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 don'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 $2$



Finite non-commutative ring with few invertible (unit) elements
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](https://en.wikipedia.org/wiki/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](https://arxiv.org/pdf/1302.3192.pdf) of Rodney Coleman (2013) in which OP's question was both asked and answered.




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[1] P. M. Cohn, ["Rings of zero-divisors"](http://www.ams.org/journals/proc/1958-009-06/S0002-9939-1958-0103202-2/S0002-9939-1958-0103202-2.pdf), 1984.  

[2] M. Henriksen, ["Rings with a unique regular element"](https://www.math.hmc.edu/~henriksen/publications/1989_Henriksen_Rings_with_a_unique_regular_element.pdf), 1989. 


