TITLE: Is $2\mathbb{Z}/4\mathbb{Z}$ NOT a field?!
QUESTION [8 upvotes]: According to wikipedia:

If $R$ is a unital commutative ring with an ideal $m$, then $k = R/m$ is a field if and only if $m$ is a maximal ideal. In that case, $R/m$ is known as the residue field. This fact can fail in non-unital rings. For example, $4\mathbb{Z}$ is a maximal ideal in $2\mathbb{Z}$ , but $2\mathbb{Z}/4\mathbb{Z}$ is not a field.

Is $2\mathbb{Z}/4\mathbb{Z}$ NOT a field?! Isn't it isomorphic to $\mathbb{Z_2}$?
I mean every element in $2\mathbb{Z}$ is of the form $2k$ for some $k \in \mathbb{Z}$. Now suppose that I define $\psi: \mathbb{2Z} \to \mathbb{Z}_2$ where $2k \mapsto \bar{0}$ if $k$ is even and $2k  \mapsto \bar{1}$ if $k$ is odd. 
This map is well-defined because we have only one representation as $2k$ for each element of $2\mathbb{Z}$. This map is surjective because at least $0$ is mapped to $\bar{0}$ and $2$ is mapped to $\bar{1}$. 
Is $\psi$ a homomorphism? Well, to me it sounds like it must be. Because the factor of $2$ is playing no role in here and only the parity of $k$ matters for the mapping so I think it is obvious that it is a homomorphism. Even though it can be checked directly by checking all 4 cases separately as well.
So, what is the kernel? Isn't the kernel equal to $4\mathbb{Z}$? Doesn't that mean $2\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}_2$? 
It makes sense that these two are isomorphic to me, where am I wrong?
Also, is the ideal correspondence theorem wrong for non-unital rings???

REPLY [7 votes]: Hint $\ 2\Bbb Z/4\Bbb Z\,$ is not a field since $\,2^2 = 0.\,$ Your map $\psi$ is not a ring hom since they must map the nilpotent $\,2\,$ to a nilpotent ($= 0$ in a field), but you have $\,\psi(2)= 1\,$ by definition. Said in further detail $\ 0 = \psi(0) = \psi(2^2)= \psi(2)^2,\, $ so $\,\psi(2) = 0,\,$ being in a field (or domain).
Remark $\ $ Properties like nilpotent, idempotent, invertible etc. that are purely "ring-theoretic" are always preserved by ring homomorphisms. As above, generally this implies constraints on where these elements may be mapped when defining ring homs.