Show that $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic as rings. One can find various proofs on this site showing that $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic by supposing there is an isomorphism and computing what happens to certain elements and deriving a contradiction. 
However, I am wondering if my proof is also valid? I have not found a posted question that uses this proof: 
Note that $2\mathbb{Z}$ and $3\mathbb{Z}$ are both ideals of $\mathbb{Z}$ because they are subrings that are closed under multiplication. 
So, then $\mathbb{Z}$/$2\mathbb{Z}$ and $\mathbb{Z}$/$3\mathbb{Z}$ are both quotient rings. However, $\mathbb{Z}$/$2\mathbb{Z} = \{\bar{0}, \bar{1}\}$ while $\mathbb{Z}$/$3\mathbb{Z} =  \{\bar{0}, \bar{1}, \bar{2}\}$, so the quotient rings are not isomorphic because they have a different cardinality. Can I conclude that $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are not isomorphic, because otherwise their quotient rings would be isomorphic? 
 A: You are implicitly claiming the following: if two ideals $I_1,I_2$ in a ring $R$ are isomorphic as rings, then the quotient rings $R/I_1$ and $R/I_2$ are isomorphic.
To see that this is false, consider the following example, let $\mathbb Z ^0$ be the ring whose underlying additive group is $(\mathbb Z, +)$, but with trivial multiplication, i.e. $\forall a,b \in \mathbb Z^0: ab = 0$. Then every subgroup of $(\mathbb Z, +)$ is an ideal of $\mathbb Z^0$. The subgroups $2\mathbb Z^0$ and $3 \mathbb Z^0$ are both isomorphic to $\mathbb Z^0$ itself, but the quotients $\mathbb Z^0 / 2 \mathbb Z^0$ and $\mathbb Z^0 / 3\mathbb Z^0$ are not.
If you want a unital example, consider $R = \displaystyle \prod_{n=1}^{\infty}\mathbb Z$, then $ I =\{0\} \times \displaystyle \prod_{n=2}^{\infty}\mathbb Z$, is an ideal which is isomorphic to $R$ itself as a ring (the isomorphism is a simple right shift). But we have $R/R = 0$, whereas $R/I \cong \mathbb Z$
A: These are rings without unity. In $2\Bbb Z$, $x=2$ solves $x^2=x+x$.
What about in $3\Bbb Z$?
