Exercise 5.1.13

Proof. $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2})\subset L=\mathbb{Q}(\sqrt{2},\sqrt{3})$.

$f=x^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$. Indeed, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$, and $f$ has no root in $\mathbb{Q}(\sqrt{2})$, otherwise $\sqrt{3}=a+b\sqrt{2},a,b\in \mathbb{Q}$. But then $3=a^2+2b^2+2\mathit{ab}\sqrt{2}$. If $\mathit{ab}\ne 0$, then $\sqrt{2}=(3-a^2-2b^2)∕(2\mathit{ab})\in \mathbb{Q}$, which is false, thus $\mathit{ab}=0$. If $b=0$, then $\sqrt{3}\in \mathbb{Q}$, and if $a=0$, $\sqrt{3∕2}\in \mathbb{Q}$ : the two cases are impossible. Consequently $f=x^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$.

(This gives an alternative proof of $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$.)

As $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $x^2-3$ over $\mathbb{Q}(\sqrt{2})$, by Proposition 5.1.8 there exists a field isomorphism $\sigma :L\to L$ which is the identity on $\mathbb{Q}(\sqrt{2})$ and which takes $\sqrt{3}$ to $-\sqrt{3}$. As $\sigma$ is the identity on $\mathbb{Q}(\sqrt{2})$, we have also $\sigma (\sqrt{2})=\sqrt{2}$. □