Let u,v be polynomials in two variables x and y over the field C of complex numbers.
Assume that the following 6 conditions are satisfied for all f(x) in C[x], g(y) in C[y]; the notation is of fields of fractions:
(1) C(u,f(x)) is strictly contained in C(x,y).
(2) C(u,g(y)) is strictly contained in C(x,y).
(3) C(v,f(x)) is strictly contained in C(x,y).
(4) C(v,g(y)) is strictly contained in C(x,y).
(5) C(u,v,f(x)) = C(x,y).
(6) C(u,v,g(y)) = C(x,y).
For convenience, call such u,v 'special' or 'a special pair'.
Denote by a the exchange involution (x,y) -> (y,x) and the set of symmetric elements w.r.t. a by S_a.
Question: Is it true that either C(u,v)=C(x,y) or C(u,v) is a subset of C(S_a)?
In other words, is it true that for a special pair u,v, we must
C(u,v)=C(x,y) or C(u,v) is contained in C(S_a)?
I have asked this question with further details in MO: https://mathoverflow.net/questions/454286/are-special-u-v-in-mathbbcx-y-must-be-symmetric-polynomials
Thank you very much!
Thank you. C(u,v) denotes the field of fractions of C[u,v].
I have later edited MO's question, adding a second weaker version question for which I have found and posted a counterexample. The question that appears here still has no answer in MO.
u is assumed to be an element of C[x,y]. We could have assumed it belongs to C(x,y), but this seems much more difficult.
The importance of C(u,f(x)) being not the whole field is because counterexamples arising when it is, for example u=y,v=x^2, hence if we allow C(u,f(x)) to be C(x,y) (here for f(x)=x we obtain C(y,x)), then the conclusion does not hold: C(y,x^2) is not C(x,y) and is not contained in C(x+y,xy) (indeed, y is not symmetric w.r.t. (x,y) -> (y,x)).
Thank you for your comments!
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