If $L:=\mathbb{R}(X,Y)$ and $G=\{\sigma_{1},\sigma_{2}\}$ where
$\sigma_{1}(X)=X\;\;,\;\;\;\sigma_{2}(X)=-X$
$\sigma_{1}(Y)=Y\;\;,\;\;\;\sigma_{2}(Y)=-Y$
Find $L^{G}$
So by def of $L^{G}:=\{l\in L\;\;\displaystyle\forall_{\sigma\in G}\sigma(l)=l\}$ i.e I need to find elements of $L$ that will be invariant under $K$-automorphism of L.
$\sigma_{i}(X^2)=\sigma_{i}(X)^2=(\pm X)^2=X^2\Rightarrow X^2\in L^{G}$
$\sigma_{i}(Y^2)=\sigma_{i}(Y)^2=(\pm Y)^2=Y^2\Rightarrow Y^2\in L^{G}\;,\;i=1,2$
Also
$\sigma_{1}(XY)=XY$ and
$\sigma_{2}(XY)=(-X)(-Y)=XY\Rightarrow XY\in L^{G}$
So If we let $K:=\mathbb{R}(X^2,Y^2,XY)$ then $K\subseteq L^{G}$
And to show other inclusion I will use the tower law. But my book says that $L^{G}=\mathbb{R}(X^2,XY)$
Why $Y^2$ is not taken in the smaller field?
Thanks
You're right about your calculations. But, the fields in fact agree, ie. $$\Bbb R(X^2,XY,Y^2) = \Bbb R(X^2,XY)$$ This is because $$Y^2 = \frac{(XY)^2}{X^2}$$ and is therefore already included.