Is the ring $3\mathbb Z$ a ring homomorphic image of the ring $2\mathbb Z$. Is the ring $2\mathbb Z$ isomorphic to the ring $3\mathbb Z$ ?
Solution: Let if possible,$\phi:\mathbb {2Z\to 3Z}$ be a ring isomorphism.
Then $\phi$ is a group isomorphism between the additive groups $2\mathbb Z$ and $3\mathbb Z$.
Now,$\phi(2)$ completely determines $\phi$.Let $\phi(2)=3x$ where $x\in \mathbb Z$.
Then $\phi(2y)=3xy $ for all $y\in \mathbb Z$.
Now,$\phi$ is surjective,so $3\in 3\mathbb Z$ has a preimage.So,$\phi(2y)=3$ for some $y\in \mathbb Z$.
Which implies,$3xy=3\implies xy=1$.So,$x,y\in \{-1,1\}$.But then $\phi(2)=\pm3$ and hence $\phi(4)=\phi(2.2)=\phi(2).\phi(2)=9$.
Again,$\phi(4)=\phi(2+2)=\phi(2)+\phi(2)=\pm 6$,but that is not possible.
So,$2\mathbb Z$ and $3\mathbb Z$ are not ring isomorphic.
Note: $3\mathbb Z$ is not a ring homomorphic image of $2\mathbb Z$,as we did not make use of injectivity of $\phi$ in the above proof.
Is the above conclusion correct?
 A: Your argument is correct.
But if you strip it of all inessential parts, you get a shorter, in my opinion also clearer, argument that actually proves something more.
The key point is, as you did, to look at $\phi(4)$, where $\phi$ is an arbitrary rng-homomorphism from $2\mathbb{Z}$ to some sub-rng of $\mathbb{Z}$. Since $4 = 2\cdot 2 = 2 + 2$ it follows that
$$0 = \phi(4) - \phi(4) = \phi(2\cdot 2) - \phi(2+2) = \phi(2)^2 - \bigl(\phi(2) + \phi(2)\bigr) = \phi(2)\cdot \bigl(\phi(2) - 2\bigr)\,.$$
Thus either $\phi(2) = 0$ or $\phi(2) = 2$. Hence either $\phi = 0$ or $\phi$ is the inclusion of $2\mathbb{Z}$ in $\mathbb{Z}$, or $\phi$ is the identity map on $2\mathbb{Z}$.
In particular, the only rng-homomorphism $2\mathbb{Z} \to 3\mathbb{Z}$ is the zero map.
A: Your argument seems correct. But there's an easier way: if $2\Bbb Z$ and $3\Bbb Z$ were isomorphic, then the quotient rings $\Bbb Z/3\Bbb Z$ and $\Bbb Z/2\Bbb Z$ would be isomorphic too. But this can't be possible since $\Bbb Z/3\Bbb Z$ has precisely 3 elements and $\Bbb Z/2\Bbb Z$ has precisely 2 elements.
