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The following question is a direct continuation of this elaborate question; it is mentioned there at the end:

Let $u,v \in \mathbb{C}(x,y)$ or $u,v \in \mathbb{C}[x,y]$, if it is easier to answer in this case.

Assume that the following condition, call it $D(f)$, is satisfied for every $f \in \mathbb{C}[x]-\mathbb{C}$:

$\mathbb{C}(u,v,f)=\mathbb{C}(x,y)$.

Question: Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

The notation is of fields of fractions.

Remarks:

  1. It seems that $y \in \mathbb{C}(u,v)$, but actually I am not sure about this, since I only have the following argument:

We know that $y \in \mathbb{C}(u,v,f)$, so there exist $F,G \in \mathbb{C}[r,s,t]$ such that $y=\frac{F(u,v,f)}{G(u,v,f)}$.

For some $\alpha \in \mathbb{C}$, $f(\alpha)=0$, and then $y=\frac{F(u(\alpha,y),v(\alpha,y),f(\alpha))}{G(u(\alpha,y),v(\alpha,y),f(\alpha))}= \frac{F(u(\alpha,y),v(\alpha,y))}{G(u(\alpha,y),v(\alpha,y))}$, which just shows that $y \in \mathbb{C}(u(\alpha,y),v(\alpha,y))$.

  1. I do not know how to show that $x \in \mathbb{C}(u,v)$, if this is true.

  2. I guess such $u,v$ must be algebraically independent.

Thank you very much!

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1 Answer 1

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The answer to the Question is ``no". As an example, let $u=xy$, $v=x+y$ be the elementary symmetric functions in $x$ and $y$. It is well-known that $[\mathbb{C}(x,y): \mathbb{C}(u,v)]=2$, so those two fields are not equal. On the other hand, consider $K_f:=\mathbb{C}(u,v,f(x))$ for any nonconstant polynomial $f(x)\in \mathbb{C}[x]$. Note that $\mathbb{C}(u,v)$ already contains the functions $f(x)+f(y)$ and $\frac{f(x)-f(y)}{x-y}$, since indeed those are symmetric functions in $x$ and $y$. So successively we find that $K_f$ contains the elements $(f(x)+f(y))-f(x)=f(y)$, thus also $f(x)-f(y)$, and thus also $\frac{f(x)-f(y)}{(f(x)-f(y))/(x-y)}=x-y$. Finally, it also contains $x+y$ and thus $x$ and $y$, showing $K_f=\mathbb{C}(x,y)$.

(Edit: One could have of course argued non-constructively even quicker: Since the field of symmetric functions is of index $2$ in $\mathbb{C}(x,y)$ and doesn't contain any nonconstant function $f(x)$, adjoining any such function must give the whole thing by lack of intermediate fields)

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  • $\begingroup$ Thank you very much for your nice answer! So now I am curious about the following question: Let $u,v \in \mathbb{C}[x,y]$. Assume that the following condition, call it $L_{a,b,c,n}$, is satisfied for every $a,b,c \in \mathbb{C}$ and every $1 \leq n \in \mathbb{N}$: $\mathbb{C}(u,v,(ax+by+c)^n)=\mathbb{C}(x,y)$. Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? $\endgroup$
    – user237522
    Commented Aug 25, 2023 at 14:06
  • $\begingroup$ Another variation is: Let $u,v \in \mathbb{C}[x,y]$. Assume that the following condition, call it $M_{a,b,c,d,n,m}$, is satisfied for every $a,b,c,d \in \mathbb{C}$ and every $1 \leq n,m \in \mathbb{N}$: $\mathbb{C}(u,v,(ax+b)^n(cy+d)^m)=\mathbb{C}(x,y)$. Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? $\endgroup$
    – user237522
    Commented Aug 25, 2023 at 14:14
  • $\begingroup$ Excluding the three cases: (i) $a=b=0$ ; (ii) $c=d=0$ ; (iii) $a=c=0$. (I really apologize if my comments are bothering you. This is the last one in this topic). $\endgroup$
    – user237522
    Commented Aug 25, 2023 at 15:51
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    $\begingroup$ @user237522 These problems may require some separate effort for each new variant, but at least the first one above still has a negative answer, and if I'm not mistaken $u=x^3+y$, $v=y^3+x$ would be an example, the very brief reason being that $\mathbb{C}(x,y)/\mathbb{C}(u,v)$ is then a nonsolvable ($S_9$)extension without any intermediate extensions, and it also does not contain any $ax+by$; but then it also cannot contain any $\zeta:=(ax+by+c)^n$, since that would lead to a solvable(!) intermediate extension. Since there are no nontrivial intermediate fields at all, $C(u,v,\zeta)=C(x,y)$. $\endgroup$ Commented Aug 26, 2023 at 6:51
  • $\begingroup$ Thank you very much; seems very nice counterexample, though I am not sure if I understand all the details now. I have similar ideas in mind, but maybe I should post them in new questions. Thanks again. $\endgroup$
    – user237522
    Commented Aug 26, 2023 at 17:40

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