Let $p $ and $q$ be two distinct primes.
$\mathbb Q(\sqrt p)$ is isomorphic to $\mathbb Q(\sqrt q)$ as fields.
$\mathbb Q(\sqrt p)$ is isomorphic to $\mathbb Q(\sqrt {-q})$ as vector spaces over $\mathbb Q$.
my try:
2nd option is correct because $\mathbb Q(\sqrt p)$ and $\mathbb Q(\sqrt {-q})$ both are $ 2$ degree extension over $\mathbb Q$. So as a vector space $\mathbb Q(\sqrt p)\cong $$\mathbb Q^2\cong \mathbb Q(\sqrt {-q})$. Is it correct, please check$?$
Yes, correct. The option (a) is false because if $\sigma:\Bbb Q(\sqrt p)\to\Bbb Q(\sqrt q)$ is an isomorphism, then let $\alpha=\sigma^{-1}(\sqrt q)$. Then $\sigma(\alpha^2)=q$, so $\alpha^2=q$, but there is no square root of $q$ in $\Bbb Q(\sqrt p)$.