Galois group of $\mathbb{C}(x(1-y),y(1-x)) \subset \mathbb{C}(x,y)$

Question

I would like to compute the Galois group $ G $ of the extension $\mathbb{C}(x(1-y),y(1-x)) \subset \mathbb{C}(x,y)$.

So far I have no idea how to determine $ G $ though I think it should be the Klein group $ C_{2} \times C_{2} $. What I know is that $ x-y \in L $ as it is simply the difference $ x(1-y)-y(1-x) $. Any help would be appreciated. Thank you!

Answer 1

Let $K=\Bbb{C}(x(1-y),y(1-x))$, $L=\Bbb{C}(x,y)$.

Consider the endomorphism $\tau$ of $\Bbb{C}$-algebras of $\Bbb{C}[x,y]$ defined by $$ \tau:x\mapsto 1-y,\quad y\mapsto 1-x. $$ We easily see that $\tau$ is its own inverse, and maps both $x(1-y)$ and $y(1-x)$ to themselves. Therefore $\tau$ induces a $K$-automorphism of $L$. Let's look at the polynomial $$ p(T):=(T-x)(T-\tau(x))=T^2-(x+\tau(x))T+x\tau(x)=T^2-(x-y+1)T+x(1-y). $$ Your crucial observation $x(1-y)-y(1-x)=x-y\in K$ (from an earlier version of the question) implies that $p(T)\in K[T]$. But $p(x)=0$, so we see that $[K(x):K]\le 2$. This settles pretty much everything:

• Because $y-x\in K$, we have $y\in K(x)$, so $L=K(x)$ and $[L:K]\le2$.
• Because $\tau$ is non-trivial, its fixed field is a proper subfield. As $K$ is containted in the fixed field, we conclude that $[L:K]\ge2$, and $p(T)$ is the minimal polynomial of $x$ over $K$.
• Therefore $[L:K]=2$, and the Galois group is generated by $\sigma$.