Let F = $\mathbb{Q}(2^{1/2}$,$3^{1/2}$) be a field extension of $\mathbb{Q}$. How do I show that the map determined by $f(2^{1/2}$) = -$2^{1/2}$ and f($3^{1/2}$)=$3^{1/2}$ is a $\mathbb{Q}$-automorphism?
I assume it suffices to show that the function is well defined, and other properties just inherit from field multiplication distributivity? I need to show any two equivalent ways of expressing an element in F will be mapped to the same element. But, this seems quite difficult.
On
We have $\mathbb{Q}(\sqrt{2},\sqrt{3}) \cong \mathbb{Q}[x,y]/(x^2-2,y^2-3)$. This isomorphism is induced by the ring homomorphism $\mathbb{Q}[x,y] \to \mathbb{Q}(\sqrt{2},\sqrt{3})$ induced by $x \mapsto \sqrt2$ and $y\mapsto \sqrt3$.
Now the ring homomorphism $\mathbb{Q}[x,y] \to \mathbb{Q}[x,y]$ induced by $x \mapsto -x$ and $y\mapsto y$ is an automorphism of $\mathbb{Q}[x,y]$.
Combining these two isomorphisms gives an automorphism of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ such that $\sqrt2 \mapsto -\sqrt2$ and $\sqrt3 \mapsto \sqrt3$.
One way is to see that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a Galois extension of $\mathbb{Q}$ of degree $4$. Therefore, there are 4 $\mathbb{Q}$ automorphisms of this field. You know that any $\mathbb{Q}$ automorphism must send $\sqrt{2}$ to $\pm \sqrt{2}$ and $\sqrt{3} \to \pm \sqrt{3}$. These exhaust the 4 possibilities!