Showing that $\mathbb{Q}(\sqrt{3})$ isomorphic to $\mathbb{Q(\sqrt{-3})}$. (or possibly disprove it) How can I show that $\mathbb{Q}(\sqrt{3})$ is isomorphic to $\mathbb{Q(\sqrt{-3})}$. (or possibly disprove it)?
What I know:
I don't know how to begin if it is the case that they are not isomorphic.
In the case that they are isomorphic:
I can show that they are isomorphic to one another by producing a bijective homomorphism.
I consider $\phi:\mathbb{Q}(\sqrt{3})\to\mathbb{Q}(\sqrt{-3})$ however I am having trouble thinking of a mapping for the elements that works.
$a+\sqrt{3}b$ maps to ??
 A: It seems that your problem is in checking that your map is a homomorphism. 
I claim that no map $\varphi: \mathbb Q(\sqrt{-3}) \to \mathbb Q(\sqrt 3)$ can be a field homomorphism.  
For assume such a map existed, then it would have to fix $\mathbb Q$ and we have that $-3 = \varphi(-3) = \varphi(\sqrt{-3} \sqrt{-3}) = \varphi(\sqrt{-3}) \varphi(\sqrt{-3})$ implying that $\varphi(\sqrt{-3}) \in \mathbb Q(\sqrt 3)$ squares to $-3$ which is a contradiction as $\mathbb Q(\sqrt 3) \subset \mathbb R$. 
A: Note that $3 = \phi(3) = \phi((0+\sqrt{3})(0 + \sqrt{3}))$, but:
$-3 = (0 + \sqrt{-3})(0 + \sqrt{-3}) = \phi((0 + \sqrt{3}))\phi((0 + \sqrt{3}))$, so the $\phi$ you exhibited is not a ring-homomorphism.
To put this situation a different way: $x^2 + 3$ splits in one field, but not the other, so how can they be isomorphic?
A: The two are not isomorphic, suppose there is an ring isomorphism $\phi:\mathbb{Q}(\sqrt{3})\to\mathbb{Q}(\sqrt{-3})$,then $\phi(1)=1, \phi(0)=0$, which implies that $\phi(x)=x$ for all $x\in\mathbb Q$. then since $\sqrt{3}^2-3=0$, we must have $(\phi(\sqrt{3}))^2-3=0$, but there is no solution of $x^2-3=0$ in $\mathbb{Q}(\sqrt{-3})$ (every element in $\mathbb{Q}(\sqrt{-3})$ is of form $a+b\sqrt{-3}$ with $a,b\in\mathbb Q$, if $b\neq0$, the corresponding minimal polynomial is $(x-a-b\sqrt{-3})(x-a+b\sqrt{-3})$, which has no real root).
