Finding number of automorphisms?

Question

Can anyone please help to find the total number of automorphisms of $\mathbb{Q}(\sqrt{2},\sqrt{3})$? Prime field of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is $\mathbb{Q}$. Can anyone please explain how to see it?

Answer 1

Let $\phi$ be an automorphism of $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then $\phi(\sqrt{2})^2=\phi(\sqrt{2}^2)=\phi(2)=2$ since $\phi_{\mid \mathbb{Q}}=Id_{\mathbb{Q}}$. Thus $\phi(\sqrt{2})$ is a root of the equation $x^2-2=0$. Thus either $\phi(\sqrt{2})=\sqrt{2}$ or $-\sqrt{2}$. Similarly, $\phi(\sqrt{3})=\sqrt{3}$ or $-\sqrt{3}$.

It is very straightforward to show that determining the values of $\phi(\sqrt{2})$ and $\phi(\sqrt{3})$ completely defines $\phi$. We thus found that there are at most $4$ such automorphisms. Moreover each of these four lead to well-defined automorphisms. Hence the automorphism group $\text{Aut}(\mathbb{Q}(\sqrt{2},\sqrt{3}))\cong \mathbb{Z}_2\oplus \mathbb{Z}_2$ since non of the automorphism has order $4$.

In case you are wondering why $\phi_{\mid \mathbb{Q}}=Id_{\mathbb{Q}}$, notice that if $\phi(1)=0$, then $\phi=0$. This is impossible since $\phi$ is an automorphism and thus bijective. Since $\phi(1)=\phi(1^2)=\phi(1)^2$, we have that $\phi(1)=1$. But then $\phi(n)=\phi(\sum_{i=1}^n1)=\sum_{i=1}^n\phi(1)=n$ for all $n\in \mathbb{N}$. It follows that $\phi(-n)=n$ since $\phi$ is a group homomorphism for the sum. Notice that $n\frac{1}{n}=1$, applying $\phi$ to this equality yields $\phi(n)\phi(\frac{1}{n})=\phi(1)$, thus $\phi(\frac{1}{n})=\frac{1}{n}$. It follows that $\phi$ is the identity on $\mathbb{Q}$.