Finite non-commutative ring with few invertible (unit) elements

for a ring $R$ with unity , let $U(R)$ denote the group of units of $R$ . Now there are lots of finite commutative rings, of arbitrarily high order, with exactly one unit ; indeed $U(R)=1$ for a finite commutative ring $R$ iff $a^2=a , \forall a \in R$ . Incidentally , I couldn't find any finite non-commutative ring with exactly one unit ; matrix rings doesn't seem to work.

So my question is : Does there exist a finite non-commutative ring with unity having exactly one invertible (unit) element ?

Small remark : Note that such a ring must have characteristic $2$

• It should be noted that such rings cannot have non-zero nilpotents, because if $u\in R$ is a such a nilpotent, then $1+u$ has $1-u+u^2-\ldots\in R$ as its inverse. The question is then, whether non-commutative reduced rings in char 2 exist.. (IDK) – kneidell Oct 29 '17 at 7:46
• @kneidell : true indeed ... – user111524 Oct 29 '17 at 7:51
• @kneidell : I have posted an answer – user111524 Oct 29 '17 at 8:26
• Late comment: the question was answered in arxiv.org/pdf/1302.3192.pdf, 2013. – Luc Guyot Feb 27 '18 at 21:12
• @LucGuyot : I had a look at the paper; it doesn't prove anything more general, and the techniques are same as that of mine. In addition I also noticed in my answer that a finite non-commutative ring with zero Jacobson radical has at least 6 invertible elements. – user111524 Feb 28 '18 at 15:33

This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity must be a Boolean ring, hence commutative. The proof given below is based on a result by Melvin Henriksen. It doesn't rely on the Artin-Wedderburn Theorem and turns out to be fully elementary.

Following Melvin Henriksen, we call $R$ a UI-ring if $R$ has an identity element $1$ and $ab = ba = 1$ for $a,b \in R$ implies $a = b = 1$.

We have

Claim. A finite ring $R$ with identity is a UI-ring if and only if $R$ consists only of idempotent elements, i.e., $R$ is a Boolean ring. In particular, a finite UI-ring is commutative.

Proof. Assume that $R$ is a UI-ring. Then $R$ is reduced and $2x = 0$ for every $x \in R$. As $R$ is a finite dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$, every element of $R$ is algebraic over $\mathbb{Z}/2\mathbb{Z}$ by the Cayley-Hamilton theorem. Thus $R$ is a Boolean ring by [2, Corollary 2.10], which shows that $R$ is commutative. Assume now that $R$ is a Boolean ring. As any element $x \neq 1$ satisfies $x(1 - x) = 0$, the identity $1$ is the only unit of $R$.

The commutative case mentioned in OP's question was solved by P. M. Cohn [2, Theorem 3], should $R$ be finite or infinite:

Cohn's Theorem Let $R$ be an algebra over a field $F$ without nontrivial units, i.e., the units of $R$ are those of $F$. Then $R$ is a subdirect product of extension fields of $F$, and every element $x$ of $R$ which is not in $F$ is transcendental over $F$, unless $F = GF(2)$ and $x$ is idempotent. If, moreover, $R$ has finite dimension over $F$, then either $R = F$ or $R$ is a Boolean algebra.

Addendum. I discovered this preprint of Rodney Coleman (2013) in which OP's question was both asked and answered.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.

• I like your argument . Note that the commutative case , for which you quote Cohn's theorem , is very easy to prove as follows : Let $R$ be a finite commutative ring with exactly one unit . Then $J(R)=0$ . Let $m_1,...,m_n$ be the distinct maximal ideals of $R$ ; then by CRT ; $R \cong R/J(R) \cong \prod_{i=1}^n R/m_i$ ; so $R$ is a finite direct product of fields and since $R$ has exactly one unit , so do all those fields , hence $R \cong \mathbb Z_2^n$ , where $n=|Spec(R)|$ – user111524 Oct 30 '17 at 14:12

I think I have it ; there can be no such non-commutative ring.

Let $x \in J(R)$ , then $1-x$ is a unit of $R$ , so $x=0$ i.e. $J(R)=0$ . Thus $R$ is an artinian semisimple ring , so by Artin-Wedderburn , $R \cong \prod_{i=1}^m M(n_i , D_i)$ , where $D_i$'s are division rings . But $R$ is finite , hence so are $D_i$'s , hence by Wedderburn little theorem, $D_i$'s are fields , so $R \cong \prod_{i=1}^m M(n_i , k_i)$ , where $k_i$'s are fields . Now since $R$ is not-commutative , at least one $n_i$ is more than $1$ , say w.l.o.g. $n_1 \ge 2$ , but then $M(n_1 , k_1)$ has at least $q^{n_1}-1 \ge q^2-1 >1$ many units (where $q=|k_1|$) , so $R$ has more than one unit .

In fact , since $M(n_1,k_1)$ has $\prod_{j=0}^{n_1-1}(q^{n_1} - q^j)$ many units , where $q=|k_1|$ and for $n_1 \ge 2$ , $\prod_{j=0}^{n_1-1}(q^{n_1} - q^j)\ge (2^2-1)(2^2-2)=6$ ; so we get that :

Any finite non-commutative ring with unity and with zero Jacobson radical has at least $6$ units .

• We note that this answers the question " Is every ring with unity with a unique regular element commutative ? " , of Melvin Henriksen as described in one of the answers here mathoverflow.net/questions/100265/… and which Henriksen comments on the introduction of his paper math.hmc.edu/~henriksen/publications/… , in the affirmative for finite rings with unity , since for finite rings , an element is regular iff it is a unit . – user111524 Oct 29 '17 at 14:48