Let $k[x_1,y_1,\ldots,x_n,y_n]$ be the polynomial ring in $2n$ variables over a field $k$ of characteristic zero. Let $\alpha_n: x_i \leftrightarrow y_i$, $1 \leq i \leq n$; clearly, $\alpha_n$ is an involution (=automorphism of order two). Denote the set of symmetric elements with respect to $\alpha_n$ by $S_{\alpha_n}$.
When $n=1$, it is not difficult to see that every element of $k[x_1,y_1]$ can be written $u+v(x-y)$, where $u,v \in S_{\alpha_1}$. Clearly, $[k(x_1,y_1):k(S_{\alpha_1})]=[k(S_{\alpha_1})(x-y):k(S_{\alpha_1})]=2$.
($(x-y)^2 \in k(S_{\alpha})$). If I am not wrong, the $k$-subalgebra of symmetric elements is generated by $\{x+y,x^2+y^2\}$ or by $\{x+y,xy\}$.
When $n \geq 2$: (1) What is $S_{\alpha_n}$? (2) what is $[k(x_1,y_1,\ldots,x_n,y_n):k(S_{\alpha_n})]$?
Notice that $S_{\alpha_n}$ contains not only $n$ copies of $S_{\alpha_1}$ (each copy with variables $x_i,y_i$), but also elements such as $x_1x_2+y_1y_2$, so things are getting quite complicated (at least for me). Also, I suggest to consider the $n=2$ case before considering a general $n \geq 2$.
Any hints are welcome. Thanks!
Easier to see this after changing variables, $u_i=x_i+y_i, v_i=x_i-y_i$, Then the involution is $u_i\mapsto u_i, v_i\mapsto -v_i$. Then, it is easy to see that the invariants are generated by $u_i, v_iv_j, 1\leq i,j\leq n$ ($i=j$ included).
For the corresponding field extension, we have $L=k(u_i, v_iv_j)\subset k(u_i,v_j)$. Take one of the $v_i$s, say $v_1$. Then we have both $v_1v_j$ and $v_1^2$ in $L$ and so, $v_j/v_1\in L$. Note that $$k(u_i,v_j)=k(u_i, v_1,v_2/v_1,\ldots, v_n/v_1).$$ Thus, we see that the fixed field contains all the generators but $v_1$. So, the extension is of the form $k(u_i,v_1^2,v_2/v_1,\ldots, v_n/v_1)=L\subset L(v_1)$. Since $v_1^2\in L$ we are done.