TITLE: $2\mathbb{Z}$ is not ring isomorphic to $3\mathbb{Z}$, But Why? (Verify Me)
QUESTION [6 upvotes]: I want to prove the ring of even integers $2\mathbb{Z}$ is not isomorphic to $3\mathbb{Z}$. I do this by contradiction, assume $\phi$ is such an isomorphism from $2\mathbb{Z}$ to $3\mathbb{Z}$. Observe this equality:
$$\phi(2)+\phi(2)=\phi(2+2)=\phi(2\cdot2)=\phi(2)\cdot\phi(2)$$
Let $x=\phi(2)$ and we have the equation $x^2-2x=0$ in $3\mathbb{Z}$. This factors into $x(x-2)=0$.
$\phi(2)=0$ is not possible because any isomorphism will have $\phi(0)=0$.
Thus we are left with $x-2=0$ which has no solution in $3\mathbb{Z}$.
Question. Is this a valid argument? What does the $2$ symbol in the last equation denote, if not an element of $3\mathbb{Z}$?

REPLY [2 votes]: The argument is valid, because $2\mathbb{Z}$ and $3\mathbb{Z}$ are subrings of $\mathbb{Z}$ and nothing prevents you from using this larger ring for doing computations in.
A simpler argument is that the equation $x+x=6$ has solution in $3\mathbb{Z}$, but no solution in $2\mathbb{Z}$.