Prove that $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(i \sqrt{5})$ are not isomorphic. The question is : 

Prove that $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(i \sqrt{5})$ are not isomorphic (I'm talking about ring isomorphism).

What I have done : suppose there is an isomorphism $f:\mathbb{Q}(\sqrt{5})\to \mathbb{Q}(i\sqrt{5})$.
We have $f(1)=1$ and 
$$f(n+1)=f(n)+1, \qquad f(n-1)=f(n)-1.$$
Which leads to $f(x)=x$ for all $x\in\mathbb{Z}$, but $f(\sqrt{5})^2 =f(5)=5.$ then $f(\sqrt{5}) \in \{\sqrt{5},-\sqrt{5}\}$, which is not possible since $\pm \sqrt{5} \notin \mathbb{Q}(i\sqrt{5})$. Then, there is no such morphism $f$.
Am I right?

They are not isomorphic as fields: While $\mathbb{Q}(i \sqrt{5})$ contains a square root of $-5$ (namely $i \sqrt{5}$), the field $\mathbb{Q}(\sqrt{5})$ does not (as it is a subfield of $\mathbb{R}$).
They are however isomorphic as $\mathbb{Q}$-vector spaces, as they are both two-dimensional over $\mathbb{Q}$ (because $\mathbb{Q}(\sqrt{5})$ has $\{1, \sqrt{5}\}$ as a $\mathbb{Q}$-basis, and $\mathbb{Q}(i\sqrt{5})$ has $\{1, i\sqrt{5}\}$ as a $\mathbb{Q}$-basis).
