Any algebraic closure of $\mathbb{Q}(\sqrt{2})$ is isomorphic to any algebraic closure of $\mathbb{Q}(\sqrt{17})$

Question

Any algebraic closure of $\mathbb{Q}(\sqrt{2})$ is isomorphic to any algebraic closure of $\mathbb{Q}(\sqrt{17})$

I know that $\sqrt{2}$ and $\sqrt{17}$ are algebraic in $\mathbb{Q}$, so if $A$ is an algebraic closure of $\mathbb{Q}(\sqrt{2})$ then this would be isomorphic to a closure of $\mathbb{Q}$ which in turn is isomorphic to an algebraic closure of $\mathbb{Q}(\sqrt{17})$? Who could I help to make a more convincing argument please? Thank you very much.

Answer 1

More generally, let $K$ be algebraic over $\mathbb Q$. We may suppose for convenience that $K \subseteq \mathbb C$.

Now, $\mathbb Q \subseteq K$ implies $\overline{\mathbb Q} \subseteq \overline{K}$. On the other hand, $\overline{K} \subseteq \overline{\mathbb Q}$ because $K$ is algebraic over $\mathbb Q$.

Therefore, $\overline{K} = \overline{\mathbb Q}$.

Apply this to $K = \mathbb{Q}(\sqrt{2})$ and $L = \mathbb{Q}(\sqrt{17})$ to conclude that $\overline{K} = \overline{L}$.