Are $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ isomorphic? I would like to show whether $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ are isomorphic.
I tried noting that $\pm\sqrt2$ and $\pm i\sqrt2$ are the roots of $x^4-4=\left(x^2+2\right)\left(x^2-2\right)$ and that if $x^4-4$ were irreducible over $\mathbb Q$, then
$$\mathbb Q\left[x\right]/\left(x^4-4\right)\cong\mathbb Q\left(\sqrt2\right)\cong\mathbb Q\left(i\sqrt2\right),$$
i.e., $\sqrt2$ and $i\sqrt2$ would be algebraically indistinguishable. Therefore, $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ are not isomorphic since $x^4-4$ is not irreducible over $\mathbb Q$.
Is this approach valid? Is there a better one? Thanks in advance.
 A: The answer is no. Note that for this isomorphism each rational number must be mapped to itself: let $\phi$ be the required field isomorphism. Then a property of field homomorphism is that $\phi(q)=q$ $\forall q\in\mathbb{Q}$. More explicitly, given a field $F$, a field homomorphism must satisfy $\phi(ax+b)=\phi(a)\phi(x)+\phi(b)$ for $a,x,b\in F$. Therefore, any rational number $m/n$ ends up \begin{equation}\phi\left(\frac{m}{n}\right)=\frac{\phi(m)}{\phi(n)}=\frac{m\phi(1)}{n\phi(1)}=\frac{m}{n}\end{equation}
Therefore, we are left with $\phi(\pm\sqrt{2})=\pm i\sqrt{2}$. For example, suppose $\phi(\sqrt{2})=i\sqrt{2}$. Then
\begin{equation}\phi(\sqrt{2})^{2}=\phi(2)=2=(i\sqrt{2})^{2}=-2\end{equation}
a contradiction. You can easily show that any combination of $\pm$ signs lead to similar contradiction. Thus the supposed isomorphism cannot exist.
The reason I show this explicitly is because this is actually a standard method of eliminating certain isomorphisms, which become important when you reach Galois theory even!
As for your method, I think you are right that if a polynomial is irreducible in one field and not in the other field, then they are not isomorphic. I think you need to explicitly prove that indeed it is the case, because I think you may have "begged the question": the statement that they are algebraically indistinguishable is precisely the question.
A: This might be seen as just an elaboration of Slade's answer, but the idea is very useful.
Let $L,K$ be to field extensions of $\mathbb Q$. As eri argues, any homomorphism between $L$ and $K$ fixes $\mathbb Q$. Now, let $\varphi\colon L\to K$ be a homomorphism and $p\in\mathbb Q[x]$ a polynomial. Note that for any $\alpha\in L$ we have $\varphi(p(\alpha)) = p(\varphi(\alpha))$ precisely because $\varphi$ is ring homomorphism and fixes $\mathbb Q$. In particular, it means that if $\alpha$ is a root of polynomial $p$, $\varphi(\alpha)$ is a root of the same polynomial as well.
Now, if there were a homomorphism $\varphi\colon\mathbb Q[i\sqrt 2]\to\mathbb Q[\sqrt 2]$, it would map $i\sqrt 2$ to a root of $x^2 + 2$ (i.e. $\pm i\sqrt 2$), but those are not elements of $\mathbb Q[\sqrt 2]$. There is no $\varphi\colon\mathbb Q[\sqrt 2]\to\mathbb Q[i\sqrt 2]$ for the same reason.
A: $\mathbb{Q}(i\sqrt{2})$ contains a number whose square is $-2$.
But every element of $\mathbb{Q}(\sqrt{2})$ is a real number, and therefore has a nonnegative square.
A: This can be proved in the same way as here. For this reason, one might add some different arguments here: for square-free $d$, the discriminant $D$ of $\mathbb{Q}(\sqrt{d})$ is an invariant of the field. We have $d=8$ for $\mathbb{Q}(\sqrt{2})$ and $D=-8$ for
$\mathbb{Q}(\sqrt{-2})$. 
Also the unit group of the ring of integers is different. In case of $K=\mathbb{Q}(\sqrt{2})$ we have $\mathcal{O}_K^{\times}\cong \pm 1\times \mathbb{Z}$, whereas for $K=\mathbb{Q}(\sqrt{-2})$ we have $\mathcal{O}_K^{\times}\cong \pm 1$ according to Dirchlet's unit theorem.
