Perhaps the following question is not at the MO question level, but has not received comments on MSE, so I also ask it here:

Leave $ beta: mathbb {C} (x, y) to mathbb {C} (x, y) $ be the involution in $ mathbb {C} (x, y) $

defined by $ (x, y) mapsto (x, -y) $.

Leave $ s_1, s_2, s_3 in mathbb {C} (x, y) $ be three symmetric elements with respect to $ beta $.

It is not difficult to see that a symmetric element w.r.t. $ beta $ It has the following form:

$ a_ {2n} y ^ {2n} + a_ {2n-2} y ^ {2n-2} + cdots + a_2y ^ 2 + a_0 $, where $ a_ {2j} in mathbb {C} (x) $.

Assume that the following two conditions are met:

**(one)** Every two of $ {s_1, s_2, s_3 } $ are algebraically independent about $ mathbb {C} $.

Note that the three $ s_1, s_2, s_3 $ are algebraically dependent upon $ mathbb {C} $, from the degree of transcendence of $ mathbb {C} (x, y) $ finished $ mathbb {C} $ are two

**(two)** $ mathbb {C} (s_1, s_2, s_3, y) = mathbb {C} (x, y) $; this notation means the fraction fields of $ mathbb {C} (s_1, s_2, s_3, y) $ Y $ mathbb {C} (x, y) $respectively.

**Example:**

$ s_1 = x ^ 2 + x ^ 5 + A (y), s_2 = x ^ 5y ^ 2 + B (y), s_3 = x ^ 3y ^ 2 + C (y) $, where $ A (y), B (y), C (y) in mathbb {C} (y ^ 2) $.

**Question 1:** Is it possible to find a specific & # 39; & # 39; of at least one of $ {s_1, s_2, s_3 } $?

A plausible answer can be: one of $ {s_1, s_2, s_3 } $ it's the way

$ Î» x ^ n and 2m + D (y) $ for some $ D (y) in mathbb {C} (y ^ 2) $, $ lambda in mathbb {C} ^ { times} $, $ n geq 1 $, $ m geq 0 $; Is it possible to find a counterexample to my plausible answer?

It may be better to first consider two (easier) questions that replace the conditions **(one)** Y **(two)** by:

**(1 & # 39;)** $ {s_1, s_2 } $ are algebraically independent about $ mathbb {C} $ +

**(2 & # 39;)** $ mathbb {C} (s_1, s_2, y) = mathbb {C} (x, y) $; call this **Question 1 & # 39;**.

**(1 & # 39; & # 39;)** $ s_1 neq 0 $ +

**(2 & # 39; & # 39;)** $ mathbb {C} (s_1, y) = mathbb {C} (x, y) $; call this **Question 1 & # 39; & # 39;**.

I guess the answer to question 1 & # 39; & # 39; it is: $ s_1 = lambda xE (y) + F (y) $, where

$ lambda in mathbb {C} ^ { times} $ Y $ E (y), F (y) in mathbb {C} (y ^ 2) $.

**Observations:**

**(I)** In the previous example we already have

$ mathbb {C} (s_2, s_3, y) = mathbb {C} (x, y) $ Y $ mathbb {C} (s_1, s_2, y) = mathbb {C} (x, y) $.

**(ii)** We can write $ x = frac {u (s_1, s_2, s_3, y)} {v (s_1, s_2, s_3, y)} $ for some $ u, v in mathbb {C} (X, Y, Z, W) $. So, if I'm not wrong, taking $ y = 0 $ (if possible?) we get that $ x = frac {u (s_1 (x, 0), s_2 (x, 0), s_3 (x, 0))} {v (s_1 (x, 0), s_2 (x, 0), s_3 (x , 0))} $Thus $ mathbb {C} (s_1 (x, 0), s_2 (x, 0), s_3 (x, 0)) = mathbb {C} (x) $.

**Question 2:** Is there any example where everyone $ s_1, s_2, s_3 $ are necessary to obtain $ mathbb {C} (s_1, s_2, s_3, y) = mathbb {C} (x, y) $? That is, it is not possible to skip one of $ {s_1, s_2, s_3 } $ and still get $ mathbb {C} (x, y) $.

I guess the answer is positive.

Thank you!