why $\mathbb Z[\sqrt 2] \ncong \mathbb Z[\sqrt 3]$? The following is a question from section $3.11$ of the book An introduction to abstract algebra by Allenby:

Explain intuitively why $\mathbb Z[\sqrt 2] \ncong \mathbb Z[\sqrt 3]$.back your intuition with proof.  
Note:this example not only says that $\theta: a+b\sqrt 2 \mapsto a+b\sqrt 3$ is not isomorphism .It says no isomorphism can be found at all - no matter how clever choice of mapping you try to make ..  

I can't see what's the intution behind this ..can anyone provide some hint on this...
 A: Hint: $2$ is a square in the first ring. Is it a square in the second?
A: let $\sqrt {2}\in \mathbb {Z}[\sqrt{2}]$
$\phi(\sqrt {2 })^2= \phi (2) =2 $
$\phi (\sqrt {2 })= \pm \sqrt {2} \not\in \mathbb {Z} \sqrt{3} $
A: Proof: Suppose that there exists an isomorphism $\psi : \mathbb{Z}[\sqrt{2}] \to \mathbb{Z}[\sqrt{3}]$. Consider $x = 0 + 1\sqrt{2} = \sqrt{2} \in \mathbb{Z}[\sqrt{2}]$
Note that since $\psi$ is an isomorphism, $\psi(1) = 1$. Hence $\psi(2) = \psi(1 + 1) = \psi(1) + \psi(1) = 2$. Now note also that $\psi(2) = \psi(\sqrt{2}\cdot \sqrt{2}) = \psi(\sqrt{2})\cdot \psi(\sqrt{2}) = \left(\psi(\sqrt{2})\right)^2 = 2$. Since $\psi(\sqrt{2}) \in \mathbb{Z}[\sqrt{3}]$ we must have $\psi(\sqrt{2}) = a+b\sqrt{3}$ for some $a, b \in \mathbb{Z}$, but then we have $\left(\psi(\sqrt{2})\right)^2 = a^2 + 2ab\sqrt{3} + b^2\sqrt{3}$ Since $\left(\psi(\sqrt{2})\right)^2 = 2$ we must have $b = 0$ and thus have $\left(\psi(\sqrt{2})\right)^2 = a^2$. Thus $a = \pm \sqrt{2}$, hence $\psi(\sqrt{2}) = \pm\sqrt{2} \not\in \mathbb{Z}[\sqrt{3}]$ a contradiction. Thus no such isomorphism exists. $\square$
