Is $\mathbb{Q}(\sqrt{2}) $ isomorphic to $\mathbb{Q}(\sqrt{-2}) $?

Question

I think the answer is no, and my reasoning is as follows: any isomorphism $ f: \mathbb{Q}(\sqrt{-2}) \to \mathbb{Q}(\sqrt{2}) $ will send $ \sqrt{-2} \mapsto p $ and therefore $ -2 \mapsto p^2 $. But $ f $ will fix $ \mathbb{Q}$, hence $ p^2 = -2 $. But $ \mathbb{Q}(\sqrt{2}) \subset \mathbb{R} $ and every square is nonnegative. Hence no such isomorphism exists. Is this correct?

Answer 1

Everything looks fine to me. You may just want to justify why $f$ fixes $\mathbb{Q}$ and mention that they are not isomorphic as fields.

Answer 2

They are isomorphic as vector spaces over $\mathbb{Q}$ for they have the same dimension, but they're not isomorphic as fields because $2$ is a square in one of them but not in the other one, and the same goes for $-2$.

Answer 3

Another approach is to consider the two fields as vector spaces over $\mathbb Q$. Then if $\alpha\in\mathbb Q[\sqrt{-2}]$, show that the determinant of the linear operation of multiplying by $\alpha$ always has non-negative determinant.

But $\sqrt{2}$ has negative determinant in its field. Since determinants are independent of the basis chosen for the vector space, there can't be any element of $\mathbb Q[\sqrt{-2}]$ that is a square root of $2$.

(This still requires the knowledge that such an isomorphism would have to fix $\mathbb Q$.)

Another way to look at this is that the minimal monic polynomial for $a+b\sqrt{-2}$, with $b\neq 0$, is $x^2-2ax +(a^2+2b^2)$, and thus has positive constant term, but the minimal monic polynomial for $\sqrt{2}$ is $x^2-2$.

Answer 4

No. I will exapnd @Chris Dugale answer: Consider the equation $x^2=-2$ and set $x_0=\sqrt{-2}$. Clearly, $x_0\in\mathbb{Q}(\sqrt{-2})$ and $x_0^2=-2$. If $\varphi:\mathbb{Q}(\sqrt{-2})\rightarrow\mathbb{Q}(\sqrt{2})$ is an isomorphism, then $\varphi(x_0^2)=\varphi(-2)$, that is $\varphi(x_0)^2=-2$. Thus, $\varphi(x_0)$ is a solution of $x^2=-2$ in $\mathbb{Q}(\sqrt{2})$. But this is impossible since $x^2=-2$ is not solvable in $\mathbb{Q}(\sqrt{2})$.