# Find the Galois group $G= Gal(L/K)$ where $L= \mathbb{R}(x,y)$ and $K=\mathbb{R}(x^2,y^2)$

Find the Galois group $$G= Gal(L/K)$$ where $$L= \mathbb{R}(x,y)$$ and $$K=\mathbb{R}(x^2,y^2)$$

My attempt :The minimial polynomial $$f(a)$$ is $$(a^2-x^2)(a^2-y^2)$$

$$f(a)=(a^2-x^2)(a^2-y^2) \in \mathbb{R}(x^2,y^2)(a)=K(a)$$

The polynomial $$f(a)$$ has a degree of $$4$$

$$|Gal(L/K)|=|Aut(L/K)|=4 \implies \mathbb{Z}_4$$ or $$\mathbb{Z}_2 \times \mathbb{Z}_2$$

I think $$Gal(L/K)=\mathbb{Z}_4$$

• Try to determine some elements of the Galois group. Commented Aug 1 at 14:12
• This is very confused. One speaks of the minimal polynomial of an element, not a pair of elements. You've chosen the strange name $a$ for the polynomial variable, and so the minimal polynomial of $x$ will be $a^2-x^2\in \mathbb{R}(x,y)[a]$ (not in $K$ as you assert. Commented Aug 1 at 14:16
• But surely you know that the polynomial ring is $K[a]$ not $K(a)$. Commented Aug 1 at 14:34
• And you still have the irrelevant polynomial $(a^2-x^2)(a^2-y^2)$. Commented Aug 1 at 14:35
• Judging from your latest questions you may be a bit out of your depth. Here you could think of this extension as being of the type you get by adjoining two square roots - quite like $\Bbb{Q}(\sqrt2,\sqrt3)/\Bbb{Q}$. True, this example wanders away from the realm of number fields, the usual playground of first examples in Galois theory. Here the algebra is actually made a bit simpler due to algebraic independence of $x$ and $y$, saving you from the counterpart of having to prove that $\sqrt3\notin\Bbb{Q}(\sqrt2)$. Commented Aug 3 at 3:19

Here is how I would tackle this question.

Let $$L=\mathbb{R}(x,y)$$, and let $$K=\mathbb{R}(x^2,y^2)$$.

That is $$L$$ is the field of rational functions in $$x,y$$ and $$K$$ is the subfield of those elements of $$L$$ which are functions of $$x^2,y^2$$. To avoid doubt: it is intended that $$x,y$$ are algebraically independent, in the sense that the only polynomial in $$x,y$$ which is zero is the zero polynomial.

Now consider the polynomial $$f(X)\in K[X]$$ given by $$f(X):=(X^2-x^2)(X^2-y^2)$$. It is clear that $$f$$ splits completely into linear factors in $$L[X]$$, as $$f(X)=(X-x)(X+x)(X-y)(X+y)$$. As $$L=K[x,y]$$ it is clear no smaller extension of $$K$$ splits $$f$$, and so $$L$$ is the splitting field of $$f$$ over $$K$$. Hence $$L$$ has a Galois Group over $$K$$, call it $$G$$.

How big is $$|L:K|$$?

Well we have that $$L=K(x)(y)$$, and so $$|L:K|=|L:K(x)||K(x):K|$$. Now $$|L:K(x)|=|K(x)(y):K(x)|\leqslant |K(y):K|$$. So we have that $$|L:K|\leqslant|K(y):K||K(x):K|$$. Now notice that $$x$$ satisfies $$m_x(X)=X^2-x^2\in K[X]$$; as neither of $$(X\pm x)$$ has coefficients in $$K$$ we see that $$m_x$$ is the minimal polynomial of $$x$$ over $$K$$, and hence $$|K(x):K|=2$$.

That is $$|L:K|\leqslant 4$$.

We now wish to find the $$K$$-automorphisms of $$L$$. Note that each of these is determined by its effects on $$x,y$$; and since $$x$$ must map to one of the roots of $$m_x$$ we see that there are at most $$4$$ automorphisms, given by $$\sigma_0:(x,y)\mapsto (x,y)$$, $$\sigma_1:(x,y)\mapsto (-x,y)$$, $$\sigma_2:(x,y)\mapsto (x,-y)$$, $$\sigma_3:(x,y)\mapsto (-x,-y)$$. As $$x,y$$ are algebraically independent always get a homomorphism when we substitute any $$(\phi(x,y),\psi(x,y))$$ for $$(x,y)$$; but the ones we have specified are clearly onto, so are automorphisms. Moreover, as they do different things to $$(x,y)$$ they are all different, so that the order of the Galois group $$G$$ is at least $$4$$.

Putting this together we have that $$|L:K|=|G|=4$$. Moreover $$G=\{\sigma_0,\sigma_1,\sigma_2,\sigma_3\}$$, and as each $$\sigma$$ squares to the identity we have that $$G\simeq C_2\times C_2$$.