I'm attempting a question as follows:
Let $K = \mathbb Q(x,y)$, where $x,y$ are independent transcendentals, and consider the group $G$ of automorphisms generated by $$\sigma: \quad x \mapsto y,\quad y \mapsto -x$$ $$\tau: \quad x \mapsto x,\quad y \mapsto -y$$ Show that the fixed field $K^G = L = \mathbb Q(x^2 + y^2, x^2y^2)$.
I get that $G = \langle \sigma, \tau : \sigma^4 = 1, \tau ^2 = 1, \tau\sigma\tau = \sigma^{-1}\rangle \cong D_8$, so by Artin's Theorem, $[K : K^G] = 8$.
However, since $x,y$ are roots of $$f = (T^2 - x^2)(T^2 - y^2) = T^4 - (x^2 + y^2)T^2 + x^2y^2 \in L[T]$$ This gives
$$ [K : L] \le 4 < [K:K^G]$$
which contradicts $K^G = L$. What am I doing wrong?
EDIT: Seems I assumed "if all elements have degree $\le 4$ then degree of extension is $\le 4$", but this isn't true? Instead, can we say:
$$ [\mathbb Q(x,y) : L] = [\mathbb Q(x,y) : L(x)] [L(x) : L] \le 8 $$
since $[L(x) : L] \le 4$ using $f$, and then $[\mathbb Q(x,y) : L(x)] = 2$ since $y$ is a root of $T^2 - (x^2y^2)/(x^2)$.
And clearly $K^G \subseteq L$, so by dimensions $K^G = L$.