There are at least three mutually non-isomorphic rings with $4$ elements? Is the following statement is true?

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

I have no idea or counterexample at the moment. Please help. So far I know about that a group of order $4$ is abelian and there are two non isomorphic groups of order $4$ like $K_4(non cyclic)$ and $\mathbb Z_4(cyclic)$.
 A: 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 $R \not\cong S$. 
A: These can be exhaustively enumerated using alg.  To enumerate the number of non-isomorphic rings of orders $1,2,\ldots,8$, we enter:
./alg theories/ring.th --size 1-8 --count

which outputs:
# Theory ring

    Constant 0.
    Unary ~.
    Binary + *.

    Axiom plus_commutative: x + y = y + x.
    Axiom plus_associative: (x + y) + z = x + (y + z).
    Axiom zero_neutral_left: 0 + x = x.
    Axiom zero_neutral_right: x + 0 = x.
    Axiom negative_inverse: x + ~ x = 0.
    Axiom negative_inverse: ~ x + x = 0.
    Axiom zero_inverse: ~ 0 = 0.
    Axiom inverse_involution: ~ (~ x) = x.

    Axiom mult_associative: (x * y) * z = x * (y * z).
    Axiom distrutivity_right: (x + y) * z = x * z + y * z.
    Axiom distributivity_left: x * (y + z) = x * y + x * z.

    size | count
    -----|------
       1 | 1
       2 | 2
       3 | 2
       4 | 11
       5 | 2
       6 | 4
       7 | 2
       8 | 52

Check the numbers [2, 2, 11, 2, 4, 2, 52](http://oeis.org/search?q=2,2,11,2,4,2,52) on-line at oeis.org

If we want it to print out the addition and multiplication tables, we can remove --count from the command line:
./alg theories/ring.th --size 4

If we want to work with unital rings, we can use:
./alg theories/unital_ring.th --size 1-8 --count

which gives:
# Theory unital_ring

    Theory unital_ring.

    Constant 0 1.
    Unary ~.
    Binary + *.

    Axiom plus_commutative: x + y = y + x.
    Axiom plus_associative: (x + y) + z = x + (y + z).
    Axiom zero_neutral_left: 0 + x = x.
    Axiom negative_inverse: x + ~ x = 0.
    Axiom mult_associative: (x * y) * z = x * (y * z).
    Axiom one_unit_left: 1 * x = x.
    Axiom one_unit_right: x * 1 = x.
    Axiom distrutivity_right: (x + y) * z = x * z + y * z.
    Axiom distributivity_left: x * (y + z) = x * y + x * z.

    # Consequences of axioms that make alg run faster:

    Axiom zero_neutral_right: x + 0 = x.
    Axiom negative_inverse: ~ x + x = 0.
    Axiom zero_inverse: ~ 0 = 0.
    Axiom inverse_involution: ~ (~ x) = x.
    Axiom mult_zero_left: 0 * x = 0.
    Axiom mult_zero_right: x * 0 = 0.


    size | count
    -----|------
       1 | 0
       2 | 1
       3 | 1
       4 | 4
       5 | 1
       6 | 1
       7 | 1
       8 | 11

Check the numbers [1, 1, 4, 1, 1, 1, 11](http://oeis.org/search?q=1,1,4,1,1,1,11) on-line at oeis.org

A: 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 $\begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_2 \\ 0 & \mathbb{F}_2 \end{pmatrix} \subseteq M_2(\mathbb{F}_2)$.
