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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 $\mathbb Z_4(cyclic)$.

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How many ring structures are there on $K_4$ (as the additive group)? –  Jyrki Lahtonen Jan 15 '13 at 17:24
    
I have no idea, I know that additive group of the ring is abelian. thats all.please teach me. –  El Angel Exterminador Jan 15 '13 at 17:26
3  
I think that $a^2$ can be any of the four elements. The choices $a^2=1$ and $a^2=0$ seem to lead to isomorphic rings (appears in BenjaLim's +1 list). The choice $a^2=a$ leads to another ring on his list. The choice $a^2=1+a$ leads to a finite field of four elements (not on his list). Anyway, do satisfy yourself about the fact that when the additive group is $K_4$, knowing $a^2$ gives you everything about the ring operations. –  Jyrki Lahtonen Jan 15 '13 at 17:36
1  
Hint: There are, up two isomorphism, exactly two groups with four elements. Find these groups and check whether they are rings. –  Johannes Kloos Jan 29 '13 at 10:07
3  
It is not a dublicate, because the other question didn't ask for a classification. @Johannes: It doesn't make sense to ask whether a group is a ring. Of course you know this, but other readers do not. –  Martin Brandenburg Jan 29 '13 at 10:30

3 Answers 3

up vote 8 down vote accepted

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:

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

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Hm? I have posted this answer elsewhere. Did someone merge it? –  Martin Brandenburg Jan 29 '13 at 20:50
    
PS: The same classification also works for rings with $p^2$ elements, where $p$ is any prime. See George's answer here: math.stackexchange.com/questions/305512 –  Martin Brandenburg Feb 16 '13 at 15:26

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
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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 $R \not\cong S$.

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@JyrkiLahtonen I have edited my answer above. –  user38268 Jan 16 '13 at 8:33
    
@YACP I have added some explanations. –  user38268 Jan 16 '13 at 8:33

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