There are at least three mutually non-isomorphic rings with $4$ elements?

By ring, I always mean unital ring. Each of the following rings has four elements:

$R_1 = \mathbb{Z}/4~, ~R_2 = \mathbb{F}_2 \times \mathbb{F}_2 = \mathbb{F}_2[x]/(x^2+x)~, ~R_3 = \mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)~, ~R_4 = \mathbb{F}_2[x]/(x^2)$

They are non-isomorphic because only $R_3$ is a field, only $R_1$ has characteristic $\neq 2$, and only $R_2,R_3$ are reduced.

Conversely, let $R$ be a ring with four elements. If $a \in R \setminus \{0,1\}$, then the centralizer of $a$ is a subgroup of $(R,+)$ with at least three elements $0,1,a$, so by Lagrange also the fourth element has to commute with $a$. Thus, $R$ is commutative. If $R$ is reduced, then it is a finite product of local artinian reduced rings, i.e. fields, so that $R \cong R_2$ or $R \cong R_3$. If $R$ is not reduced, there is some $a \in R \setminus \{0\}$ such that $a^2=0$. Since $0,1,a,a+1$ are pairwise distinct, these are the elements of $R$. If $2=0$, then we get an injective homomorphism $\mathbb{F}_2[x]/(x^2) \to R, x \mapsto a$. Since both sides have four elements, it is an isomorphism. If $2 \neq 0$, the characteristic has to be $4$, i.e. we get an embedding $\mathbb{Z}/4 \to R$, which again has to be an isomorphism.

Of course, this classification can also be obtained by more elementary methods. For other orders, see:

• http://oeis.org/A037291: Number of rings with $n$ elements
• http://oeis.org/A127707: Number of commutative rings with $n$ elements
• http://oeis.org/A127708: Number of non-commutative rings with $n$ elements
• http://oeis.org/A027623: Number of rngs with $n$ elements
• http://oeis.org/A037289: Number of commutative rngs with $n$ elements

The smallest non-commutative ring has $8$ elements and is given by $\begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_2 \\ 0 & \mathbb{F}_2 \end{pmatrix} \subseteq M_2(\mathbb{F}_2)$.

Consider the ring $R = \Bbb{Z}/2\Bbb{Z}[i]$. Alternatively $R$ can be constructed as a quotient

$$R \cong \Bbb{Z}[x]/(2,x^2+1).$$ As a ring $R$ is not isomorphic to either $S = \Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$ or $ T = \Bbb{Z}/4\Bbb{Z}$.

Edit: Perhaps I should add why the ring $R$ is not isomorphic to either $S$ or $T$. Firstly by counting orders of elements $R$ cannot be isomorphic to $T$; $T$ has an element of order 4 while $R$ does not. So now the penultimate question is why is $R$ is not isomorphic to $S$? As groups they are certainly isomorphic but as rings they can't be. The reason is because the presence of $i$ means that the multiplication in $\Bbb{Z}/2\Bbb{Z}[i]$ is not the same as the multiplication in $S$ which is the usual one coming from the product ring structure.

In view of this we see that $R$ has non-trivial nilpotent elements, $(1+i)^2 = 1 - 2i +i^2 = 1 - 1 = 0$ while obviously $\text{nilrad}$ $S = 0$. Thus $R \not\cong S$.