A two-dimensional rationality problem and intersections of two quadrics.

Let $k$ be a field with char $k\neq 2$ and $k$ be not algebraically closed. Let $a\in k\setminus k^2$ and $L=k(\sqrt{a})(x,y)$ be a field extension of $k$ where $x,y$ are algebraically independent over $k$. Assume that $\sigma$ is a $k$-automorphism on $L$ defined by \[ \sigma: \sqrt{a}\mapsto -\sqrt{a},\ x\mapsto \frac{b}{x},\ y\mapsto \frac{c(x+\frac{b}{x})+d}{y} \] where $b,c,d \in k$, $b\neq 0$ and at least one of $c,d$ is non-zero. Let $L^{\langle\sigma\rangle}=\{u\in L:\sigma(u)=u\}$ be the fixed subfield of $L$ and $(a,b)_{2,k}$ be the Hilbert symbol over $k$. Theorem. Assume that $d^2-4bc^2\neq 0$. Then the following statements are equivalent: (i) $L^{\langle\sigma\rangle}$ is rational over $k$; (ii) $L^{\langle\sigma\rangle}$ is stably rational over $k$; (iii) $L^{\langle\sigma\rangle}$ is unirational over $k$; (iv) there exist $\alpha, \beta, \gamma, \delta\in k$ such that $\alpha ^2-a \beta ^2=b$, $\gamma^2-a \delta ^2= 2c \alpha + d$; (v) there exist elements $\alpha, \beta \in k$ such that $\alpha ^2-a \beta ^2=b$, and either $2c\alpha + d =0$, or $2c\alpha + d \neq 0$ with $(a, 2c\alpha + d)_{2, k(\sqrt{d^2-4bc^2})}=0$; (vi) $(a,b)_{2, k}=0$, and for any elements $\alpha, \beta \in k$ such that $\alpha ^2-a \beta ^2=b$ we have either $2c\alpha + d =0$, or $2c\alpha + d \neq 0$ with $(a, 2c\alpha + d)_{2, k(\sqrt{d^2-4bc^2})}=0$. The same question when $d^2-4bc^2=0$ is solved also.