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Let $f=f(x,y), g=g(x,y) \in \mathbb{C}[x,y]$.

Which $f,g$ satisfy $\mathbb{C}(fy,gy)=\mathbb{C}(x,y)$?

Examples:

(1) $f=2x$, $g=1-2x$. It is easy to see that $\mathbb{C}(2xy,(1-2x)y)=\mathbb{C}(2xy,y-2xy)=\mathbb{C}(2xy,y)=\mathbb{C}(x,y)$.

(2) $f=x$, $g=xy$.

Attempts to generalize the two examples:

(1) $af+bg \in \mathbb{C}[x]-\{0\}$, for some $a,b \in \mathbb{C}$, similarly to example (1). But if $f=x^2$ and $g=x^4$, then this does not work since $\mathbb{C}(x^2y,x^4y)=\mathbb{C}(x^2,y)$. So perhaps we should require $af+bg \in \mathbb{C}-\{0\}$? This will imply that $\mathbb{C}(fy,gy) \ni afy+bgy=y(af+bg)=\lambda y$, and then $f,g \in \mathbb{C}(fy,gy)$, which still does not guarantee that $x \in \mathbb{C}(fy,gy)$.

(2) The following seems to be a false claim: $f$ and $g$ themselves already generate $\mathbb{C}$, namely: $\mathbb{C}(f,g)=\mathbb{C}(x,y)$; although in example (2) it works, but in other cases it may not work.

Edit: After receiving a comment that the resultant may be relevant, I add the following links: Theorem 2.1 and this question.

If I am not wrong, the criterion in the second link implies that, in our case: If $\gcd(f,g)=1$, considering $f,g \in (\mathbb{C}(x))[y]$, then $\mathbb{C}(x)(fy,gy)=(\mathbb{C}(x))(y)=\mathbb{C}(x,y)$.

Thank you very much!

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  • $\begingroup$ I’m a bit confused. In the second of your “attempts”: what does “$fy$” mean? Isn’t $\,f$ an expression in $x$ and $y$? $\endgroup$ – Lubin Aug 13 '18 at 16:31
  • $\begingroup$ Yes, thanks, I meant: "namely, $\mathbb{C}(f,g)=\mathbb{C}(x,y)$". My mistake. I will correct this now. $\endgroup$ – user237522 Aug 14 '18 at 3:17
  • $\begingroup$ The resultant seems relevant here. $\endgroup$ – Servaes Aug 14 '18 at 9:46
  • $\begingroup$ Thank you for your comment! I do know that for two polynomials $F,G \in \mathbb{C}[x,y]$, we have: $\mathbb{C}(F,G)=\mathbb{C}(x,y)$ if and only if the d-resultant of $F$ and $G$ is a non-zero polynomial (if it happens to be a scalar, then $\mathbb{C}[F,G]=\mathbb{C}[x,y]$). But the problem is that I do not see how to calculate the d-resultant of two arbitrary $fy$ and $gy$; but perhaps I am missing something, and the 'specific' $fy$ and $gy$ may yield something interesting. $\endgroup$ – user237522 Aug 14 '18 at 11:23
  • $\begingroup$ Actually, I was not accurate in my above comment, since the criterion I meant to mention deals with one variable polynomials: $\mathbb{C}(u(t),v(t))=\mathbb{C}(t)$ if and only if the d-resultant of $u(t),v(t)$ is a non-zero polynomial. It is possible to adjust that criterion to our $fy$ and $gy$, thought of as polynomials in one variable $y$ over $\mathbb{C}(x)$. $\endgroup$ – user237522 Aug 14 '18 at 11:49

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