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Denote the set of prime numbers by $P$, $P=\{2,3,5,7,\ldots\}$. Let $F \subseteq \mathbb{C}(x,y)$ be a subfield of $\mathbb{C}(x,y)$, and for $w \in \mathbb{C}[x,y]$ denote by $F(w)$ the subfield of $\mathbb{C}(x,y)$ generated by $F$ and $w$.

Assume that $F$ satisfies the following three properties:

(p1) $F=\mathbb{C}(u,v)$ with $u,v \in \mathbb{C}[x,y]$ algebraically independent over $\mathbb{C}$.

(p2) For every $p \in P$, $F(x^p+y^p)=\mathbb{C}(x,y)$.

(p3) $F(x+y)=\mathbb{C}(x,y)$.

Question: Is it true that for such $F$ necessarily $x \in F$ or $y \in F$? If not, is there something interesting that can be said about $F$?

For example: Take $F=\mathbb{C}(xf,y)$, where $f \in \mathbb{C}[x]$ and $\gcd(f,x)=1$. $F$ satisfies the three properties; indeed, for every $n \in \{1\} \cup P$, $F(x^n+y^n)=\mathbb{C}(xf,y,x^n+y^n)=\mathbb{C}(xf,y,x^n)$. By the result presented here, $\mathbb{C}(xf,x^n)=\mathbb{C}(x)$ (because $\langle xf,x^n \rangle = \langle x \rangle$ is a maximal ideal of $\mathbb{C}[x]$, by our choice of $f$), so $F(x^n+y^n)=\mathbb{C}(xf,x^n,y)=\mathbb{C}(x,y)$.

But maybe there are examples of $F$ such that $x \notin F$ and $y \notin F$?

Remarks:

  • An 'opposite' question is this question, in which $x^n+y^n$ is already in the subfield, for some fixed $n$, not a generator. Here $x^n+y^n$ are generators for prime numbers $n$ or $n=1$.

  • Also asked here without comments.

Thank you very much!

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  • $\begingroup$ If we change the second property to: For every $p \in P$, $F(x^p+ay^p)=\mathbb{C}(x,y)$, for some $a \in \mathbb{C}$ (fixed for all $p \in P$), then the question has a negative answer. Symmetric functions is a counterexample. $\endgroup$
    – user237522
    Commented Jul 9 at 19:24

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Not much can be said about $F$. The second conditions should hold for $\mathbb C(u,v)$ for $u$ and $v$ two "generic" algebraically independent polynomials.

For an explicit construction, take $u=(x-1)^2$ and $v= (y-1)^2$. Then $F=\mathbb C(u,v)$ is the fixed field of the group of automorphisms generated by $(x,y )\mapsto (2-x, y)$ and $(x,y)\mapsto (x,2-y)$. By Galois theory $F(t)=\mathbb C(x,y)$ if and only if $t$ is not fixed by any nontrivial automorphism in this group. By plugging in, we see that $x^n+y^n$ is not fixed by any nontrivial automorphism in this group for any $n>0$, so indeed these all generate $\mathbb C(x,y)$.

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  • $\begingroup$ Thank you! (I have written a comment, with a slight change of the second property, finding a counterexample, not for my original question, but to the new version presented in my comment). $\endgroup$
    – user237522
    Commented Jul 9 at 19:30
  • $\begingroup$ Truly, I thought about the above question when thinking about another question: mathoverflow.net/questions/472877/… I will explain later what is the connection between the two questions. $\endgroup$
    – user237522
    Commented Jul 9 at 19:36
  • $\begingroup$ The connection between the two questions: In the former, linked question, I have asked if $R=\mathbb{C}(u,v)$ with $u,v$ algebraically independent over $\mathbb{C}$, such that $C_{i,j}$ holds for every $(i,j) \in \mathbb{N} \times \mathbb{N} - (0,0)$, then $R=\mathbb{C}(x,y)$? The answer: $[\mathbb{C}(x,y):R]=2$. (Actually. I have asked also about an additional condition $R(x+y)=\mathbb{C}(x,y)$, so maybe with this additional condition we have $R=\mathbb{C}(x,y)$). Where $C_{i,j}$ says: $R(x^iy^j)=\mathbb{C}(x,y)$, namely, $x^iy^j$ is a primitive element for the extension. $\endgroup$
    – user237522
    Commented Jul 9 at 20:18
  • $\begingroup$ @user237522 The idea behind answering both questions is the same: Take $R$ to be the fixed points of a group action on $\mathbb C(x,y)$, and just check no nontrivial element of the group fixes the additional elements you are adding (e.g. $x^i y^j$ in that case). $\endgroup$
    – Will Sawin
    Commented Jul 9 at 20:20
  • $\begingroup$ Thank you, nice observation! Please, if we combine the edit of the linked question + the above question to the following: Given $R=\mathbb{C}(u,v)$ as usual, such that (1) $C_{i,j}$ holds for every $(i,j) \in \mathbb{N} \times \mathbb{N} - D$, $D=\{(n,n)\}_{n \in \mathbb{N}}$, (2) $E_{p,p}$ holds for every $p \in P \cup \{1\}$, (3) $xy \in R$. Where $C_{i,j}$ says $R(x^iy^j)=\mathbb{C}(x,y)$ and $E_{i,j}$ says $R(x^i+y^j)=\mathbb{C}(x,y)$. Remark: (3) is to guarantee that $x^ny^n \in R$, so $C_{n,n}$ is not assumed anymore (to exclude $R=R(x^my^m)=\mathbb{C}(x,y)$, if $xy \in R$). $\endgroup$
    – user237522
    Commented Jul 9 at 20:45

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