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

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.

• What's your definition of 'an' algebraic closure? If it's 'a set containing the specified set which is algebraically closed' then this statement is false: $\bar{\mathbb{Q}}$ can't be isomorphic to $\mathbb{R}$ (for one thing, they have different cardinalities). If it's the minimal algebraic closure, then this would seem to be basically trivial somewhat along the lines you outline. – Steven Stadnicki Sep 21 '17 at 18:16
• @StevenStadnicki: Presumably, the OP means the usual definition. Algebraic closures aren't unique in the absolute sense; they are only unique up to isomorphism of field extensions. (incidentally, you probably meant $\mathbb{C}$ for your example) – Hurkyl Sep 21 '17 at 18:18
• Oh! Of course, $\mathbb{R}$ isn't algebraically closed. Apparently the coffee hasn't kicked in this morning. :P But I'm not sure what you mean by 'not unique in the absolute sense' here; at least over $\mathbb{Q}$ it's certainly unique as a subset of $\mathbb{C}$, no isomorphism needed, no? – Steven Stadnicki Sep 21 '17 at 18:39

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 $\overline{K} = \overline{L}$.