$ would be $0$, and $f(2,2,4)=24$, $g(2,2,4)=-4$.

Bernard
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