Understanding that $\mathbb{R}(X^2 + Y^2, XY)(x) \supset \mathbb{R}(Y)$?

I'm trying to show that $\mathbb{R}(x^2 + y^2, xy)(x) = \mathbb{R}(x,y)$, where the $\mathbb{R}(x)$ denotes the field of rational functions in $x$. I have an idea of how I might go about proving this, but there's something I'm trying to understand first.

Starting with - how can I show that $\mathbb{R}(x^2 + y^2, xy)(x) \supset \mathbb{R}(y)$? I've tried to write out what $\mathbb{R}(x^2 + y^2, xy)(x)$ and $\mathbb{R}(y)$ means in sigma notation, and I've got

$\mathbb{R}(x^2 + y^2, xy)(x) = \sum_{i,a,b,c = 0}k_i(x^2+y^2)^a(xy)^b(x)^c $ and $\mathbb{R}(y) = \sum_{j,d = 0} = p_j(y)^d$.

But how do you obtain $\mathbb{R}(y)$ from this? Can I just set $x=1$? Then how would I show that $\mathbb{R}(x^2 + y^2, xy)(x) \supset \mathbb{R}(x)$... would I just pick $y=1$?

Solution 1:

What you're doing isn't really the right way to go about this. Your expressions cannot build the element $\frac{1}{x^2+y^2+xy}$, for instance. (You're also claiming that a field is equal to an element, which is problematic, but I'm taking a guess at what you really mean.)

The right way to do this is to show that $y\in\Bbb R(x^2+y^2,xy)(x)$: since $\Bbb R(y)$ is the smallest field extension of $\Bbb R$ containing $y$, the presence of $y$ is all you need to conclude the containment you're after. Since $y=\frac{xy}{x}$, you're done. $\Bbb R(x)$ is easier: $x\in \Bbb R(x^2+y^2,xy)(x)$ because we adjoined $x$.