Let $p $ and $q$ be two distinct primes.

  1. $\mathbb Q(\sqrt p)$ is isomorphic to $\mathbb Q(\sqrt q)$ as fields.

  2. $\mathbb Q(\sqrt p)$ 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})$. Is it correct, please check$?$

  • $\begingroup$ I removed the finite-field tag as all the fields appearing in the question are extensions of $\Bbb{Q}$ and therefore have infinitely many elements. $\endgroup$ – Jyrki Lahtonen Jan 2 '17 at 11:10

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 $\Bbb Q(\sqrt p)$.

  • $\begingroup$ Because $\sqrt{q} \in \mathbb{Q}(\sqrt{p})$ means $\sqrt{q} = a\sqrt{p}+b$ and $q = (a\sqrt{p}+b)^2 = a^2p+b^2+2a\sqrt{p}b$ a contradiction if $a,b \in \mathbb{Q}$ $\endgroup$ – reuns Jan 2 '17 at 11:58

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