Find the smallest non-commutative ring with unity. (By smallest it means it has the least cardinal.)

I tried rings of size 4 and I found no such ring.

  • $\begingroup$ Rumor has it that $\{0\}$ is a perfectly-defined commutative ring with unity XD $\endgroup$
    – Hui Yu
    Dec 25 '13 at 16:28
  • $\begingroup$ It's easy finding one with 16 elements. $\endgroup$
    – egreg
    Dec 25 '13 at 16:33
  • $\begingroup$ @HuiYu but I nees a non commutative ring. $\endgroup$
    – pardis
    Dec 25 '13 at 16:36
  • $\begingroup$ By Wedderburn's little theorem, it must have zero divisors. I don't know how much that helps. $\endgroup$ Dec 25 '13 at 17:16
  • 1
    $\begingroup$ The smallest one is of cardinality $8$. $\endgroup$ Dec 25 '13 at 23:01

In an earlier version of this post I caused an accident by giving a wrong answer. Thanks to those who pointed out the error! Here is a little update:

The ring $M_2(\Bbb F_2)$ of $2 \times 2$-matrices with entries in $\Bbb F_2$ is a non-commutative ring with 16 elements, because $$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$ It has a subring of order $8$, namely the upper triangular matrices which are non-commutative by the example above.

We show that $8$ is minimal:

Let $R$ be a finite ring with unity with $n$ elements.

We need two preliminaries:

1.) If the additive group is cyclic, then $R$ is commutative.

Proof: If the additive group of $R$ is cyclic, we can choose $1$ as a generator: If we have $0 = 1+\dots+1 = m \cdot 1$ for some $m\in \Bbb N$, then $0=(m\cdot 1) \cdot g = m\cdot(1\cdot g)=m\cdot g$ so the additive order of $1$ is maximal. Thus, the multiplication table of $R$ is determined by $1 \cdot 1 = 1$, showing $R \cong \Bbb Z/n \Bbb Z$ and $R$ is commutative.

2.) All rings of order $4$ are commutative. Proof: As a general result, all rings with order equal to a squared prime are commutative: Ring of order $p^2$ is commutative.

Thus any ring of order $1,2,3,5,6,7$ is ruled out by 1.) using the Sylow-theorems and $4$ is ruled out by 2.).

So $8$ is the minimal cardinality a non-commutative ring can have.

  • 2
    $\begingroup$ Could you explain why in the proof of 2), you say $R^{\times}$ is a group of order $n-1$? This would be true if all nonzero elements are units. But how do we know this a priori? (I am assuming that $R^{\times}$ stands for the group of units of $R$) $\endgroup$
    – Prism
    Dec 25 '13 at 22:39
  • $\begingroup$ There is a ring of order $8$. See. math.stackexchange.com/questions/618332/… . $\endgroup$
    – studying
    Dec 25 '13 at 22:55
  • $\begingroup$ This is incorrect, there is a subgroup even of $M_2(\mathbb{F}_2)$ which has smaller cardinality and is also a noncommutative ring with unity. Take the subring of upper triangular matrices of $M_2(\mathbb{F}_2)$. $\endgroup$ Dec 25 '13 at 22:56
  • 1
    $\begingroup$ I am so sorry, this is absolutely emberassing. $\endgroup$
    – benh
    Dec 25 '13 at 23:01
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    $\begingroup$ If I were you, I would just delete your original text, making a note to the OP about the original error. Then you need only keep your first point about the additivity of $+$ to for $R$ to be commutative. This eliminates cases of order $1,2,3,4,5,6,7$. Finding the case of order $8$ and showing that it is the smallest is as simple as running through the construction of $M_2(\mathbb{F}_2)$, which has order $16$ and then explaining how to find the subring of upper triangular matrices of this ring and showing it has order $8$. Then your solution is correct and the ring is as it need be. $\endgroup$ Dec 25 '13 at 23:41

When one thinks of a noncommutative ring with unity (at least I), tend to think of how I can create such a ring with $M_n(R)$, the ring of $n \times n$ matrices over the ring $R$. The smallest such ring you can create is $R=M_2(\mathbb{F}_2)$. Of course, $|R|=16$. Now it is a matter if you can find a even smaller ring than this. Of course, the subring of upper/lower triangular matrices of $R$ is a subring of order $8$ which is a noncommutative ring with unity. This is indeed the smallest such rings.

In fact, there is a noncommutative ring with unity of order $p^3$ for all primes $p$. See this paper for this and many other interesting/useful constructions.


Here is another approach: Let $R$ be a non-zero ring with identity detone its center by $Z(R)$. It can be easily proved that: If $\frac {R}{Z(R)}$ is a cyclic group (with additive structure) then $R$ is a commutative ring.

Now since $0,1 \in Z(R)$ then for any ring $R$ with $|R|< 8$ we have $|\frac{R}{Z(R)}| \leq 3$ which implies that $R$ is commutative (since any group of order $1,2$ or $3$ is cyclic) Therefore the smallest non-commutative with identity must have at least $8$ elements and there are such rings of course as it is mentioned in other solutions. The thing I would like to add is that there is no other example other than the ring of upper triangular matrices over $\Bbb{Z}_2$ scince we have the following theorem:

Let $p$ be a prime number and let $R$ be a non-commutative ring with identity and suppose $|R| = p^3$ then $R$ is isomorphic to the ring of upper triangular matrices over $\Bbb{Z}_p$.

Also I would like to add that in a similar way we can show that the smallest non-commutative ring (not necessary with identity) has order $4$. As an example consider $R = \{ \begin{pmatrix}a & b \\ 0 & 0 \end{pmatrix} \; | \; a, b \in \Bbb{Z}_2\}$.


As others have noted, the upper triangular $2\times 2$ matrices with entries in $\mathbb{F}_2$ is a smaller ring. Here is one way to see that it is the smallest: cyclicity of the additive group, as remarked above, rules out the orders $1,2,3,5,6,7$.

As for $4$, it is not hard to see that any ring with unity of order $p^2$ is commutative: either it is cyclic or the additive subgroup generated by $1$ is of order $p$; hence there is an element $x$ that is not in this subgroup, whence $R$ must be a quotient of $\mathbb{Z}/p[t]$ via $t\mapsto x$ since the $x$ element necessarily commutes with all elements of the additive subgroup $\langle 1\rangle$.


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