# Smallest non-commutative ring with unity

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.

• Rumor has it that $\{0\}$ is a perfectly-defined commutative ring with unity XD Commented Dec 25, 2013 at 16:28
• It's easy finding one with 16 elements. Commented Dec 25, 2013 at 16:33
• @HuiYu but I nees a non commutative ring. Commented Dec 25, 2013 at 16:36
• The smallest one is of cardinality $8$. Commented Dec 25, 2013 at 23:01
• What type of rings do you know that are not commutative? Thinking linear algebra might help. Now, how can you be sure you are dealing with only finitely many elements in your ring? Commented Jan 18, 2014 at 20:46

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.

• 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 $R$) Commented Dec 25, 2013 at 22:39
• 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 $M_2(\mathbb{F}_2)$. Commented Dec 25, 2013 at 22:56
• I am so sorry, this is absolutely emberassing.
– benh
Commented Dec 25, 2013 at 23:01
• I am very glad though, the review process is working here.
– benh
Commented Dec 25, 2013 at 23:02
• 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. Commented Dec 25, 2013 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.

• Let $M$ be a noncommutative monoid of order 3. Then the monoid ring $\mathbb{F}_p[M]$ with $p^3$ elements is an example. Commented Apr 13 at 12:40

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 $\langle 1\rangle$.