Counting solutions to equations in $\mathbb{Q}(q)$

Question

Consider $\mathbb{Q}(q)$, the field of rational functions in an indeterminate $q$. This field has unique $\mathbb{Q}$-ring involution $\psi$ sending $q$ to $q^{-1}$. Consider elements $a,b\in \mathbb{Q}(q)$ with $a\neq 0$. I am looking for solutions $x\in\mathbb{Q}(q)$ to the equation $$x^2-\psi(x^2)+\psi(x)a+b=0.$$ My question is the following: How many solutions does this equation have? My hope is that this equation has a unique solution, in the sense that it looks like a quadratic equation with extra conditions. Does anyone know how to this problem?

Answer 1

Indeed you have in general at most 2 solutions, given by intersecting a conic and a line. But there are degenerate cases when you have infinitely many solutions.

Notation: Write $y=y_++\frac{q-q^{-1}}2y_-$ where $y_+=\frac12(y+\psi(y))$, $y_-=\frac{y-\psi(y)}{q-q^{-1}}$. Note that we have similar decomposition algebraically as decomposing real and imaginary parts of a complex numbers, and $y_{\pm}\in\mathbb{Q}(q)^{\langle\psi\rangle}=\mathbb{Q}(q+q^{-1})=:k$. Finally, \begin{align*} (yz)_+&=y_+z_++\frac{(q+q^{-1})^2-4}4y_-z_-\\ (yz)_-&=y_+z_-+y_-z_+ \end{align*}

So our equation: $$x^2-\psi(x^2)+\psi(x)a+b=0$$ decomposes into a pair of equations $$ \begin{aligned} 0&=[x^2-\psi(x^2)+\psi(x)a+b]_+\\ &=x_+a_+-\frac{(q+q^{-1})^2-4}4x_-a_-+b_+\\ 0&=[x^2-\psi(x^2)_- + \psi(x)a+b]_-\\ &=2x_+x_- + x_+a_- + x_-a_++b_- \end{aligned} $$ The first equation is a line, and the second equation is a conic (or two lines $x_\pm=\frac12 a_{\mp}$ if $b_-=\frac12a_+a_-$) in $k^2$. So there are at most 2 intersection points in the $(x_+,x_-)$-plane unless the conic is degenerate and the first line coincide with one of the lines.