- $\mathbb{Z}$ and $\mathbb{2Z}$
My solution: To prove they are isomorphic as groups, I take the mapping $f: \mathbb{Z} \rightarrow \mathbb{2Z}$ defined by $f(x)=2x$. I prove it's a homomorphism and surjective and I am done.
To prove they are not isomorphic as rings, I take the equation $x^2=1$. It has solutions $x=1,-1$ in $\mathbb{Z}$ but no solutions in $\mathbb{2Z}$ and are hence not isomorphic.
- $\mathbb{Z}[\sqrt2]$ and $\mathbb{Z}[\sqrt5]$
My solution: To prove they are isomorphic as groups, I take the mapping
$f: \mathbb{Z}[\sqrt2] \rightarrow \mathbb{Z}[\sqrt5]$ defined by $f(a+b\sqrt2)=a+b\sqrt5$. I prove it's a homomorphism and surjective and I am done.
Here, to prove they are not isomorphic as rings, I take the equation $x^2=2$, which has solutions $x=\sqrt2, -\sqrt2$ in $\mathbb{Z}[\sqrt2]$ but no solution in $\mathbb{Z}[\sqrt5]$.
Is this the correct approach to proving non-isomorphism as rings? That an equation has a solution in one ring but not in another?