# If $\mathbb{C}(u(x,y),v(x,y),f(x))=\mathbb{C}(x,y)$, for every $f(x) \in \mathbb{C}[x]-\mathbb{C}$, then already $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

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!

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)
• 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 $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? Commented Aug 25, 2023 at 14:06
• 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 $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? Commented Aug 25, 2023 at 14:14
• 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). Commented Aug 25, 2023 at 15:51
• @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, $C(u,v,\zeta)=C(x,y)$. Commented Aug 26, 2023 at 6:51