Generators of fixed function fields under involutions

I am trying to prove something for function fields in two generators and only one involution but I don't know if it is true, if true, I would like to generalize, here it is.

Let $K=k(\eta_1,\eta_2)$ be a function field, and let $\sigma$ be an involution such that $\sigma|_k=id$, how can I describe the generators of $K^{\langle \sigma \rangle}$?

If $\tau_1=\eta_1+\sigma\eta_1\neq 0$ and $\tau_2=\eta_2+\sigma\eta_2\neq 0$, When is true that $K^{\langle\sigma\rangle}=k(\tau_1,\tau_2)$ ?

What I can tell is that if $\lambda_1=\eta_1\cdot\sigma\eta_1$ and $\lambda_2=\eta_2\cdot\sigma\eta_2$ and of course $\eta_i^2-\tau_i\eta_i+\lambda_i=0$ for $i\in\lbrace 1,2\rbrace$ then $k(\tau_1,\tau_2,\lambda_1,\lambda_1)\subset k(\eta_1,\eta_2)$ is a field extension of degree $1$ or $2$ or $4$, so, Can I get a description of $K^{\langle\sigma\rangle}$?

• It follows from the classification of algebraic surfaces that $K^{\sigma}$ is of the form $k(\rho_1,\rho_2)$ for elements $\rho_1,\rho_2\in K^{\sigma}$. However, the elements $\tau_1$, $\tau_2$ above may not generate, e.g., if $\sigma(\eta_i)=-\eta_i$, then $\tau_i$ is zero. – Jason Starr Jan 11 '16 at 14:42
• I should have said: the classification of surfaces gives this result if $k$ is algebraically closed. If $k$ is not algebraically closed, then there may be counterexamples (for other finite groups than $\langle \sigma \rangle \cong \mathbb{Z}/2\mathbb{Z}$ there are certainly counterexamples, but I need to think about this for $\mathbb{Z}/2\mathbb{Z}$). – Jason Starr Jan 11 '16 at 14:44
• What I am interested is in fact is in the shape of $\rho_1,\rho_2$ , thanks for the comment. – Eduardo R. Duarte Jan 11 '16 at 14:45
• Unfortunately, I doubt that there is any completely general result. In the example I suggested above, you can take $\rho_1 = \tau_1 = -\eta_1^2$, and you can take $\rho_2 = \eta_1\cdot \eta_2$. Note, for instance, it is impossible to choose both $\rho_1$ and $\rho_2$ to be $k$-linear combinations of $\lambda_i$ and $\tau_j$. – Jason Starr Jan 11 '16 at 14:53
• @JasonStarr , can you tell me more about what you mean by "It follows from the classification of algebraic surfaces" exacly , or which classification? – Eduardo R. Duarte Jan 11 '16 at 15:10