so I know how to prove that $\mathbb{Z}\left[\sqrt{2}\right]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ and $\mathbb{Z}\left[\sqrt{3}\right]=\{a+b\sqrt{3}:a,b\in\mathbb{Z}\}$ are subrings of $\mathbb{R}$. However, I don't understand the following proofs. Thanks a lot for your help! :)
Question: Give a one sentence intuitive reason why the ring $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$ cannot be isomorphic.
Answer: $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$ are not isomorphic because $2$ is not a square in $\mathbb{Z}[\sqrt{3}]$ but it is in $\mathbb{Z}[\sqrt{2}]$.
What I don't understand: What does it mean by "$2$ is not a square in $\mathbb{Z}[\sqrt{3}]$ but it is in $\mathbb{Z}[\sqrt{2}]$"?
And,
Question: Convert your intuition into a rigorous proof.
Answer: If $\Psi:\mathbb{Z}[\sqrt{2}] \to \mathbb{Z}[\sqrt{3}]$ then $\Psi(1)=1, so \Psi(2)=\Psi(1+1)=\Psi(1)+\Psi(1)=1+1=2$ and hence $\Psi(\sqrt{2})=x$ has the property that $x^2=\Psi(\sqrt{2})^2=\Psi((\sqrt{2})^2)=\Psi(2)=2$.
Let $x=a+b\sqrt{3}$. We deduce $(a^2+3b^2)+2ab\sqrt{3}=2$ $$\implies\sqrt{3}=\frac{2-a^2-3b^2}{2ab}\in\mathbb{Q}\text{ (since $\sqrt{3}$ is irrational),}$$
provided $ab\ne 0$.
If $a=0$, then $x=b\sqrt{3}$ so $x^2=3b^2=2$, which is impossible.
And if $b=0$, then $x^2=a^2=2$, which is also impossible.
What I don't understand: what is $\Psi$? I don't understand how it gives those results for $1$ and $\sqrt{2}$.
