Question about algebraic closures Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. Let $K = \mathbb{Q}(\sqrt{d})$ and $\overline{K}$ defined to be the algebraic closure of $K$. Is it true that $\overline{K} \cong \overline{\mathbb{Q}}$?
 A: Algebraic closures are only defined up to isomorphism, so it doesn't make sense to ask if $\overline K$ and $\overline{\Bbb{Q}}$ are equal. It does make sense to ask if they are isomorphic, which they are: $\overline K$ is an algebraic extension of $\Bbb Q$ (since it's an algebraic extension of an algebraic extension) and is algebraically closed, and hence is an algebraic closure of $\Bbb Q$.
A: If $d \in \mathbb{Q}$, $K$ will be a subfield of $\overline{\mathbb{Q}}$, since $\sqrt{d} \in \overline{\mathbb{Q}}$. Since $\overline{\mathbb{Q}}$ is algebraically closed, it suffices to show that $\overline{\mathbb{Q}}$ is algebraic over $K$. But if $\alpha \in \overline{\mathbb{Q}}$, choose a polynomial in $\mathbb{Q}[x]$ with $\alpha$ as a root. But then this polynomial also has coefficients in $K$, and so $\alpha$ is algebraic over $K$, too. Hence $\overline{\mathbb{Q}}$ is an algebraically closed field that is algebraic over $K$. These are the defining properties of an algebraic closure, so $\overline{K} = \overline{Q}$. Up to isomorphism, of course.
